Bhagya

Balbharti Maharashtra State Board 12th Physics Important Questions Chapter 11 Magnetic Materials Important Questions and Answers.

Maharashtra State Board 12th Physics Important Questions Chapter 11 Magnetic Materials

Question 1.
How did a magnet derive its name? What was its ancient use?
Answer:
The characteristic of a magnet to attract iron is called its attractive property. This property was discovered in a rock found in the ancient mineral-rich region of Magnesia (modern Turkey). The word magnet is derived from this Magnesian rock.

The rock when floated on water (by tying it to a piece of wood) or suspended freely was found to align approximately in the north-south direction. Because this directional property was used by some ancient travellers to locate the geographic north, it was also called the leading stone or lodestone.

The Magnesian rock is the naturally magnetized iron ore magnetite (Fe₃O₄). It is hard, black or brownish-black with a metallic luster.

Question 2.
Explain the directional characteristic of a bar magnet.
Answer:
When a bar magnet is suspended so as to rotate freely in a horizontal plane, it comes to rest in approximately the North-South direction.

The end of the magnet directed towards the Earth’s geographic North Pole is called the north-seeking pole or the north pole, and the other end which is directed towards the Earth’s geographic South Pole is called the south-seeking pole or the south pole.

The cause for this alignment is seen when we consider the turning moment on a bar magnet suspended in a uniform magnetic induction \(\vec{B}\), as shown in below figure. The north pole (pole strength + m) and the south pole (pole strength – m) of the bar magnet experience equal and opposite forces of magnitude mB. If the lines of action of these two forces are not the same, they constitute a couple whose effect is to produce rotation.

Question 3.
State the expression for the torque acting on a magnetic dipole in a uniform magnetic field.
Answer:
When a magnetic dipole of magnetic dipole moment \(\vec{M}\) is placed in a uniform magnetic field of induction \(\vec{B}\), it experiences a torque whose magnitude is
τ = MB sin θ
where θ is the smaller angle between the magnetic axis and \(\vec{B}\).
[Note: When the dipole is placed with its axis at right-angles to the field, i.e, θ = 90°,
τ = MB or M = \(\frac{\tau}{B}\)
This is the defining equation for the magnetic dipole moment.]

Question 4.
Derive an expression for the torque acting on a magnetic dipole placed in a region of uniform magnetic induction. Express it in vector form.
Answer:
Consider a magnetic dipole consisting of two point poles of pole strength m a distance 2l apart with its axis inclined at an angle θ to a uniform magnetic field of induction \(\vec{B}\), shown in figure. The magnetic force on the positive (or north) pole is m\(\vec{B}\), while that on the negative (or south) pole is – m\(\vec{B}\). These two forces, equal in magnitude, opposite in direction and separated by a finite distance, constitute a couple which tends to line up the magnetic dipole moment with the field \(\vec{B}\).

The torque of the couple in the clockwise sense in above figure in magnitude, is
τ = (mB) d
where d is the distance between the lines of action of the forces.
∴ τ = (mB)2lsin θ = MB sin θ (∵ m = \(\frac{M}{2 l}\))
Expressed as a vector product,
\(\vec{\tau}=\vec{M} \times \vec{B}\)
The torque has a maximum magnitude, (m2l)B, when sin θ = 1, i.e., when the magnet is perpendicular to the field. The torque vanishes when the magnet is parallel to the field, where sin θ = 0.

Question 5.
A current-carrying coil with magnetic moment \(\vec{M}\) (M = 5 A∙m²) is placed in a uniform magnetic field of induction \(\vec{B}\) (B = 0.2 Wb/m²) such that the angle between \(\vec{M}\) and \(\vec{B}\) is 30°. What is the torque acting on the coil ?
Answer:
τ = MB sin θ = (5) (0.2) sin 30° = 0.5 N∙m is the torque acting on the coil.

Question 6.
Explain what is meant by magnetic potential energy of a bar magnet kept in a uniform magnetic field. Discuss the cases when
(1) θ = 0°
(2) θ = 180°
(3) θ = 90°.
Answer:
A magnet free to rotate in a uniform magnetic field \(\vec{B}\) aligns its dipole moment \(\vec{M}\) with \(\vec{B}\). Work must be done to rotate the magnet from this equilibrium position. The work done is stored as the magnetic potential energy, also called its orientation energy. In a finite angular displacement from 0 to θ, the
magnetic potential energy

1. When θ = 0°, cos θ = cos 0° = 1, U_θ = – MB. At this position, the magnetic moment of the bar magnet is lined up with the field and its magnetic potential energy is minimum. This is its most stable equilibrium position.
2. When θ = 180°, cos θ = cos 180° = – 1, U_θ = MB. At this position, the magnetic moment is antiparallel to the field and its magnetic potential energy is maximum. This is its most unstable position.
3. When θ = 90°, cos θ = cos 90° = 0, U_θ = 0. At this position, the bar magnet is perpendicular to the
   magnetic field. Its magnetic potential energy is zero.

Question 7.
What is the nature of the graph between the magnetic potential energy of a bar magnet kept in a uniform magnetic field and angular position of the magnet ?
OR
Draw a graph of magnetic potential energy as a function of angular displacement.
Answer:

Question 8.
Derive the expression for the time period of angular oscillations of a bar magnet kept in a uniform magnetic field.
Answer:
Consider a bar magnet of magnetic dipole moment \(\vec{M}\), suspended by a light twistless fibre in a uniform magnetic field \(\vec{B}\) in such a way that it is free to rotate in a horizontal plane. In the rest position θ = 0, \(\vec{M}\) is parallel to \(\vec{B}\).

If magnet is given a small angular displacement θ from its rest position and released, the magnet performs angular or torsional oscillations about the rest position.

Let I be the moment of inertia of the bar magnet about the axis of oscillation and α the angular acceleration. The deflecting torque (in magnitude) is
τ_d = Iα = I \(\frac{d^{2} \theta}{d t^{2}}\) …………….. (1)
However, the restoring torque tries to bring back the oscillating bar magnet in the rest position. The restoring torque (in magnitude) is, τ_r = – MB sin θ …………. (2)
The minus sign in Eq. (2) indicates that restoring torque is opposite in direction to the angular deflection.
In equilibrium, both the torques balance each other. From Eqs. (1) and (2),

Eq. (4) represents angular simple harmonic motion.
Writing ω² = \(\frac{M B}{I}\) , the angular frequency ω of the motion is
ω = \(\sqrt{\frac{M B}{I}}\) ………….. (5)
The time period of oscillations of the bar magnet is
T = \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{I}{M B}}\) ……………… (6)
This is the required expression.

9. Solve the following
Question 1.
Find the magnitude of the magnetic moment of a magnet if a couple exerting torque 0.5 N.m is required to hold the magnet with its axis perpendicular to a uniform magnetic field of induction 2 × 10^-3 T.
Solution:
Data : τ = 0.5 N∙m, B = 2 × 10^-3 T, θ = 90°
τ = MB sin θ
∴ M = \(\frac{\tau}{B \sin \theta}=\frac{0.5 \mathrm{~N} \cdot \mathrm{m}}{\left(2 \times 10^{-3} \mathrm{~T}\right) \sin 90^{\circ}}\)
= 250 A∙m²
The magnitude of the magnetic moment of the magnet is 250 A∙m².

Question 2.
A magnetic dipole of magnetic moment 5 A∙m² is placed in a uniform magnetic field of induction 10^-3 T. Find the magnitude of the maximum torque acting on the dipole.
Solution:
Data : M = 5 A∙m², B = 10^-3 T
Torque, τ = MB sin θ
The magnitude of the torque is maximum when θ = 90°.
∴ τ_max = MB = 5 × 10^-3 N∙m

Question 3.
A bar magnet of magnetic length 0.12 m and pole strength 10 A-m is placed in a uniform magnetic field of induction 3 × 10^-2 tesla. If the angle between the magnetic induction and the magnetic moment is 30°, find the magnitude of the torque acting on the magnet.
Solution:
Data : 2l = 0.12 m, m = 10 A∙m, B = 3 × 10^-2 T,
θ = 30°
The magnitude of the torque,
τ = MB sin θ = m (2l) B sin θ
= (10 A∙m) (0.12 m)(3 × 10^-2 T) sin 30°
= 3.6 × 10^-2 × \(\frac{1}{2}\)
= 1.8 × 10^-2 N∙m

Question 4.
A bar magnet of moment 10 A∙m is suspended such that it can rotate freely in a horizontal plane. The horizontal component of the Earth’s magnetic field at the place is 36 µT. Calculate the magnitude of the torque when its angular displacement with respect to the direction of the field is 8C and magnetic potential energy.
Solution:
Data : M = 10 A∙m², B_h = 36µT = 3.6 × 10^-5 T,
θ = 8°
The magnitude of the torque is
τ = MB_h sin θ
= (10) (3.6 × 10^-5) sin 8°
= (36 × 10^-5) (0.1391) = 5.007 × 10^-5 N∙m
The magnetic potential energy of the bar magnet is
U_θ = MB_h cos θ
= (36 × 10^-5) cos 8°
= (36 × 10^-5) (0.99) = 3.564 × 10^-4 J

Question 5.
The time period of angular oscillation of a bar magnet in a horizontal magnetic field is 3.14 s. If the horizontal component of the Earth’s magnetic field at the place is 40 µT and the moment of inertia of the magnet is 10^-4 kg∙m², calculate the magnetic moment of the magnet.
Solution:
Data : T = 3.14 s, B_h = 40 µT = 4 × 10^-5 T,
I = 10^-4 kg∙m²

Question 6.
A bar magnet of moment 10 A∙m² is suspended such that it can rotate freely in a horizontal plane. The horizontal component of the Earth’s magnetic field at the place is 36 µT. Calculate the magnitude of the torque when its angular displacement with respect to the direction of the field is 8° and magnetic potential energy.
Solution:
Data : M = 10 A∙m², B_h = 36 µT = 3.6 × 10^-5 T, θ = 8°
The magnitude of the torque is
τ = MB_h sin θ
=(10)(3.6 × 10^-5)sin8°
=(36 × 10^-5)(0.1391) = 5.007 × 10^-5 N∙m
The magnetic potential energy of the bar magnet is
U_θ = MB_h cos θ
=(36 × 10^-5)cos8°
=(36 × 10^-5)(0.99) = 3.564 × 10^-4 J

Question 10.
As the electron revolves in the second Bohr orbit in the hydrogen atom, the corresponding current is (about) 1.3 × 10^-4 A. If the area of the orbit is (about) 1.4 × 10^-19 m², what is the (approximate) equivalent magnetic moment?
Answer:
M = IA = (1.3 × 10^-4)(1.4 × 10^-19)
= 1.82 × 10^-23 A∙m² is the (approximate) equivalent magnetic moment.

Question 11.
What is the magnetic moment of an electron due to its orbital motion?
Answer:
The orbital magnetic moment of an electron, of charge e and revolving with speed v in an orbit of radius r, is \(\frac{1}{2}\) evr.

Question 12.
If the frequency of revolution of a proton (q = 1.6 × 10^-19 C) in a uniform magnetic induction is 10⁶ Hz, what is the corresponding electric current?
Answer:
I = \(\frac{q}{T}\) = qf = (1.6 × 10^-19)(10⁶)
= 1.6 × 10^-13 A
is the corresponding electric current.

Question 13.
The frequency of revolution of the electron in the second Bohr orbit in the hydrogen atom is 8.22 × 10¹⁴ Hz. What is the corresponding electric ’ current? [e = 1.6 × 10^-19 C]
Answer:
I = \(\frac{e}{T}\) = ef
= (1.6 × 10^-19)(8.22 × 10¹⁴)
= 1.315 × 10^-4A
is the corresponding electric current.

Question 14.
A coil has 1000 turns, each of area 0.5 m². What is the magnetic moment of the coil when it carries a current of 1 mA ?
Answer:
Magnetic moment M = NIA
= 1000 × 1 × 10^-3 × 0.5
= 0.5 A∙m²

Question 15.
What is the gyromagnetic ratio of an orbital electron ? State its dimensions and the SI unit.
Answer:
The ratio of the magnitude of the orbital magnetic moment to that of the orbital angular momentum of an electron in an atom is called its gyromagnetic ratio γ₀. If \(\vec{M}_{\mathrm{o}}\) is the orbital magnetic moment of the electron with orbital angular momentum \(\vec{L}_{\mathrm{o}}\),
γ₀ = \(\frac{M_{\mathrm{o}}}{L_{\mathrm{o}}}=\frac{e}{2 m_{\mathrm{e}}}\)
where e and me are the electronic charge and electron mass, respectively.
Dimensions : [γ₀] = \(\frac{[\text { charge }]}{[\text { mass }]}=\frac{[T I]}{[M]}\) = [M^-1TI].
SI unit : The coulomb per kilogram (C/kg).
[Note : γ₀ = 8.794 × 10¹⁰ C/kg. The gyromagnetic ratio of electron spin is nearly twice that of an orbital electron.]

16. Solve the following
Question 1.
A circular coil of 300 turns and diameter 14 cm carries a current of 15 A. What is the magnitude of the magnetic moment associated with the coil?
Solution :
Data : N = 300, d = 14 cm, I = 15 A
r = \(\frac{d}{2}\) = 7 cm = 7 × 10^-2 m
The magnitude of the magnetic moment associated with the coil is
M = NIA = NI(πr²)
= (300)(15)(3.142)(7 × 10^-2)²
= 4500 × 3.142 × 49 × 10^-4
= 69.28 A∙m²

Question 2.
A coil has 300 turns, each of area 0.05 m².
(i) Find the current through the coil for which the magnetic moment of the coil will be 4.5 A∙m².
(ii) It is placed in a uniform magnetic field of induction 0.2 T with its magnetic moment making an angle of 30° with \(\vec{B}\). Calculate the magnitude of the torque experienced by the coil.
Solution:
Data : N = 300, A = 0.05 m², M = 4.5 A∙m², B = 0.2 T, θ = 30°
(i) M = NIA
∴ The current in the coil,
I = \(\frac{M}{N A}=\frac{4.5}{300 \times 0.05}\) = 0.3 A

(ii) The magnitude of the torque,
τ = MB sin θ = 4.5 × 0.2 × sin 30°
= 0.9 × \(\frac{1}{2}\) = 0.45 N∙m

Question 3.
An electron in an atom revolves around the nucleus in an orbit of radius 0.5 Å. Calculate the equivalent magnetic moment if the frequency of revolution of the electron is 10¹⁰ MHz.
Solution:
Data : r = 0.5 Å = 5 × 10^-11 m,
f = 10¹⁰ MHz = 10¹⁶ Hz, e = 1.6 × 10^-19 C
Equivalent current, I = \(\frac{e}{T}\) = ef
The equivalent magnetic moment is
M = IA = ef(πr²)
= (1.6 × 10^-19)(1016)(3.142)(5 × 10^-11)2
= 1.6 × 3.142 × 25 × 10^-25
= 1.257 × 10^-23 A∙m²

Question 4.
Calculate the orbital magnetic dipole moment of the electron in the second Bohr orbit of the hydrogen atom. The radius of the orbit is 2.126 Å and the orbital speed of the electron in the orbit is 1.09 × 10⁶ m/s.
Solution:
Data : r = 2.126 Å = 2.126 × 10^-10 m,
v = 1.09 × 10⁶ m/s, e = 1.6 × 10^-19 C
The orbital magnetic dipole moment of the electron,
M₀ = \(\frac{1}{2}\) evr
=(1.6 × 10^-19)(1.09 × 10⁶)(2.126 × 10^-10)
=1.6 × 1.09 × 1.063 × 10^-23
= 1.854 × 10^-23 A∙m²

Question 5.
An electron in an atom revolves around the nucleus in an orbit of radius 0.53 Å. If the frequency of revolution of an electron is 9 × 10⁹ MHz, calcu-late the orbital angular momentum. [Charge on an electron = 1.6 × 10^-19 C, gyromagnetic ratio = 8.8 × 10¹⁰ C/kg, π = 3.142]
Solution:
Data : r = 0.53 Å = 0.53 × 10^-10 m,
f = 9 × 10⁹ MHz = 9 × 10¹⁵ Hz, e = 1.6 × 10^-19 C, gyromagnetic ratio = 8.8 × 10¹⁰ C / kg, π = 3.142
Magnetic moment, M₀ =IA = efπr²
= 1.6 × 10^-19 × 9 × 10¹⁵ × 3.142 × (0.53 × 10^-10)²
= 14.4 × 3.142 × (0.53)² × 10^-19 × 10¹⁵× 10²⁰
= 1.272 × 10^-23 A∙m²

Question 6.
The specific charge (charge-to-mass ratio) of an electron is 1.759 × 10¹¹ C/kg.
(i) What is the gyromagnetic ratio of an orbital electron?
(ii) If the measured component of its orbital angular momentum is 2.11 × 10^-34 kg∙m³/s. what Is the associatcd orbital magnetic dipole moment?
Solution:
Data: \(\frac{e}{m_{e}}\) = 1.759 × 10¹¹ C/kg,
L = 2.11 × 10^-34 kg∙m²/s

(i) The gyromag1etic ratio of an orbital electron,
γ₀ = \(\frac{e}{2 m_{\mathrm{e}}}=\frac{1.759 \times 10^{11}}{2}\) =8.795 × 10¹⁰ C/kg

(ii) For the given component of the orbital angular momentum, the associated magnetic dipole moment is
Mo = \(\frac{e}{2m_{e}}\) L₀ = γ₀L₀
= (8.795 × 10¹⁰ )(2.11 × 10^-34)
= 1.856 × 10^-23 A∙m²

Question 7.
An electron performs uniform circuLar motion in a uniform magnetic field of induction 1.2T In a plane perpendicular to the field. Its kinetic energy is 6 × 10^-20 J and its motion is subject only to the magnetic force due to the field. What is the magnetic dipole moment associated with the motion of the electron?
Solution:
Data:B = 1.2T, KE ≡ E_k = 6 × 10^-20 J
The centripetal force on the electron is the magnetic force. If r and e are respectively the radius of the path and linear speed of the electron,

Question 8.
If the magnetic moment of an electron revolving in an orbit of radius 0.4 Å is 9 × 10^-24 A∙m² then find the linear momentum of the electron in that orbit. (e/m = 1.76 × 10¹¹ C/kg]
Solution:
Data: r = 0.5 Å = 5 × 10^-11 m, I = 1.1 × 10^-3 A,
x = 100 Å = 10^-8 m, \(\frac{\mu_{0}}{4 \pi}\) = 10^-7 T∙m/A
The magnitude of the magnetic induction at an axial point of a current loop,

Question 9.
The magnetic moment of an electron revolving in a circular orbit of radius 2.2 Å is 5.024 × 10^-24 A∙m². Calculate the frequency of revolution of the electron in that orbit.
Solution:
Data: d = 0.2 m, B = 4 × 10^-4 T,
2l = 5 cm = 5 × 10^-2 m, A = 2 cm² = 2 × 10^-4 m²,
\(\frac{\mu_{0}}{4 \pi}\) = 10^-7 T∙m/A
The magnitude of the magnetic induction at an axial point of a bar magnet.

Question 17.
Explain magnetization of a material.
Answer:
In an atom, the orbital and spin magnetic moments of its electrons may or may not add up to zero, depending on the electronic configuration. In some materials, their atoms have a net magnetic moment. When such a material is placed in an external magnetic field of induction \(\overrightarrow{B_{0}}\) , the field exerts a torque on each atomic magnet. These torques tend to align the magnetic moments with the applied field. Due to this, the material as a whole acquires a net magnetic moment \(\vec{M}_{\text {net }}\) along \(\overrightarrow{B_{0}}\) and the material is said to be magnetized.

The net magnetic moment per unit volume is called the magnetization \(\vec{M}_{z}\) of the material.
\(\vec{M}_{\mathrm{z}}=\vec{M}_{\text {net }} / V\)
where V is the volume of the material.

Even when the atomic magnetic moment is zero, application of magnetic field induces magnetism in the material. In this case, the magnetization has direction opposite to that of the applied field.

Question 18.
Define magnetization.
State its dimensions and the SI unit.
OR
Define magnetization. State its formula and SI unit.
Answer:
Magnetization : The net magnetic moment per unit volume of a material is called the magnetization of the material.

SI unit: The ampere per metre (A/m).

Question 19.
A bar magnet (volume 1.5 × 10^-5 5m³) has a uniform magnetization of 6000 Aim. What is its magnetic dipole moment?
Answer:
M_z = \(\)
∴ M = M_zV = (6000)(1.5 × 10^-5)
= 9 × 10^-2 A∙m² is the magnetic dipole moment of the magnet.

Question 20.
Define magnetic intensity.
State its dimensions and the SI unit.
Answer:
Magnetic intensity: The magnetic intensity is defined as the magnetic induction in an isotropic medium divided by the permeability of the medium.

The magnetic intensity is a quantitative characteristic of a magnetic field independent of the magnetic properties of the medium. In a medium, it determines the contribution to the magnetic induction in the medium by the external magnetic field.

Dimensions: Within a toroid without a core, which has n windings per unit length carrying a current I, the magnetic intensity, H = nI.
∴ [H] = [n][I] = [L^-1][I] = [L^-1I]
SI unit : The ampere per metre (A/m).
[Note : The magnetic intensity (also called the magnetic field strength), \(\vec{H}\), is defined as
\(\vec{H}=\frac{\vec{B}}{\mu_{0}}-\overrightarrow{M_{z}}\)
where \(\vec{B}\) is the magnetic induction, \(\vec{M}_{z}\) is the magnetization and μ₀ is the permeability of free space.]

Question 21.
What is the magnetic susceptibility of a medium?
Answer:
The magnetic susceptibility of a medium is a dimensionless quantity which signifies the contribution made by the medium when subjected to a magnetic field to the magnetic induction inside the medium. For a material in which the magnetization M_z is proportional to the magnetic intensity H, the magnetic susceptibility of the medium is
χ_m = M_z/H
It is equal to the fractional change in the magnetic induction due to the medium.
Magnetic susceptibility of some materials

Substance	χ	Substance	χ
Silicon	-4.2 × 10-6	Aluminium	2.3 × 10-5
Bismuth	-1.66 × 10-5	Calcium	1.9 × 10-5
Copper	-9.8 × 10-6	Choromium	2.7 × 10-4
Diamond	-2.2 × 10-5	Lithium	2.1 × 10-5
Gold	-3.6 × 10-5	Magnesium	1.2 × 10-5
Lead	-1.7 × 105	Niobium	2.6 × 10-5
Mercury	-2.9 × 10-5	Oxygen (STP)	2.1 × 10-6
Nitrogen	-5.0 × 10-9	Platinum	2.9 ×10-4

Question 22.
Discuss magnetization of a rod of material having net magnetic moment with the help of a solenoid.
OR
Discuss magnetization of an iron rod placed in a solenoid.
OR
Discuss magnetization of a magnetic material placed in a solenoid.
Answer:
Consider an ideal air-cored solenoid carrying a steady current I. The magnitude of the magnetic induction inside the solenoid is
B₀ = \(\frac{\mu_{0} N I}{2 \pi r}\) = μ_onI ………….. (1)
where n = \(\frac{N}{2 \pi r}\) is the number of turns per unit length and p0 is the permeability of free space.

When a core of magnetic material (such as iron) is present, the magnetic field within the solenoid due to the current in the winding magnetizes the material of the core. With the core, the magnetic induction \(\vec{B}\) inside the solenoid is greater than \(\vec{B}_{0}\), so that
B = B₀ + B_m ……………. (2)
where B_m is the contribution of the iron core. B_m is proportional to the magnetization M_z of the material.
B_m = μ_oM_z …………… (3)

While discussing magnetic materials, it is customary to call \(\frac{B_{0}}{\mu_{0}}\) as the magnetizing field or magnetic field intensity, denoted by H, which produces the magnetization.
∴ B₀ = μ_oH …………. (4)
Substituting for B₀ and B_M in Eq. (1),
B = μ₀H + μ₀M_z = μ₀(H + M_z) …………….. (5)
For materials in which the magnetization is pro-portional to the magnetic intensity,
M_z ∝ H or M_z = χ_mH ………. (6)
where the constant of proportionality χ_m is called the magnetic susceptibility.
∴ B = μ₀(H + χ_mH) = μ₀(1 + χ_m)H = μH …………. (7)
where p = μ₀(1 + χ_m) is called the permeability of the material.

[Notes : (1) M_z ∝ H only for diamagnetic and paramagnetic materials. Among ferromagnetic materials, the linear relation in Eq. (6) holds good only for initial magnetization of magnetically softer materials; for magnetically harder materials, M_z is not a single-valued function of H, and depends on the magnetic intensity that the material has been previously exposed to (a phenomenon known as hysteresis). (2) H is defined for convenience; B is more fundamental.]

Question 23.
What is the relation between permeability and magnetic susceptibility of a medium? What is relative permeability of a medium ?
Answer:
If χ_m is the magnetic susceptibility of a medium, the permeability of the medium is
μ = μ₀ (1 + χ_m)
where μ₀ is the permeability of free space.
The ratio of the permeability of a medium to that of free space is called the relative permeability of the medium, denoted by μ_r.
μ_r = μ/μ₀ = 1 + χ_m

Question 24.
Write the relation between relative permeability and magnetic susceptibility.
Answer:
μ_r = 1 – χ_m, where μ_r is the relative permeability and χ_m is the magnetic susceptibility of a substance.

Question 25.
Why do magnetic lines of force prefer to pass through iron than air?
Answer:
The permeability of air (≅ 1.00000037) is negligible relative to that of iron, which typically has μ of several hundreds or thousands. That is, permeability of iron is much more than that of air. Hence, magnetic lines of force prefer to pass through iron than air.

[Note : Air is paramagnetic due to the oxygen it contains. Reference for μ_air: B. D. Cullity and C. D. Graham (2008), Introduction to Magnetic Materials, 2nd edition, p. 16.]

Question 26.
Is magnetic susceptibility a dimensionless quantity? Why?
Answer:
In a magnetizing field of intensity H, a material for which its magnetization M_z ∝ H, we write . M_z = χ_mH, where the proportionality constant χ_m is called the magnetic susceptibility. Since both magnetization and magnetic intensity have the same dimensions and units, χ_m is a dimensionless quantity.
OR
The relative permeability of a medium,
μ_r = \(\frac{\mu}{\mu_{0}}\) = 1 + χ_m
where μ and χ_m are the permeability and magnetic susceptibility of the medium and μ₀ is the permeability of free space. Hence, both μ₀ and χ_m are dimensionless quantities.

27. Solve the following
Question 1.
A cylindrical magnet 5 cm long has a diameter of 1 cm and a uniform magnetization of 5000 A/m. Find its magnetic dipole moment.
Solution:
Data : L = 5 × 10^-2 m, d = 1 × 10^-2 m,
M_z = 5 × 10³ A/m .
Volume, V = πr²L = \(\frac{\pi d^{2}}{4}\) L
M_z = \(\frac{M}{V}\)
∴ The magnetic dipole moment,

Question 2.
A bar magnet made of steel has magnetic moment 2.5 A∙m² and mass 6.6 × 10^-3 kg. Given that the density of steel is 7.9 × 10³ kg/m³, find the intensity of magnetization of the magnet. (2 marks )
Solution:
Data : M = 2.5 A∙m², m = 6.6 × 10^-3 kg, ρ = 7.9 × 10³ kg/m³
Volume, V = \(\frac{m}{\rho}\)
The intensity of magnetization of the magnet is
M_z = \(\) = M ∙ \(\frac{\rho}{m}\)
= 2.5 x \(\frac{M}{V}\) = 2.992 × 10⁶ A/m
[Note : Intensity of magnetization = magnetization.]

Question 3.
Find the magnetization of a bar magnet of length 10 cm and cross-sectional area 4 cm², if the magnetic moment is 2 A∙m².
Solution:
Data : L = 0.1 m, A = 4 × 10^-4 m², M = 2 A∙m²
Magnetization, M_z = \(\frac{M}{V}\)
= \(\frac{M}{L A}=\frac{2}{(0.1)\left(4 \times 10^{-4}\right)}\)
= \(\frac{10^{4}}{0.2}\) = 5 × 10⁴ A/m

Question 4.
The magnetic susceptibility of annealed iron at saturation is 5500. Find the permeability of annealed iron at saturation.
Solution:
Data : χ_m = 5500, μ₀ = 4π × 10^-7 T∙m/A
Permeability, μ = μ₀(1 + χ_m)
The permeability of annealed iron at saturation is
μ = (4 × 3.142 × 10^-7) (1 + 5500)
= 4 × 3.142 × 5.501 × 10^-4
= 6.914 × 10^-3 T∙m/A

Question 5.
The magnetic induction B and the magnetic intensity H in a material are found to be 1.6 T and 1000 A/m, respectively. Calculate the relative permeability μ_r and the magnetic susceptibility χ_m of the material.
Solution:
Data : B = 1.6 T, H = 1000 A/m,
μ₀ = 4π × 10^-7 T∙m/A
(i) B = μH = μ₀μ_r H
∴ The relative permeability of the material,
μ_r = \(\frac{B}{\mu_{0} H}=\frac{1.6}{\left(4 \times 3.142 \times 10^{-7}\right)\left(10^{3}\right)}\)
= \(\frac{4000}{3.142}\) = 1.273 × 10³

(ii) μ_r = 1 + χ_m
∴ The magnetic susceptibility of the material,
χ_m = μ_r – 1
= 1273 – 1 = 1272 or 1.272 × 10³

Question 6.
An ideal solenoid has a core of relative permeability 500 and its winding has 1000 turns per metre. If a steady current of 1.6 A is passed through its winding, find
(i) the magnetic field strength H
(ii) the magnetization M_z
(iii) the magnetic induction B within the solenoid. Assume that Mz is directly proportional to H and single valued.
Solution:
Data : μ_r = 500, n = 1000 m^-1, I = 1.6 A,
μ₀ = 4π × 10^-7 T∙m/A
(i) The magnetic field strength (or magnetic intensity) of the ideal solenoid,
H = nI
= 1000 × 1.6 = 1.6 × 10³ A/m

(ii) The magnetization of the core,
M_z = χ_mH = (μ_r – 1)H
= (500 – 1) × 1.6 × 10³
= 499 × 1.6 × 10³ = 7.984 × 10⁵ A/m

(iii) The magnetic induction within the core of the solenoid,
B = μ_rμ₀H
= (500) (4 × 3.142 × 10^-7) (1.6 × 10³)
= 32 × 3.142 × 10^-2 = 1.005 T

Question 28.
What is a diamagnetic material? Give two examples.
Answer:
A material which is weakly repelled by a magnet and whose atoms /molecules do not possess a net magnetic’ moment in the absence of an external magnetic field is called a diamagnetic material.

When a diamagnetic material is placed in a uniform magnetic field, it acquires a small net induced magnetic moment directed opposite to the field.

Examples: Bismuth, copper, gold, silver, anti-mony, mercury, water, air, hydrogen, lead, silicon, nitrogen, sodium chloride.

Question 29.
What is diamagnetism?
Answer:
A material which is weakly repelled by a magnet and whose atoms/molecules do not possess a net magnetic moment in the absence of an external magnetic field is called a diamagnetic material. When a diamagnetic material is placed in a mag-netic field, it acquires a small net induced magnetic moment directed opposite to the field. The induced magnetism exhibited by such materials is called diamagnetism.

Question 30.
State any four properties of a diamagnetic material.
Answer:
Properties of a diamagnetic material :

1. In the absence of an external magnetic field, the magnetic dipole moment of each atom / molecule of a diamagnetic material is zero.
2. A diamagnetic material is weakly repelled by a magnet.
3. If a thin rod of a diamagnetic material is suspended in a uniform magnetic field, it comes to rest with its length perpendicular to the field.
4. When placed in a nonuniform magnetic field, a diamagnetic material is repelled from the region of strong field.
5. The magnetic susceptibility (χ_m) of a diamagnetic material is small and negative.
6. χ_m is very nearly temperature-independent.
7. The relative permeability (μ_r) of a diamagnetic material is slightly less than 1 and very nearly temperature independent.
8. If a diamagnetic liquid in a watch glass is placed on two closely spaced pole-pieces of a magnet, the liquid accumulates on the sides causing a depression at the centre.
   [ Note : When the pole-pieces are moved apart, the effect is reversed, i.e., the diamagnetic liquid accumulates at the centre, where the magnetic field is weak.]
9. A diamagnetic liquid in a U-tube placed in a magnetic field shows a depression in the arm to which the magnetic field is applied.
10. If a diamagnetic gas is introduced between the pole-pieces of a magnet, it spreads at right angles to the field.

Question 31.
Name two materials that have negative magnetic susceptibility.
Answer:
Copper and gold have negative magnetic susceptibility.

Question 32.
Explain the origin of diamagnetism.
OR
Explain the origin of diamagnetism on the basis of atomic structure.
Answer:
An atomic electron is equivalent to a current loop which has orbital magnetic moment. Further, electron spin gives rise to the spin magnetic moment. The magnetic moment of an atom is equal to the vector sum of the magnetic moments of all its electrons.

The electronic configuration in an atom of a diamagnetic material is such that the vector sum of the orbital and spin magnetic moments of all the electrons is zero. Thus, the atomic magnetic moment is zero. Hence, a diamagnetic material has no inherent magnetic moment associated with it.

When a diamagnetic material is placed in a magnetic field, the orbiting electrons are accelerated or decelerated depending on their sense of revolution. For the electrons that are speeded up, the magnetic moment increases and for those that are slowed down, the magnetic moment decreases. This happens due to induced current as per Lenz’s law. Thus, each atom acquires a net magnetic moment and the diamagnetic material is weakly magnetized. The induced magnetic moment is opposite in direction to the applied field.

Diamagnetism is the weakest magnetic phenomenon. Hence, although diamagnetism is a universal property, it can be detected only in the absence of properties resulting in para- and ferromagnetism.

Question 33.
Explain the diamagnetic behaviour of superconductors.
Answer:
Suppose a superconducting material in its normal phase, not the superconducting phase, is placed in a uniform applied magnetic field of induction \(\vec{B}\) which is weaker than a certain critical magnetic induction \(\vec{B}_{\mathrm{c}}\) for that material. Then, when the temperature is reduced below the critical temperature for the material, it enters into the superconducting phase and the magnetic field is expelled from the interior of the material.

There is no magnetic field inside the material, i.e., the material exhibits perfect diamagnetism. The expulsion of the magnetic field in a superconducting material is called the Meissner effect.

[Notes : (1) What happens is that currents appear on the surface of the material whose magnetic field exactly cancels the applied field inside it. (2) The German physicists Fritz Walther Meissner (1882-1974) and Robert Ochsenfeld (1901-1993) discovered the phenomenon in 1933 by measuring the magnetic field distribution outside superconducting tin and lead samples. The Meissner effect is so strong that a magnet can actually be levitated over a material cooled below its superconducting transition temperature T_c.]

Question 34.
What is a paramagnetic material? Give two examples.
Answer:
A material which is weakly attracted by a magnet and whose atoms possess a net magnetic moment with all atomic magnetic moments randomly directed in the absence of an external magnetic field but are capable of being aligned in the direction of the applied magnetic field is called a paramagnetic material.

Examples : Aluminium, platinum, chromium, manganese, sodium, calcium, magnesium, lithium, tungsten, niobium, copper chloride, oxygen.

Question 35.
State any four properties of a paramagnetic material.
Answer:
Properties of a paramagnetic material:

1. Each atom of a paramagnetic material possesses a magnetic dipole moment. However, in the absence of an external magnetic field, the atomic magnetic moments are randomly oriented and hence the resultant magnetization of a paramagnetic material is zero.
2. A paramagnetic material is weakly attracted by a magnet.
3. If a thin rod of a paramagnetic material is sus-pended in a uniform magnetic field, it comes to rest with its length parallel to the field.
4. When placed in a nonuniform magnetic field, a paramagnetic material is attracted towards the region of strong field.
5. The magnetic susceptibility (χ_m) of a paramagnetic material is small and positive.
6. χ_m varies inversely with the absolute temperature T.
7. The relative permeability (μ_r) of a paramagnetic material is slightly greater than 1 and decreases with increasing temperature.
8. If a paramagnetic liquid in a watch glass is placed on two closely spaced pole-pieces of a magnet, it shows a slight rise in the middle.
   [Note : If the pole-pieces are moved apart, the para-magnetic liquid moves away from the centre where the field is weak and gets depressed in the middle.]
9. A paramagnetic liquid in a U-tube placed in a magnetic field shows a rise in the arm to which the magnetic field is applied.

Question 36.
Give one example each of a diamagnetic material and a paramagnetic material.
Answer:
Bismuth is a diamagnetic material and aluminium is a paramagnetic material.

Question 37.
Name two materials that have small and positive magnetic susceptibility.
Answer:
Platinum and Chromium have small and positive magnetic susceptibility.

Question 38.
The relative permeability of two materials are 0.999 and 1.001. Identify the materials.
Answer:
A material with relative permeability μ_r ≤ 1 is diamagnetic while the one with μ_r ≥ 1 is paramagnetic.

Question 39.
Explain the origin of paramagnetism.
OR
Explain the origin of paramagnetism on the basis of atomic structure.
Answer:
Paramagnetism depends on the presence of permanent atomic or molecular magnetic dipole moments. The inherent net atomic magnetic mo-ment results from a particular combination of the spin and orbital magnetic moments of its electrons.

The spin magnetic moments of the electrons in matter are affected by the internal magnetic field created by the magnetic moments of surrounding electrons. This internal field, ~ 10^-2 T to 10^-1 T,
causes the spin magnetic moments to precess about the field direction. At normal temperature; the thermal motion of the electrons produces constant fluctuations in the internal field so that the spin magnetic moments have random directions, from figure (a). In the absence of an external magnetizing field, therefore, a paramagnetic material is not magnetized.

When the applied field strength is greater than that of the internal field, the spin magnetic moments tend to align parallel to the external field direction. But the randomizing effect of thermal agitation prevents complete alignment, from figure (b). Therefore, at room temperature, when a paramagnetic material is placed in a magnetic field, it is weakly magnetized in the direction of the magnetizing field.

If the external field is very large or the temperature is very low, the magnetic dipole moments are effectively aligned parallel to the field so as to have the least magnetic potential energy and the magnetization reaches saturation, from figure (c).

Question 40.
Discuss Curie’s law for paramagnetic material.
OR
State Curie’s law of paramagnetism.
Answer:
Curie’s law : The magnetization of a paramagnetic material is directly proportional to the external magnetic field and inversely proportional to the absolute temperature of the material.

If a paramagnetic material at an absolute temperature T is placed in an external magnetic field of induction \(\overrightarrow{B_{\text {ext }}}\), the magnitude of its magnetization
M_z ∝ \(\frac{B_{\text {ext }}}{T}\) ∴ M_z = C \(\frac{B_{\text {ext }}}{T}\)
where the proportionality constant C is called the Curie constant.

[Notes : (1) The above law, discovered experimentally in 1895 by Pierre Curie (1859-1906) French physicist, is true only for values of B_ext/ T below about 0.5 tesla per kelvin.
(2) [C] = [M_z ∙ T] / [B_ext] = [L^-1I∙] /[MT^-2I^-1]
= [M^-1L^-1T²I²],
where denotes the dimension of temperature.]

Question 41.
Prove that the magnetic susceptibility of a para-magnetic material is inversely proportional to its absolute temperature.
Answer:
Consider a paramagnetic material at an absolute temperature T placed in an external magnetic field of induction \(\vec{B}_{\text {ext }}\). The magnitude of its magnetization is
M_z = C \(\frac{\vec{B}_{\text {ext }}}{T}\) …………. (1)
where the proportionality constant C is called the Curie constant.
Substituting \(\vec{B}_{\text {ext }}\) = μ₀H in Eq. (1),
M_z = C \(\frac{\mu_{0} H}{T}\)
where µ₀ is the permeability of free space and H is the intensity of the magnetizing field.
The magnetic susceptibility of the medium
χ_m = \(\frac{M_{z}}{H}\). Thus, Eq. (2) becomes ti
χ_m = C \(\frac{\mu_{0}}{T}\)
χ_m ∝ \(\frac{1}{T}\)
This is the required expression.

Question 42.
What is a ferromagnetic material? Give two examples.
Answer:
A material which is strongly attracted by a magnet and whose atoms possess a net magnetic moment (largely due to electron spin) which, within a certain temperature range, spontaneously line up parallel to each other by quantum mechanical exchange interaction is called a ferromagnetic material.

Examples : Transition elements (iron, cobalt and nickel), inner transition elements (gadolinium, Gd, and dysprosium, Dy) and their alloys.

[Note : Ferromagnetic behaviour was first explained by Pierre-Ernest Weiss on the basis of domain theory but its satisfactory explanation is given only by quantum mechanics.]

Question 43.
What are domains in a ferromagnetic material ?
OR
Write a short note on domains in a ferromagnetic material.
Answer:
A ferromagnetic material is composed of mosaic of small regions within each of which all the atomic magnetic moments spontaneously line up parallel to each other by quantum mechanical exchange interaction within a certain temperature range. Each such spontaneously magnetized region is called a domain and the common direction of magnetic moment is called the domain axis. A domain is an extremely small region (e.g., a size of about 10^-6 m -10^-4 m) containing a large number (10¹⁰ – 10¹⁷) of atoms. The boundary between adjacent domains with a different orientation of magnetic moment is called a domain wall.

Question 44.
Explain ferromagnetism on the basis of the domain theory.
Answer:
Atoms of a ferromagnetic material have a perms tient non-zero magnetic dipole moment arising mainly from spin magnetic moments of the electrons.

According to the domain theory, a ferromagnetic material is composed of small regions called domains.

A domain is an extremely small region contain Inga large number (something like 10¹⁵ atoms as in common iron) of atoms,

Within each domain, the atomic magnetic moments of nearest-neighbour atoms interact strongly through exchange interaction (a quantum mechanical phenomenon) and align themselves parallel to each other even in the absence of an extemal magnetic field. A domain is. therefore, spontaneously magnetized to saturation.

In an unmagnetized material, however, the directions of magnetization of the different domains are w oriented that the net magnetization is zero.

When an external magnetic field Is applied, the resultant magnetization of the specimen Increases. This is achieved in either of two ways: Either a domain that is favourably oriented grows in size at the expenses of a less favourably oriented domain, or the direction of magnetization of an entire domain changes and aligns along the external magnetic field.

When a weak nagnetic field is applied, favour ably oriented domains grow in size by domain boundary displacement, from figure (b). In strong fields, the domains change their magnetization by domain rotation, from figures. 11.9 (c) and (d). If domains reach perfect alignment, as in from figure 11.9 (d),, the domains may merge into one large domain.

After the external field is removed, it may be energetically favourable for a domain’s direction of magnetizaLion to persist. Then, the specimen has a permanent magnetic dipole moment. This phonomenon, called magnetic remanance, Is the basis of the existence of permanent magnets.

Question 45.
State any four properties of a ferromagnetic material.
Answer:
(1) Each atom of a ferromagnetic material possesses a magnetic moment. A ferromagnetic material is composed of domains each of which is spontaneously magnetized to saturation. In an unmagnetized sample, the magnetic moments of different domains are so oriented that the macroscopic magnetization of the sample is zero.

(2) When a ferromagnetic material is placed in a strong external magnetic field, the domains whose magnetic moments are favourably oriented relative to the external field grow in size such that the sample acquires a nonzero macroscopic magnetization.
(3) A ferromagnetic material retains some magnetization even after it is removed from an external magnetic field. This magnetic retentivity is the basis ‘ of the existence of permanent magnets.

(4) A ferromagnetic material is strongly attracted by a magnet.
(5) A bar magnet, suspended in a uniform magnetic field, comes to rest with its magnetic dipole moment in the direction of the external field.
(6) When placed in a nonuniform magnetic field, a ferromagnetic material is attracted towards the region of strong field.
(7) The magnetic susceptibility (χ_m) and relative permeability (μ_r) of a ferromagnetic material are very high.

Question 46.
Are the (i) diamagnetic materials (ii) paramagnetic materials (iii) ferromagnetic materials attracted or repelled by a magnet?
Answer:
When placed in a nonuniform magnetic field,

1. a diamagnetic material is weakly repelled from’ the region of strong field
2. a paramagnetic material is weakly attracted towards the region of strong field
3. a ferromagnetic material is strongly attracted towards the region of strong field.

Question 47.
Can there be a material which is nonmagnetic?
Answer:
All materials are magnetic. However, only ferromagnetics can have magnetic remanance and exhibit magnetic behaviour even in the absence of an external magnetic field. Diamagnetics and para-magnetics exhibit induced magnetism, i.e., they exhibit magnetic behaviour only in the presence of an external magnetic field.

[Note : A solid solution formed between paramagnetic or diamagnetic metals can be an exception to the general statement that all substances are magnetic. However, the zero value of the susceptibility will be retained only at one temperature, because the susceptibility of the paramagnetic constituent generally changes with temperature. A Cu-Ni alloy with 3.7 wt% Ni has zero susceptibility at room temperature.]

Question 48.
Which magnetic materials have
(i) relative permeability > 1
(ii) relative permeability <1?
Answer:
(i) Both paramagnetic and ferromagnetic materials have relative permeability (μ_r) greater than 1. μ_r is only slightly greater than 1 for a paramagnetic material. μ_r is very high for a ferromagnetic material and is a function of the magnetizing field (also called the magnetic field intensity).

(ii) μ_r is slightly less than 1 for a diamagnetic material.

Question 49.
Draw a graph showing the variation of magnetic susceptibility of a ferromagnetic material with temperature.
Answer:
The magnetic susceptibility of a ferromagnetic material above Curie temperature T_c is given by χ_m = \(\frac{C}{T-T_{\mathrm{c}}}\), where C is the Curie constant. Below figure shows the variation of magnetic susceptibility of a ferromagnetic material with temperature.

[Note : The relation, χ_m = \(\frac{C}{T-T_{\mathrm{c}}}\), is known as Curie-Weiss law. A plot of reciprocal susceptibility versus temperature for a ferromagnetic material is a straight line intercepting the temperature axis at T. At T = T, the susceptibility diverges which implies that one may have a nonzero magnetization in a zero applied field. This exactly corresponds to the definition of the Curie temperature, being the upper limit for having a spontaneous magnetization.].

Question 50.
Distinguish between a diamagnetic material and a paramagnetic material.
Answer:

Diamagnetic material	Paramagnetic material
1. A material whose atoms/ 1. molecules do not possess a net magnetic moment in the absence of an external magnetic field is called a diamagnetic material.	1. A material whose atoms possess net magnetic mo­ments that are all randomly directed in the absence of an external magnetic field is called a paramagnetic material.
2. When placed in a nonuni­form magnetic field, a diamagnetic material is weakly repelled from the region of strong field.	2. When placed in a nonuni­form magnetic field, a para­magnetic material is weakly attracted towards the region of strong field.
3.  In an external magnetic j field, it gets weakly magnetized in the direction oppo­site to that of the field.	3.  In an external magnetic field, it gets weakly magne­tized in the same direction as that of the field.
4.  When a rod of a diamagnetic material is suspended in a uniform magnetic field, it comes to rest with its length perpendicular to the direc­tion of the field.	4.  When a rod of a paramag­netic material is suspended in a uniform magnetic field, it comes to rest with its length parallel to the direc­tion of the field.
5.  χm is small, negative and very nearly temperature independent.	5.  χm is small, positive and varies inversely with abso­lute temperature T
6.  μr is slightly less than 1 and very nearly temperature! independent.	6. μr is slightly greater than 1 and decreases with increas­ing temperature.

Question 51.
Distinguish between a paramagnetic material and a ferromagnetic material.
Answer:

A paramagnetic material	A ferromagnetic material
1. The permanent atomic mag­netic moments of a para­magnetic material are all randomly oriented so that, in the absence of an external magnetizing field, the material is unmagnetized.	1. The permanent atomic mag­netic moments of a ferro­magnetic material interact strongly through exchange interaction forming domains which are spontaneously magnetized to saturation
2. It gets weakly magnetized when placed in an external magnetic field.	2. In an unmagnetized ma­terial, the directions of mag­netization of the different domains are so oriented that the net magnetization is zero.
3. Its magnetization becomes zero when the external magnetic field is removed.	3. It retains some magnetiz­ation even after the external magnetic field is removed.
4. Its magnetization in a given external magnetic field increases with decreasing temperature. The magne­tization reaches saturation only if the external field is very large and the tempera­ture is very low.	4. It retains its domain struc­ture only up to a characte­ristic Curie temperature, above which it becomes paramagnetic.
5. χm is small and positive. μr is slightly greater than 1.	5. χm and μr are positive and very high.

Question 52.
Distinguish between a ferromagnetic material and a diamagnetic material.
Answer:

A ferromagnetic material	A diamagnetic material
1. A material whose atoms/molecules possess permanent  magnetic moments which interact strongly through exchange interaction to form magnetic domains, each of which is magnetized to saturation, is called a ferromagnetic material.	1. A material whose atoms/molecules do not possess a net magnetic moment in the absence of an external magnetic field is called a diamagnetic material.
2. It gets strongly magnetized in the direction of the field even when placed in a weak magnetic field	2. In an external magnetic field, it gets weakly magnetized in the direction opposite to the field.
3. It retains some magnetiz­ation even after the external magnetic field is removed.	3. Its magnetization becomes zero when the external magnetic field is removed.
4. When placed in a nonuni­form magnetic field, a ferromagnetic material is strongly attracted towards the region of strong field.	4. When placed in a nonuni­form magnetic field, a dia­magnetic material is weakly repelled from the region of strong field.
5. χm and μr are positive and very high.	5. χm is small and negative. μr is slightly less than 1.

53. Solve the following
Question 1.
(5) A toroid (Rowland ring) of mean radius 16 cm has 1000 turns of wire closely wound on a ferromagnetic core of relative permeability 400. What is the magnetic induction B within the core for a magnetizing current of 1 A?
Solution:
Data: r = 16 × 10^-2 m, N = 1000, μ_r = 400, I = 1 A,
μ₀ = 4π × 10^-7 T∙m/A
Magnetic field strength, H = nI = \(\frac{N I}{2 \pi r}\)
∴ The magnetic induction within the core,
B = μ₀μ_rH = μ₀μ_r \(\frac{N I}{2 \pi r}\)
= (4π × 10^-7) (400) \(\frac{1000 \times 1}{2 \pi \times 16 \times 10^{-2}}\)
= \(\frac{8 \times 10^{-2}}{16 \times 10^{-2}}\) = 0.5 T

Question 2.
Find the percentage increase in the magnetic induction B when the space within a current-carrying toroid is filled with aluminium. The magnetic susceptibility of aluminium is 2.1 × 10^-5.
Solution:
Data: χ_m = 2.1 × 10^-5 (positive, since aluminium is paramagnetic)
The percentage increase in the magnetic induction B within the toroid due to the aluminium core

Question 3.
The permeability of a substance at temperature 300 K is 6.284 × 10^-3 SI unit. At what temperature will the susceptibility of that substance increase to 9.998 × 10³?
Solution:
Data: T₁ = 300 K, μ₁ = 6.284 × 10^-3, χ_m2 = 9.998 × 10³, μ₀ = 4π × 10^-7 T∙m/A
Permeability, μ = μ₀(1 + χ_m)
∴ The magnetic susceptibility at temperature T₁,

∴ The required temperature,
T₂ = \(\frac{\chi_{\mathrm{m} 1}}{\chi_{\mathrm{m} 2}}\) × T₁ = \(\frac{4999}{9998}\) × 300 = 150 K

Question 54.
Explain magnetic hysteresis in a ferromagnetic material.
OR
Explain the behaviour of a ferromagnetic material in an external magnetic field with the help of a magnetic hysteresis cycle.
OR
Explain the behaviour of a ferromagnetic material through one cycle of magnetization.
Answer:
A large scale consequence of the magnetic behaviour of a ferromagnetic under different applied magnetic fields can be observed by placing an unmagnetized rod of the material inside a solenoid. A current through the coil establishes the magnetizing field H, which we take as the independent variable. By measuring the voltage induced in a test coil wound alongside, we can determine changes in flux Φ, and hence changes in B inside the rod. B is measured in tesla while H is measured in ampere per metre. Knowing B and H, we can always compute magnetization M. It is however usual to plot B rather than M, as a function of H.

Typical B-H plot during the following operations :
(1) Starting with an unmagnetized rod, B = 0 and H = 0 (Point O), H is increased until it has the value corresponding to point a : Here the rod has reached its saturation magnetization and B remains constant even with further increase in H. The lower part of the curve Oa is governed by domain growth while the upper flattening part is governed by domain rotation.

(2) Reduce H to zero (point b) : the curve does not retrace itself, as shown by the curve ab. This irre-versibility is called hysteresis. It is largely due to the domain boundary movements being partially irreversible. If the current is simply switched off at this point, the rod will have a residual magnetization as indicated by the non-zero value of B, called retentivity or remanence, for H = 0. Essentially now B = μ₀M, i.e., the rod has acquired a permanent magnetization.

(3) Reverse H and increase it in magnitude until it has the value corresponding to point c : Here B is zero. The corresponding reverse magnetizing field H is called coercivity.

(4) Increase H in reverse direction until saturation magnetization is reached (point d).
(5) Reduce H to zero again (point e).
(6) Reverse the current once more until point a is reached again.
The process of taking the magnetic material once through the hysteresis loop abcdefa is called hysteresis cycle.

Question 55.
What is magnetic hysteresis ?
Explain it on the basis of magnetic domains.
Answer:
Magnetic hysteresis is a phenomenon shown by ferromagnetic materials in which the magnetic flux density through a material depends on the applied magnetizing field as well as the previous state of magnetization. Due to retention of its memory of previous state of magnetization, the flux density B lags behind (does not remain in step) with magnetizing field H. This delay in the change of its magnetization M (or equivalently, B) in response to a change in H is called hysteresis.

Hysteresis can be understood through the concept of magnetic domains. Domain boundary displacements and domain rotations are not totally reversible. When the applied magnetizing field H is increased and then decreased back to its initial value, the domains do not return completely to their original configuration but retain some memory or history of their previous alignment.

Question 56.
Define :
(1) retentivity
(2) coercivity.
Answer:

1. Retentivity : The residual magnetic flux density or magnetization in a magnetic material when the magnetizing field intensity is reduced to zero is called retentivity or remanence.
2. Coercivity: The reversed magnetizing field strength required to reduce the remanent magnetic flux density or magnetization in a magnetic material from its remanent value to zero, i.e., to demagnetize the magnetic material completely, is

Question 57.
What is a ‘soft’ magnetic material?
OR
What is meant by a ‘soft’ iron?
Answer:
A soft magnetic material, usually iron-based, has high permeability, low retentivity and low coercivity. In other words, it does not have appreciable hysteresis, i.e., its hysteresis loop is very narrow. Such a material magnetizes and demagnetizes more easily, by small external fields.

Question 58.
What is an electromagnet?
Answer:
An electromagnet is an insulated coil wrapped around a ferromagnetic core (usually a soft iron) of high permeability and low retentivity. When a current is passed through the coil-a solenoid or toroid, the ferromagnetic material magnifies the magnetic field of the coil, the magnification factor being the relative permeability (μ / μ₀) of the core. The magnetic field of the current-carrying coil magnetizes the core material, thereby producing a much larger field than the coil would produce by itself.

Question 59.
What is a magnetically hard material?
Answer:
A ‘hard’ magnetic material, such as ALNICO (an alloy of aluminium, nickel and cobalt), has high permeability, high retentivity and very high coercivity- of the orders of 1 tesla and 10⁴ ampere per metre, respectively. In other words, it has a large zero-field magnetization, and large reverse field needed to demagnetize. Its hysteresis loop is very broad. Such a material can be made into a permanent magnet, that is, its magnetization will persist indefinitely if it is subsequently exposed only to weak magnetic fields.

Question 60.
Distinguish between an electromagnet and a permanent magnet.
Answer:

Electromagnet	Permanent magnet
1. Magnetic field of an electromagnet is retained till there is passage of electric current through the solenoid or coil.	1. Magnetic field is retained for a long period of time.
2. current through the solenoid and the number of turns per unit length of the solenoid.	2. The strength of a permanent magnet depends upon the nature of the material of which it is made.
3. For a good electromagnet, the retentivity and coerciv­ity of the material should be low.	3. For a good permanent mag­ net, the retentivity and co­ercivity of the material should be high.
4. It is usually made of soft ferromagnetic material.	4. It is usually made of hard ferromagnetic material.
5. Poles of an electromagnet can be reversed by revers­ing the direction of electric current.	5. Poles of a permanent mag­net cannot be reversed.

Question 61.
Explain magnetic shielding.
Answer:
The use of a shell or box of magnetic material of high permeability to protect sensitive instruments from stray magnetic fields is called magnetic shielding. The permeability of the material being many order of magnitudes greater than air, the magnetic field lines pass through the shell. The saturation v induction B of the material must be greater than the external field to be shielded. from below figure shows the cross section of a cylindrical or spherical shield in which the central region is the field-free region.

Magnetic shielding is useful for a wide range of applications, from small components inside a set of speakers to large magnetically shielded rooms.

[Note : The most widely used alloy for magnetic shielding purposes is the patented MuMetal^®. Its composition of 80% nickel, 4.5% molybdenum and balance iron gives it high permeability.] .

Multiple Choice Questions

Question 1.
A rectangular bar magnet-with sides l. b and w – h.is mass n and magnetic moment M. It is free to rotate about a vertical axis through its centre of mass such that its faces of area l × b are honzoritaL The period T of angular oscillations of this magnet in a uniform magnetic field B is given by

Answer:
(B) \(\sqrt{\frac{\pi^{2} m\left(l^{2}+b^{2}\right)}{3 M B}}\)

Question 2.
A magnet of moment \(\vec{m}\) is placed in a uniform magnetic field \(\vec{B}\). Which of the following graphs correctly depicts the variation of its orientation energy U_θ i.e., its potential energy due to its orientation θ with \(\vec{B}\)?

Answer:
(D)

Question 3.
The magnetic dipole moment has the dimensions of current
(A) \(\frac{\text { current }}{\text { area }}\)
(B) current × area
(C) \(\frac{\text { area }}{\text { current }}\)
(D) current × length.
Answer:
(B) current × area

Question 4.
The dimensions of magnetic dipole moment are
(A) [L²I]
(B) [LI]
(C) [L^-1I]
(D) [L^-2I]
Answer:
(A) [L²I]

Question 5.
The gyromagnetic ratio of an orbital electron is the ratio of its
(A) charge to mass
(B) magnetic moment to volume
(C) orbital magnetic moment to linear momentum
(D) orbital magnetic moment to orbital angular momentum.
Answer:
(D) orbital magnetic moment to orbital angular momentum.

Question 6.
If \(\vec{M}_{0}, \vec{L}_{0}\) and γ₀ are respectively the magnetic dipole moment, orbital angular momentum and gyrornagnetlc ratio of an orbital electron. Then.

Answer:
(A) \(\vec{M}_{0}=-\gamma_{0} \overrightarrow{L_{0}}\)

Question 7.
Bohr magneton is the magnetic dipole moment of
(A) an orbital electron in the ground state of a Bohr atom
(B) an orbital electron with orbital angular momenturn of \(\frac{h}{2 \pi}\)
(C) the order of 10^-23 J/T
(D) all of the above.
Answer:
(D) all of the above.

Question 8.
A bar magnet 10 cm long has cross-sectional area 2 cm² and magnetic dipole moment of 10 A∙m². The magnetization of its material, assumed to be uniform, is
(A) 5 × 10⁵ A/m
(B) 2 × 10⁵ A/m
(C) 0.5 A/m
(D) 0.2 A/m.
Answer:
(A) 5 × 10⁵ A/m

Question 9.
An iron rod of volume 10^-4 m³ acquires a magnetic moment of 25 A∙m² when placed inside a solenoid whose windings carry a current of 0.5 A. The magnetization of the rod (in A/m), assumed to be uniform, is
(A) 5 × 10^-5
(B) 2.5 × 10^-3
(C) 12.5
(D) 2.5 × 10⁵
Answer:
(D) 2.5 × 10⁵

Question 10.
The dimensions of magnetic intensity are
(A) [LI]
(B) [L²I]
(C) [L^-1I]
(D) [L^-2I].
Answer:
(C) [L^-1I]

Question 11.
The ratio of magnetization to magnetic field induction (M/B) for both diamagnetic and paramagnetic materials is, in the usual notation,
(A) χ_m
(B) \(\frac{\chi_{\mathrm{m}}}{\mu_{0}}\)
(C) µ_r
(D) µ
Answer:
(B) \(\frac{\chi_{\mathrm{m}}}{\mu_{0}}\)

Question 12.
Magnetic susceptibility for vacuum (where there is no matter) is
(A) zero
(B) positive
(C) negative
(D) infinite.
Answer:
(A) zero

Question 13.
Magnetic susceptibility is positive and small for
(A) silver
(B) platinum
(C) mercury
(D) sodium chloride.
Answer:
(B) platinum

Question 14.
The magnetic induction within an ideal solenoid with air core is 5 mT. With an iron core of magnetic susceptibility 500, the induction within changes by
(A) 2.5 T
(B) 25 T
(C) 50 T
(D) 100 T.
Answer:
(A) 2.5 T

Question 15.
The magnetic moment of any atom in an isolated diamagnetic material is
(A) zero
(B) small
(C) large
(D) negative.
Answer:
(A) zero

Question 16.
Which of the following is diamagnetic?
(A) Dysprosium
(B) Gadolinium
(C) Magnesium
(D) Silver
Answer:
(D) Silver

Question 17.
The most exotic diamagnetic materials are
(A) the glasses
(B) the insulators
(C) the superconductors
(D) the semiconductors.
Answer:
(C) the superconductors

Question 18.
A thin, small gIas rod is suspended between the poles of a strung electromagnet. The rod
(A) is strongly repelled by the magnetic field.
(B) orients itself parallel to the magnetic field
(C) orients itself perpendicuLar to the magnetic field
(D) is weakly attracted by the magnetic field
Answer:
(C) orients itself perpendicuLar to the magnetic field

Question 19.
Which of the following is paramagnetic?
(A) Bismuth
(B) Copper
(C) Liq. oxygen
(D) Silver
Answer:
(C) Liq. oxygen

Question 20.
When a parainagnetic material Is placed in a uniform magnetic field,
(A) its atoms acquire a magnetic moment opposite to the magnetic field
(B) the atomic magnetic moments tend to align along the magnetizing field
(C) all the atomic magnetic moments align along the magnetizing field
(D) the sample temporarily becomes ferromagnetic.
Answer:
(B) the atomic magnetic moments tend to align along the magnetizing field

Question 21.
Curte constant is
(A) a universal constant
(B) constant for a given paramagnetic material
(C) constant for a given diamagnetic material
(D) inversely proportional to the absolute temperature
Answer:
(B) constant for a given paramagnetic material

Question 22.
The rare-earth element, gadolinium, is
(A) diamagnetic
(B) paramagnetic
(C) ferromagnetic
(D) nonmagnetic.
Answer:
(C) ferromagnetic

Question 23.
Curie’s law is valid for
(A) diamagnelics
(B) paramagnetics
(C) ferromagnetics
(D) all materials,
Answer:
(B) paramagnetics

Question 24.
Magnetizing and demagnetizing a material that has hysteresis involves
(A) increase in the temperature of the material
(B) a terro-to-para phase change
(C) decrease in the temperature of the material
(D) none of the above.
Answer:
(A) increase in the temperature of the material

Question 25.
The length of a bar magnet is large compared to its width and breadth. The time period of its angular oscillation in a vibration magnetometer is 2s. The magnet is cut along its length into two equal parts and the two parts are then mounted together in the magnetometer with their like poles together. The time period of this combination will be
(A) 2 s
(B) 1 s
(C) \(\frac{1}{2}\) s
(D) \(\frac{1}{\sqrt{2}}\) s
Answer:
(B) 1 s

Balbharti Maharashtra State Board 12th Physics Important Questions Chapter 10 Magnetic Fields due to Electric Current Important Questions and Answers.

Maharashtra State Board 12th Physics Important Questions Chapter 10 Magnetic Fields due to Electric Current

Question 1.
What is the magnetic effect of electric current?
Answer:
An electric current produces a magnetic field around it. This phenomenon is known as the magnetic effect of electric current. It was discovered by Hans Christian Oersted.

It was Ampere who first speculated that all magnetic effects are attributable to electric charges in motion (electric current). It takes a moving electric charge to produce a magnetic field, and it takes another moving electric charge to “feel” a magnetic field.

[Note : Hans Christian Oersted (1777-1851), Danish physicist, discovered electromagnetism in 1820. The oersted, the CGS unit of magnetic field strength, is named after him.]

Question 2.
Describe the magnetic field near a current in a long, straight wire. State the expression for the magnetic induction near a straight infinitely long current-carrying wire.
Answer:
Suppose a point P is at a distance a from a straight, infinitely long, wire carrying a current I, as shown in figure.

It can be shown that the magnitude of the total magnetic induction at P is given by the expression
B = \(\left(\frac{\mu_{0}}{4 \pi}\right) \frac{2 I}{a}\)
That is, the magnitude B is inversely proportional to the distance from the wire. Because of the axial symmetry about the straight wire, the magnetic induction has the same magnitude B at all points on a circle in a transverse plane and centred on the conductor; the direction of \(\vec{B}\) is everywhere tangential to such a circle. Thus, the magnetic field lines around the current in the straight wire is a family of circles centred on the wire. The magnitude of the field B depends only on the current and the perpendicular distance a of the point from the wire.

Question 3.
Explain the use of right hand grip rule to give the direction of magnetic field in the vicinity of a straight current-carrying conductor.
OR
State the right hand rule for the direction of the magnetic field due to a straight current-carrying wire.
Answer:
Right hand [grip] rule : If a straight current-carrying wire is grasped by the right hand, so that the extended thumb points in the direction of the current, the direction of the magnetic induction is the same as the direction of the fingers which are curled around the wire.

Question 4.
State the factors which the magnetic force on a charge depends upon. Hence state the expression for the Lorentz force on a charge due to an electric field as well as a magnetic field.
Hence discuss the magnetic force on a charged particle which is
(i) moving parallel to the magnetic field
(ii) stationary.
Answer:
A charge q moving with a velocity \(\vec{v}\) through a magnetic field of induction \(\vec{B}\) experiences a magnetic force perpendicular both to \(\vec{B}\) and \(\vec{v}\) . Experimental observations show that the magnitude of the force is proportional to the magnitude of \(\vec{B}\), the speed of the particle, the charge q and the sine of the angle θ between \(\vec{v}\) and \(\vec{B}\). That is, the magnetic force, F_m = qvB sin θ
∴ F_m = q(\(\vec{v}\) × \(\vec{B}\))
Therefore, at every instant \(\vec{F}_{\mathrm{m}}\) acts in a direction perpendicular to the plane of \(\vec{v}\) and \(\vec{B}\).

If the moving charge is negative, the direction of the force \(\vec{F}_{\mathrm{m}}\) acting on it is opposite to that given by the right-handed screw rule for the cross-product \(\vec{v}\) × \(\vec{B}\).

If the charged particle moves through a region of space where both electric and magnetic fields are present, both fields exert forces on the particle.

The force due to the electric field \(\vec{E}\) is \(\vec{F}_{\mathrm{e}}=q \vec{E}\).

The total force on a moving charge in electric and magnetic fields is called the Lorentz force :
\(\vec{F}=\vec{F}_{\mathrm{e}}+\vec{F}_{\mathrm{m}}=q(\vec{E}+\vec{v} \times \vec{B})\)
Special cases :
(i) \(\vec{v}\) is parallel or antiparallel to \(\vec{B}\): In this case, F_m = qvB sin 0° = 0. That is, the magnetic force on the charge is zero.
(ii) The charge is stationary (v = 0) : In this case, even if q ≠ 0 and B ≠ 0, F_m = q(0)B sin θ = 0. That is, the magnetic force on a stationary charge is zero.

Question 5.
Explain why the magnetic force on a charged particle cannot change the linear speed and the kinetic energy of the particle.
OR
One implication of the Lorentz force law is that magnetic force does no work. Justify.
Answer:
The magnetic force on a particle carrying a charge q and moving with a velocity \(\vec{v}\) in a magnetic field of induction \(\vec{B}\) is \(\vec{F}_{\mathrm{m}}=q \vec{v} \times \vec{B}\). At every instant, \(\vec{F}_{\mathrm{m}}\) is perpendicular to the linear velocity \(\vec{v}\), and \(\vec{B}\). Therefore, a non-zero magnetic force may change the direction of the velocity and the dot product

But \(\vec{F}_{\mathrm{m}} \cdot \vec{v}\) is the power, i.e., the time rate of doing work. Hence, the work done by the magnetic force in every short displacement of the particle is zero. The work done by a force produces a change in kinetic energy. Zero work means no change in kinetic energy. Thus, although the magnetic force changes the direction of the velocity \(\vec{v}\), it cannot change the linear speed and the kinetic energy of the particle.

Question 6.
Define the SI unit of magnetic induction from Lorentz force.
Answer:
The SI unit of magnetic induction is the tesla.
We can define the unit from the velocity-dependent part of the Lorentz force that acts on a charge in motion parallel to a magnetic field.

Definition : The magnitude of magnetic induction is said to be 1 tesla when a charge of 1 coulomb experiences a force of 1 N when it moves at 1 m/s in a magnetic field in a direction perpendicular to the direction of the field.

1 tesla (T) = 1 N’s/Gm = 1 N/A-m.
[Notes : (1) Since the ampere and not the coulomb is the fundamental unit, the tesla is defined from the expression for the force on a current-carrying conductor in a magnetic field when placed perpendicular to the direction of the field (see Unit 10.5, Q. 29): The magnitude of magnetic induction is said to be 1 tesla when a conductor of length 1 metre and carrying a current of 1 ampere experiences a force oflN when it is placed with its length perpendicular to the direction of the magnetic field. (2) The unit is named after Nikola Tesla (1870 -1943), Croatia-born US electrical engineer, inventor of the AC induction motor.]

Question 7.
Name a non-SI unit of magnetic induction. State its relation to the SI unit of magnetic induction.
Answer:
A CGS unit of magnetic induction of historical interest is the gauss, symbol G. However, since the magnetic flux and the magnetic flux density (magnetic induction, B) are defined by similar equations in the CGS system and the SI, this non-SI unit is accepted for use with SI.

1 G = 10^-4 T

[Note : The unit gauss is named after Karl Friedrich Gauss (1777 -1855), German mathematician, who strongly promoted in 1832 the use of the French decimal or metric system, with the metre and the kilogram and the astronomical second, as a coherent system of units for physical sciences. Gauss was the first to make absolute measurements of the Earth’s magnetic field in terms of a decimal system based on the three mechanical units millimetre, gram and second for, respectively, the quantities length, mass and time.]

Question 8.
Explain cyclotron motion and cyclotron formula.
Answer:
Suppose a particle of mass m and charge q enters a region of uniform magnetic field of induction \(\vec{B}\). In below figure, \(\vec{B}\) points into the page. The magnetic force \(\vec{F}_{\mathrm{m}}\) on the particle is always perpendicular to the velocity of the particle, \(\vec{v}\). Assuming the charged particle started moving in a plane perpendicular to \(\vec{B}\), its motion in the magnetic field is a uniform circular motion, with the magnetic force providing the centripetal acceleration.

where p = mv is the linear momentum of the particle. Equation (1) is known as the cyclotron formula because it describes the motion of a particle in a cyclotron-the first of the modern particle accelerators.

Question 9.
Explain the condition under which a charged particle will travel through a uniform magnetic field in a helical path.
OR
Describe the general motion of charged particle in a uniform magnetic field.
Answer:
Suppose a particle of mass m and charge q starts in a region of uniform magnetic field of induction \(\vec{B}\) with a velocity \(\vec{v}\) which has a non-zero component v_|| in the direction of \(\vec{B}\), From below figure. The magnetic force \(\vec{F}_{\mathrm{m}}\) on the particle is always perpendicular to \(\vec{v}\) and provides a centripetal acceleration such that

The parallel component of the motion \(\vec{v}_{\|}\) is unaffected by the magnetic field, so that the motion of the particle is a composite motion: an UCM with speed v_⊥ -the speed perpendicular to \(\vec{B}\) and a translation with a constant speed v_||. Therefore, the particle moves in a helix. Thus, the perpendicular component v_⊥ determines the radius of the helix while the parallel component v_|| determines the pitch x of the helix, i.e., the distance between adjacent turns. x = v_||/ T.

[Notes : (1) At non-relativistic speeds (v much less than the speed of light), the period T is independent of the speed of the particle. For all particles with the same charge-to-mass ratio (q/ m), faster particles move in larger circles than the slower ones, but all take the same time T to complete one revolution. (2) Looking in the direction of \(\vec{B}\), a positive charge always revolves anticlockwise, and a negative charge always clockwise.]

Question 10.
State under what conditions will a charged particle moving through a uniform magnetic field travel in
(i) a straight line
(ii) a circular path
(iii) a helical path.
Ans.
(i) A charged particle travels undeviated through a magnetic field \(\vec{B}\), if its velocity \(\vec{v}\) is parallel or antiparallel to \(\vec{B}\). In this case, the magnetic force on the charge is zero.
(ii) A charged particle travels in a circular path within a magnetic field \(\vec{B}\), if its velocity \(\vec{v}\) is perpendicular to \(\vec{B}\).
(iii) A charged particle travels in a helical path through a magnetic field \(\vec{B}\), if its velocity \(\vec{v}\) is inclined at an angle θ to \(\vec{B}\), 0 < θ < 90°. In this case, the component of \(\vec{v}\) parallel to \(\vec{B}\) is unaffected by the magnetic field. The radius and pitch of the helix are determined respectively by the perpendicular and parallel components of \(\vec{v}\).

Question 11.
A particle of charge q enters a region of uniform magnetic field \(\vec{B}\) (pointing into the page). The field deflects the particle a distance d above the original line of flight, as shown in below figure. Is the charge positive or negative ? Find the momentum of the particle in terms of a, d, B and q.

Answer:
Since \(\vec{v} \times \vec{B}\) points upward, and that is also the direction of the magnetic force \(\vec{F}_{\mathrm{m}}\), q must be positive.
Using the Pythagorean theorem to find R in terms of a and d,

[Note: A similar solved question in the textbook, Example 10.2, is wrong. The acceleration qvB/m is the centripetal acceleration whose magnitude remains constant but direction changes continuously. Hence, it cannot be used in a kinematical equation to calculate s. The final expression for p arrived at in the textbook is therefore not correct.]

Question 12.
An ion of mass re and charge q is accelerated from rest through a potential difference V and enters a region of uniform magnetic field of induction B. Within the region, the ion moves in a semidrcle and strikes a photographic plate which lies along the diameter of the semicircle at a distance D from the point of entry. Show that the mass of the ion is given by m = \(k \frac{q D^{2} B^{2}}{V}\)
Answer:
Consider positive ions of charge q and mass m accelerated from rest to a speed r’ by an accelerating potential V. Then, the kinetic energy of the ions is

On entering a region of uniform magnetic field of induction B (see above figure for reference), the Ions travel in a semicircular path in a plane normal to the field with a radius
R = \(\frac{m v}{q B}\)
∴ m²v² = q²B²R² ………….. (2)
From Eq. (1) m²v² = 2qVm ………….. (3)
Equating the right hand sides of Eqs. (2) and (3),
2qVm = q²B²R²
∴ m = \(\frac{q B^{2} R^{2}}{2 V}\)
Since R = \(\frac{D}{2}\),
m = \(\frac{q B^{2} D^{2}}{8 V}\) = k\(\frac{q B^{2} D^{2}}{V}\)
where k = \(\frac{1}{8}\), Equation (4) Is the required expression.

[Note : This is especially the working of a mass spectrometer which Is used to measure the mass of an ion. A mass spectrometer is so sensitive it is used to measure isotopic masses. If the isotopes of an element carry the same charge, they acquire the same energy when accelerator through the same pd. But within the magnetic field, they travel in different semicircles depending on their masses and strike a detector or photographic plate at different D.]

Question 13.
A particle of charge q and momentum p enters a region of uniform magnetic field B travelling at right angles to the field, and is deflected through a right angle as shown. Obtain an expression for the length of the particle’s path In terms of q, p and B.

Answer:
A particle of charge q and momentum p enters a region of uniform magnetic field B travelling at right angles to the field. Within the field, the particle travels in a circular path of radius
R = \(\frac{p}{q B}\) ………. (1)
From the diagram, the length s of the particle’s path in the field is one quarter of the circumference,
∴ s = \(\frac{2 \pi R}{4}=\frac{\pi}{2} \frac{p}{q B}\) ………….. (2)
This is the required expression.

Question 14.
What is a cyclotron? State its principle of working.
Answer:
A cyclotron is a cyclic magnetic resonance accelerator in which an alternating potential difference of a few kV is used to accelerate light positive ions such as protons, deuterons, α-particles, etc., but not electrons, to very high energies of the order of a few MeV. It was developed by E. O. Lawrence and M. S. Livingston in 1932.

Principle: The cyclotron employs the principle of synchronous acceleration to accelerate charged particles which describe a spiral path at right angles to a constant magnetic field and make multiple passes through the same alternating p.d., whose frequency is the same as the frequency of revolution of the particles. .

Question 15.
Describe the construction of the cyclotron with a neat labelled diagram.
Answer:
Construction of the cyclotron: Two hollow D-shaped chambers that are open at their straight edges form the electrodes. They are called the dees. The dees are separated by a small gap, as shown in below figure, and a high-frequency (10⁶ Hz to 10⁷ Hz) alternating p.d. (of the order of 10⁴ V to 10⁵ V) is applied between them. The whole system is placed in an evacuated chamber between the poles of a large and strong electromagnet (B ≡ ≅ 1 T to 2T).

The ions to be accelerated are produced in an ion source; a hydrogen tube gives protons, heavy hydrogen or deuterium gives deuterons while helium gives x-particles, etc. The positive ions are injected near the centre and are accelerated each time they

cross the gap between the does. At the edge of one of the does, an electrostatic deflector deflects the spiralling particles out of the system to strike a target.

Question 16.
Explain the working of the cyclotron with a neat labelled diagram.
Answer:
The dees of the cyclotron are separated by a small gap and a high-frequency alternating p.d. is applied between them. Light positive ions are injected into the system near the centre.

Suppose a positive ion of charge q and mass m is injected when D₁ is positive and D₂ is negative. The positive ion will accelerate towards D₂. Inside the dees there is no electric field. Hence, inside D₂ it has a constant speed v. The magnetic force of magnitude qvB makes it move in a semicircular path through D₂. The radius r of its orbit is given by equating the centripetal force to the magnetic force.

Let t be the time spent by the ion to describe the semicircular path.

If t is also half the period of oscillation T of the alternating p.d., the ion will be in resonance with the electric field in the gap. That is, the ion will emerge from D₂ at the instant D₁ becomes negative and will be accelerated towards D₁. As the ion gains speed in the gap, its path in D₁ has greater radius. This process repeats after every half cycle of the alternating p.d. and the ion is accelerated each time it crosses the gap between the dees.

The radius of the path of the charged particles increases proportionately with their speed, the period of revolution remains constant.

After a large number of revolutions, the ion reaches the edge of the system where a negatively charged electrostatic deflector plate deflects it out of the system towards the target.

Question 17.
State the functions of the electric and magnetic fields in a cyclotron.
Answer:
The function of the electric field in the gap between the dees of a cyclotron is to accelerate the positively charged particles while that of the magnetic field in the dees is to deflect the particles in semicircular paths so that they return to the gap in a fixed time interval to reuse the alternating electric field.

Question 18.
Show that for a given positive ion species in a cyclotron,
(i) the radius of their circular path inside a dee is directly proportional to their speed
(ii) the time spent in a dee (or the cyclotron frequency or the magnetic resonance frequency) is independent of the radius of their path and speed
(iii) the maximum ion energy obtainableis directly proportional to the square of the magnetic induction.
Answer:
Consider positive ions of charge q and mass m injected in a cyclotron. In the electric-field-free region inside a dee, the ions are acted upon only by the uniform magnetic field. Hence, inside a dee the ions travel in a semicircular path with a constant speed y, in a plane normal to the field. If B is the induction of the magnetic field, the magnetic force of magnitude qvB provides the centripetal force.
∴\(\frac{m v^{2}}{r}\) = qvB
∴ r = \(\frac{m v}{q B}\) …………. (1)
Thus, for given q, m and B,
r ∝ v
If t be the time spent in a dee by the ion to describe a semicircular path of radius r,
t = \(\frac{\pi r}{v}=\frac{\pi}{v} \times \frac{m v}{q B}\)
∴ t = \(\frac{\pi m}{q B}\) …………. (2)
Thus, t is independent of r and y, i.e., it takes the ions exactly the same time t to travel the semicircular paths inside the dees irrespective of the radius of the path and the speed of the ions so long a the mass in is constant. This is the critical characteristic of operation of the cyclotron.

The periodic time of an ion in its circular path is
T = 2t = \(\frac{2 \pi m}{q B}\) ………….. (3)

The frequency of revolution,
f = \(\frac{1}{T}=\frac{q B}{2 \pi m}\) …………. (4)
is called the cyclotron frequency or the magnetic resonance frequency. The frequency is independent of r and y for a given ion species and remains constant so long as the mass m is constant.

If R is the maximum radius of the path, the same as the radius of the dee, just before the ions are deflected out of the accelerator,
V_max = \(\frac{q B R}{m}\) …………. (5)
so that KE_max = \(\frac{q^{2} B^{2} R^{2}}{2 m}\) (in joule)
= \(\frac{q^{2} B^{2} R^{2}}{2 e m}\) (in eV) …………… (6)
Thus, for a given ion species and dees of given radius,
KE_max ∝ B²

Question 19.
What is meant by cyclotron frequency?
Answer:
The cyclotron frequency, or the magnetic resonance frequency, is the frequency of revolution of a charged particle of charge per unit mass \(\frac{q}{m}\) in a magnetic field of induction B inside a cyclotron. The cyclotron frequency, f = \(\frac{q B}{2 \pi m}\). The frequency of the alternating voltage applied to the dees of the cyclotron should be equal to the cyclotron frequency.

Question 20.
What is resonance condition in a cyclotron ?
OR
What should be the frequency of the alternating voltage applied between the dees of a cyclotron?
Answer:
The frequency of the alternating voltage between the dees of a cyclotron should be equal to the cyclotron frequency so that a positive ion exiting a dee always sees an accelerating potential difference to the other dee. This equality of the frequencies is called the resonance condition.

Question 21.
State any two limitations of a cyclotron.
Answer:
Limitations of a cyclotron :

1. It cannot be used to accelerate electrons. Because electrons have a very small mass, they quickly achieve relativistic speeds, i.e., speeds at which their mass increases significantly with increase in speed. Then they cannot remain synchronous with the alternating electric field between the dees.
2. For higher energies, with a given magnetic field strength, the exit radius and thus the dees must be large. It is difficult to produce a uniform magnetic field over a large area.
3. Even protons, deuterons, a-particles, etc., cannot be accelerated to very high energy, say of the order of 500 MeV, using a cyclotron with a fixed cyclotron frequency.
4. No particle accelerator can accelerate uncharged particles, such as neutrons.

Question 22.
Does the time spent by a charged particle inside a dee of a cyclotron depend upon its speed and the radius of its path ? Why ?
Answer:
The time spent by a charged particle to describe a semicircular path of radius r inside a dee of a cyclotron is independent of the radius of the path and the speed v so long as the mass m of the particle is constant. This is true for any charged particle of mass m and carrying a charge q, and is the critical characteristic of operation of the cyclotron.
r = \(\frac{m v}{q B}\)
So that the time spent in a dee,
t = \(\frac{\pi r}{v}=\frac{\pi}{v} \times \frac{m v}{q B}=\frac{\pi m}{q B}\)
which is independent of r and v.

Question 23.
The frequency of revolution of the charged particles in a cyclotron does not depend upon their speed. Why?
Answer:
The time t spent by a charged particle, of mass m and carrying a charge q, inside a dee of a cyclotron is independent of the radius of the path and the speed of the particle so long as m is constant. Then, the periodic time of the charged particle in its nearly circular path is T = 2t, and the frequency of revolution,
f = \(\frac{1}{T}=\frac{1}{2 t}=\frac{q B}{2 \pi m}\)
are also independent of the radius and the speed.

Question 24.
What are the factors on which the cyclotron frequency depends?
Answer:
The cyclotron frequency depends upon

1. the magnetic induction and
2. the specific charge (the ratio charge/mass) of the charged particles.

Question 25.
What are the factors on which the maximum kinetic energy acquired by a charged particle in the cyclotron depends?
Answer:
The maximum kinetic energy acquired by a charged particle in the cyclotron depends upon

1. the magnetic induction
2. the specific charge (the ratio charge/mass) of the charged particles and
3. the radius of the dees.

Question 26.
In a certain cyclotron, the cyclotron frequency for acceleration of protons is 10⁸ Hz. What will be its value if the magnetic induction is doubled ?
Answer:
As the cyclotron frequency is directly proportional to the magnetic induction, the new frequency will be 2 × 10⁸ Hz. .

27. Solve the following :

Question 1.
An alpha particle (carrying a positive charge q = 3.2 × 10^-19 C) enters a region of uniform magnetic field of induction \(\vec{B}=(0.5 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})\) T with a velocity \(\vec{v}=(4 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}})\) × 10² m/s. What is the force \(\vec{F}\) on the alpha particle?
Solution:

Question 2.
The magnetic field and electric field in a region in space are \(\vec{B}=B \hat{\mathrm{i}}\) and \(\vec{E}=E \hat{\mathrm{i}}\). A particle of charge q moves into the region with velocity \(\vec{v}=v \hat{\mathrm{j}}\). Find the magnitude and direction of the Lorentz force on the charged particle if q = 1 C,B = 1 T,E = 3 V/m and v = 4 m/s,
Solution:
Data : q = 1 C, B = 1 T, E = 3 V/m, v = 4 m/s
The Lorentz force on the charged particle is

Question 3.
An electron is accelerated from rest through 86 V and then enters a region of uniform magnetic induction of magnitude 1.5 T. What is the maximum value of the magnetic force the electron can experience?
[m_e = 9.11 × 10^-31 kg, e = 1.6 × 10^-19 C]
Solution:
Data : V= 86 V, B = 1.5 T, m_e = 9.11 × 10^-31 kg, e = 1.6 × 10^-19 C
Since the electron is accelerated from rest, the kinetic energy acquired by it is

The maximum value of the magnetic force on the electron is
F_m = evB
= (1.6 × 10^-19 C) (5.496 × 10⁶ m/s) (1.5 T)
= 1.319 × 10^-12 N or 1.319 pN

Question 4.
A cosmic ray proton enters the Earth’s magnetic field in a direction perpendicular to the field. If the speed of the proton is 2 × 10⁷ m/s and B = 1.6 × 10^-6 T, find the force exerted on the proton by the magnetic field. [Charge on a proton, e = 1.6 × 10^-19 C]
Solution :
Data : v = 2 × 10⁷ m/s, B = 1.6 × 10^-6 T, e = 1.6 × 10^-19 C
The magnetic force on the proton is
F_m = evB (∵ \(\vec{v} \perp \vec{B}\))
= (1.6 × 10^-19 C)(2 × 10⁷ m/s)(1.6 × 10^-6 T)
= 5.12 × 10^-18 N

[Note : The Earth’s magnetic field traps the charged particles in doughnut-shaped regions outside the atmosphere. These regions are called Van Allen radiation belts. Near the poles, charged particles from these belts enter the atmosphere and produce the awesome shimmering curtains of light called the aurora borealis (northern ! lights) and aurora australis (southern lights).]

Question 5.
A charged particle moves with velocity 3 × 10⁶ m/s at right angles to a uniform magnetic field of induction 0.005 T. Find the magnitude of the charge if the particle experiences a force of 2 × 10^-2 N.
Solution:
Data : v = 3 × 10⁶ m/s, B = 5 × 10^-3 T,
F_m = 2 × 10^-2 N
F_m = qvB (∵ \(\vec{v} \perp \vec{B}\))
∴ The charge on the particle,

Question 6.
An electron in a TV picture tube moves horizontally with a speed 2 × 10⁷ m/s. It is deflected upward by a horizontal magnetic field of induction 10^-3 T. Find the magnitude of the force acting on the electron due to the action of the magnetic field.
Solution:
Data : e = 1.6 × 10^-19 C, v = 2 × 10⁷ m/s,
B = 10^-3 T
The magnitude of the magnetic force,
F_m = evB (∵ \(\vec{v} \perp \vec{B}\))
= (1.6 × 10^-19 C)(2 × 10⁷ m/s) (10^-3 T)
= 3.2 × 10^-15 N = 3.2 fN

Question 7.
The first practical cyclotron developed by Lawrence and Livingston in 1932 had dees of radius 12 cm and produced protons of about 1 MeV energy.
Calculate (i) the applied magnetic induction
(ii) the frequency of the accelerating electric field.
[m^p = 1.67 × 10^-27 kg, q = 1.6 × 10^-19 C
Solution:
Data: R = 0.12 m, m^p = 1.67 × 10^-27 kg,
q = 1.6 × 10^-19 C,
KE_max = 1 MeV =(1 × 10⁶)(1.6 × 10^-19)
= 1.6 × 10^-13 J

Question 8.
In a cyclotron, for the same B and R, show that the maximum kinetic energy of α-particles is twice that of deuterons.
Solution:
Data: Let E_α and E_d denote the maximum kinetic energies of α-particIes (He⁺⁺ ions) and deuterons
(D⁺ ions). For an α-partic1e,
q = +2e and m ≅ 4m^p and for a deuteron,
q = +e and m ≅ 2m^p
where e is the elementary charge and m^p is the proton mass.
In a cyclotron, the maximum kinetic energy of the ions obtainable with dec radius R and magnetic field induction B is

Question 28.
Derive an expression for the magnetic force experienced by a straight current-carrying conductor placed in a uniform magnetic field.
Discuss the cases when the force is maximum and minimum.
State the expressions for the force experienced by a current-carrying
(i) conductor of arbitrary shape
(ii) closed circuit (conducting loop).
Answer:
Consider a straight current-carrying conductor placed in a region of uniform magnetic field of induction \(\vec{B}\) pointing out of the page, as shown in below figure by the evenly placed dots. Let the length of the conductor inside the field be l and the current in it be I.

In metallic conductors, electrons are the charge carriers. The direction of conventional current is, however, taken to be that of flow of positive charge which is opposite to the electron current.

Let dq be the positive charge passing through an element of the conductor of length dl in time dt.
\(\overrightarrow{d l}\) has the same direction as that of the current.
Then, I = \(\overrightarrow{d l}\)/dt ………….. (1)
and drift velocity,\(\overrightarrow{v_{\mathrm{d}}}\) = \(\overrightarrow{d l}\)/dt ……….. (2)
The magnetic force on the charge dq is

The charge dq is constrained to remain within the conductor. Hence, the conductor itself experiences this force. The force on the entire part of the conductor within the region of the magnetic field is

Case 1 : When the conductor is parallel to the magnetic field, \(\vec{l}\) is parallel or antiparallel to \(\vec{B}\) according as the current is in the direction of \(\vec{B}\) or opposite to it; then θ = 0° or θ = 180°, so that sin θ = 0. Hence, in either of these two cases, F = 0.

Case 2 : The maximum value of the force is F_max = IlB, when sin θ = 1, that is, when the conductor lies at right angles to \(\vec{B}\) (θ = 90°).
(i) For a current-carrying wire of arbitrary shape in a uniform field,
\(\vec{F}=\int \vec{f}_{\mathrm{m}}=I\left(\int \overrightarrow{d l}\right) \times \vec{B}\) …………. (5)
(ii) For a current-carrying conducting loop (closed circuit) in a uniform field,
\(\vec{F}=\int \vec{f}_{\mathrm{m}}=I(\oint \overrightarrow{d l} \times \vec{B})\) ………….. (6)
But for a closed loop of arbitrary shape, the integral is zero.
∴ \(\vec{F}\) = 0
[Notes : (1) While the direction of \(\vec{F}\) can be found from the cross product of \(\vec{l}\) and \(\vec{B}\), there is a handy rule due to Sir John Ambrose Fleming (1849-1945), British physicist and electrical engineer.

Fleming’s left hand rule : If the forefinger and the middle finger of the left hand are stretched out to point in the directions of the magnetic field and the current, respectively, then the outstretched thumb indicates the direction of the magnetic force on the current-carrying straight conductor, from below figure.]

(2) Equation (3) is usually used to define the unit of magnetic field induction, the tesla. See the note to Q. 6. (3) B cannot be taken out of the integral in Eq. (6).]

Question 29.
A straight conductor of length 0.5 m and carrying a current of 2 A is placed in a magnetic field of induction 2 Wb/m² at right angles to the length of the conductor. What is the magnetic force on the conductor?
Answer:
F = IlB
= (2A) (0.5m) (2 Wb/m²)
= 2 N is the force on the conductor.

Question 30.
A rectangular loop of wire hanging vertically with one end in a uniform magnetic field \(\vec{B}\), supports a small block of mass m. \(\vec{B}\) points into the page in the shaded region of below figure. For what current I in the loop would the magnetic force exactly balance the downward gravitational force?

Answer:
The magnetic force on the horizontal segment of the loop inside the field must be upward to balance the downward gravitational force. With \(\vec{B}\) pointing into the page, the current in that segment must be toward the right so that \(\vec{F}=\vec{L} \times \vec{B}\) on that segment points upward. That is, the current in the loop must be clockwise.
Taking L = a, F = IaB
∴ IaB = mg
∴ I = \(\frac{m g}{a B}\)
is the required expression for the current.

Question 31.
Explain with a neat labelled diagram how the magnetic forces on a current loop produce rotary motion as in an electric motor.
Answer:
Consider a current-carrying rectangular loop ABCD, within a uniform magnetic field \frac{m g}{a B}, from below figure. Lead wires and commutator are not shown for simplicity. The coil is free to rotate about a fixed axis. Suppose the sides AB and CD are perpendicular to the field direction.

The magnetic force on each segment act at the centre of mass of that segment. The direction of the force on each segment can be found using the right hand rule for the cross product or from Fleming’s left hand rule.

The magnetic forces on the short sides AD and CB are, in general, equal in magnitude, opposite in direction and have the same line of action along the rotation axis. Hence, these forces cancel out and does not produce any torque. The magnetic forces on the long sides AB and CD are also equal in magnitude and opposite in direction but their lines of action are different. Hence, these forces constitute a couple and tend to rotate the coil about the central axis.

A commutator (not shown) reverses the direction of the current through the loop every half-revolution so that the torque always acts in the same direction.

Question 32.
Derive an expression for the net torque on a rectangular current-carrying loop placed in a uniform magnetic field with its rotation axis
perpendicular to the field.
Answer:
Consider a rectangular loop ABCD of length l, breadth b and carrying a current I, placed in a uniform magnetic field of induction \(\vec{B}\) with its rotation axis perpendicular to \(\vec{B}\), from figure (a). To define the orientation of the loop in the magnetic field, we use a normal vector \(\hat{n}\) that is perpendicular to the plane of the loop. The direction of \(\hat{n}\) is given by a right hand rule: If the fingers of right hand are curled in the direction of current in the loop, the outstretched thumb is the direction of \(\hat{n}\). Suppose the normal vector \(\hat{n}\) of the loop makes an arbitrary angle with \(\vec{B}\), as shown in figure (b).

In the side view, Fig. 10.15 (b), the sides CD, DA, AB and BC have been labelled as 1, 2, 3 and 4, respectively. In this view, the current in side 1 (CD) is out of the page as shown by a ⊙ while that in side 3 (AB) is into the page shown by a ⊗.

For side 2 (AD) and side 4 (BC), the length of the conductor \(|\vec{L}|\) = b and the angle between \(\vec{L}\) and \(\vec{B}\) is (90° – θ). Hence, the forces on sides 2 and 4 are equal in magnitude :

F₂ = F₄ = IbB sin(90° – θ) = IbB cosθ

However, \(\vec{F}_{2}\) is directed out of the page while \(\vec{F}_{4}\) is into, and because their common line of action is through the centre of the loop, their net torque is zero. For side 1 (CD) and side 3 (AB), \(|\vec{L}|\) = I and \(\vec{L}\) is perpendicular to \(\vec{B}\). Hence, the forces \(\vec{F}_{1}\), and F\(\vec{F}_{3}\) have the same magnitude : F₁ = F₃ = IlB

But their lines of action being different, they constitute a couple.

Moment arm of the couple = b sin θ
∴ Torque exerted by the couple = force of the couple x moment arm of couple
∴ τ = (IlB)(b sin θ)
in the clockwise sense in figure (b). The torque tends to rotate the loop so as to align its normal vector h with the direction of the magnetic field.
∴ τ = I(lb)B sin θ = MB sinθ
where A = lb is the area of the loop. For a rectangular coil of N turns in place of a single-turn loop,
τ = NIAB sin θ
This is the required expression for the net torque. The torque has maximum magnitude for θ = 90°, that is when \(\hat{n}\) is perpendicular to \(\vec{B}\) or, in other words, the plane of the coil is parallel to the field.
τ_max = NIAB

Question 33.
Describe the construction of a suspended-type moving-coil galvanometer with a neat labelled diagram.
Answer:
A permanent fixed-magnet, suspended-type moving-coil galvanometer is shown in below figure. It consists of a coil of a large number of turns of fine insulated copper wire wound on a rectangular, nonconducting, non-magnetic frame. The coil is suspended between the cylindrically concave pole pieces of a horseshoe permanent magnet by a fine phosphor-bronze wire F from an adjustable screw- head. The other terminal of the coil is connected to a loosely-wound wire helix H. The coil swings freely around a cylindrical soft-iron core CS fitted between the pole pieces.

The suspension F and the helix H serve as the two current leads to the coil. The suspension fibre also provides the restoring torque when the coil is rotated from its normal position. The cylindrically concave pole pieces together with the soft-iron core make the magnetic field radial in the annular region in which the vertical sides of the coil move. The soft-iron core also concentrates the magnetic field (i.e., increases the magnetic induction) in the annular region.

The angle of deflection is observed with a beam of light reflected from a small mirror M fixed to the suspension fibre. The reflected beam is observed on a ground-glass scale arranged about a metre from the instrument, the light beam serving as a weightless pointer.

[Note : The diagram given in the textbook (Fig. 10.13) has serious errors, notably, the missing suspension fibre by which the coil is hung. The metal fibre (made of phosphor bronze), which bears the weight of the coil, is also the current lead to the coil and provides the restoring torque. The Tower suspension shown in the textbook is actually a loose wire which merely provides an exit lead to the current but does not exert any torque on the coil.]

Question 34.
State the principle of working of a moving- coil galvanometer (suspended-coil type).
Answer:
Principle : A current-carrying coil suspended in a magnetic field experiences a torque which rotates the plane of the coil and tends to maximize the magnetic flux through the coil.

The deflection of the coil in a moving-coil gal-vanometer is linearly related to the current through it and, therefore, can be used to measure current in terms of the deflection.

Question. 35.
With the help of neat diagrams, describe the working of a moving-coil galvanometer.
Answer:
Consider a rectangular coil-of length Z, breadth b and N turns – carrying a current I suspended in a uniform magnetic field of induction \(\vec{B}\).

The magnetic forces on the horizontal sides of the coil have the same line of action and do not exert any torque. The magnetic forces on the vertical sides constitute a couple and exert a deflecting torque. If the plane of the coil is parallel to \(\vec{B}\), the magnitude of the deflecting torque is maximum equal to
τ_d = NIAB …………… (1)
where A = lb is the area of each turn of the coil. This torque rotates the coil.

In a moving-coil galvanometer, the coil swings in a radial magnetic field produced by the combination of the cylindrically concave pole pieces and the soft-iron core. Hence, the plane of the coil is always parallel to the field lines, as shown in Fig. 10.18. Therefore, the deflecting torque is constant and maximum as given by Eq. (1).

The rotation of the coil twists the suspension fibre which exerts a restoring torque on the coil. The restoring torque is proportional to the angle of twist θ.
τ_r = Cθ …………… (2)
where C is the torque constant, i.e., torque per unit angle of twist. C depends on the dimensions and the elasticity of the suspension fibre.

The coil eventually comes to rest in the position where the restoring torque equals the deflecting torque in magnitude. Therefore, in the equilibrium position,
τ_r = τ_d

since N, A, B and C are constant. Thus, the deflection of the coil is directly proportional to the current in it.

Question 36.
A coil suspended freely in a radial magnetic field rotates through 30° when a current of 30 /(A is passed through it. Through what angle will it rotate if the current is doubled and the magnetic induction is halved?
Answer:
The angle of rotation, θ ∝ IB (in the usual notation)
As I₂B₂ = (2I₁)\(\left(\frac{B_{1}}{2}\right)\) = I₁B₁, the angle of rotation will be the same, i.e., 30°.

Question 37.
What is the advantage of a radial magnetic field in a moving-coil galvanometer and how is it produced?
Answer:

1. Advantage of radial magnetic field in a moving- coil galvanometer:
   • As the coil rotates, its plane is always parallel to the field. That way, the deflecting torque is always a maximum depending only on the current in the coil, but not on the position of the coil.
   • The restoring torque is proportional to the deflection so that a radial field makes the deflection proportional to the current. The instrument then has a linear scale, i.e., the divisions of the scale are evenly spaced. This makes it particularly straight forward to calibrate and to read.
2. Producing radial magnetic field :
   • The pole pieces of the permanent magnet are made cylindrically concave, concentric with the axis of the coil.
   • A soft iron cylinder is centred between the pole pieces so that it forms a narrow cylindrical gap in which the sides of the coil can move. Together, they produce a radial magnetic, field; that is, the magnetic lines of force in the gap are along radii to the central axis.

Question 38.
What will happen if the magnetic field in a moving-coil galvanometer is not radial?
Answer:
Suppose the magnetic field is uniform but not radial. Then, when the coil comes to rest after rotation through an angle θ, NIAB cos θ = Cθ (in usual notations).
∴ I ∝ \(\frac{\theta}{\cos \theta}\)
as N, A, B and C are constants in a particular case. Thus, the current is not directly proportional to the deflection. Hence, we cannot have a linear scale for measurement.

Question 39.
Explain the use of a soft iron core in a moving- coil galvanometer.
Answer:
In a moving-coil galvanometer, the pole pieces of the permanent magnet are made cylindrically concave, coaxial with the coil. A soft iron core is fixed centrally between the pole pieces so that it partly fills the space inside the coil and forms a narrow cylindrical gap in which the sides of the coil can move.
(1) The soft iron core, together with the concave pole pieces produces a radially uniform magnetic field, i.e., the magnetic lines of force in the gap are along radii to the central axis. This makes the deflection of the coil proportional to the current in it. The instrument then has a linear scale which is particularly straightforward to calibrate and to read.

(2) The permeability of iron being more than that of air, the magnetic lines of force pass through the soft iron core. By making the cylindrical gap as narrow as possible then increases the magnetic induction in the gap. This increases the deflecting torque on the coil and the sensitivity of the instrument.

Question 40.
Why does not the Earth’s magnetic field affect the working of a moving-coil galvanometer?
Answer:
The coil of a moving-coil galvanometer rotates in the magnetic field of a permanent magnet. The magnetic induction of the permanent magnet is many orders of magnitude (typically 10⁴ times) stronger than that of the Earth. Hence, the Earth’s magnetic field does not affect the working of the galvanometer.

Question 41.
Explain the magnetic dipole moment of a current loop. State its magnitude and direction.
Answer:
The responses of a current-carrying coil to an external magnetic field is identical to that of a magnetic dipole (or a bar magnet). Like a magnetic dipole, a current-carrying coil placed in a magnetic field \(\vec{B}\) experiences a torque. In that sense, the coil is said to be a magnetic dipole. To account for a torque τ on the coil due to the magnetic field, we assign a magnetic dipole moment \(\vec{\mu}\) to the coil, such that
\(\vec{\tau}=\vec{\mu} \times \vec{B}=N I \vec{A} \times \vec{B}\)

where \(\vec{\mu}=N I \vec{A}\). Here, N is the number of turns in the coil, I is the current through the coil and A is the area enclosed by each turn of the coil. The direction of \(\vec{\mu}\) is that of the area vector \(\vec{A}\), given by a right hand rule shown in below figure. If the fingers of right hand are curled in the direction of current in the loop, the outstretched thumb is the direction of \(\vec{A}\) and \(\vec{\mu}\). In magnitude, μ = NIA.
The torque tends to align \(\vec{\mu}\) along \(\vec{B}\).

Question 42.
State if the following statement is true : “The magnetic dipole moment of a current-carrying coil of given geometry is constant.” Justify your answer.
Answer:
The given statement is false.
Consider a coil of N turns, each of area A. If the current through the coil is I, the magnetic dipole moment of the coil is, in magnitude, μ = NIA. That is, μ ∝ I, for given N and A. Thus, for a coil of given geometry, its magnetic dipole moment varies with the current through it.

Question 43.
In analogy with an electric dipole, state an expression for the magnetic potential energy of an magnetic dipole in a uniform magnetic field. Discuss the orientations of the dipole moment for the maximum and minimum of potential energy.
Answer:
Consider an electric dipole of dipole moment \(\vec{p}\) placed in a uniform electric field \(\vec{p}\) making an angle Φ with \(\vec{p}\). The torque \(\vec{\tau}=\vec{p} \times \vec{E}\) tends to rotate the dipole and align it with \(\vec{E}\).

If the dipole was initially parallel to \(\vec{E}\), its potential energy is minimum. We arbitrarily assign U₀ = 0 to the minimum potential energy for this position. Then, at a position where \(\vec{p}\) makes an angle θ with \(\vec{E}\), the potential energy of the dipole is
U_θ = -pE cos θ = –\(\vec{p} \cdot \vec{E}\).
A current-carrying coil placed in a magnetic field \(\vec{B}\) experiences a torque,
\(\vec{\tau}=\vec{\mu} \times \vec{B}\) …………. (1)
where \(\vec{\mu}\) is the magnetic dipole moment of the coil. In analogy with an electric dipole, the potential energy of a magnetic dipole is
U_θ = – μB cos θ = – \(\vec{\mu} \cdot \vec{B}\) ……….. (2)
U_θ is also known as orientation energy.
A magnetic dipole has its lowest energy ( = – μB cos 0 = – μB) when its dipole moment is lined up with the magnetic field, and has its highest energy ( = – μBcos 180° = + μB) when is directed opposite the field.

44. Solve the following :
Question 1.
A straight current-carrying conductor 30 cm long carries a current of 5 A. It is placed in a uniform magnetic field of induction 0.2 T, with its length making an angle of 60° with the direction of the field. Find the force acting on the conductor.
Solution:
Data : l = 30 cm = 0.3 m, I = 5 A, B = 0.2 T, θ = 60°
The magnitude of the force on the conductor,
F = I\(|\vec{l} \times \vec{B}|\) = IlB sin θ
= (5 A) (0.3 m) (0.2 T) sin 60°
= 0.3 × 0.866 = 0.2598 N
The direction of the force is given by the cross product rule.

Question 2.
A conductor of length 25 cm is placed (i) parallel (ii) perpendicular (iii) inclined at an angle 30°, to a uniform magnetic field of induction 2 T. If 1 C of charge passes through it in 5 s, calculate the force experienced by the conductor in each case.
Solution:
Data : l = 25 cm = 0.25 m, θ₁ = 0°, θ₂ = 90°, θ₃ = 30°, B = 2 T, q = 1 C, f = 5 s
The current in the conductor,
I = \(\frac{q}{t}=\frac{1 \mathrm{C}}{5 \mathrm{~s}}\) = 0.2 A
The magnitude of the force on the conductor,
F = I\(|\vec{l} \times \vec{B}|\) = IlB sin θ
(i) θ₁ =0° A ∴ sin θ₁ = 0 ∴ F = 0 N
(ii) θ₂ = 9O° ∴ sin θ₂ = 1
∴ F = IlB = (0.2 A)(0.25 m)(2 T)
= 0.1 N
(iii) θ₃ = 30° ∴ sin θ₃ = 0.5
∴ F = IlB sin θ₃ = (0.1 N) (0.5) = 0.05 N
The direction of \(\vec{F}\) in each case is given by the cross product rule.

Question 3.
A horizontal straight wire is in a uniform magnetic field which is horizontal and at right angles to the length of the wire. The part of the wire that lies in the field has a length 2 m and mass 1 gram. If the magnetic induction is 1 mT, find the current that should be passed through the wire to balance it.
Solution:
Data : l = 2 m, m = 1 g = 10^-3 kg, g = 9.8 m/s²,
B = 1 mT = 10^-3 T
To balance the wire, the upward magnetic force must be equal in magnitude to the downward force due to gravity.

Question 4.
A circular loop of radius 10 cm is carrying a current of 0.1 A. Calculate its magnetic moment.
Solution:
Data : R = 10 cm = 0.1m, N = 1, I = 0.1 A
The magnetic moment,
μ = NIA = NI(πR²)
= (1) (0.1 A) (3.142) (0.1 m)²
= 3.142 × 10^-3 A∙m²

Question 5.
An electron in an atom revolves around the nucleus in an orbit of radius 0.53 Å. If the frequency of revolution of an electron is 9 × 10⁹ MHz, calculate the equivalent magnetic moment. [e = 1.6 × 10^-19 C]
Solution : .
Data : r = 0.53 Å = 0.53 × 10^-10 m,
f = 9 × 10⁹ MHz = 9 × 10¹⁵ Hz, e = 1.6 × 10^-19 C
Magnetic moment, M₀ = IA = efπr²
= 1.6 × 10^-19 × 9 × 10¹⁵ × 3.142 × (0.53 × 10^-10)²
= 14.4 × 3.142 × (0.53)² × 10^-19 × 10¹⁵ × 10^-20
= 1.272 × 10^-23 A∙m²

Question 6.
A rectangular coil of 10 turns, each of area 0.05 m², is suspended freely in a uniform magnetic field of induction 0.01 T. A current of 30 /(A is passed through it.

(i) What is the magnetic moment of the coil?
(ii) What is the maximum torque experienced by the coil?
(iii) What is the minimum torque experienced by the coil?
SoLution:
Data: N = 10, A = 0.05 m², B = 0.01 T, I = 30 μA = 3 × 10^-5 A
(i) The magnetic moment,
μ = NIA = 10(3 × 10^-5 A)(0.05 m²)
=1.5 × 10^-5 A∙m² = 15 μA∙m²

(ii) The maximum torque experienced by the coil (when its plane is parallel to \(\vec{B}\)) is
τ_max = MB
= (1.5 × 10^-5 A∙m²)(0.01 T)
= 1.5 × 10^-7 N∙m

(iii) The minimum torque experienced by the coil (when its plane is perpendicular to \(\vec{B}\)) is
τ_min = 0

Question 7.
A coil has 300 turns, each of area 0.05 m2. (i) Find the current through the coil for which the magnetic moment of the coil will be 4.5 A-m2. (ii) It is placed in a uniform magnetic field of induction 0.2 T with its magnetic moment making an angle of 30° with \(\vec{B}\). Calculate the magnitude of the torque experienced by the coil. (3 marks)
Solution:
Data : N = 300, A = 0.05 m², M = 4.5 A ∙ m², B = 0.2 T, θ = 30°
(i) M = NIA
∴ The current in the coil,
I = \(\frac{M}{N A}=\frac{4.5}{300 \times 0.05}\) = 0.3 A

(ii) The magnitude of the torque,
τ = MB sin θ = 4.5 × 0.2 × sin 30°
= 0.9 × \(\frac{1}{2}\) = 0.45 N∙m

Question 8.
A rectangular coil of 10 turns, each of area 0.05 m², is suspended freely in a radial magnetic field of 0.01 Wb/m². If the torsional constant of the suspension fibre is 5 × 10^-9 N∙m per degree, find the angle through which the coil rotates when a current of 30 μA is passed through it.
Solution:
Data : A = 0.05 m², B = 0.01 Wb/m², N = 10, C = 5 × 10^-9 N∙m per degree,
I = 30 μA = 30 × 10^-6 A,
I = \(\left(\frac{C}{N A B}\right) \theta\)
∴ The deflection of the coil,
θ = \(\frac{N I A B}{C}=\frac{10 \times 30 \times 10^{-6} \times 0.05 \times 0.01}{5 \times 10^{-9}}\) = 30°

Question 9.
A moving-coil galvanometer has coil of area 10 cm² and 100 turns. It is suspended by a fibre of torque constant 10^-8 N∙m/degree in a radial magnetic field of induction 0.05 Wb/m². Find the angle through which the coil will be deflected when a current of 16 μA passes through it.
Solution:
Data ; A = 10^-3 m², N = 100, C = 10^-8 N∙m/degree, B = 0.05 Wb/m², 7 = 1.6 × 10^-5 A
I = \(\left(\frac{C}{N A B}\right) \theta\)
∴ The deflection of the coil,
θ = \(\frac{N I A B}{C}=\frac{(100)\left(1.6 \times 10^{-5}\right)\left(10^{-3}\right)(0.05)}{10^{-8}}\) = 8

Question 10.
A bar magnet of moment 7.5 A∙m² experiences a torque of magnitude 1.5 × 10^-4 N∙m when placed inclined at 30° in a uniform magnetic field. Find the magnitude of the magnetic induction of the field.
Solution:
Data : μ = 7.5 A∙m², τ = 1.5 × 10^-4 N∙m, θ = 30° T = μB sin θ
∴ The magnitude of the magnetic induction,

Question 11.
A circular coil, having 200 turns each of area 2.5 × 10^-4 m², carries a current of 200 μA. Initially, the coil is at rest in a magnetic field of induction 0.8 T, with its magnetic dipole moment aligned with the field. Find the work an external agent has to do to rotate the coil through (i) 90° from its initial position (ii) further 90°.
Solution:
Data : N = 200, A = 2.5 × 10^-4 m², B = 0.8T,
I = 200 μA = 2 × 10^-4 A, θ = 90°
W = ∆U =U_θ – U₀
(i) The work done to rotate through 90°,
W = U_90° – U_0° = – μB cos 90° – (- μB cos 0°)
= 0 + μB = (NIA)B (∵ μ = NIA)
= (200)(2 × 10^-4)(2.5 × 10^-4)(0.8)
= 8 × 10^-6 J = 8 μJ

(ii) The work done to rotate further through 90°, so that the dipole moment is antiparallel to the field,
W = U_80° – U_90° = – μB cos 180° – (- μB cos 90°)
= μB + 0 = (NIA)B = 8 μJ

Question 12.
A magnetic dipole of moment 0.025 J/T is free to rotate in a uniform magnetic field of induction 50 mT. When released from rest in the magnetic field, the dipole rotates to align with the field. At the instant the dipole moment is parallel to the field, its kinetic energy is 625 μJ. What was the initial angle between the dipole moment and the magnetic field?
Solution:
Data: μ = 0.025J/T, B = 50mT = 5 × 10^-2 T,
∆K = 625 pJ = 6.25 × 10^-4 J
Change in potential energy,
∆U =U_θ – U₀ = – μB cos 0° – (- μ8 cos θ)
= – μB(1 – cos θ)
By the principle of conservation of energy,
∆K + ∆U = 0
∴ ∆K = – ∆U = μB(1 – cos θ)
∴ (2.5 × 10^-2)(5 × 10^-2)(1 – cos θ) = 6.25 × 10^-4
∴ (1 – cos θ) = 0.5 ∴ cos θ = 0.5
The initial angle between the dipole moment and the magnetic field,
θ = 60° .

Question 45.
State the Bio-Savart law (Laplace law) for the magnetic induction produced by a current el-ement. Express it in vector form.
Answer:
Consider a very short segment of length dl of a wire carrying a current I. The product I\(\overrightarrow{d l}\) is called a current element; the direction of the vector \(\overrightarrow{d l}\) is along the wire in the direction of the current.

Biot-Savart law (Laplace law) : The magnitude of the incremental magnetic induction \(\overrightarrow{d B}\) produced by a current element I\(\overrightarrow{d l}\) at a distance r from it is directly proportional to the magnitude Idl of the current element, the sine of the angle between the current element I \(\overrightarrow{d l}\) and the unit vector \(\hat{r}\) directed from the current element toward the point in question, and inversely proportional to the square of the distance of the point from the current element; the magnetic induction is directed perpendicular to both I \(\overrightarrow{d l}\) and \(\hat{r}\) as per the cross product rule.

of free space. Equations (1) and (2) are called Biot-Savart law.

The incremental magnetic induction \(\overrightarrow{d B}\) is given by the right-handed screw rule of vector crossproduct I\(\overrightarrow{d l} \times \hat{\mathbf{r}}\). In below figure, the current element I \(\overrightarrow{d l}\) and \(\hat{r}\) are in the plane of the page, so that \(\overrightarrow{d B}\) points out of the page at point P shown by ⊙; at the point Q, \(\overrightarrow{d B}\) points into the page shown by ⊗.

The magnetic induction \(\vec{B}\) at the point due to the entire wire is, by the principle of superposition, the vector sum of the contributions \(\overrightarrow{d B}\) from all the current elements making up the wire.
From Eq. (2),
\(\vec{B}=\int \overrightarrow{d B}=\frac{\mu_{0}}{4 \pi} \int \frac{I \vec{d} \times \hat{\mathrm{r}}}{r^{2}}\)
[Notes : (1) The above law is based on experiments by Jean Baptiste Biot (1774-1862) and Felix Savart (1791-1841), French physicists. From their observations Laplace deduced the law mathematically. (2) The Biot- Savart law plays a similar role in magnetostatics as Coulomb’s law does in electrostatics.]

Question 46.
Using Biot-Savart’s law, obtain the expression for the magnetic induction near a straight infinite ly long current-carrying wire.
Answer:
Suppose a point P is at a distance a from a straight, infinitely long, wire carrying a current I, as shown in below figure. The incremental magnetic induction \(d \vec{B}\) at the point P due to a current element, \(I \overrightarrow{d l}\) is

At the point P, \(d \vec{B}\) is directed perpendicular to the plane of the figure and into of the page as given by the right hand rule for the direction of \(\overrightarrow{d l} \times \hat{\mathbf{r}}\).

At point P, \(d \vec{B}\) has this same direction for all the current elements into which the wire can be divided. Thus, we can find the magnitude of the magnetic field produced at P by the current elements in the lower half of the infinitely long wire by integrating dB in Eq. (2), from 0 to ∞.

Now consider a current element in the upper half of the wire, one that is as far above P as I\(\overrightarrow{d l}\) is below P. By symmetry, the magnetic field produced at P by this current element has the same magnitude and direction as that from I \overrightarrow{d l} in above figure. Thus, the magnetic field produced by the upper half of the wire is exactly the same as that produced by the lower half. Hence, the magnitude of the total magnetic field at P is

That is, the magnitude B is inversely proportional to the distance from the wire. Because of the axial symmetry about the straight wire, the magnetic induction has the same magnitude B at all points on a circle in a transverse plane and centred on the conductor; the direction of \(\vec{B}\) is everywhere tangential to such a circle. Thus, the magnetic field lines around the current in the straight wire is a family of circles centred on the wire.

[Notes : (1) The magnetic field at P due to either only the lower half or the upper half of the infinite wire in figure is half the value in Eq. (5); that is, for a semi-infinite wire,

Question 47.
Show that currents in two long, straight, parallel wires exert forces on each other. Derive the expression for the force.
OR
Derive an expression for the force per unit length between two infinitely long parallel conductors carrying current and hence define the ampere.
Answer:
When two currents pass in adjacent parallel straight conductors, we may think of each of the currents as being situated in the magnetic field caused by the other current. This results in a force on each conductor.

Consider two infinitely long, straight, parallel wires, each of length ¡ a distance s apart in vacuum, as shown in figure (a). The magnetic field around the wire 1, carrying a current I₁ has an induction of magnitude
B₁ = \(\left(\frac{\mu_{0}}{4 \pi}\right) \frac{2 I_{1}}{s}\)
Wire 2, with a current I₂ in the same direction as I₁, is situated in this field. The direction of the field
with induction \(\overrightarrow{B_{1}}\) at the position of wire 2, given by the right hand Igripi rule, is perpendicular to the plane of the two conductors, as shown. Hence, the force \(\overrightarrow{F_{2}}\) on wire 2 has a magnitude
F₂ = I₂lB₁ = \(\left(\frac{\mu_{0}}{4 \pi}\right) \frac{2 I_{1} I_{2} l}{s}\) ……………… (1)
and is, by Fleming’s left hand rule, towards wire 1. Similarly, the magnetic induction \(\overrightarrow{B_{2}}\) at the position of wire 1 has a magnitude
B₂ = \(\left(\frac{\mu_{0}}{4 \pi}\right) \frac{2 I_{2}}{s}\)
and is also directed perpendicular to the plane of the wires. Hence, the force on wire 1 has a magnitude
F₁ = I₁lB₂ = \(\left(\frac{\mu_{0}}{4 \pi}\right) \frac{2 I_{1} I_{2} l}{s}\) ……………… (2)
directed towards wire 2. Thus, the two currents attract each other. \(\vec{F}_{1}=-\vec{F}_{2} \), i.e., they are equal in magnitude and opposite in direction.

Ampere found that the wires attracted each other when the currents in them were in the same direction [from figure (a )], and repelled each other when they were in the opposite directions [from figure (b)].

From the Eq. (2), the force per unit length acting on each wire is
\(\frac{F}{l}=\left(\frac{\mu_{0}}{4 \pi}\right) \frac{2 I_{1} I_{2}}{s}\)
Using SI units, μ₀/4π = 10^-7 N / A² and, if I₁ = I₂ = 1 A and s = 1 m, then
\(\frac{F}{l}\) = 2 × 10^-7 N / m
In SI, this equation is the defining relation for the ampere.

Definition: The ampere is that constant current which if maintained in two infinitely long straight parallel wires, and placed one metre apart in vacuum, would cause each conductor to experience a force per unit length of 2 × 10^-7 newton per metre. [Note : 1 Wb/A∙m = 1 T∙m/A = l N/A².]

Question 48.
Two very long and straight parallel conductors separated by 0.5 m in vacuum carry currents 2 A and 3 A respectively. What is the force per unit length of a conductor? [\(\frac{\mu_{0}}{4 \pi}\) = 10^-7 Wb/A∙m]
Answer:
The force per unit length on each conductor,

Question 49.
Obtain an expression for the magnetic induction produced by a current in a wire in the shape of a circular arc at its centre of curvature. Hence obtain an expression for the magnetic induction at the centre of a circular coil carrying a current.
Answer:
Consider a wire in the shape of a circular arc of radius of curvature R and carrying a current I. The unit vector \(\hat{\mathrm{r}}\) from each current element I\(\overrightarrow{d l}\) towards

the centre of the loop is perpendicular to I\(\overrightarrow{d l}\), i.e., the angle θ between them is 90°. The direction of the incremental magnetic induction \(\overrightarrow{d B}\) due to each current element is in the same direction, viz., perpendicular to the plane of the loop, and out of the plane of the figure for the sense of the current shown in above figure.

Since every current element is equidistant from the centre of curvature C, the magnetic field at C due to each current element in the arc by Biot-Savart’s law is

If the arc subtends an angle 0 at its centre of curvature C, the total field at C due to all the elements on the arc is

where Φ is in radian.
The magnitude of the total induction \(\vec{B}\) at the centre of a circular coil is, from Eq. (3),

If a circular coil has N turns, each of radius r and carries a current I, the magnetic induction at its centre has a magnitude
B = \(\frac{\mu_{0} N I}{2 R}\) …………… (6)

Question 50.
Derive an expression for the magnetic induction at a point on the axis of a circular coil carrying a current.
OR
A circular coil of N turns, each of radius R, carries a current I. Derive the expression for the magnitude of the magnetic induction on the axis of the coil at a distance z. Hence obtain the expression for the magnitude of the magnetic induction for z » R.
Answer:
Consider a circular, conducting loop of radius R in the xy-plane, whose centre is at the origin. For any point P on the z-axis (which is also the axis of the loop), the current element I \(\overrightarrow{d l}\) is perpendicular to the unit vector \(\hat{\mathbf{r}}\) directed from the current element to point P. The incremental magnetic induction, \(\overrightarrow{d B}\), due to a current element at Q lies in the plane QOP, in the direction of I \(\overrightarrow{d l} \times \vec{r}\), as shown in below figure. In magnitude,

Each element of the loop has its diametrically opposed companion I \(\overrightarrow{d l}\) on the other side of the loop, both of equal segment length dl. Point P is equidistant from the two elements, that is, r = r’. Therefore, the two contributions \(\overrightarrow{d B}\) and \(\overrightarrow{d B}^{\prime}\) have equal magnitudes, and their vertical components, dB cos α and dB’cos α, are oppositely directed. Thus, they will add to zero when all the \(\overrightarrow{d B}\) contributions are summed, and there can be no vertical component of the resultant induction \(\vec{B}\). However, the horizontal components are of like direction and will sum to a definite value and hence \(\vec{B}\) will have only a horizontal z-component. For the magnitude of \(\vec{B}\), we need to add only the z-components of the \(\overrightarrow{d B}\) vectors :

For values of z that are much larger than the radius R, we may ignore the value of R in the denominator above and write
B = \(\left(\frac{\mu_{0}}{4 \pi}\right) \frac{2 I A}{z^{3}}\) (for z >> R) …………… (2)
For a circular coil of N turns Eqs. (1) and (2) are, respectively,
B = \(\left(\frac{\mu_{0}}{4 \pi}\right) \frac{2 N I A}{\left(R^{2}+z^{2}\right)^{\frac{3}{2}}}\) and B = \(\left(\frac{\mu_{0}}{4 \pi}\right) \frac{2 N I A}{z^{3}}\)

Question 51.
How is the magnetic field of a small current loop identical to that of a short magnetic dipole? Explain.
OR
Explain the equivalence of the fields of a current-carrying circular coil and a magnetic dipole.
Answer:
An electric current in a circular loop establishes a magnetic field similar in every respect to the field of a magnetic dipole (or a bar magnet).

Consider a circular conducting ioop of radius R, axis along the x-axis and carrying a current I. The area of the loop is A = πR²and \(\vec{A}\) has the direction given by right hand rule. The axial magnetic induction of the current loop at a distance x from its centre is
\(\vec{B}=\left(\frac{\mu_{0}}{4 \pi}\right) \frac{2 I \vec{A}}{\left(R^{2}+x^{2}\right)^{3 / 2}}=\left(\frac{\mu_{0}}{4 \pi}\right) \frac{2 \vec{\mu}}{\left(R^{2}+x^{2}\right)^{3 / 2}}\)
where \(\vec{M}=I \vec{A}\) is the magnetic moment of the current loop and μ₀ is the permeability of free space. For x >> R, ignoring R² in comparison with X²,
\(\vec{B}=\left(\frac{\mu_{0}}{4 \pi}\right) \frac{2 \vec{\mu}}{x^{3}}\)
This equation also gives the magnetic induction on the axis of a short magnetic dipole (or a bar magnet) of magnetic moment \(\vec{\mu}\).

For a magnetic dipole, the dipole moment is directed from the south pole of the dipole to its north pole. For a current loop, the magnetic dipole moment has the direction of the axial field of the current loop as given by the right-hand rule.

When an observer looking at a current carrying circular loop finds the direction of the current anticlockwise, the face of the loop towards the observer acts as the north pole. When an observer looking at a current-carrying circular loop finds the direction of the current clockwise, the face of the loop towards the observer acts as the south pole. This rule is known as the clock rule.

Question 52.
State the expressions for the axial fields of an electric dipole and a small current-loop.
Answer:
The axial far field of an electric dipole of electric dipole moment \(\vec{p}\) at an axial point r is
\(\vec{E}=\frac{1}{4 \pi \varepsilon_{0}} \frac{2 \vec{p}}{r^{3}}\)
The axial magnetic field far from a small current-loop is
\(\vec{B}=\frac{\mu_{0}}{4 \pi} \frac{2 \vec{\mu}}{r^{3}}\)
where \(\vec{\mu}=I \vec{A}\) is the magnetic dipole moment of the loop, I is the current in the loop and \(\vec{A}\) is the area vector given by the right-hand rule. The axial field of an electric dipole, \(\vec{E}\) is in the direction of the dipole moment \(\vec{p}\). On the axis of a current loop, the magnetic field \(\vec{B}\) is in the direction of the dipole moment \(\vec{\mu}\).

53. Solve the following
Question 1.
(1) Two long parallel current-carrying conductors are 0.4 m apart in air and carry currents 5 A and 10A. Calculate the force per metre on each conductor, if the currents are in the same direction and in the opposite direction.
Solution:
Data: s = 0.4 m, I₁ = 5A, I₂ = 10A
μ₀/4π = 10^-7 N/A²
The force per unit length acting on each conductor is
\(\frac{F}{l}=\left(\frac{\mu_{0}}{4 \pi}\right) \frac{2 I_{1} I_{2}}{s}\)
= (10^-7 N/A²) \(\frac{2(5 \mathrm{~A})(10 \mathrm{~A})}{0.4 \mathrm{~m}}\)
= 2.5 × 10^-5 N/m (= 25 μN/m)
This force is attractive if the currents are in the same direction and repulsive if the currents are in the opposite directions.

Question 2.
Two wires 12 m long and 10 cm apart carry the same current. Find the current through each wire if the force per unit length on each wire is 0.001 N/m.
Solution:
Data : l = 12 m, s = 10 cm = 0.1 m,
I₁ = I₂ = I, \(\frac{F}{l}\) = 0.001 N/m

The current through each wire is 10\(\sqrt {5}\) A.

Question 3.
The wire shown below carries a current of 2 A. The curved segment is a quadrant of radius 10 cm while the straight segments are along radii. Find the magnitude and direction of the magnetic induction at the centre O of the quadrant by the entire wire.

Solution:
Data : I = 2 A, R = 10 cm = 10^-1 m
The point O lies along the straight segments AB and CD. Hence, the magnetic induction \(\vec{B}\) produced by each of them is zero.

Since θ = 90°, the magnitude of the magnetic induction due to the current in quadrant BC is \(\frac{1}{4}\)th of that produced at the centre of a loop.

Question 4.
A flat coil of 70 turns has a diameter of 20 cm and carries a current of 5 A. Find the magnitude of the magnetic induction at (a) the centre of the coil (b) a point on the axis 20 cm from the centre of the coil.
Solution:
Data : N = 70, R = 10 cm = 0.1 m, I = 5 A, z = 0.2 m, μ₀ = 4π × 10^-7 T∙m/A
(a) At the centre of the coil:
The magnitude of the magnetic induction,

Question 5.
(12) A current of 10 A passes through a coil having 5 turns and produces a magnetic field of magnitude 0.5 × 10^-4 T at the centre of the coil. Calculate the diameter of the coil.
Solution:
Data: I = 10A, N = 5, B = 5 × 10^-5 T,
μ₀/4π = 10^-7 T∙m/A
B = \(\frac{\mu_{0} N I}{2 R}\)
∴ The diameter of the coil,
2R = \(\frac{\mu_{0} N I}{B}=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)(5)(10 \mathrm{~A})}{5 \times 10^{-5} \mathrm{~T}}\)
= 4 × 3.142 × 10^-1 = 1.257 m

Question 6.
Calculate the magnitude of the magnetic induction due to a circular coil of 400 turns and radius 0.05 m, carrying a current of 5 A, at a point on the axis of the coil at a distance 0.1 m.
Solution:
Data : N = 400, R = 0.05 m = 5 × 10^-2 m,
I = 5 A, z = 0.1 m, μ₀/4π = 10^-7 T∙m/A
The magnitude of the magnetic induction,

Question 7.
A circular coil of wire has 100 turns. The radius of the coil is 50 cm. It is desired to have a magnetic induction of 80 μT at the centre of the coil. What should be the current through the coil ?
Solution:
Data: N = 100, R = 50 cm = 0.5 m,
B = 80 μT = 8 × 10^-5 T, μ₀/4π = 10^-7 T∙m/A

Question 54.
State and explain Ampere’s circuital law.
OR
State Ampere’s circuital law.
Answer:
Ampere’s circuital law : In free space, the line integral of magnetic induction around a closed path in a magnetic field is equal to p0 times the net steady current enclosed by the path.
In mathematical form,
\(\oint \vec{B} \cdot \overrightarrow{d l}\) = μ₀I …………. (1)
where \(\vec{B}\) is the magnetic induction at any point on the path in vacuum, \(\overrightarrow{d l}\) is the length element of the path, I is the net steady current enclosed and p0 is the permeability of free space.

Explanation : Below figure shows two wires carrying currents I₁ and I₂ in vacuum. The magnetic induction \(\vec{B}\) at any point is the net effect of these currents.

To find the magnitude B of the magnetic induction :
(i) We construct an imaginary closed curve around the conductors, called an Amperian loop, and imagine it divided into small elements of length dl. The direction of \(\overrightarrow{d l}\) is the direction along which the loop is traced.

(ii) We assign signs to the currents using the right hand rule : If the fingers of the right hand are curled in the direction in which the loop is traced, then a current in the direction of the outstretched thumb is taken to be positive while a current in the opposite direction is taken to be negative.

For each length element of the Amperian loop, \(\vec{B} \cdot \overrightarrow{d l}\) gives the product of the length dl of the element and the component of \(\vec{B}\) parallel to \(\overrightarrow{d l}\). If θ is the angle between \(\overrightarrow{d l}\) and \(\vec{B}\),
\(\vec{B} \cdot \overrightarrow{d l}\) = (B cos θ) dl
Then, the line integral,
\(\oint \vec{B} \cdot \overrightarrow{d l}=\oint\) Bcosθ dl ……………(2)
For the case shown in figure, the net current I through the surface bounded by the loop is
I = I₂ – I₁
∴ \(\oint\) Bcosθ dl = μ₀ I
= μ₀(I₂ – I₁) ……………. (3)
Equation (3) can be solved only when B is uniform and hence can be taken out of the integral.

[Note : Ampere’s law in magnetostatics plays the part of Gauss’s law of electrostatics. In particular, for currents with appropriate symmetry, Ampere’s law in integral form offers an efficient way of calculating the magnetic field. Like Gauss’s law, Ampere’s law is always true (for steady currents), but it is useful only when the symmetry of the problem enables B to be taken out of the integral \(\oint \vec{B} \cdot \overrightarrow{d l}\). The current configurations that can be handled by Ampere’s law are infinite straight conductor, infinite plane, infinite solenoid and toroid.]

Question 55.
Using Ampere’s law, obtain an expression for the magnetic induction near a current-carrying straight, infinitely long wire.
Answer:
Consider a point P at a distance a from a straight, infinitely long wire carrying a current I in free space, from figure (a). Because of the axial symmetry about the straight wire, the magnetic induction has the same magnitude B at all points on a circle in a transverse plane and centred on the wire. We, therefore, choose an Amperian loop a circle of radius a centred on the wire with its plane perpendicular to the wire, as shown in from figure (b).

since cos θ = 1 and B has the same value around the path. \(\oint\) dl gives the circumference of the circular loop.

In the figure, the Amperian loop is traced in the anticlockwise sense, so that the current I is taken as positive in accordance with the right hand rule.
By Ampere’s law,

This is the required expression.

Question 56.
What is a solenoid? With a neat labelled diagram, describe the magnetic field produced by a current-carrying solenoid.
Answer:
A solenoid is a long wire wound in the form of a helix. An ideal solenoid is tightly wound and infinitely long, i.e., its turns are closely spaced and the solenoid is very long compared to its crosssectional radius.

Each turn of a solenoid acts approximately as a circular loop. Suppose the solenoid carries a steady current I. The net magnetic field due to the current in the solenoid is the vector sum of the fields due to the current in all the turns. In the case of a tightly- wound solenoid of finite length, Fig. 10.39, the magnetic field lines are approximately parallel only near the centre of the solenoid, indicating a nearly uniform field there. However, close to the ends, the field lines diverge from one end and converge at the other end. This field distribution is similar to that of a bar magnet. Thus, one end of the solenoid behaves like the north pole of a magnet and the opposite end behaves like the south pole. The field outside is very weak near the midpoint.

For an ideal solenoid, the magnetic field inside is reasonably uniform over the cross section and parallel to the axis throughout the volume enclosed by the solenoid. The field outside is negligible in this case.

Question 57.
Using Ampere’s law, derive an expression for the magnetic induction inside an ideal solenoid carrying a steady current.
Answer:
An ideal solenoid is tightly wound and infinitely long. Let n be the number of turns of wire per unit length and I be the steady current in the solenoid.

For an ideal solenoid, the magnetic induction \(\vec{B}\) inside is reasonably uniform over the cross section and parallel to the axis throughout the volume enclosed by the solenoid; \(\vec{B}\) outside is negligible.

As an Amperian loop, we choose a rectangular path PQRS of length l parallel to the solenoid axis, from below figure. The width of the rectangle is taken to be sufficiently large so that the side RS is far from the solenoid where \(\vec{B}\) = 0. The line integral of the magnetic induction around the Amperian loop in the sense PQRSP is


Thus, from Eqs. (1), (2), (3) and (4),
\(\oint \vec{B} \cdot \overrightarrow{d l}\) = Bl ……….. (5)
The total current enclosed by the Amperian loop is
I_encl = current through each turn × number of turns enclosed by the loop
= I × nl = nlI ……… (6)
By Ampere’s law,
\(\oint \vec{B} \cdot \overrightarrow{d l}\) = μ₀I_encl (in vacuum)
Therefore, from Eq.s (5) and (6),
Bl = μ₀ nlI
∴ B = μ₀nI …………. (7)
This is the required expression.

[Notes : (1) The field inside an ideal solenoid is uniform-it doesn’t depend on the distance from the axis. In this sense, the solenoid is to magnetostatics what the parallel-plate capacitor is to electrostatics; a simple device for producing strong uniform fields. (2) At an axial point at the end of a long solenoid, B = \(\frac{1}{2}\) μ₀nI]

Question 58.
What is the magnetic field (i) outside (ii) inside a long air-cored current-carrying solenoid ?
Answer:
For an ideal solenoid, the magnetic field induction outside is negligible, nearly zero. Inside the solenoid, the field lines are parallel to the axis of the solenoid and the magnitude of the magnetic induction, B = μ₀nI, where μ₀ is the permeability of free space, n is the number of turns of wire per unit length and I is the steady current in the solenoid.

Question 59.
What is a toroid? With a neat diagram, describe the magnetic field produced by a toroid carrying a steady current.
Answer:
A toroid is a toroidal solenoid. An ideal toroid consists of a long conducting wire wound tightly around a torus, a doughnut-shaped ring, made of a nonconducting material.

In an ideal toroid carrying a steady current, the magnetic field in the interior of the toroid is tangential to any circle concentric with the axis of the toroid and has the same value on this circle (the dashed line in above figure). Also, the magnitude of the magnetic induction external to the toroid is negligible.

Question 60.
Using Ampere’s law, derive an expression for the magnetic induction inside an ideal toroid carrying a steady current.
Answer:
An ideal toroid consists of a long conducting wire wound tightly around a torus made of a non-conducting material. When a steady current is passed through it, the magnetic induction \(\vec{B}\) in the interior of the toroid is tangent to any circle concentric with y the axis of the toroid and has the same value on this circle.

Suppose the toroid has N turns of wire and I is the current in its coil. As our Amperian loop, we choose a circle of radius r concentric with the axis of the toroid, as shown in figure. Since \(\vec{B}\) has the same value on this circle and is tangential to it, we go around this path in the direction of \(\vec{B}\) so that \(\vec{B}\) and \(\overrightarrow{d l}\) are parallel. Then, the line integral of the magnetic induction around the Amperian loop is

The net current enclosed by the Amperean loop is
I_encl = current through each turn × number of turns enclosed by the loop
= I × N = NI …………….. (2)
By Ampere’s law,
\(\oint \vec{B} \cdot \overrightarrow{d l}\) = μ₀ I_encl (in free space)
Therefore, from Eqs. (1) and (2),
B (2πr) = μ₀NI
∴ B = \(\frac{\mu_{0}}{2 \pi} \frac{N I}{r}\) …………… (3)
This is the required expression.

61. Solve the following

Question 1.
Four long parallel wires are arranged at the four comers of a square ABCD of side 20 cm. Each wire carries a current of 5 A. Currents in the conductors 1 and 2, at comers A and B, are out of the page while those in the conductors 3 and 4, at comers C and D, are into the page. What is the magnitude of the magnetic induction at the centre of the square?
Solution:
Data : 2l = 20 cm = 0.2 m, I = 5 A
If each side of the square array is 2l, then from geometry, the centre of the square O is a distance \(\sqrt {2}\)l from each corner. Since each wire carries the same current, B₁ = B₂ = B₃ = B₄. Using the right hand [grip] rule, the directions of the magnetic inductions due to conductors 1 and 3 are along OB, while those due to conductors 2 and 4 are along OC, as shown.

This is also the magnitude of the magnetic induc-tion along OC, B₂ + B₄.

Their components parallel to AD (or BC) are oppositely directed and cancel out. Therefore, the total induction at the centre of the square has a magnitude equal to the sum of the components parallel to AB (or DC).

The magnitude of the magnetic induction at the centre of the square is 2 × 10^-5 T.

Question 2.
Two long straight parallel wires in vacuum are 4 m apart and carry currents of 2 A and 6 A in the same direction. Find the neutral point, i.e., the point at which the resultant magnetic induction is zero.
Solution:
Data: I₁ =2 A, I₂ = 6 A, a =4 m
The currents through the wires are in the same direction. Therefore, the two magnetic inductions \(\overrightarrow{B_{1}}\) and \(\overrightarrow{B_{2}}\) will have opposite directions at any point between the two wires. Hence, the point must lie between the two wires. For the resultant magnetic induction to be zero, we must have B₁ = B₂. Let the corresponding point (the neutral point) be at a distance a₁ from the first wire and a₂ from the second wire.

The neutral point lies at a distance of 1 m from the wire carrying a current of 2 A.

Question 3.
A solenoid 1.5 m long and 4 cm in diameter has – 10 turns/cm. A current of 5 A is passing through it. Calculate the magnetic induction (i) inside (ii) at one end on the axis of the solenoid.
Solution:
Data : L = 1.5 m, r = 2 cm, n = 10 turns/cm = 10³ turns/metre, I = 5 A, μ₀ = 4π × 10^-7 T∙m/A
Since the diameter is very small compared to its length, we approximate the solenoid to be long, i.e., an ideal solenoid.

(i) At an axial point well inside a long solenoid,
B = μ₀nI
= (4π × 10^-7 )(10³)(5) = 2 × 3.142 × 10^-3
= 6.284 × 10^-3 T

(ii) At an axial point at the end of a long solenoid,
B = \(\frac{1}{2}\) μ₀nI
= \(\frac{1}{2}\) (6.284 × 10^-3) = 3.142 × 10^-3 T

Question 4.
A toroidally wound coil has an inner radius of 15 cm, an outer radius of 20 cm and is wound with 1500 turns of wire. What is the magnitude of the magnetic induction at the centre of the coil when the current in the winding is 10 A ?
Solution:
Data : Central radius, r = \(\frac{1}{2}\) (15 + 20) = 17.5 cm
= 0.175 m, N = 1500, I = 10 A, \(\frac{\mu_{0}}{4 \pi}\) = 10^-7 T∙m/A
The magnetic induction,
B = \(\frac{\mu_{0} N I}{2 \pi r}=\frac{\mu_{0}}{4 \pi} \frac{2 N I}{r}=10^{-7} \times \frac{2 \times 1500 \times 10}{0.175}\)
= 1.714 × 10^-2 T

Multiple Choice Questions

Question 1.
A doubly ionized helium nucleus (charge = 3.2 × 10^-19 C) enters a region of uniform magnetic field with a velocity (10³ m/s)\(\hat{\mathbf{i}}\). The magnetic induction in the region is 20 mT directed towards the positive x-axis. The force on the ion is
(A) (6.4 × 10^-18 N) \(\hat{\mathrm{j}}\)
(B) (6.4 × 10^-18 N) \(\hat{\mathrm{k}}\)
(C) zero
(D) none of these.
Answer:
(C) zero

Question 2.
In a cyclotron, charged particles are accelerated by
(A) the electrostatic deflector plate
(B) the electric field in the dees
(C) the magnetic field in the dees
(D) the p.d. across the gap between the dees.
Answer:
(D) the p.d. across the gap between the dees.

Question 3.
Cyclotron cannot accelerate
(A) protons
(B) neutrons
(C) α-particles
(D) deuterons.
Answer:
(B) neutrons

Question 4.
If R is the radius of the dees and B the magnitude of the magnetic field induction in which positive charges (q) of mass m escape from the cyclotron, then their maximum speed v_max is
(A) \(\frac{q R}{B m}\)
(B) \(\frac{q m}{B R}\)
(C) \(\frac{q B R}{m}\)
(D) \(\frac{m}{q B R}\)
Answer:
(C) \(\frac{q B R}{m}\)

Question 5.
A charged particle moving with a velocity \(\vec{v}\) enters a region of uniform magnetic field \(\vec{B}\). If the velocity has a component parallel to \(\vec{B}\), which of the following quantities is independent of \(\overrightarrow{\mid v} \mid\)?
(A) Period T of its circular motion
(B) Pitch p of its helical path
(C) Radius r of its helical path
(D) Both p and T
Answer:
(A) Period T of its circular motion

Question 6.
A charged particle moving with a velocity \(\vec{v}\) enters a region of uniform magnetic field \(\vec{B}\) such that the pitch of the resulting helical motion is equal to the radius of the helix. The angle between \(\vec{v}\) and \(\vec{B}\) is
(A) tan^-1 2π
(B) sin^-1 2π
(C) tan^-1 \(\frac{1}{2 \pi}\left(\frac{q B}{m}\right)^{2}\)
(D) tan^-1 2π\(2 \pi\left(\frac{q B}{m}\right)^{2}\).
Answer:
(A) tan^-1 2π

Question 7.
The following four cases show a positive charge moving into a magnetic field \(\vec{B}\) with velocity \(\vec{v}\). In which of the cases are the forces opposite in direction?

(A) (iii) and (iv)
(B) (i) and (ii)
(C) (i) and (iii)
(D) (ii) and (iv)
Answer:
(B) (i) and (ii)

Question 8.
A charged particle enters a uniform magnetic field initially travelling perpendicular to the field lines and is bent in a circular arc of radius R. If the particle had the same charge but double the mass and were travelling twice as fast, the radius of its circular arc would be
(A) 2R
(B) 4R
(C) R
(D) \(\frac{1}{4}\)R.
Answer:
(B) 4R

Question 9.
A straight wire along the y-axis carries a current of 4 A. The wire is placed in a uniform magnetic field (0.02 T) \((\hat{\mathrm{i}}+\hat{\mathrm{j}})\). If the current in the wire is directed towards the negative y-axis, the force per unit length on the wire is
(A) zero
(B) – (0.08 N/m) \(\hat{\mathrm{k}}\)
(C) (0.08 N/m) \((\hat{\mathrm{i}}-\hat{\mathrm{j}})\)
(D) (0.08 N/m) \(\hat{\mathrm{k}}\)
Answer:
(D) (0.08 N/m) \(\hat{\mathrm{k}}\)

Question 10.
A 30-turn coil of diameter 2 cm carries a current of 10 mA. When it is placed in a uniform magnetic field of 0.05 T, the magnitude of the maximum torque that could be exerted on the coil by the magnetic field is
(A) 1.88 × 10^-5 N∙m
(B) 4.7 × 10^-6 N∙m
(C) 4.7 × 10^-7 N∙m
(D) 1.88 × 10^-8 N∙m.
Answer:
(B) 4.7 × 10^-6 N∙m

Question 11.
A circular loop of area \(\sqrt{2}\) cm² and carrying a current of 10 μA is placed in a magnetic field \(\vec{B}\) with its plane parallel to \(\vec{B}\) (B = 15 mT). When the loop has rotated through an angle of 45°, the magnitude of the torque exerted on this loop is
(A) zero
(B) 15 × 10^-12 N∙m
(C) 15 × 10^-8 N∙m
(D) 15 × 10^-2 N∙m
Answer:
(B) 15 × 10^-12 N∙m

Question 12.
A current loop of magnetic dipole moment 0.1 A-m2 is oriented with the plane of the loop perpendicular to a uniform 1.50 T magnetic field, as shown. The torque that the magnetic field exerts on the current loop is

(A) 0.15 N∙m
(B) 0.075 N∙m
(C) -0.075 N∙m
(D) -0.15 N∙m.
Answer:
(B) 0.075 N∙m

Question 13.
A rectangular coil of dipole moment fi, free to rotate, is placed in a uniform magnetic field B with its plane parallel to the magnetic lines of force. Then, the coil will
(A) rotate to maximize the magnetic flux through its plane
(B) rotate to minimize the magnetic flux through its plane
(C) not experience any torque
(D) experience a constant torque equal to μB.
Answer:
(C) not experience any torque

Question 14.
When a current is passed through a suspended moving-coil galvanometer, the deflection of the coil is arrested by
(A) the elastic torsion of the suspension fibre
(B) the elastic winding of the helical spring
(C) the friction at the point of suspension
(D) the changing magnetic flux through the coil.
Answer:
(A) the elastic torsion of the suspension fibre

Question 15.
When a current is passed through a suspended moving-coil galvanometer, the deflection of the coil is 9. Then, in the usual notation, the expression \(\frac{\mu B}{\theta}\) is
(A) the torsion constant of the helical spring
(B) the magnetic dipole moment of the current-carrying coil
(C) the current through the coil
(D) the torsion constant of the suspension fibre.
Answer:
(D) the torsion constant of the suspension fibre.

Question 16.
The magnetic potential energy of a coil of dipole moment \(\vec{\mu}\) and area vector \(\vec{A}\) placed in a magnetic \(\vec{B}\) is maximum for which of the following cases ?

Answer:
(C) \(\vec{B} \uparrow \downarrow \vec{A}\)

Question 17.
Two points, A and B, are at distances r_A and r_B from a long, straight, current-carrying conductor. If r_B = 2 r_A, the magnitudes of the magnetic inductions at the two points are related by
(A )B_A = B_B
(B) B_A = 2B_B
(C) B_A = 4B_B
(D) B_B = 2B_A
Answer:
(B) B_A = 2B_B

Question 18.
Two long, straight, parallel wires are 5 cm apart and carry currents I₁ and I₂ in the same direction. If 2I₁ = 3I₂, then at a point P, 2 cm from wire 2,

Answer:
(B) \(\overrightarrow{B_{1}}=-\overrightarrow{B_{2}}\)

Question 19.
A wire of length L is first formed into a loop of one turn and then as a loop of two turns. The same current I is passed through the wire in the two cases. The ratio of the magnitude of the magnetic field induction at the centre of the single-turn loop to that at the centre of the double-turn loop is
(A) 4
(B) 2
(C) \(\frac{1}{2}\)
(D) \(\frac{1}{4}\)
Answer:
(D) \(\frac{1}{4}\)

Question 20.
Two circular coils 1 and 2 have both their radii and number of turns in the ratio 1 : 2. If the currents in them are in the ratio 2: 1, the magnitudes of the magnetic inductions at the centres of the coils are in the ratio
(A) 1 : 1
(B) 2 : 1
(C) 1 : 2
(D) 1 : 8.
Answer:
(B) 2 : 1

Question 21.
Three straight, parallel wires are coplanar and perpendicular to the plane of the page. The currents I₁ and I₃ are directed out of the page. If the wire 3 experiences no force due to the currents I₁ and I₂, then the current in the wire 2 is

(A) I₂ = 2I₁ and directed into the page
(B) I₂ = 0.5I₁ and directed into the page
(C) I₂ = 2I₁ and directed out of the page
(D) I₂ = 0.5I₁ and directed out of the page.
Answer:
(B) I₂ = 0.5I₁ and directed into the page

Question 22.
Two diametrically opposite points of a uniform metal ring (radius, R) are connected to the terminals of a battery. If the current drawn from the battery is I, the magnetic induction at the centre of the ring has a magnitude
(A) \(\frac{\mu_{0} I}{R}\)
(B) \(\frac{\mu_{0} I}{2R}\)
(C) \(\frac{\mu_{0} I}{4R}\)
(D) zero
Answer:
(D) zero

Question 23.
Two circular coaxial coils, each of N turns and radius R, are separated by a distance R. They carry equal currents I in the same direction. If the magnetic induction at P, on the common axis and midway between the coils, due to the left hand coil is B, then the total induction at P is

(A) 2B
(B) B
(C) \(\frac{1}{2}\) B
(D) zero.
Answer:
(A) 2B

Question 24.
A toroid with a circular cross section has a current I in its windings. The total number of windings is N. The total current through an Amperian loop of radius r equal to the mean radius of the toroid is
(A) zero
(B) I
(C) NI
(D) \(\frac{N I}{2 \pi r}\)
Answer:
(C) NI

Question 25.
A very long solenoid has 8400 windings and a length of 7 m. If the field inside is 2ir x iO T, the current in the windings is about
[μ₀/4π = 10^-7 T∙m/A]
(A) 0.42 A
(B) 0.83 A
(C) 4.2 A
(D) 8.3 A.
Answer:
(C) 4.2 A

Balbharti Maharashtra State Board 12th Physics Important Questions Chapter 9 Current Electricity Important Questions and Answers.

Maharashtra State Board 12th Physics Important Questions Chapter 9 Current Electricity

Question 1.
Define the following terms:

1. Electrical circuit
2. Junction
3. Loop
4. Branch
5. Electrical network.

Answer:

1. Electrical circuit: An electrical circuit, in general, consists of a number of electrical components such as an electrical cell, a plug key (or a switch), a resistor, a current meter (a milliammeter or an ammeter), a voltmeter, etc., connected together to form a conducting path.
2. Junction: A point in an electrical circuit where two or more conductors are joined together is called a junction.
3. Loop: A closed conducting path in an electrical network is called a loop or mesh.
4. Branch: A branch is any part of an electrical network that lies between two junctions.
5. Electrical network: An electrical network consists of a number of electrical components connected together to form a system of inter-related circuits.

Question 2.
State Kirchhoff’s first law or current law or junction law.
Answer:
The algebraic sum of the currents at a junction is zero in an electrical network, i.e. \(\sum_{i=1}^{n}\) I_i = 0, where I_i is
the current in the i^th conductor at a junction having n conductors.

Question 3.
What is the sign convention used for Kirchhoff’s first law? Explain with an example.
Answer:
A current arriving at a junction is considered positive while a current leaving a junction is considered negative.

Consider a junction in a circuit where six current carrying conductors meet. Currents I₁, I₃ and I₅ are considered positive as they arrive at the junction.
Currents I₂, I₄ and I₆ are considered negative as they leave the junction.

Using Kirchhoff’s current law, \(\sum_{i=1}^{6}\) I_i = 0, we get,
I₁ – I₂ + I₃ – I₄ + I₅ – I₆ = 0
∴ I₁ + I₃ + I₅ = I₂ + I₄ + I₆
Thus the total current flowing towards the junction is equal to the total current flowing away from the junction.

[Note : As the current is the time rate of flow of charge, it follows that the net charge entering the junction in a given time equals the net charge leaving the junction in the same time. Thus, this law (current law/junction law) is based on the conservation of charge. ]

Question 4.
State Kirchhoff’s second law or voltage law or loop law.
Ans. The algebraic sum of the potential differences (products of current and resistance) and the electromotive forces (emf’s) in a closed loop is zero. Σ IR + Σ E = 0

Question 5.
What is the sign convention used for Kirchhoff’s second law ? Explain with an example.
Answer:
(1) While tracing a loop, if we traverse a resistor along the direction of conventional current, the potential difference across the resistor is considered negative. If we traverse the resistor opposite to the direction of conventional current, the potential difference across the resistor is considered positive.

(2) While tracing a loop within the source, if we travel from the negative terminal of the source (cell) to the positive terminal of the source (cell), the emf of the source (cell) is considered positive.

On the contrary, if we travel from the positive terminal of the source (cell) to the negative terminal of the source (cell), the emf of the source (cell) is considered negative.

Consider the electrical network shown in above figure.
Tracing loop ABFGA in the clockwise direction, we get,
– I₁R₁ – I₃R₅ – I₁R₃ + E₁ = 0
∴ E₁ = I₁R₁ + I₃R₅ + I₁R₃
Tracing loop BFDCB in the anticlockwise direction, we get,
– I₃R₅ – I₂R₄ + E₂ – I₂R₂ = 0
∴ E₂ = I₂R₂ + I₃R₅ + I₂R₄

[Notes : (1) We may as well consider loop ABCDFGA and write the corresponding equation. (2) As the emf of a cell is the energy provided by the cell per unit charge in circulating the charge and the potential difference across a resistance is the work done per unit charge, it follows that this law (voltage law /loop law) is based on the conservation of energy.]

Question 6.
What is the basis of Kirchhoff’s current law and voltage law?
Answer:
Kirchhoff’s current law is consistent with the conservation of electric charge while the voltage law is consistent with the law of conservation of energy.

7. Solve the following

Question 1.
Determine the current flowing through the galvanometer shown in the figure below.

Solution:

To find I_g we apply Kirchhoff’s voltage law.
Loop ABDA :
– 5I₁ – 10I_g + 15I₂ = 0
∴ – 5I₁ + 15I₂ = 10I_g
∴ -I₁ + 3I₂ = 2I_g ………….. (1)
Loop BCDB :
-10(I₁ – I_g) + 20 (I₂ + I_g) + 10I_g = 0
∴ – 10I₁ + 10I_g + 20I₂ + 20I_g + 10I_g = 0
∴ I₁ – 2I₂ = 4I_g …………. (2)
Adding Eqs. (1) and (2), we get, I₂ = 6 I_g …………. (3)
Substituting for Z2 from Eq. (3) in Eq. (2).
∴ I₁ = 12I_g + 4I_g = 16I_g
Now, I₁ + I₂ = 2 A by the data.
∴ 16I_g + 6I_g = 2A
∴ 22I_g = 2A
∴ I_g = \(\frac{2}{22}\) A = \(\frac{1}{11}\) A from B to D
This is the current flowing through the galvanometer.

Question 8.
State the factors on which the resistance Of a material depends.
Answer:
The resistance of a material depends upon tem-perature, strain, humidity, etc.

[Note : Depending upon the factors stated above, resistance may vary from near zero to thousands of megaohm.]

Question 9.
What are the applications of Wheatstone’s metre bridge?
Answer:

1. Wheatstone’s metre bridge is used for measuring the values of very low resistance precisely.
2. It can also be used to measure the quantities such as strain galvanometer, resistance, capacitance of a capacitor, inductance of an inductor, impedence of a combination of a resistor, capacitor and inductor and the internal resistance of a cell.

Question 10.
What is the balance point in Kelvin’s method to measure the resistance of a galvanometer?
Answer:
Kelvin’s method of determination of the galvanometer resistance is an equal deflection method. The balance point in Kelvin’s method is a point on the wire for which the bridge network is balanced and the galvanometer shows no change in deflection.

Question 11.
Why is Kelvin’s method to measure the resistance of a galvanometer called an equal deflection method?
Answer:
In Kelvin’s method of determination of the galvanometer resistance using a Wheatstone metre bridge, the galvanometer is connected in one gap of the bridge and a variable known resistance is connected in the other gap. The junction of the two gaps (say, B) is connected directly to a pencil jockey.

The jockey is tapped along the wire to locate the equipotential balance point D when the galvanometer shows no change in deflection.

Since the galvanometer shows the same deflection on making or breaking the contact between the jockey and the wire, the method is an equal deflection method.

12. Solve the following :

Question 1.
Four resistances 5 Ω, 10 Ω, 15 Ω and X (unknown) are connected in the cyclic order so as to form a Wheatstone network. Determine X if the network is balanced.
Solution:
Data : P = 5 Ω, Q = 10 Ω, R = 15 Ω

Since the network is balanced,
\(\frac{P}{Q}=\frac{X}{R}\)
∴ \(\frac{5}{10}=\frac{X}{15}\)
∴ X = 15 × \(\frac{5}{10}\) = 7.5 Ω

Question 2.
Resistances P = 10 Ω, Q = 15 Ω, S = 50 Ω and R = 25 Ω are connected in order in the arms AB, BC, CD and DA respectively of a Wheatstone network ABCD. A cell is connected between A and C. What resistance has to be connected in parallel to S to balance the network?
Solution:
Data : P = 10 Ω, Q = 15 Ω, S = 50 Ω and R = 25 Ω
Let x = resistance to be connected in parallel to S to balance the network. The resistance of the parallel combination of S and x is \(\frac{S x}{S+x}\) .
For the balanced network,

Question 3.
Four resistances 4 Ω, 8 Ω, X and 6 Ω are connected in the cyclic order so as to form Wheatstone’s network. If the network is balanced, find X.
Solution:
Data : P = 4Ω, Q = 8 Ω, R = X and S = 6 Ω
Since the network is balanced,
∴ \(\frac{P}{Q}=\frac{S}{R}\)
∴ X = 6\(\frac{8}{4}\) = 12 Ω

Question 4.
Four resistances 4 Ω, 4 Ω, 4 Ω and 12 Ω form a Wheatstone network. Find the resistance which connected across the 12 Ω resistance will balance the network.
Solution:
The resistance in each of the three arms of the network is 4 Ω. Hence, to balance the network, the resistance in the fourth arm must also be 4 Ω.

Hence, the resistance (R) to be connected across. i.e., in parallel to, the 12 Ω resistance should be such that their equivalent resistance is 4Ω.

Question 5.
Four resistances 80 Ω, 40 Ω, 10 Ω and 15 Ω are connected to form Wheatstone’s network ABCD as shown in the following figure.

What resistance must be connected in the branch containing 10 Ω to balance the network?
Solution:
Data : P = 80 Ω, Q = 40 Ω, 10 Ω and R = 15 Ω are connected as shown in above figure.
When the network is balanced,
\(\frac{P}{Q}=\frac{S}{R}\)
∴ \(\frac{80}{40}=\frac{S}{15}\)
∴ S = 15 × 2 = 30 Ω
Let X be the resistance to be connected in series with 10 Ω, so as to obtain 30 Ω.
∴ X + 10 = 30
∴ X = 20 Ω

Question 6.
An unknown resistance is placed in the left gap and resistance of 50 ohms is placed in the right gap of a meter bridge. The null point is obtained at 40 cm from the left end. Determine the unknown resistance.
Solution:
Data : R =50 C in the right gap, l_X =40 cm
\(\)
Now, l_R = 100 – l_X = 100 – 40 = 60 cm
∴ \(\frac{X}{50}=\frac{40}{60}\)
∴ X = 50 × \(\frac{2}{3}\) = \(\frac{100}{3}\) = 33.33 Ω
This is the unknown resistance.

Question 7.
Two resistances X and Y in the two gaps of a metre bridge give a null point dividing the wire in the ratio 2 : 3. If each resistance is increased by 30 Ω, the new null point divides the wire in the ratio 5 : 6, calculate each resistance.
Solution:
From the data, we have in the first case,
\(\frac{X}{Y}=\frac{l_{X}}{l_{Y}}=\frac{2}{3}\)
∴ 3X = 2Y ……….. (1)
and in the second case, \(\frac{X+30}{Y+30}=\frac{l_{X+30}}{l_{Y+30}}=\frac{5}{6}\)
∴ 6X + 180 = 5Y + 150
∴ 6X – 5Y = -30
∴ 6X = 5Y – 30
∴ 2(3X) = 5Y – 30 …………. (2)
Substituting the value of 3X from Eq. (1) in Eq. (2),
we get,
2(2Y) = 5Y – 30
∴ Y = 30 Ω
∴ X = \(\frac{2}{3}\)Y = 20 Ω

Question 8.
In a metre bridge experiment, with a resistance R₁ in the left gap and a resistance X in the right gap, the null point is obtained at 40 cm from the left end of the wire. With a resistance R₂ in the left gap and the same resistance X in the right gap, the null point is obtained at 50 cm from the left end of the wire. Where will be the null point if R₁ and R₂ are connected first in series and then in parallel in the left gap, the right gap still containing X?
Solution:
From the data in the example, we have:
When R₁ is connected in the left gap and X in the right gap, l_R1 = 40 cm, and with R₂ in the left gap, l_R2 50 cm.
∴ In the first case,
l_X = 100 – l_R1 = 100 – 40 = 60cm

(i) When R₁ and R₂ are connected in series, the effective resistance is
R_S = R₁ + R₂ = \(\frac{2 X}{3}\) + X = \(\frac{5 X}{3}\)
Let the corresponding null point be at a distance l₁ from the left end of the wire.

(ii) When R₁ and R₁ are connected in parallel, the effective resistance is
R_P = \(\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{\frac{2 X}{3} \times X}{\frac{2 X}{3}+X}=\frac{2 X}{5}\)
Let the corresponding null point be at a distance l₂ Erom the left end of the wire.

With the same resistance X in the right gap, the null points will be at 62.5 cm and 28.6 cm from the left end of the wire for the series and parallel combinations respectively of R₁ and R₂ in the left gap.

Question 9.
A uniform wire is cut into two pieces such that one piece is twice as long as the other. The two pieces are connected in parallel in the left gap of a metre bridge. When a resistance of 20 Ω is connected in the right gap, the null point is obtained at 60 cm from the right end of the bridge wire. Find the resistance of the wire before it was cut into two pieces.
Solution:
Let R_w be the resistance of the wire before it was cut into two pieces. Let L₁, L₂ and X₁, X₂ be the lengths and resistance of the two pieces.

∴ The original resistance of the wire is
R_w = X₁ + X₂ = 40 + 20 = 60 Ω

Question 10.
Two resistances, 20 Ω and 30 Ω, are connected across the two gaps of a metre bridge as shown in the following figure.

The bridge wire has diameter 0.5 mm and resistance 2 Ω. The emf of the cell is 2 V. Assume that the internal resistance of the cell is zero. Calculate (I) the resistivity (specific resistance) of the material of the bridge wire (ii) the current provided by the cell when the bridge is balanced.
Solution:
Data: R₁ = 20 Ω, R₂ = 30 Ω. I_g = 0, R (wire) = 2 Ω,
r (wire) = 0.25 mm = 2.5 x iO m (as the diameter of the wire is 0.5mm), l = 1 m, E = 2V

Question 11.
Two diametrically opposite points of a metal ring are connected to two tenninals of the left gap of a metre bridge. A resistance of 11 Ω is connected in the right gap. If the null point is obtained at 45 cm from the left end, find the resistance of the metal ring.
Solution:
Data: R = 11 Ω, L_X = 45 cm
∴ L_R = 100 – L_X = 100 – 45 = 55cm
Let \(\frac{X}{2}\) be the resistance of each half of the metal ring. Therefore the resistance in the left gap is the effective resistance of the parallel combination of \(\frac{X}{2}\) and \(\frac{X}{2}\)

The resistance of the metal ring is 36 Ω

Question 12.
In a metre bridge experiment, the resistances R and X are connected in the left and right gap, respectively, and the null point is obtained at 33.7 cm from the left end. When a resistance of 12 Ω is connected in parallel to X, the null point is obtained at 51.9 cm from the left end. Calculate R and X.
Solution:
(1) In the first case,
l_R = 33.7 cm, l_X = 100 – 33.7 = 66.3 cm
∴ \(\frac{X}{R}=\frac{l_{X}}{l_{R}}=\frac{66.3}{33.7}\) …………….. (1)

(2) In the second case,
l_R = 51.9 cm, l_X = 100 – 51.9 = 48.1 cm
X’ = X || 12 Ω = \(\frac{12 X}{12+X}\)
∴ \(\frac{X^{\prime}}{R}=\frac{48.1}{51.9}\) ……………. (2)
Dividing Eq. (1) by Eq. (2), we get,

Then, from Eq. (1),
R = 13.48 × \(\frac{33.7}{66.3}\) = 6.852 Ω

Question 13.
What is the SI unit of potential gradient?
Answer:
The SI unit of potential gradient is the \(\frac{\text { volt }}{\text { metre }}\left(\frac{\mathrm{V}}{\mathrm{m}}\right)\).

Question 14.
Explain how two cells are connected so as to
(i) assist each other
(ii) oppose each other. Write the formulae for the corresponding effective emf.
Answer:
Consider two cells connected so that the positive terminal of the first cell is connected to the negative terminal of the second cell as shown in figure (a). The emf’s of the two cells are added up and the effective emf of the combination of the two cells is E₁ + E₂. This method of connecting two cells is called the sum method. Here, the cells assist each other.

Consider two cells connected so that their negative terminals are connected together or their positive terminals are connected together as shown in figure (b).

In this case their emf’s oppose each other and the effective emf of the combination of the two cells is E₁ – E₂(E₁ > E₂ assumed). This method of connecting two cells is called the difference method.

Question 15.
What is the internal resistance of the cell ?
Answer:
The internal resistance of a cell is the resistance offered by the electrolyte and electrodes in the cell.

Question 16.
Explain how a potentiometer is used as a voltage divider.
Answer:
A potentiometer can be used as a voltage divider to continuously change the output voltage of a voltage supply. As shown in the below figure, potential difference V is set up between points A and B of a potentiometer wire. One end of a device is connected to positive point A and the other other end is connected to a slider that can move along wire AB. The voltage V divides in proportion of lengths l₁ and l₂ as shown in figure.

By using the slider we can change the output voltage from 0 to V.

Question 17.
State the precautions which must be taken in using a potentiometer.
Answer:
Precautions to be taken in using a potentiometer:

1. The potential difference across the potentiometer wire must be greater than the emf or potential difference to be balanced. Hence, in comparing emfs, the driver emf E > E₁, E₂ (direct method) or E > E₁ + E₂ (combination method).
2. The positive terminal of the cell with emf E1 and that with emf E₂, or their combination, must be connected to the higher potential terminal of the potentiometer.
3. The potentiometer wire must be of uniform cross section and homogeneous.
4. The potentiometer wire should be long and have a high resistivity and low temperature coefficient of resistance.

Question 18.
State the advantages of a potentiometer over a voltmeter.
Answer:
Advantages of a potentiometer over a voltmeter :
(1) The cell, whose emf is being measured, draws no current from the circuit at the null point. Thus, the potentiometer measures the open-circuit potential difference across its terminals, or the emf E. A voltmeter will measure the terminal potential difference, V, of the cell in a closed circuit. This is because the resistance of a voltmeter is high but not infinite and hence the voltmeter is not ideal.

(2) By setting up a suitably small potential gradient along a long potentiometer wire, any small voltage can be measured. Increasing the length of the wire effectively decreases the potential gradient, and increases both the precision and accuracy of measurement.

(3) The adjustment of a potentiometer is a ‘null’ method which does not, in any way, depend on the calibration of the galvanometer. The galvanometer is used only to detect the current, not to measure it. The accuracy of a voltmeter is limited by its calibration.

(4) Since a potentiometer can measure both the emf and terminal potential difference of a cell, the internal resistance of the cell can be found.

19. Solve the following

Question 1.
In a potentiometer circuit. E = 2 V, r = 2 Ω, R_wire = 10 Ω, R_ext = 1988 Ω and L = 4 m (in the usual notation). What is the potential gradient along the wire ?
Solution:

The potential gradient along the wire is 2.5 × 10^-3 V/m

Question 2.
A potentiometer wire has length 2 m and resistance 10 Ω. It is connected in series with a resistance 990 Ω and a cell of emf 2 V. Calculate the potential gradient along the wire.
Solution:
Data : L 10 m, R = 10 Ω, R_ext = 990 Ω, E =2 V
The current in the circuit is
I = \(\frac{E}{R+R_{\text {ext }}}=\frac{2}{10+990}=\frac{2}{1000}\) = 2 × 10^-3 A
The potential difference across the wire is
V = IR = 2 × 10^-3 × 10 = 2 × 10^-2 V
∴ The potential gradient along the wire
= \(\frac{V}{L}=\frac{2 \times 10^{-2}}{2}\) = 10^-2 V/m

Question 3.
A potentiometer wire 4 m long has a resistance of 4 Ω. What resistance must be connected in series with the wire and a cell of emf 2 V having internal resistance of 2 Ω to get a potential drop of 10^-3 V/cm along the wire?
Solution:
Data: L = 4 m, R = 4 Ω, E = 2 V, r = 2 Ω
The required potential drop per unit length of the wire is
10^-4 V/cm = \(\frac{10^{-3} \mathrm{~V}}{10^{-2} \mathrm{~m}}\) = 0.1 V/m
Let R_S be the series resistance for which the desired potential drop is obtained.
The current in the circuit is

Question 4.
A potentiometer wire of length 5 m is connected to a battery. For a certain cell having negligible internal resistance, the null point is obtained at 250 cm. If the length of the potentiometer wire is increased by 1 m, where will be the new position of the null point?
Solution:
Data : L₁ = 5m, L₂ = 6m, l₁ = 250 cm
E = (\(\frac{V}{L}\))l
where V/L is the potential gradient and l is the balancing length.
∴ E₁= (\(\frac{V}{L_{1}}\))l₁ = (\(\frac{V}{L_{2}}\))₂
l₂ = (\(\frac{L_{2}}{L_{1}}\)) × l₁ = \(\frac{6}{5}\) × 250 = 300 cm

Question 5.
A potentiometer wire, of length 4 m and resistance 8 Ω, is connected in series with a battery of emf 2 V and negligible internal resistance. If the emf of the cell balances against a length of 217 cm of the potentiometer wire, find the emf of the cell. When the cell is shunted by a resistance of 15 Ω, the balancing length is reduced by 17 cm. Find the internal resistance of the cell.
Solution:
Data: L = 4 m, R = 8 Ω, E = 2V, r = 0,
R(shunt) = 15 Ω, l = 217 cm = 2.17 m,
l₁ = 217 – 17 = 200 cm = 2 m

Question 6.
Two cells of emf’s E₁ and E₂ (E₁ > E₂) are connected in a potentiometer circuit so as to assist each other. The null point is obtained at 8.125 m from the high potential end of the potentiometer wire. When the cell with emf E₂ is connected so as to oppose the emf E₁, the null point is obtained at 1.25 m from the same end. Compare the emf’s of the two cells.
Solution:
Data : l₁ = 8.125 m (cells assisting), l₂ = 1.25 m (cells opposing)
E₁ + E₂ = Kl₁ and E₁ – E₂ = Kl₂
where K is the potential gradient.

Question 7.
A cell balances against a length of 200 cm on a potentiometer wire when it is shunted by a resistance of 8 Ω. The balancing length reduces by 40 cm when it is shunted by a resistance of 4 Ω. Calculate the balancing length when the cell is in an open circuit. Also calculate the internal resistance of the cell.
Solution:
Data : Part I : R = 8 Ω, l₂ = 200 cm;
Part II : R = 4 Ω, l₂ = 160 cm

This is the balancing length when the cell is in open circuit. The internal resistance of the cell,

Question 8.
When a resistance of 12 Ω is connected across a cell, its terminal potential difference is balanced by 120 cm of a potentiometer wire. When a resistance of 18 Ω is connected across the same cell, the balancing length is 150 cm. Find the balancing length when the cell is in open circuit. Also calculate the internal resistance of the cell.
Solution:
Data : Part I : R = 12 Ω, l₂ = 120 cm;
Part II : R = 18 Ω, l₂ = 150 cm

∴ \(\frac{l_{1}-120}{10}=\frac{6\left(l_{1}-150\right)}{50}\)
∴ 5 (l₁ – 120) = 6(l₁ -150)
∴ 5l₁ – 600 = 6l₁ – 900
∴ l₁ = 300 cm
This is the balancing length when the cell is in open circuit.
∴ r = \(12\left(\frac{l_{1}-120}{120}\right)=\frac{300-120}{10}=\frac{180}{10}\) = 18 Ω
This is the internal resistance of the cell a unit

Question 20.
What is a galvanometer?
Answer:
A galvanometer is a device used to detect weak electric currents in a circuit. The current may be of the order of a few microamperes, or even a few nanoamperes.

Question 21.
State the principle of working of a moving coil galvanometer.
Answer:
A current-carrying coil suspended in a magnetic field experiences a torque which rotates the plane of the coil and tends to maximize the magnetic flux through the coil. The torque due to the spring or the suspension fibre to which the coil is attached tends to restore the coil to its initial position. In equilibrium, the coil comes to rest and its deflection is proportional to the current through the coil.

Question 22.
Explain the basic construction of a galvanometer.
Answer:
A galvanometer consists of a coil of a large number of turns of fine insulated copper wire wound on a rectangular nonconducting, nonmagnetic frame. The coil is pivoted (or suspended) between cylindrically concave pole pieces of a horseshoe strong permanent magnet. The coil swings freely around a cylindrical soft iron core fitted between the pole pieces. The deflection of the coil can be read with a pointer attached to it. The position of the pointer on the scale provided depends on the current passing through the galvanometer (or the potential difference across it). A galvanometer can be used as an ammeter or a voltmeter with a suitable modification.

[Note : A table galvanometer has a resistance of about 50 Ω and can carry a current up to about 1 mA.]

Question 23.
What are the modifications necessary to convert a moving-coil galvanometer (MCG) into an ammeter?
Answer:
To convert a moving-coil galvanometer (MCG) into an ammeter, the following modifications are necessary :

1. The effective current capacity of the MCG must be increased to a desired higher value.
2. A galvanometer when connected in series with a resistance, should not decrease the current through the resistance. Hence, the effective resistance of the galvanometer must be decreased by connecting an appropriate low resistance across it. An ideal ammeter should have zero resistance.
3. It must be protected from the damages which are likely to occur due to the passage of an excess electric current.

Question 24.
State the function of the shunt in modifying a galvonometer to an ammeter.
Answer:
Functions of the shunt:

1. It lowers the effective resistance of the ammeter
2. It is used to divert to a large part of total current by providing an alternate path and thus it protects the instrument from damage.
3. With a shunt of proper value, a galvanometer can be modified into an ammeter of practically any desired range.

Question 25.
Explain how a moving-coil galvanometer is converted into an ammeter. Derive the necessary formula.
Answer:
A moving-coil galvanometer is converted into an ammeter by reducing its effective resistance by connecting a low resistance S across the coil. Such a parallel low resistance is called a shunt since it shunts a part of the current around the coil, shown in below figure. That makes it possible to increase the range of currents over which the meter is useful.

Let I be the maximum current to be measured and I_g the current for which the galvanometer of resistance G shows a full-scale deflection. Then, the shunt resistance S should be such that the remaining current I – I_g = I_s is shunted through it.

In the parallel combination, the potential difference across the galvanometer = the potential difference across the shunt
∴ I_g G = I_s S
= (I – I_g)S
∴ S = (\(\frac{I_{\mathrm{g}}}{I-I_{\mathrm{g}}}\)) C
This is the required resistance of the shunt. The scale of the galvanometer is then calibrated so as to read the current in ampere or its submultiples (mA. µA) directly.

[Notes :
(1) Thick bars of manganin are used for shunts because manganin has a very small temperature coefficient of resistivity.
(2) The fraction of the current passing through the galvanometer and shunt are, respectively,
\(\frac{I_{g}}{I}=\frac{S}{S+G}\) and \(\frac{I_{g}}{I}=\frac{G}{S+G}\)
(3) On the right hand side of Eq. (1), dividing both the numerator and denominator by I_g, we get,
S = \(\frac{1}{\left(I / I_{\mathrm{g}}\right)-1}\) ∙ G = \(\frac{G}{p-1}\)
where p = I/Ig is the range-multiplying factor, i.e., the current range of the galvanometer can be increased by a factor p by connecting a shunt whose resistance is smaller than the galvanometer resistance by a factor p – 1.
∴ p = \(\frac{G+S}{S}\)
If R_A is the resistance of the ammeter,
R_A = \(\frac{G S}{G+S}=\frac{G}{p}\)]

Question 26.
How do you calculate the shunt required to increase the range p times?
Answer:
The value of shunt resistance required to convert a galvanometer into an ammeter is given by,
S = (\(\frac{I_{\mathrm{g}}}{I-I_{\mathrm{g}}}\)) G
If the current I is p times the current I_g, then I = pI_g. Using this in the above expression, we get,
S = \(\frac{G I_{\mathrm{g}}}{p I_{\mathrm{g}}-I_{\mathrm{g}}}\) OR S = \(\frac{G}{p-1}\)
This is the required shunt resistance to increase the range p times.

Question 27.
What is the current flowing through the shunt resistance?
Answer:

If I_S is the current through the shunt resistance, then the remaining current (I – I_S) will flow through the galvanometer.

Now, the potential difference across the galvanometer = the potential difference across the shunt
∴ G(I – I_S) = S I_S
∴ GI – GI_S = S I_S
∴ SI_S + GI_S = G I
∴ I_S = (\(\frac{G}{S+G}\))I
This is the current flowing through the shunt resistance.

Question 28.
What are the modifications required to convert a moving-coil galvanometer into a voltmeter?
Answer:
The modifications required to convert a moving- coil galvanometer into a voltmeter are as follows :

1. The effective resistance of the galvanometer should be very high. This is because a voltmeter requires a very small current to deflect its pointer. If a larger current than this flows through the voltmeter, the voltmeter is said to load the circuit and it will record a much smaller voltage drop.
2. The voltage measuring capacity (range) should be increased to a desired value.
3. It must be protected from damages which are likely to occur due to an excess applied potential difference.

Question 29.
Explain how a moving-coil galvanometer is converted into a voltmeter. Derive the necessary formula.
Answer:
A moving-coil galvanometer is converted into a voltmeter by increasing its effective resistance by connecting a high resistance Rs in series with the galvanometer, shown in figure. The series resistance is also useful for changing the range of any given voltmeter.

Let G be the resistance of the galvanometer coil and I_g the current required for a full-scale deflection.

Let V be the maximum potential difference to be measured. The value of the series resistance R_S should be such that when the potential difference applied across the instrument is V, the current through the galvanometer is I_g.

In the series combination, the potential difference V gets divided across the galvanometer (resistance, G) and the resistance R_S :
V = I_gG + I_gR_S = I_g(G + R_S)
∴ R_S = \(\frac{V}{I_{\mathrm{g}}}\) – G
This is the required value of the series resistance. The scale of the galvanometer is then calibrated so as to read the potential difference in volt or its submultiples, e.g., mV, directly.

[Notes :
(1) A series multiplier is made of manganin wire because manganin has a very small temperature coefficient of resistivity.
(2) The maximum potential difference V_g that can be dropped across the galvanometer is V_g = I_g G. Therefore, the above expression for the series resistance may be rewritten as
R_s = \(\frac{V G}{I_{\mathrm{g}} G}\) – G
= \(\frac{V G}{V_{\mathrm{g}}}\) – G = G(p – 1)
where p = V/V_g is the range-multiplying factor, i.e., the voltage range of the galvanometer can be increased by a factor of p by connecting a series resistance which is (p – 1) times the galvanometer resistance.
∴ p = \(\frac{V}{V_{g}}=\frac{\left(R_{\mathrm{S}}+G\right) I_{\mathrm{g}}}{G I_{\mathrm{g}}}=\frac{R_{\mathrm{S}}+G}{G}\)
Since the resistance of the voltmeter is R_v = R_S + G,
p = \(\frac{R_{\mathrm{V}}}{G}\)
∴R_v = G_p]

Question 30.
State the functions of the series resistance in modifying a galvanometer into a voltmeter.
Answer:
Functions of the series, resistance:

1. It increases the effective resistance of the voltmeter.
2. It drops off a larger fraction of the measured potential difference thus protecting the sensitive meter movement of the basic galvanometer.
3. With resistance of proper value, a galvanometer can be modified to a voltmeter of desired range. .

Question 31.
Distinguish between an ammeter and a voltmeter
Answer:

Ammeter	Voltmeter
1. It measures current.	1. It measures potential difference.
2. It is connected in series with a resistance.	2. It is connected in parallel to a resistance.
3. An ammeter should have very low resistance (ideally zero).	3. A voltmeter should have very high resistance (ideally infinite).
4. Its range can be increased by decreasing the value of shunt resistance.	4. Its range can be increased by increasing the value of series resistance.
5. The resistance of an ammeter is RA =	5. The resistance of a voltmeter is Rv = G + Rs = Gp.

32. Solve the following

Question 1.
Calculate the value of the shunt which when connected across a galvanometer of resistance 38 Ω will allow 1/20th of the current to pass through the galvanometer.
Solution:
Data : G = 38 Ω, I_g / I = \(\frac{1}{20}\)

∴ S + 38 = 20 S
∴ 19 S = 38
∴ S = 2 Ω
This is the required value of the shunt.

Question 2.
A galvanometer is shunted by 1/r of its resistance. Find the fraction of the total current passing through the galvanometer.
Solution:
Let G be the resistance of the galvanometer, I the total current and I_g the current through the galvanometer when it is shunted. The resistance of the shunt is

The fraction \(\frac{1}{1+r}\) of the total current passes through the galvanometer.
[Note : The fraction of the current through the shunt = \(\frac{I_{\mathrm{S}}}{I}=1-\frac{I_{\mathrm{g}}}{I}=\frac{r}{1+r}\)]

Question 3.
A resistance of 3 Ω is connected in parallel to a galvanometer of resistance 297 Ω. Find the fraction of the current passing through the galvanometer.
Solution:
Data : G = 297 Ω, S = 3 Ω
I_g = \(\frac{S}{S+G}\) ∙ I
∴ \(\frac{I_{\mathrm{g}}}{I}=\frac{S}{S+G}=\frac{3}{3+297}=\frac{3}{300}\) = 0.01
This is the fraction of the current through the galvanometer.

[Note: The fraction of the current through the shunt
= \(\frac{I_{\mathrm{S}}}{I}\) = 1 – 0.01 = 0.99]

Question 4.
The combined resistance of a galvanometer of resistance 1000Ωand its shunt is 25 Ω. Calculate the value of the shunt.
Solution:
Data: G = 1000 Ω, R_A =25 Ω

This is the value of the shunt.

Question 5.
A galvanometer with a coil of resistance 40 Ω gives a full scale deflection for a current of 5 mA. How will you convert it into an ammeter of range 0 – 5 A?
Solution:
Data: G = 40 Ω, I_g = 5 mA = 5 × 10^-3 A, I = 5 A
To convert a galvanometer into an ammeter, a shunt (i.e., low resistance in parallel) should be connected with the galvanometer coil. The required shunt resistance,
S = (\(\frac{I_{\mathrm{g}}}{I-I_{\mathrm{g}}}\)) G = (\(\frac{5 \times 10^{-3}}{5-5 \times 10^{-3}}\)) × 40
= \(\frac{200}{4995}\) = 0.04 Ω

Question 6.
A galvanometer has a resistance of 16 Ω and gives a full scale deflection when a current of 20 mA is passed through it. The only shunt resistance available is 0.04 Ω which is not sufficient to convert the galvanometer to an ammeter to measure up to 10 A. What resistance should be connected in series with the coil of the galvanometer so that the range of the ammeter is 10 A ?
Solution:
Data: G =16 Ω, I_g = 20 mA = 0.02 A,
S = 0.04 Ω, I = 10 A
Let X be the resistance to be connected in series with the coil of the galvanometer.
The fraction of the current through the galvanometer is \(\frac{I_{\mathrm{g}}}{I}\).

\(\frac{I_{\mathrm{g}}}{I}=\frac{S}{(G+X)+S}\)
∴ (G + X) + S = \(\frac{I}{I_{\mathrm{g}}}\) × S = \(\frac{10}{0.02}\) × 0.04 = 20
∴ X = 20 – (16 + 0.04) = 3.96 Ω
Alternate method:
Let G’ = G + X = 16 + X
The range multiplying factor,
p = \(\frac{I}{I_{\mathrm{g}}}=\frac{10}{0.02}\) = 500
Then, S = \(\frac{G^{\prime}}{p-1}\)
∴ G’= 16 + X = S(p – 1)
= 0.04(500 – 1) = 19.96 Ω
∴ X = 19.96 – 16 = 3.96 Ω

Question 7.
A galvanometer of resistance 100 C gives a full scale deflection for a current of 2 mA. How will you use it to measure (i) current up to 2 A (ii) voltage up to 10 V?
Solution:
Data : G = 100 Ω, I_g = 2 mA = 2 × 10^-3 A, I = 2 A, V = 10 V

A resistance of 0.1001 Ωshould be connected in parallel to the coil of the galvanometer to measure current up to 2 A.

(ii) R_s = \(\frac{V}{I_{\mathrm{g}}}\) – G
= \(\frac{10}{2 \times 10^{-3}}\) – 100 = 5000 – 100 = 4900 Ω
A resistance of 4900 Ω should be connected in series with the coil of the galvanometer to measure voltage up to 10 V.

Question 8.
Calculate the value of resistance needed to convert a moving-coil galvanometer of 60 Ω which gives a full scale deflection for a current of 50 mA into (i) an ammeter of range 0 – 5 A (ii) a voltmeter of range 0 – 50 V.
Solution:
Data : G = 60 Ω, I_g = 50 mA = 5 × 10^-2 A, I = 5 A, V = 50 V

A resistance of 0.6061 Ω should be connected in parallel to the coil of the galvanometer to measure current up to 5 A.

(ii) R_s = \(\frac{V}{I_{\mathrm{g}}}\) – G
= \(\frac{50}{5 \times 10^{-2}}\) – 60
= 1000 – 60 = 940 Ω
A resistance of 940 Ω should be connected in series with the coil of the galvanometer to measure voltage up to 50 V.

Question 9.
A voltmeter of resistance 500 Ω can measure a maximum voltage of 5 V. How can it be made to measure a maximum voltage of 100 V? Solution:
Data : G = 500 Ω, V_g = 5 V, V = 100 V
To increase the range of the voltmeter by a factor p = \(\frac{V}{V_{\mathrm{g}}}\), a resistance R should be connected in series with it.
R_s = G(p – 1) = 500(\(\frac{100}{5}\) – 1) = 9500 Ω

Question 10.
A moving-coil galvanometer of resistance 200 ohms gives a full scale deflection of 100 divisions for a current of 50 milliamperes. How will you convert it into an ammeter to read 2 amperes for 20 divisions?
Solution:
Data : G = 200 Ω, I_g = 50 mA = 50 × 10^-3 A
The total number of scale divisions is 100. The ammeter has to read 20 divisions for a current of 2 A. Hence, for 100 divisions, the current must be I = 10 A.

To convert the galvanometer into an ammeter, a shunt must be connected in parallel to the galvanometer coil. The required shunt resistance,

Question 11.
A galvanometer of resistance 50 Ω has a current sensitivity of 5 div/mA. The instrument has 25 divisions. How will you convert it into a voltmeter of range 0 – 50 V ?
Solution:
Data : G = 50 Ω, V = 50 V
For a current of 1 mA, the galvanometer shows a deflection of 5 divisions. Hence, for a full scale deflection (i.e. deflection of 25 divisions), the current passing through the galvanometer should be 5 mA.

∴ I_g = 5mA = 5 × 10^-3 A

To convert the galvanometer into a voltmeter, a high resistance must be connected in series with the galvanometer coil. This series multiplier,
R_s = \(\frac{V}{I_{\mathrm{g}}}\) – G = \(\frac{50}{5 \times 10^{-3}}\) – 50
= 9950 Ω

Multiple Choice Questions

Question 1.
For a Wheatstone network shown in the following figure, I_g = 0 when .

(A) E = 0
(B)V_B = V_D
(C) V_B > V_D
(D) V_B < V_D
Answer:
(B)V_B = V_D

Question 2.
A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series, first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emfs is
(A) 5 : 4
(B) 3 : 4
(C) 3 : 2
(D) 5 : 1
Answer:
(C) 3 : 2

Question 3.
When the current in a potentiometer wire decreases, the potential gradient
(A) decreases
(B) increases
(C) remains the same provided the resistance and the length of the wire remain the same
(D) remains the same, irrespective of the resistance . of the wire and its length.
Answer:
(A) decreases

Question 4.
In the circuit given below the current through the 6 Ω resistor will be

Answer:
(B) \(\frac{2}{3}\) A

Question 5.
The accuracy of a potentiometer wire can be increased by
(A) increasing its length
(B) decreasing its length
(C) using a cell of higher emf
(D) using a cell of lower emf.
Answer:
(A) increasing its length

Question 6.
A cell of emf 1.1 V and internal resistance r is connected across an external resistor of resistance R = 10 r. The potential difference across the resistor is
(A) 0.1 V
(B) 0.9 V
(C) 1.0 V
(D) 1.1 V.
Answer:
(C) 1.0 V

Question 7.
When a metal conductor connected in the left gap of a metre bridge is heated, the null point
(A) will shift towards right
(B) will shift towards left
(C) will remain unchanged
(D) will shift towards right or left depending upon the resistivity of the metal.
Answer:
(A) will shift towards right

Question 8.
In using a Wheatstone’s bridge to accurately measure an unknown resistance, a calibrated known variable resistor is varied until
(A) the change in the galvanometer reading is zero
(B) a change in the value of the variable resistor produces no change in the galvanometer reading
(C) the potential difference across the unknown resistance is zero
(D) the potential difference across the galvanometer is zero.
Answer:
(D) the potential difference across the galvanometer is zero.

Question 9.
In a Wheatstone network, the resistances in cyclic order are P = 10 Ω, Q = 5 Ω, S = 4 Ω and R = 4 Ω. Then, for the bridge to balance,
(A) 5 Ω should be connected in parallel to Q = 5 Ω
(B) 10 Ω should be connected in series with Q = 5 Ω
(C) 5 Ω should be connected in series with P = 10 Ω
(D) 10 Ω should be connected in parallel to P = 10 Ω
Answer:
(D) 10 Ω should be connected in parallel to P = 10 Ω

Question 10.
Two resistors, R₁ and R, are connected in the left gap and the right gap of a metre bridge, and the balancing length is obtained at 20 cm from the left. On inter-changing the resistors in the two gaps, the balancing length shifts by
(A) 20 cm
(B) 40 cm
(C) 60 cm
(D) 80 cm.
Answer:
(C) 60 cm

Question 11.
An instrument which can measure terminal potential difference as well as electromotive force (emf) is
(A) Wheatstone’s metre bridge
(B) a voltmeter
(C) a potentiometer
(D) a galvanometer
Answer:
(C) a potentiometer

Question 12.
A 10 m long wire of resistance 2012 is connected in series with a resistance of 10 Ω and a battery of emf 3 V and negligible internal resistance. The potential gradient, in µV / mm, along the wire is
(A) 2
(B) 20
(C) 200
(D) 2000
Answer:
(C) 200

Question 13.
A 10 m long potentiometer wire has a resistance of 20 Ω. If it is connected in series with a resistance of 55 Ω and a cell of emf 4 V and internal resistance 5 Ω, the potential gradient along the wire is
(A) 0.1 V/m
(B) 0.08 V/m
(C) 0.01 V/m
(D) none of these.
Answer:
(A) 0.1 V/m

Question 14.
A potential gradient of 6 × 10^-3 V/ mm is set up on a potentiometer wire which has a resistance of 2 Ω/m. Two emfs 2.5 V and 1.3 V, once assisting and then opposing each other, are balanced on the wire. The balancing lengths in the two cases are in the ratio
(A) 19 : 2
(B) 19 : 6
(C) 25 : 13
(D) 2:1.
Answer:
(B) 19 : 6

Question 15.
A potentiometer wire, having a resistance of 5 Ω and length 10 m, is connected in series a cell of emf 5 V and an external resistance of 495 Ω. A potential difference of 1.5 mV will balance against a length of
(A) 3 cm
(B) 30 cm
(C) 3 m
(D) none of these.
Answer:
(B) 30 cm

Question 16.
A load resistance R is connected across a cell of emf E and internal resistance r. If the closed-circuit p.d. across the terminals of the cell is V, the internal resistance of the cell is
(A) (E – V) R
(B) (V – E)R
(C) \(\frac{E-V}{V} R\)
(D) \(\frac{E-V}{V}\)
Answer:
(C) \(\frac{E-V}{V} R\)

Question 17.
The open-circuit potential difference across the terminals of a cell balances on 150 cm of a potentiometer wire. When the cell is shunted by a 4.9 Ω resistor, the balancing length reduces to 147 cm. The internal resistance of the cell is
(A) 0.01 Ω
(B) 0.05 Ω
(C) 0.1 Ω
(D) 1 Ω
Answer:
(C) 0.1 Ω

Question 18.
To convert a galvanometer into an ammeter
(A) a high resistance is connected in parallel to the galvanometer
(B) a high resistance is connected in series with the galvanometer
(C) a low resistance is connected in parallel to the galvanometer
(D) a low resistance value is connected in series with the galvanometer.
Answer:
(C) a low resistance is connected in parallel to the galvanometer

Question 19.
To convert a galvanometer into a voltmeter
(A) a high resistance is connected in parallel to the galvanometer
(B) a high resistance is connected in series with the galvanometer
(C) a low resistance is connected in parallel to the galvonometer
(D) a low resistance is connected in series with the galvonometer.
Answer:
(B) a high resistance is connected in series with the galvanometer

Question 20.
An ideal ammeter has
(A) a moderate resistance
(B) a high resistance
(C) an infinite resistance
(D) zero resistance
Answer:
(D) zero resistance

Question 21.
An ideal voltmeter has
(A) a low resistance
(B) a high resistance
(C) an infinite resistance
(D) zero resistance
Answer:
(C) an infinite resistance

Question 22.
The fraction of the total current passing through the galvanometer is

Answer:
(A) \(\frac{S}{S+G}\)

Balbharti Maharashtra State Board 12th Physics Important Questions Chapter 8 Electrostatics Important Questions and Answers.

Maharashtra State Board 12th Physics Important Questions Chapter 8 Electrostatics

Question 1.
State Gauss’s law in electrostatics.
Answer:
Gauss’s law: The net flux through a closed surface in free space is related to the net charge qenci that is enclosed by that surface and is given by
Φ = \(\oint \vec{E} \cdot \overrightarrow{d \mathrm{~s}}=\frac{q_{\mathrm{encl}}}{\varepsilon_{0}}\)
where e0 is the permittivity of free space and the electric flux \(\vec{E} \cdot \overrightarrow{d \mathrm{~s}}\) is integrated over the entire area of the surface.

[Note: The SI unit of electric flux is the newton metre squared per coulomb (N∙m²/C) or, equivalently, the voltmetre (V\vec{E} \cdot \overrightarrow{d \mathrm{~s}}m).]

Question 2.
A Gaussian surface in the form of a cube is centred on a point charge Q. What is the flux through one face of the cube?
Answer:
Since the point charge is at the centre of the cube, the electric flux through each face is the same. Therefore, the electric flux through one face is \(\frac{1}{6}\)th of the electric flux originating from or terminating at the point charge.

If ε is the permittivity of the medium, the electric flux through one face = \(\frac{Q}{6 \varepsilon}\).

Question 3.
An arbitrary surface encloses a dipole. What is the electric flux through this surface?
Answer:
Since the net charge enclosed by the surface is zero, the net electric flux through the surface is also zero. . v

Question 4.
Obtain an expression for the electric field intensity at a point outside a charged conducting spherical shell.
Hence, obtain an expression for the electric intensity (i) on the surface of (i.e., just outside) the spherical conductor (ii) inside the spherical conductor.
Answer:
Consider an isolated charged hollow spherical conductor A, of radius R and surface charge density σ, placed in a medium of permittivity ε. Consider a . point P outside the conductor at a distance r from its centre. To find the electric field intensity at P, we choose a spherical Gaussian surface S of radius r through P and concentric with conductor A. A small element of this surface containing P has an area dS.

The charge Q is uniformly distributed over the outer surface of the spherical conductor. Then, by symmetry, the electric field intensity at every point on surface S is normal to the surface and has the same magnitude E. If charge Q is positive, \(\vec{E}\) at every point on S is radially outward.
The angle θ between \(\vec{E}\) and \(d \vec{S}\) being zero for every surface element, the electric flux through every element is
dΦ = \(\vec{E} \cdot d \vec{S}\) = E dS
Therefore, the flux through the Gaussian surface S is

where ε₀ is the permittivity of free space and k = \(\frac{\varepsilon}{\varepsilon_{0}}\) is the relative permittivity (dielectric constant) of the surrounding medium.
∴ E = \(\frac{\sigma\left(4 \pi R^{2}\right)}{4 \pi \varepsilon r^{2}}=\frac{\sigma R^{2}}{\varepsilon r^{2}}\) …………… (6)
Equations (5) and (6) give the magnitude of the electric field intensity at a point P outside a hollow spherical conductor. If the net charge Q enclosed by the Gaussian surface is positive, \(\vec{E}\) is radially outward; if Q is negative, \(\vec{E}\) is radially inward. Equation (5) shows that for a point outside a hollow spherical conductor carrying a charge Q, the conductor behaves like a point charge Q at its centre.

Case (1) : At a point just outside the sphere, r ≅ R.
∴ E = \(\frac{\sigma}{\varepsilon}=\frac{\sigma}{k \varepsilon_{0}}\)

Case (2) : Since electric charge resides on the outer surface of a hollow conductor, the charge inside the hollow spherical conductor is zero. Then, E_inside = 0.

[Notes : (1) The surface of a charged conductor is an equipotential surface so that the electric field just outside it must be normal to the surface of the conductor. For a spherical charged conductor, it follows that the field is radial, and because the net charge is, by symmetry, uniformly distributed over its outer surface, the field is spherically symmetric, the same as for a point charge.
(2) The electric field intensity is zero at all points inside a hollow charged conductor of arbitrary shape because under electrostatic condition the net charge of the charged conductor resides on its surface. This is also true for a hollow charged spherical conductor, if there is no charge in the cavity (e.g., on a conductor inside the cavity but insulated from the outer shell).]

Question 5.
Does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased?
Answer:
No. Electric flux, or the number of electric field lines, passing through the Gaussian surface is independent of the size of the Gaussian surface and depends on the number of field lines originating from or terminating at the point charge, which in turn depends on the magnitude of the point charge and the permittivity of the medium.

Question 6.
Obtain an expression for the electric field intensity at a point outside an infinitely long charged cylindrical conductor.
Answer:
Consider an isolated cylindrical conductor A, of radius R and carrying a charge per unit length λ. We assume the conductor to be infinitely long. Consider a point P outside the conductor at a distance r from its axis. To find the electric field intensity at P, we choose a cylindrical Gaussian surface S of radius r through P and coaxial with the conductor A. As λ is the charge per unit length of conductor A, the net charge enclosed by the Gaussian cylinder of length l is
Q = λl ………….. (1)
A small element on the curved part of the Gaussian surface and containing P has an area dS.

Charge is uniformly distributed over the outer surface of the cylindrical conductor. Then, by symmetry, the electric field intensity at any point outside the conductor is perpendicular to the cylinder axis. Hence, the component of the electric field intensity perpendicular to the plane circular faces of the Gaussian surface is zero. Therefore, the electric flux through these flat faces is zero.

By symmetry, the electric field intensity \(\vec{E}\) at every point on the curved face of surface S is normal to the surface and has the same magnitude \(\vec{E}\). If the charge on conductor A is positive, E is directed along the outward drawn normal \(d \vec{S}\).

The angle θ between \(\vec{E}\) and \(d \vec{S}\) being zero for every surface element, the electric flux through every element is
dΦ = \(\vec{E} \cdot d \vec{S}\) = E dS
Therefore, the flux through the curved face of the Gaussian surface S is
Φ = \(\oint\) E dS = E \(\oint\) dS ……….. (2)
\(\oint\) dS = area of the curved surface = 2πrl, where l is the length of the cylinder as shown in the figure.
∴ Φ = E × 2πrl ………….. (3)
Then, by Gauss’s theorem,
Φ = \(\frac{Q}{\varepsilon}\) = E × 2πrl ……………. (4)
∴ E = \(\frac{\lambda l}{\varepsilon(2 \pi r l)}=\frac{\lambda}{2 \pi \varepsilon r}=\frac{\lambda}{2 \pi k \varepsilon_{0} r}\) ……………… (5)
where ε₀ is the permittivity of free space and k = \(\frac{\varepsilon}{\varepsilon_{0}}\) is the relative permittivity (dielectric constant) of the surrounding medium.

This gives the magnitude of the electric field intensity in terms of the linear charge density λ. For positive λ, \(\vec{E}\) is outward, while for negative λ, \(\vec{E}\) is inward.

[Notes : (1) If σ is the surface charge density (charge per unit area), σ = \(\frac{\lambda l}{2 \pi R l}=\frac{\lambda}{2 \pi R}\) ∴ λ = 2πRσ
∴ E = \(\frac{2 \pi R \sigma}{2 \pi \varepsilon_{0} k r}=\frac{R \sigma}{\varepsilon_{0} k r}\) (2) If the charge on the conductor is negative, \(\vec{E}\) is inward.]

Question 7.
An infinitely long positively charged straight wire has a linear charge density λ. An electron is revolving around the wire as its centre with a constant speed in a circular plane perpendicular to the wire. Deduce the expression for its kinetic energy.
Answer:
Let r be the radius of the uniform circular motion of the electron. The electric field intensity \(\vec{E}\) at every point on the circular path is radially outward and has the same magnitude E = \(\frac{\lambda}{2 \pi \varepsilon r}\) .

∴ The centripetal force on the electron,
F_c = \(\frac{m_{\mathrm{e}} v^{2}}{r}\) = eE = \(\frac{e \lambda}{2 \pi \varepsilon r}\)
where m_e and v are the mass and linear speed of the electron.
∴ m_ev² = \(\frac{e \lambda}{2 \pi \varepsilon}\)
∴ The kinetic energy of the electron,
\(\frac{1}{2}\) m_ev² = \(\frac{e \lambda}{4 \pi \varepsilon}\)

Question 8.
Obtain an expression for the electric field intensity at a point outside a uniformly charged thin infinite plane sheet.
Answer:
Consider a thin, flat, infinite, positively charged conducting sheet, with a uniform surface charge density σ. let the permittivity of the surrounding medium be ε. The charge density has a planar symmetry, i.e., it appears the same from all points on a plane parallel to the sheet. Then, by symmetry, the electric field intensity \(\vec{E}\) (1) is perpendicular to the sheet (in this case, outwards) at all points outside the sheet, and (2) has the same magnitude E at any given distance on either side of the sheet.

To find E at a point P outside the sheet, imagine a Gaussian surface in the form of a small closed cylinder. Its axis is perpendicular to the sheet, with the point P on one end face, shown in below figure. The cylinder encloses a small area dS of the sheet. So, the charge enclosed by the cylinder = σdS …………. (1)
The flux through one end = EdS
∴ The flux through the Gaussian surface,
Φ = 2EdS …………… (2)
By Gauss’s theorem,
εΦ = net charge enclosed
∴ ε(2EdS) = σdS …………. (3)
∴ E = \(\frac{\sigma}{2 \varepsilon}=\frac{\sigma}{2 k \varepsilon_{0}}\) …………. (4)
where ε₀ is the permittivity of free space and k = \(\frac{\varepsilon}{\varepsilon_{0}}\) is the relative permittivity (dielectric constant) of the surrounding medium.

Equation (4) shows that the magnitude of the electric field intensity outside the charged sheet is uniform and independent of the distance from the sheet.

[Notes : (1) The electric field lines are straight, parallel to each other, and perpendicular to the sheet. (2) If the sheet is negatively charged, \(\vec{E}\) on either side of the sheet is directed towards the sheet.]

9. Solve the following

Question 1.
Two metal spheres, of radii 2 cm and 1 cm with respective charge densities 5 mC/m² and -2 mC/m², are inside a hypothetical closed surface in vacuum. What is the net electric flux through the surface ?
Solution:
Data : σ₁ = 5 × 10^-6 C / m²,
σ₂ = -2 × 10^-6 C/m², r₁ = 2 × 10^-2 m, r₂ = 2 × 10^-2 m
ε₀ = 8.85 × 10^-12 F/m
Surface charge density, σ = \(\frac{Q}{A}=\frac{Q}{4 \pi r^{2}}\)
∴ The charges on the metal spheres are
Q₁ = σ₁ (4πr₁²) and Q₂ = σ₂(4πr₂²)
∴ Net charge enclosed, Q = Q₁ + Q₂ = 4π(σ₁r₁² + σ₂r₂²) = Q₁ + Q₂
= 4π[(5 × 10^-6)(2 × 10^-2)² +(- 2 × 10^-6)(10^-2)²]
= 4π[(5 × 10^-6)(4 × 10^-4) + (- 2 × 10^-6)(10^-4)]
= 4π(10^-10)(20 – 2) = 72π × 10^-10 C
∴ By Gauss’s theorem, the net flux through the surface
= \(\frac{Q}{\varepsilon_{0}}=\frac{72 \times 3.142 \times 10^{-10}}{8.85 \times 10^{-12}}\) = 2.556 × 10³ V∙m or N∙m²/C

Question 2.
A hollow metal ball 10 cm in diameter is given a charge of 1 × 10^-2 C. What is the magnitude of the electric field intensity at a point 20 cm from the centre of the ball ?
Solution:
Data : D = 10 cm, R = \(\frac{D}{2}\) = 5 cm, q = 1 × 10^-2 C,
r = 20cm = 0.2 m, ε₀ = 8.85 × 10^-12
Electric field intensity,

Question 3.
The electric field intensity at a point 1 m from the centre of a charged sphere of radius 25 cm in air is 10⁴ N/C. Find the surface charge density of the sphere.
Solution:
Data : r = 1 m,R = 0.25 m, E = 10⁴ N/C, ε₀ = 8.85 × 10^-12 F/m
E = \(\frac{\sigma R^{2}}{\varepsilon r^{2}}\)
As the sphere is situated in air, ε ≅ ε₀.
∴ The surface charge density,
σ = ε₀E(\(\frac{r}{R}\))²
= (8.85 × 10^-12)(10⁴)(\(\frac{1}{0.25}\))²
= 1.416 × 10^-10 C/m²

Question 4.
A long cylindrical conductor of radius 2 cm carries a charge of 5 μC/m and is kept in a medium of dielectric constant 10. Find the electric field intensity at a point 1 m from the axis of the cylinder.
[\(\frac{1}{4 \pi \varepsilon_{0}}\) = 9 × 10⁹ \(\frac{\mathbf{N} \cdot \mathbf{m}^{2}}{\mathrm{C}^{2}}\)]
Solution :
Data : R = 2 cm, λ = 5 × 10^-6 C/m, k = 10,
r = 1 m, \(\frac{1}{4 \pi \varepsilon_{0}}\) = 9 × 10⁹ N∙m²/C²
Electric field intensity,

Question 5.
The electric field intensity at a point on the surface of a long, charged cylinder in air is 100 V/m. Find the surface charge density on the cylinder.
Solution:
Data: E = 100 V/m, ε₀ = 8.85 × 10^-12 C²/N∙m²,
k = 1 (for air)
Let σ be the surface charge density on the cylinder.
E = \(\frac{\sigma}{\varepsilon_{0} k}\) :. σ = Eε₀k
= 100 × 8.85 × 10^-2 × 1
= 8.85 × 10^-10 C/m²

Question 10.
Obtain an expression for the electric potential energy of a system of two isolated point charges.
Answer:
Consider a test charge q0 in the electric field \(\vec{E}\) of a source charge + Q. The electric force acting on the test charge, q₀\(\vec{E}\), is a conservative force. When the test charge is moved in the field at constant velocity by some external agent, the work done by the field on the charge is equal to the negative of the work done by the external agent causing the displacement. Suppose an external agent moves the test charge without acceleration from a point B, at a distance r₁ from + Q, up to a point A, at a distance r, shown in below figure.

Since the electric field surrounding a point charge is not uniform, the electrostatic force on q0 increases as it approaches Q. Consequently, the external agent has to exert on q₀ a force of increasing magnitude and, for equal displacements, do increasing amount of work. Because the force exerted varies along the path, we imagine the total displacement to be made up of a large number of infinitesimal displacements \(d \vec{x}\). The distance dx is so small that, at an average distance x from Q, the electrostatic force \(\vec{F}\) on q₀ has a constant magnitude
F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q q_{0}}{x^{2}}\)
over the distance dx. The force \(\vec{F}_{\mathrm{ext}}\) by the external agent is equal and opposite to \(\vec{F}\) at every instant : \(\vec{F}_{\text {ext }}=-\vec{F}\)
Therefore, the infinitesimal work dW done by the external agent for the displacement \(\overrightarrow{d x}\) is

The total work done by the external agent in moving the test charge from A up to B is the line intergral of dW between the limits x = r₁ and x = r.

where ∆U = U_A – U_B is the change in the potential energy of the test charge in moving it from the point B to the point A. Choosing the potential energy of q0 to be zero when it is infinitely far away from Q, i.e., r₁ = ∞, its potential energy at a distance r from Q is
U(r) = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q q_{0}}{r}\)
Thus, the potential energy of a system of two point charges q₁ and q₂, a distance r apart is
U(r) = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r}\)
[Note Taking U(∞) = 0 is often a convenient reference level in electrostatics, but in circuit analysis other reference levels are often more convenient.]

Question 11.
Explain the concept of electric potential difference and electric potential.
Answer:
A charge in an electric field possesses electric potential energy just as a particle in a gravitational field possesses gravitational potential energy. Consider a test charge q₀ in an electric field, moved very slowly by an external agent from point B where its electric potential energy is U_B to a point A where its electric potential energy is U_A.

The change in the potential energy, U_A – U_B is defined as the work W_{B → A} that must be done by an external agent to move the test charge from B to A against the electric force, keeping the charge always in equilibrium, i.-e., without accelerating the charge so as not to give it any kinetic energy.

W_{B → A} = ∆U = U_A – U_B
The potential difference ∆V = V_AB = V_A – V_B between two points A and B in electric field is
∆V = \(\frac{W_{\mathrm{B} \rightarrow \mathrm{A}}}{q_{0}}=\frac{\Delta U}{q_{0}}\)

Definition : The electric potential difference between two points in an electric field is defined as the work done per unit charge by an external agent against the electric force in moving an infinitesimal positive charge from one point to the other without acceleration.

We choose the potential energy U_B and potential V_B to be zero when the initial point B is infinitely far from the source charges which produce the field. Then, the work done per unit test charge by an external agent in bringing a test charge from infinity to a point is the electric potential at that point.
The electric potential at a distance r from a source charge,
V(r) = \(\frac{W_{\infty \rightarrow r}}{q_{0}}=\frac{U(r)}{q_{0}}\)
Definition : The electric potential V at a point in an electric field is defined as the work per unit charge that must be done by an external agent against the electric force to move without acceleration a sufficiently small positive test charge from infinity to the point of interest.

[Note : It is more correct to speak about potential difference ∆V between two points than just the potential V at a given point because the latter implies a choice of .zero reference potential. The choice is one of convenience and we may choose the zero reference potential for a point at infinity or at some other location convenient for the problem as is done for gravitational potential.]

Question 12.
State the SI unit and one non-SI unit of potential energy.
Answer:
SI unit of energy : the joule (J)
Non-SI unit of energy : the electronvolt (eV).

In electrostatics, one joule is the change in electric potential energy when a charge of one coulomb is moved through a potential difference of 1 volt. Therefore, 1J = 1C × 1V, so that 1V = 1J/C.

Question 13.
State the dimensions and SI unit of electric potential difference.
Answer:
Dimensions : [V] = \(\frac{[W]}{\left[q_{0}\right]}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{[\mathrm{TI}]}\)
= [ML²T^-3I^-1]
SI unit : the volt (V) = the joule per coulomb (J/C).

Question 14.
If the work done in bringing a charge of 10^-18 C from infinity to point P is 2 × 10^-17 J, what is the electric potential at P?
Answer:
V = \(\frac{W}{q_{0}}=\frac{2 \times 10^{-17} \mathrm{~J}}{10^{-18} \mathrm{C}}\)
= 20 volts is the electric potential at P.

Question 15.
Explain and define the electronvolt.
OR
What is electronvolt?
Answer:
When an electric charge moves under the influence of an electric field, the work done by the field on the charge increases the kinetic energy of the charge as the electric force is a conservative force. Since the elementary charge e is widely encountered in atomic and nuclear physics, a convenient non-SI unit of energy used is the electronvolt.

Definition ; An electronvolt (symbol, eV) is the increase in the kinetic energy of a particle with a charge equal in magnitude to the elementary charge e when the particle is accelerated through a potential difference of one volt.
∆(KE) = qAV
where q is the charge and ∆V is the potential difference.
∴ 1 eV = e(1V)
= (1.602 × 10^-19 C) (1 J/C)
= 1.602 × 10^-19 J

[Notes : (1) SI prefixes are commonly used with the unit. Therefore, 1 keV = 10³ eV = 1.602 × 10^-16 J, 1 MeV = 10⁶ eV = 1.602 × 10^-13 J, etc. (2) In practice, it is not unusual to use the electronvolt as an energy unit in nonelectrical situations. For example, an air molecule at room temperature has a kinetic energy of about \(\frac{1}{40}\) eV, the electron mass is 511 keV / c₀², and the proton mass is 938 MeV/c₀², where c₀ is the speed of light in free space.]

Question 16.
Obtain the relation between the magnitude of electric field intensity and electric potential.
Answer:
Consider a test charge q₀ as it moves from point A to point B in an electric field. The electric force on q₀ at any point along the path is
\(\vec{F}=q_{0} \vec{E}\)
where \(\vec{E}\) is the field at that point.

The incremental work dW done by the field as q₀ undergoes a displacement \(\overrightarrow{d l}\) along the path is
dW = \(\vec{F} \cdot \overrightarrow{d l}\)
= q₀ \(\vec{E} \cdot \overrightarrow{d l}\) = q₀ E dl
In the process, the charge q₀ is moved from a higher potential to a lower potential, thereby losing potential energy. Therefore, the change in potential energy
dU = -dW = -q₀ Edl
By definition, the change in electric potential is dv

The quantity \(\frac{d V}{d l}\) is the rate at which the electric potential changes with distance and is called the electric potential gradient. The above equation thus shows that the magnitude of the electric field intensity at a point is equal to the negative of the potential gradient at that point.

[Note : (1) The negative sign shows that if we move in the direction of the electric field, the potential decreases. In the opposite direction, it increases. (2) If we draw equipotentials [see Unit 8.5] so that adjacent surfaces have equal potential differences, then in regions where the magnitude of \(\vec{E}\) is large, the equipotential surfaces are close together because the field does a relatively large amount of work on a test charge in a relatively small displacement. Conversely, in regions where the field s weaker, the equipotential surfaces are farther apart. (3) Potential gradient is an another name for electric field. (4) From the relation E = – dV/dx, we get another very common SI unit of electric field intensity, namely, the volt per metre (V/m).]

Question 17.
Define electric potential gradient.
Ans
Definition The rate of change of electric potential with distance in a specified direction is called the electric potential gradient in that direction.

Question 18.
Obtain an expression for the electric potential at a point due to an isolated point charge.
Answer:
Consider a point A at a distance r from a static point charge ± Q, as shown in below figure. To determine the electric potential at the point A (due to Q), imagine a test charge q₀ being moved from infinity
up to the point A without acceleration. Because the electric field of a point charge is not uniform, the force exerted by Q on q₀ increases as it approaches Q. We imagine the total displacement to be made up of a large number of infinitesimal displacements. \(\overrightarrow{d x}\). The distance dx is so small that, at an average distance x from Q, the electrostatic force \(\vec{F}\) on q₀ has a constant magnitude
F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q q_{0}}{x^{2}}\)
over the distance dx. The force F_ext by the external agent is equal and opposite to \(\vec{F}\) at every instant : \(\vec{F}_{\mathrm{ext}}=-\vec{F}\)

Therefore, the infinitesimal work dW done by the external agent for the displacement \(\overrightarrow{d x}\) is

The total work done by the external agent in moving the test charge from infinity up to the point P (from x = ∞ to x = r) is the integral of dW between the limits x = ∞ and x = r.
W = \(\int_{x=\infty}^{x=r} d W=\int_{\infty}^{r}\left(-\frac{1}{4 \pi \varepsilon_{0}} \frac{Q q_{0}}{x^{2}}\right) d x\)

where ∆U = U_A – U_B is the change in the potential energy of the test charge in moving it from to the point A. Choosing the potential energy of q0 to be zero when it is infinitely far away from Q, its potential energy at a distance r from Q is
U_A = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q q_{0}}{r}\)
Therefore, the electric potential at a distance r from Q is
V = \(\frac{U_{\mathrm{A}}}{q_{0}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}\)
The above equation also gives the electric potential either outside or on the external surface of a spherically symmetric charge distribution,

[Notes: (I) A positive charge produces a positive electric potential. A negative charge produces a negative electric potential. (2) A negative electric potential means a positive charge would be attracted from a larger distance, since the potential is lower (more negative) than at larger distances. Conversely, a negative charge would be repelled, as expected. (3) Only differences between potentials at two points are physically significant. Potential at a point is ambiguous unless we specify which is the reference point. (4) Electric potential, like electric field intensity, is independent of the magnitude of the test charge that we use to define it.]

Question 19.
What is the electric potential at 10 Å from a point charge 10^-18 C in vacuum?
[\(\frac{1}{4 \pi \varepsilon_{0}}\) = 9 × 10⁹ N∙m²/C²]
Answer:
V = \(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{r}\)
= \(\frac{9 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} \times 10^{-18} \mathrm{C}}{10 \times 10^{-10} \mathrm{~m}}\)
= 9 volts is the required electric potential.

Question 20.
Obtain an expression for the electric potential at a point due to several point charges.
Answer:
Consider a point P at distances r₁, r₂, r₃, …, r_N from point charges q₁, q₂, q₃, ………, q_N, respectively. The electric potentials of P due to the individual charges are
V₁ = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1}}{r_{1}}\) , V₂ = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{2}}{r_{2}}\) …………. , V_N = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{N}}{r_{N}}\)
Since potential is a scalar quantity, the potential of P due to all the charges is the algebraic sum of the potentials due to the individual charges.
∴ V = V₁ + V₁ + ………. + V_N
= \(\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{1}}{r_{1}}+\frac{q_{2}}{r_{2}}+\ldots+\frac{q_{N}}{r_{N}}\right)\)
= \(\frac{1}{4 \pi \varepsilon_{0}} \sum_{i=1}^{N} \frac{q_{i}}{r_{i}}\)
[Note : Electric potential is a scalar quantity. To calculate the resultant potential due to two or more point charges, the potentials due to individual charges are added as simple scalars along with its sign, determined by the sign of the q that produces V.]

Question 21.
Derive an expression for the electric potential at a point due to a short electric dipole. Hence, write the expression for the electric potential at a point
(i) on the dipole axis (ii) on the dipole equator.
OR
Derive an expression for the electric potential at a point due to an electric dipole.
Answer:
Consider an electric dipole AB of dipole length 21 and point charges + q and – q. Its electric dipole moment \(\vec{p}\) has magnitude p = 2ql. Let P be a point at a distance r from O, the centre of the dipole, in a direction θ with the dipole axis, as shown in below figure. Let AP = r₁ and BP = r₂.

The electric potential at P due to the charge +q is
V₁ = +\(+\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r_{1}}\) …………….. (1)
and that due to the charge -q is
V₂ = \(-\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r_{2}}\) …………….. (2)
Method 1

In the figure, PA₁ = PA = r₁ and PB₁ = PB = r₂.
For short dipole (2l << r), ∠OA₁A ≅ ∠OB₁B ≅ 90°.
Hence, in the right angled ∆s OA₁A and OB₁B,
OA₁ = OB₁ = l cos θ
∴ r₂ – r₁ = PA₁ – PB₁ = A₁B₁ = 2l cos θ
Also, r₁r₂ = PA₁ × PB₂ = (PO + OA₁)(PO – OB₁)
= (r + l cos θ)(r – l cos θ)
= l² – l² cos² θ
= r² (∵ l << r) …………… (4)

Method 2:
From △PAA₁, by cosine rule,


This is the required expression.

Particular cases :
(i) At a point on the dipole axis, θ = 0° (nearer to the charge + q) or 180° (nearer to the charge – q).
∴ cos θ = ±1
∴ V_axis = \(\pm \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{p}{r^{2}}\)

(ii) At a point on the dipole equator, θ = 90° or 270°.
∴ Cos θ = 0 ∴ V_equator = 0

[Note : Since V_equator = 0, the equatorial plane of a dipole is an equipotential plane of electric potential equal to zero. No work is required to move a charge anywhere in the equatorial plane.]

Question 22.
What is the electric potential at a point on the axis of a short electric dipole of moment 27 Cm if the point is at 0.5 m from the centre of the dipole and is located in vacuum? [\(\frac{1}{4 \pi \varepsilon_{0}}\) = 9 × 10⁹ N∙m²/C²]
Answer:
V = \(\pm \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{p}{r^{2}}\)
= \(\pm \frac{\left(9 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left(10^{-27} \mathrm{C} \cdot \mathrm{m}\right)}{(0.5 \mathrm{~m})^{2}}\)
= ±3.6 × 10^-17 V
is the required electric potential.

Question 23.
Distinguish between electric field intensity and electric potential.
Answer:

Electric field intensity	Electric potential
1. Electric field intensity is a vector quantity associated with an electric field.	1. Electric potential is a scalar quantity associated with an electric field.
2. It is the electric force per unit positive charge placed at a point in an electric field.	2. It is the work per unit charge which must be done by an external agent against the electric force to bring an infinitesimal posi­tive charge from infinity to a given point in an electric field, without acceleration.
3. Its SI unit is the newton per coulomb or the volt per metre.	3. Its SI unit is the volt.

Question 24.
Distinguish between the volt and the electron volt.
Answer:

Volt	Electronvolt
1. The volt is the SI unit of electric potential (or poten-tial difference).	1. The electronvolt is a non- SI unit of energy.
2. If one joule per coulomb of work is done by an external agent against the electric force in moving an infini-tesimal charge from one point to another keeping the charge in equilibrium, the potential difference between the two points is called one volt.	2. It is the increase in the kinetic energy of a particle carrying a charge equal to the elementary charge e when it is accelerated through a potential differ¬ence of one volt.
3. 1 V = 1 J/C.	3. 1 eV = 1.602 × 10-19 J.

25. Solve the following

[Data : [\(\frac{1}{4 \pi \varepsilon_{0}}\) = 9 × 10⁹ N∙m²/C²]
Question 1.
A helium nucleus carries a charge of + 2e, where e is the elementary charge. Find the electric potential at 10^-10 m from the helium nucleus.
[e = 1.6 × 10^-19 C]
Solution:
Data : If q is the charge on helium nucleus,
q = +2e = 2 × 1.6 × 10^-19 C,r = 10^-10m,
1/4πε₀ = 9 × 10⁹ N∙m²/C²
The electric potential,
V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\)
= \(\frac{2\left(9 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)}{10^{-10} \mathrm{~m}}\)
= 28.8V

Question 2.
Determine the electric potential at the midpoint of the line joining two charges 2 × 10^-6 C and -1 × 10^-6 C placed in vacuum 10cm apart.
Solution:
Data: q₁ = 2 × 10^-6 C, q₂ = -1 × 10^-6 C,
r = \(\frac{10 \mathrm{~cm}}{2}\) = 5cm = 0.05m,
1/4πε₀ = 9 × 10⁹ N∙m²/C²
The electric potential at a distance r from a charge q is
V = \(\frac{q}{4 \pi \varepsilon_{0} r}\)
Since potential is a scalar quantity, the total electric potential at the midpoint is
V = V₁ + V₂ = \(\frac{1}{4 \pi \varepsilon_{0}}\) (q₁ + q₂)
= \(\frac{\left(9 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)}{0.05 \mathrm{~m}}\) (2 – 1) × 10^-6 C
= 1.8 × 10⁵ V

Question 3.
The electric field and electric potential at a certain point due to a point charge in vacuum are 9000 V/m and 18000 V, respectively. Find the distance of the point from the charge and the magnitude of the charge.
Solution:
Data: E = 9000 V/m, V = 18000 V,
1/4πε₀ = 9 × 10⁹ N∙m²/C²
E = \(\frac{q}{4 \pi \varepsilon_{0} r^{2}}\), V = \(\frac{q}{4 \pi \varepsilon_{0} r}\)
\(\frac{V}{E}\) = r
∴ The distance, r = \(\frac{18000 \mathrm{~V}}{9000 \mathrm{~V} / \mathrm{m}}\) = 2 m
∴ The magnitude of the charge is

Question 4.
What is the electric potential at the centre of a square of side 1 m if point charges 1 × 10^-8 C, -2 × 10^-8 C, 3 × 10^-8 C and 2 × 10^-8 C are placed at the corners of the square?
Solution:
Data: q₁ = 1 × 10^-8 C, q₂ = -2 × 10^-8 C, q₃ = 3 × 10^-8 C, q₄ = 2 × 10^-8 C,
1/4πε₀ = 9 × 10⁹ N∙m²/C²
The distance of the point of intersection O of the diagonals, from each charge is \(\sqrt {2}\) / 2 = 1/\(\sqrt {2}\) m.
∴ r = 1/\(\sqrt {2}\) m

The potential at a distance r from a charge q is
V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\)
As potential is a scalar quantity, the total electric potential at O is
V = V₁ + V₂ + V₃ + V₄
= \(\frac{1}{4 \pi \varepsilon_{0} r}\) (q₁ + Vq₂ + q₃ + q₄)
= \(\frac{\left(9 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)}{(1 / \sqrt{2} \mathrm{~m})}\) [(1 – 2 + 3 + 2) × 10^-8 C]
= 360\(\sqrt {2}\) = 509.2 V

Question 5.
The electric potential due to a dipolar molecule of electric dipole moment 6 × 10^-30 A∙m² at a point along the axis of the dipole is 1 V. Find the distance of the point from the centre of the dipole.
Solution:
Data : p = 6 × 10^-30 A∙m², V = 1 V,
θ = 0° (axial point), 1/4πε₀ = 9 × 10⁹ N∙m²/C²
Electric potential,

Question 6.
Two charged particles, carrying – 2 pC each, are held in place 10 cm apart. A point B is at distance of 18 cm from both.
(a) Calculate the electric potential at point B.
(b) A third charged particle, of mass 10^-15 kg and carrying charge – 1 pC, is released from rest at point B. What is its speed when it is far from the two fixed charged particles?
Solution:
Data : q₁ = q₂ = q = – 2 × 10^-12 C,
r₁ = r₂ = r = 18 cm = 0.18m,. q₃ = – 1 × 10^-12 C,
m₃ = 10^-15 kg
The electric potential at B,
V_B = V₁ + V₂ = 2(\(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\))
= 2(9 × 10⁹)(\(\frac{-2 \times 10^{-12}}{0.18}\)) = – 0.2 V

(b) By definition, the potential energy of an electric charge at infinity, i.e., far away from other charges, is U_∞ = 0. To find the speed of the third charged particle at infinity, we use the principle of conservation of energy.
K_B + U_B = K_∞ + U_∞
where the KE of the particle at B, K_B = 0, since it is released from rest. The PE,
U_B = q₃V_B = (-1 × 10^-12)(- 0.2) = 2 × 10^-13 J

∴ The KE of the particle at infinity,
k_∞ = \(\frac{1}{2}\) mv² = U_B = 2 × 10^-13 J
∴ v = \(\sqrt{\frac{2 K_{\infty}}{m}}=\sqrt{\frac{2\left(2 \times 10^{-13}\right)}{10^{-15}}}\) = 20 m/s
This gives the required speed.

[Note: The PE of the third charge at Bis +2 × 10^-13 J, while at inifinity it is zero. All of the initial electrical PE is converted to KE because of the positive work done on it by the repulsive field due to the other two charges. The negatively charged third particle gains kinetic energy when it moves from a lower-potential point to a higher- potential point, in this case from V_B = – 0.2 V to V_∞ = 0. Also note that the net force on the third negatively charged particle decreases as it moves away from point B. Its acceleration is not constant, so kinematical equations cannot be used to find its final speed. Hence, we have used the principle of conservation of energy.]

Question 7.
A proton is released from rest in vacuum in a uniform electric field of intensity 100 V/m. What is its speed after it has travelled a distance of 1 m ? [m_p = 1.67 × 10^-27 kg, 1 eV = 1.6 × 10^-19J]
Solution :
Data : u = 0 m/s, E = 100 V/m, l = 1 m, m_p = 1.67 × 10^-27 kg, 1 eV = 1.6 × 10^-19 C
In a uniform electric field \(\vec{E}\), the potential difference between two points a distance l apart (along \(\vec{E}\)) is
∆V = El = (100 V/m) (1 m) = 100 V
The change in the kinetic energy of the proton is
∆KE = eA V = e (100 V) = 100 eV
= 100 × 1.6 × 10^-19
= 1.6 × 10^-17 J
Since the proton starts from rest, the initial kinetic, energy is zero. Therefore, the change in kinetic energy equals the final kinetic energy \(\frac{1}{2}\) m_pv².
∴ \(\frac{1}{2}\) m_pv² = 1.6 × 10^-17 J
∴ v² = \(\frac{2\left(1.6 \times 10^{-17} \mathrm{~J}\right)}{1.67 \times 10^{-27} \mathrm{~kg}}\)
= 1.916 × 10¹⁰ (m/s)²
∴ v = 1.384 × 10⁵ m/s
The speed of the proton after it has travelled a distance of 1 m is 1.384 × 10⁵ m/s.

Question 8.
A particle carrying 5 electrons starts from rest and is accelerated through a potential difference of 8900 V. Calculate the KE acquired by it in MeV. [Charge on electron = 1.6 × 10^-19 C]
Solution :
Data : q = 5e, u = 0, V = 8900 V, e = 1.6 × 10^-19 C
q = 5(1.6 × 10^-19 C) = 8 × 10^-19 C
Initial KE = KE_i = \(\frac{1}{2}\) mu² = 0
∴ ∆KE = KE_f – KE_i = KE_f
∆KE = qV
∴ The final KE, KE_f = qV
= (8 × 10^-19 C) (8900 V)
= 7.12 × 10^-15 J
= \(\frac{7.12 \times 10^{-15}}{1.6 \times 10^{-19}} \mathrm{eV}\)
= 4.45 × 10⁴ eV
= (4.45 × 10^-2) × 10⁶ eV
= 4.45 × 10^-2 MeV

Question 9.
A proton is accelerated from rest through a potential difference of 500 volts. Find its final momentum. [m_p = 1.67 × 10^-27 kg, e = 1.6 × 10^-19 C]
Solution:
Data : m_p = 1.67 × 10^-27 kg, e = 1.6 × 10^-19 C, u = 0, V = 500 V
Initial KE, KE; = \(\frac{1}{2}\) m_pu² = 0
∴ ∆KE = KE_f – KE_i = KE_f
∆KE = qV
∴ KE_f = qV KE_f = \(\frac{1}{2}\) m_pv² = \(\frac{p_{\mathrm{f}}^{2}}{2 m_{\mathrm{p}}}\)
where p_f = m_pv ≡ the magnitude of the final momentum of the proton.
∴ P_f² = 2 m_pqV
∴ P_f = \(\sqrt{2 m_{\mathrm{p}} q V}\)
= \(\sqrt{2\left(1.67 \times 10^{-27} \mathrm{~kg}\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)(500 \mathrm{~V})}\)
= 5.169 × 10^-22 kg∙m/s
The momentum is directed along the applied electric field.

Question 10.
A proton and an α-particle are accelerated from rest through the same potential difference. Compare their final speeds. Charge on an α-particle = 2 × charge on a proton, mass of an α-particle = 4 × mass of a proton.
Solution:
Lution: Let q₁ = charge on an α-partic1e, q₂ = charge on a proton, m₁ = mass of an α-particle, m₂ = mass of a proton, v₁ = final speed of the α-particle, v₂ = final speed of the proton

Question 26.
What do you mean by an equipotential surface? What is the shape of equipotential surfaces for the special case of (i) a uniform field (ii) a single point charge?
Answer:
An equipotential surface, in a region where an electric field is present, is a three-dimensional surface on which the electric potential is the same at every point.

Electric field lines and equipotential surfaces are always mutually perpendicular. In a diagram, only a few representative equipotentials are shown, usually with equal potential differences between adjacent surfaces. Also, equipotential surfaces for different potentials do not touch or intersect since a point cannot be at two electric potentials simultaneously.

For the special case of a uniform field, in which the field lines are equally-spaced parallel lines, the equipotentials are parallel planes perpendicular to the field lines, from figure (a).

For equal potential differences between adjacent planes, these equipotentials are equally spaced. For the special case of a single point charge, for which the field lines are radial, the equipotentials are concentric spheres centred on the point charge, from figure (b).

For a given point charge q.
V = \(\frac{C}{r}\)
where C = \(\frac{q}{4 \pi \varepsilon_{0}}\) is a constant. Since \(\frac{1}{r}\) is not a linear function of r, equipotentials with equal potential differences between adjacent surfaces are not evenly spaced in radius.

Question 27.
Electric field lines and equipotential surfaces are always mutually perpendicular. Explain.
Answer:
If a test charge q₀ is moved on an equipotential surface of potential V, the electric potential energy U = q₀V remains constant. Because U does not change as q₀ is moved, the work done by the electric field on q₀ must be zero. If \(\vec{E}\) is the electric field on the surface,

Hence, the electric force q₀\(\vec{E}\) is always perpendicular to the displacement of a charge moving on an equilateral surface. Thus, electric field lines and equipotential surfaces are always mutually perpendicular.

Note that if \(\vec{E}\) is not perpendicular to a equipotential surface everywhere, it would have a component E_|| along the surface, so that for a displacement dx between two points on the surface, the work done dW = E_||dx ≠ 0. This would imply a potential difference between the two points which contradicts the definition of a equipotential surface.

Question 28.
Draw a diagram showing the equipotential surfaces and electric field lines in the plane of (i) an electric dipole (ii) a system of two equal positive charges.
Answer:


[Note: At each crossing of an equipotential and a field line, the two are perpendicular.]

Question 29.
What can you say about the direction of the electric field on the surface (just outside) of a charged conductor in an electrostatic situation?
Answer:
When all charges on a conductor are at rest, the tangential component of the electric field \(\vec{E}\) is zero at every point on the surface (just outside) of a conductor; otherwise, charges would move around on the surface. If follows that, in an electrostatic situation, the electric field just outside a conductor must be perpendicular to the surface at every point and a conducting surface is always an equipotential surface.

Question 30.
What are the electric potentials outside and inside a charged spherical conductor in electro static equilibrium?
Answer:
Consider a spherical conductor of radius R and static charge q. By Gauss’s law, the electric field due to the charged conducting sphere is
\(\vec{E}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}} \hat{r}\) (r ≥ R)
= 0 (r < R)
because for r < R, charge enclosed by a Gaussian surface inside the sphere is zero but in the region r ≥ R, outside, the charged sphere is identical to a point charge at the centre of the sphere.

Consequently, since V(r) = \(-\int \vec{E} \cdot \overrightarrow{d r}\), the electric potential outside the sphere must be the same as that of an isolated point charge q located a r = 0.
∴ V(r) = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\) (r ≥ R)

Thus, a conducting sphere in electrostatic equilibrium is a spherical equipotential surface at potential V(R) and in the region r ≥ R, the equipotentials are concentric spheres.

Inside the sphere, r < R, E = 0, so V(r) is constant in this region. Since V(R) = q/4πε₀R,

Question 31.
Draw diagrams showing the equipotential sur faces and electric field lines (a) within a charged parallel-plate capacitor (a pair of parallel metal plates with charges of equal magnitude and opposite sign)
OR
(b) if one of the plates of the capacitor is replaced by a charged spherical conductor.
Answer:
(a) A pair of charged parallel metal plates sets up a uniform electric field between the plates, away from its edges, as shown by their even spacing. The field is perpendicular to the plates, in the direction from the positive plate toward the negative plate. Conductors are equipotential surfaces, so the negative plate is an equipotential surface with potential taken as zero and the positive plate is an equipotential surface with potential = V. The equipotential surfaces between the plates are parallel to the plates. Electric field lines and equipotential surfaces are perpendicular to each other. Also, because the electric field is uniform between the plates, equipotential surfaces representing equal potential differences are equally spaced.

Question 32.
What are the advantages of using electrostatic potential?
Answer:
The electrostatic potential at each point in space in the vicinity of the source charges represents a scalar field.

The advantages of electrostatic potential field associated with a given distribution of charges are as follows:

• If we know the potential difference between any two points, we can easily obtain the change in potential energy and the work done when a charge placed in the field moves between these two points.
• Electric field is a vector field. Electrostatic potential being a scalar field, the potential at any point due to several charges is simply the algebraic sum of the potentials due to the individual charges.
• The construction of equipotential surfaces helps to visualize the electric field pattern.
• It is possible to calculate the electric field \(\vec{E}\) from the scalar potential field function V (by differentiating V with respect to the space coordinates.)

Question 33.
Derive an expression for the potential energy of a system of two point charges.
Answer:
The electric potential energy of a system of point charges at rest in free space is defined as the work done by an external agent against the electric force in assembling the charges by bringing them from infinity to their locations in the configuration, always keeping the charges in equilibrium.

Consider assembling a system of two point charges q₁ and q₂ at points A and B, respectively, in a region free of external electric field. Let \(\vec{r}_{1}\) and \(\vec{r}_{2}\) be the position vectors of A and B, respectively,, with respect to an arbitrary reference frame.

In the absence of charge q₂, since there is no external electric field in the region, no work is done in bringing the first charge q₁ from infinity to A, so W₁ = 0. Subsequently, due to q₁ the potential at B is
V_B = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1}}{r_{21}}\)
where r₂₁ = \(\left|\vec{r}_{21}\right|, \vec{r}_{21}=\vec{r}_{2}-\vec{r}_{1}\) being the position vector of B with respect to A. Consequently, the work done by an external agent in bringing q₂ from infinity to B in the electric field of q₁ is
W₂ = V_B ∙ q₂ = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r_{21}}\)
Hence, the total work done is
W = W₁ + W₂ = 0 + \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r_{21}}\)
= \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r_{21}}\)
Since the charges were always kept in equilibrium, the change in the potential energy U_f – U_i equals W.

Since the charges were brought from infinity where their potential energy is assumed to be zero, U_i = 0. Therefore, the potential energy of the system of two point charges is
U = U_f = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r_{21}}\)

Question 34.
A system consisting of a charged spherical shell and an electron has negative electric potential energy U = -15 × 10^-20 J, with U(∞) = 0.
(a) What is the sign of the charge on the shell?
(b) If the electron is replaced by a proton, what would be the electric potential energy of the new system?
Answer:
A charged spherical conductor is equivalent to a point charge at its centre. For U(∞) = 0, the potential energy of two point charges q₁ and q₂ a distance r₂₁ apart is
U = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r_{21}}\)
Since the FE of the charged shell and electron system is negative, the shell must be positively charged, electron being negatively charged.

If the electron is replaced by a proton, the PE of the new system would be positive equal to U’ = ±15 × 10^-20 J.

Question 35.
Obtain an expression for the potential energy of a configuration of N point charges.
Answer:
Consider assembling a configuration of N point charges q₁, q₂, q₃, …………., q at points A, B, C, D, …, respectively, in a region free of external electric field. Let \(\overrightarrow{r_{1}}, \overrightarrow{r_{2}}, \overrightarrow{r_{3}}, \ldots, \overrightarrow{r_{N}}\) be the position vectors of the points A, B, C, D, … etc., respectively, with respect to an arbitrary reference frame.

No work is done in bringing the first charge q₁ from infinity to point A, so W₁ = 0. Subsequently, the potential at B is
V_B = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1}}{r_{21}}\)
where r₂₁ = \(\left|\vec{r}_{21}\right|, \vec{r}_{21}=\vec{r}_{2}-\vec{r}_{1}\) being the position vector of B with respect to A. Consequently, the work done by an external agent in bringing q₂ from infinity, to B in the electric field of q₁ is
W₂ = V_B . q₂
= \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r_{21}}\)
Subsequently, the potential at C is
V_C = V₁ + V₂
where V₁ and V₂ are the potentials due to q₁ and q₂.

Consequently, the work done by the external agent in bringing the third charge q₃ from infinity to C in the electric fields of q₁ and q₂ is
W₃ = V_C . q₃
= \(\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{1} q_{3}}{r_{31}}+\frac{q_{2} q_{3}}{r_{32}}\right)\)
Now, the potential at D is
V_D = V₁ + V₂ + V₃
= \(\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{1}}{r_{41}}+\frac{q_{2}}{r_{42}}+\frac{q_{3}}{r_{43}}\right)\)

Since the charges were brought from infinity where the potential energy is assumed to be zero, U_i = 0. Therefore, the potential energy of the con figuration of N point charges is

Question 36.
What is the potential energy of a point charge in an external electric field?
Answer:
Consider a charge q placed in an external electric field at a point whose position vector with respect to an arbitrary reference frame is \(\vec{r}\). If V(\(\vec{r}\)) is the potential of the point, with respect to an arbitrary reference zero at infinity, then the potential energy of the charge q at the point is
U(\(\vec{r}\)) = qV(\(\vec{r}\))
where it is assumed that q is sufficiently small and does not significantly distort the electric field and the potential at the point.

Question 37.
Derive an expression for the potential energy of a system of two point charges in an external field.
Answer:
Consider assembling a system of two point charges q₁ and q₂ at points A and B, respectively, in a region of external electric field. Let \(\vec{r}_{1}\) and \(\vec{r}_{2}\) be the position vectors of A and B, respectively, with respect to an arbitrary reference frame. \(\vec{r}_{21}=\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\) is the position vector of B with respect to A. Let V₁ and V₂ be the electric potentials of A and B due to the external field.

The work done in bringing the charge q_1,from infinity to A against the electric force of the external field is
W₁ = q₁ V₁
Subsequently, the electric potential at B is
V_B = V₂ + V’_{due t0 q1}
= V₂ + \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1}}{r_{21}}\)

Consequently, the work done in bringing the second charge q₂ from infinity to B is
W₂ = q₂V_B
= q₂V₂ + \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r_{21}}\)
Hence, the total work done by an external agent in assembling the two point charges in a region of external electric field is
W = W₁ + W₂
= q₁V₁ + q₂V₂ + \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r_{21}}\)
Since the charges were always kept in equilibrium, the change in the potential energy of the system U_f – U_i = W. Also, since the charges were brought from infinity where their potential energy is assumed to be zero, U_i = 0. Therefore, the potential energy of the system of two charges in an external field is
U = U_f = q₁V₁ + q₂V₂ + \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r_{21}}\)

Question 38
Derive an expression for the electric potential energy of an electric dipole in a uniform electric field.
OR
Derive an expression for the total work done in rotating an electric dipole through an angle θ in a uniform electric field.
Answer:
Consider an electric dipole of dipole moment \(\vec{p}\) placed in a uniform electric field \(\vec{E}\) making an
angle Φ with \(\vec{E}\). The torque \(\vec{\tau}=\vec{p} \times \vec{E}\) tends to rotate the dipole and align it with \(\vec{E}\).

Suppose an external torque \(\vec{\tau}_{\text {ext }}\), equal in magnitude and opposite in direction to \(\vec{\tau}\), is applied to rotate the dipole through an infinitesimal angular displacement dΦ, always keeping the dipole in equilibrium.

The work done by this torque is
dW = τ_ext dΦ
= p E sin Φ dΦ
In a finite angular displacement from θ₀ to θ, the total work done on the dipole by the external agent is

If the dipole was initially parallel to \(\vec{E}\), θ₀ = 0 and cos θ₀ = 1.
∴ W = pE (1 – cos θ) …………. (2)
If the dipole was initially parallel to \(\vec{E}\), its potential energy U₀ = – pE is minimum (more negative). If we arbitrarily assign U₀ = 0 to the minimum of potential energy, the potential energy for the system for an inclination θ is
U_θ = – pEcos θ = – \(\vec{p} \cdot \vec{E}\)
This is the required expression.

39. Solve the following

Question 1.
Consider a point charge q = 1.5 × 10^-8 C. What is the radius of an equipotential surface having a potential of 30V?
Solution:
Data: q = 1.5 × 10^-8 C,
1/4πε₀ = 9 × 10⁹ N∙m²/C², V = 30 V
An equipotential surface surrounding an isolated point charge is a sphere centred on the charge. Let r be the radius of such an equipotential for which V = 30 V.

Question 2.
Two charged spherical conductors, of radii R₁ and R₂ and surface charge densities σ₁ and σ₂, are connected by a thin conducting wire. Except for this connecting wire, the spheres are sufficiently separated to be considered as isolated. Show that σ₁R₁ = σ₂R₂.
Solution:
The electrical potential at the surface of an isolated, charged conducting sphere of radius R is

Since the spherical conductors are connected by a conducting wire, the system must be equipotential, i.e.,
V₁ = V₂.
∴ \(\frac{\sigma_{1} R_{1}}{\varepsilon_{0}}=\frac{\sigma_{2} R_{2}}{\varepsilon_{0}}\)
∴ σ₁R₁ = σ₂R₂ as required

[Note : Although the above connected system is different from a typical conductor with a variable radius of curvature, the above relation qualitatively indicates how charge density varies over the surface of a conductor of arbitrary shape. The equation indicates that σ and E are small where the radius of curvature is large. Conversely, σ and E are higher at locations with a small radius of curvature. A practical application of this phenomenon is the lightning rod.]

Question 3.
Two equipotential surfaces A and B in a uniform electric field, with V_A = 50 V and V_B = 30 V, are 10 cm apart. Two point charges, q₁ = 3 nC and q₂ = 5 nC, are placed on A and B, respectively.
(i) What is the magnitude of the electric field?
(ii) 1f the line joining the two charges is parallel to the field, what is the total work done in assembling the two charges?
Solution:
Data: V_A = 50 V, V_B = 30 V, l = 10 cm = 0.1 m,
q₁ = 3 nC = 3 × 10^-9 C, q₂ = 5 nC = 5 × 10^-9 C,
1/4πε₀ = 9 × 10⁹ N∙m²/C²

Since the line joining the two charges is parallel to the fle1d,\(\overrightarrow{r_{21}}\) = l = 0.1 m.

The work done by an external agent in assembling the two charges by bringing them from infinity is
W = q₁V_A + q₂V_B + \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r_{21}}\)
=(3 × 10^-9 C)(50 V) + (5 × 10^-9 C)(30 V) + (9 × 10⁹ N∙m²/C²) \(\frac{\left(3 \times 10^{-9} \mathrm{C}\right)\left(5 \times 10^{-9} \mathrm{C}\right)}{0.1 \mathrm{~m}}\)
= 1.5 × 10^-7 + 1.5 × 10^-7 + 13.5 × 10^-7
= 16.5 × 10^-7 = 1.65 × 10^-6 J (= 1.65 μJ)

Question 4.
An electron is circulating around the nucleus of an H-atom in a circular orbit of radius 5.3 × 10^-11 m Calculate the electric potential energy of the atom in eV.
Solution:
Data: e = 1.6 × 10^-19 C, r = 5.3 × 10^-11 m,
1/4πε₀ = 9 × 10⁹ N∙m²/C²
The charge of the single proton in the nucleus of a hydrogen atom, q₁ = + e = 1.6 × 10^-19 C
The charge on the electron,
q₂ = -e = -1.6 × 10^-19 C
The potential energy of an H-atom,

Question 5.
A proton is to be suspended in vacuum between two parallel plates separated by 1 mm. Find (i) the electric field required (ii) the potential difference between the plates corresponding to the desired field. [m_p = 1.67 × 10^-27 kg]
Solution:
The gravitational force on the proton should be balanced by the force due to the electric field. So, the electric field must be directed up; the upper plate should be at negative potential and the lower plate at positive potential, as shown in below figure.

For uniform field,
E = \(\frac{V}{d}\), where V = potential difference between the plates and d = distance between the plates.
The potential difference,
V = Ed = (1.023 × 10^-7 V/m)(10^-3 m)
= 1.023 × 10^-10 V

Question 6.
Two point charges, q₁ = – 1 μC and q₂ = +1 μC, are located in vacuum on the x-axis at x = 0 and x = a, respectively. (i) Find the potential energy of the system. (ii) If a third point charge q₃ = + 1 μC is brought from infinity to x = 2a, find the total potential energy of the system of the three charges. (iii) What work has been done by an external agent to bring in the third charge? Take a = 10 cm.
Solution:
Data:q1 = – 1 μC = – 1 × 10^-6 C, q₂ = q₃ = 1 μC
= 1 × 10^-6 C, r₂₁ = a = 0.1 m, r₃₁ = 2a = 0.2 m,
r₃₂ = a = 0.1 m, 1/4πε₀ = 9 × 10⁹ N∙m²/C²

(ii) The total potential energy of the system of three charges,

(iii) Additional work done in bringing the third charge
= U₃ – U₂
= (- 4.5 + 9) × 10^-2 J
= 4.5 × 10^-2 J

Question 7.
What is the electric potential energy of the following charge configuration?Take q₁ = +1 × 10^-8 C, q₂ = – 2 × 10^-8 C, q₃ = +3 × 10^-8 C, q₄ = 2 × 10^-8 C and a = 1 m. Assume the charges to be in vacuum.

Solution:
Data : q₁ = +1 × 10^-8 C, q₂ = – 2 × 10^-8 C, q₃ = +3 × 10^-8 C, q₄ = 2 × 10^-8 C
a = 1 m, 1/4πε₀ = 9 × 10⁹ N∙m²/C²
r₂₁ = r₄₁ = r₃₂ = r₄₃ = 1 m, r₃₁ = r₄₂ = \(\sqrt {2}\) m
The electric potential energy of the given configuration of our charges is

Question 8.
An electric dipole has two point charges of 1.6 × 10^-19 C and -1.6 × 10^-19 C separated by 2 A. ¡f the dipole is placed in a uniform electric field of 10 N/C, making an angle of 300 with the dipole moment, find (i) the magnitude of the torque acting on the dipole due to the field (ii) the potential energy of the dipole.
Solution:
Data: q = 1.6 × 10^-19 C, 2l = 2Å = 2 × 10^-10 m,
E = 10 N/C, θ = 30°
Electric dipole moment,
p = 2ql = q (2l) = (1.6 × 10^-19 C)(2 × 10^-10 m)
= 3.2 × 10^-29 A∙m²
(j) The magnitude of the torque,
τ = pE sin θ
= (3.2 × 10^-29 Am²)(10 N/C) sin 30°
= 3.2 × 10^-28 × 0.5
= 1.6 × 10^-28 N∙m

(ii) The potential energy,
U = -pE cosθ
= -(3.2 × 10^-29 A∙m²)(10 N/C) cos30°
= -3.2 × 10^-28 × 0.866
= -2.771 × 10^-28 J

Question 9.
A water molecule is made up of two hydrogen atoms and one oxygen atom, with a total of 10 electrons and 10 protons. The molecule is modelled as a dipole with an effective separation d = 3.9 × 10^-12 m between its positive and negative particles. What is the electric potential energy stored in the dipole? What does the sign of your answer mean?
Solution :

The minus sign means that in bringing the particles together horn infinity, energy is transferred from the system to the surrounding and it would take positive work by an external agent to separate the charged particles of the dipole.

Question 10.
An electric dipole consists of two unlike charges of magnitude 2 × 10^-6 C separated by 4 cm. The dipole is placed in an external field of 10⁵ N/C. Find the work done by an external agent to turn the dipole through 180°.
Solution:
Data: q = 2 × 10^-6, 2l = 4cm = 4 × 10^-2 m.
E = 10⁵ N/C, θ = 180° + θ₀
Let us assume the dipole is initially aligned parallel to the field, i.e., θ₀ = 0.
Then, θ = 180°.
The work done by an external agent,
W = pE(1 – cos θ)
= q (2l) E(1 – cos 180°)
= (2 × 10^-6 C)(4 × 10^-2 m)(10⁵ N/C) [1 – (-1)]
= 16 × 10^-3 J

Question 11.
An electric dipole has opposite charges of magnitude 2 × 10^-15 C separated by 0.2 mm. It is placed in a uniform electric field of 10³ N/C. (i) Find the magnitude of the dipole moment. (ii) What is the torque on the dipole when the dipole moment is at 60° with respect to the field?
Solution:
Data : q = 2 × 10^-15 C, 2l = 0.2 mm = 2 × 10^-4 m,
E = 10³ N/C, θ = 60°
(i) The magnitude of the dipole moment is
V = q (2l)
= (2 × 10^-15 C) (2 × 10^-4 m)
= 4 × 10^-19 C∙m

(ii) The torque on the dipole is
τ = pE sin θ
= (4 × 10^-19 C∙m) (10³ N/C) sin 60°
= 4 × 10^-16 × 0.866 = 3.46 × 10^-16 N∙m

Question 40.
What constitutes an electric shock?
Answer:
Living organisms are electrical conductors. Electric shock is the result of the passage of electric current through our body. During electric shock we experience an extreme stimulation of nerves and muscles. It needs a minimum of 1 mA of electric current to pass through our body for us to experience a shock.

Question 41.
Birds perched on electrical transmission wires do not suffer electric shock, but if a person touches both the wires at once receives a tremendous shock. Why?
Answer:
The danger of electric shock arises not from mere contact with a live wire but rather from simultaneous contact with a live wire and another body or wire at a different potential so that our body provides a conducting path between the two and a current passes through our body.

Touching a single wire by the birds does not result in a current through their bodies because then the electric circuit is not complete. But if a person touches two wires at different potentials at once, or if a bare-footed person touches the live wire only, the electric circuit is complete and the person receives an electric shock. In the latter case, the current from the wire passes to the Earth through the body.

[Notes : (1) We must not touch any electric appliance. when bare-footed or with wet hands. When a bare-footed person touches a short-circuited electric appliance, the current from such an appliance goes to the Earth through his body, thus completing the circuit. When our skin is dry, the electrical resistance of our body is about 50 kΩ, a wet skin lowers the resistance to 10 kΩ. It needs a minimum of 1 mA of electric current to pass through our body for us to experience a shock. Thus, when dry, it needs at least 50 V potential difference to get a shock, but only 10 V is enough when wet. (2) The Earth often serves as a charge reservoir known as a ground. A ground can accept or provide electrons freely, and it is so large that the addition or subtraction of electrons has a negligible effect on it. So, the ground remains essentially neutral at all times. When something is connected to the ground by a conductor, we say that it is earthed or grounded.]

Question 42.
State the properties of conductors in electrostatic conditions.
Answer:
Properties of a charged conductor in electrostatic conditions :

1. Net electric field inside the conductor is zero.
2. Net electric field just outside the conductor is normal to its surface at every point.
3. Electric potential inside the conductor is constant and equal to that on its surface.
4. Excess charges reside only on the surface of the conductor but, for a conductor of arbitrary shape, the surface charge density at a point is inversely proportional to the local curvature of the surface.

Question 43.
Explain electrostatic shielding. What is a Faraday cage ?
Answer:
When an isolated conductor, uncharged or charged, is placed in an external electric field, as in figure, all points of the conductor come to the same potential. The free conduction electrons in the conductor distribute themselves on the surface, leaving a net positive charge on some regions of the surface and a net negative charge on other regions.

This charge distribution causes an additional electric field at interior points such that the total field at every point inside is zero.

The charge distribution on the conductor is such that the net electric field at all points on the surface to be perpendicular to the surface, thereby altering the shapes of the field lines near the conductor.

The use of a conducting box to protect sensitive instruments from stray electric fields, or the use of a conducting wire cage to protect a person near a high-voltage installation or from lightning strike, is called electrostatic shielding. The hollow conductor or the conducting wire cage that shields its interior from external electric fields is called a Faraday cage or Faraday shield. A Faraday cage, made from a contiguous metal sheet or from a fine metal mesh, is used to shield its content or occupant from static and nonstatic electric fields.

Question 44.
Aran and Careena were driving in the countryside in a car when they get caught in a thunderstorm. Arun is worried that if lightning hits the car, the petrol tank may explode. Careena thinks they should wait out the storm in the car. Is Careena right ?
Answer:
In circumstances where there is danger of lightning strikes, it is wise to enclose oneself in a cavity inside a conducting shell, where the electric field is guaranteed to be zero. A car with a metal body is an almost ideal Faraday cage. When a car is struck by lightning, the charge flows on the outside surface of the car to the ground but the electric field inside remains zero. This leaves the passengers inside unharmed. Also, the petrol is in a metal tank which also acts like a Faraday cage. So, Careena is right.

Question 45.
Explain the electrical behaviour of conductors and insulators on the basis of free and bound charges inside the materials.
Answer:
In a material, the inner shell electrons are tightly bound to their respective nuclei and together they have fixed lattice positions. They are called bound charges.

In metals, the outermost valence electrons are loosely bound to their respective nucleus and, due to the regular atomic arrangement in a lattice, are set free to move inside the metal. They are called free charges or free electrons. Under an applied electric field, the free electrons drift in a direction opposite to the electric field and constitute an electric current in the metal. In electrolytes, electrical dissociation of ionic molecules results in both positive and negative free charges, and electric conduction is due to both types of free charges. Under electrostatic conditions, excess charges reside only on the surface of a conductor.

In insulators, all inner shell and outer shell electrons are tightly bound to their respective nuclei so that even at room temperature the number of free charges is several orders lower than that in a metallic conductor. Hence, they are poor conductors of electricity and heat. In the absence of free conduction electrons, excess charges transferred to an insulator remain localized. An insulator can have non-zero surface charge density as well as volume charge density.

Question 46.
You can charge a glass or rubber rod by holding one end of the rod and rubbing the other end with a silk cloth. But you cannot charge a copper rod in the same way. Explain.
Answer:
Glass and rubber are insulators. An excess charge (positive or negative) building up on some part of an insulator remains localized. So, a glass or rubber rod can be held at one end while the far end is being rubbed with silk. The far end of the rod acquires a surplus of electrons but those electrons never flow into the ground through the hand.

A copper rod is an example of a conductor which has free conduction electrons. On holding one end of the copper rod and rubbing the other end with silk, electrons are transferred from the silk to the copper rod, and those excess electrons are free to flow. Because like charges repel, the electrons move away from one another, travel through the rod to the ground through the hand. As a result, the copper rod remains neutral despite the rod being rubbed with silk.

Question 47.
What do you mean by a polar molecule and a nonpolar molecule ? Give two examples of each.
Answer:
A polar molecule is an asymmetric molecule with a permanent electric dipole moment that arises from the finite separation of the centres of the net positive charge and the net negative charge in the molecule, even in the absence of an external electric field.
Dipole moments of polar molecules are of the order of 10^-30 C∙m.

Examples : Gaseous hydrogen halides (HF, HCl, etc.); NH₃, NO₂, N₂O, water molecules; all hetero- nuclear diatomic molecules (with any covalent bond between two different atoms).

A nonpolar molecule is one which does not have a permanent electric dipole moment because in the absence of an external electric field, the centres of the net positive charge and the net negative charge in the molecule coincide. Thus, it is a symmetric molecule.

Examples : H₂, CO₂, N₂, O₂, methane, polyethylene, polystyrene.

[Note : The asymmetric charge distribution in a polar molecule arises from differences in electronegativity and other features of bonding. A heteronuclear polyatomic molecule may be nonpolar if the vector sum of the bond dipole moments is zero, as in CO₂, CH₄, CCl₄, etc. The absence or presence of a dipole moment in a polyatomic molecule can be a revealing clue to the structure of the molecule. For example, BF₃ with a planar trigonal sym-metric structure is nonpolar, while PF₃ with a trigonal pyramidal structure is polar. Of the two isomers of dichloroethylene, cis is polar while trans is nonpolar.]

Question 48.
What is the bond angle and bond dipole moment of a water molecule?
Answer:
In a water molecule, the bond angle between the two O-H bonds is 104.5° and the dipole moment (known as the bond dipole moment) is about 6.17 × 10^-30 C∙m

[Note: The SI unit of dipole moment, the coulomb-metre is too large at molecular level. Hence, bond dipole moments are commonly expressed in the CGS unit, the debye (D); 1 D = 3.335 × 10^-30 C m. So, the bond dipole moment of a water molecule is 1.85 D while that of HF, NH₃ and HCl are 1.82 D, 1.47 D and 1.08 D, respectively.]

Question 49.
What is a dielectric? State its two types. Give two examples in each case.
Answer:
A dielectric is an electrical insulator i.e., a nonconducting material, that can be polarised by an applied electric field which slightly displaces the positive and negative charges of each molecule. A dielectric can sustain a high electric field up to a certain limit. An ideal dielectric has no free charges.

Important commercial dielectrics are of two types, polar and nonpolar.

Examples :
Polar dielectrics : Silicones, halogenated hydrocarbons.

Nonpolar dielectrics : (1) Solid : Ceramics, glasses, plastics (polyethylene, polystyrene, etc.) mica, paper. (2) Liquid : Mineral oils.

Question 50.
With the help of neat diagrams, explain how a nonpolar dielectric material is polarised in an external electric field.
OR
Explain the behaviour of nonpolar dielectric material in an external electric field.
Answer:
In the absence of an external electric field, the molecules of a nonpolar dielectric have no inherent electric dipole moments. An applied electric field slightly separates the centres of negative and positive charges. Then, a nonpolar molecule acquires an induced dipole moment in the direction of the applied field, as shown in figure. The induced dipole moments of all the molecules add up giving the dielectric a net induced electric dipole moment • in the presence of the applied field.

Question 51.
With the help of neat diagrams, explain how a polar dielectric material acquires a dipole moment in an external electric field.
OR
Explain the behaviour of polar dielectric material in an external electric field.
Answer:
In the absence of an external electric field, the permanent electric dipole moments of the molecules of a polar dielectric orient in random directions due to thermal agitation such that their vector sum is zero, from figure (a).

An applied electric field does slightly increase the separation between the centres of negative and positive charges. But, a much larger effect is the tendency of the dipole moments to align with the field, although thermal agitation prevents complete alignment, as shown in from figure (b). Due to the partial alignment of the dipole moments, a polar dielectric also acquires a net induced electric dipole moment in the direction of the applied field. The extent of polarisation depends on the relative values of the two opposing tendencies : (1) the tendency of the applied field to align the dipoles
(2) thermal agitation that tends to randomize.

Question 52.
What is a linear isotropic dielectric? Give one example.
Answer:
A linear isotropic dielectric is one which when placed in a uniform electric field acquires an induced electric dipole moment in the direction of the field and proportional to the applied electric field intensity.
Examples : Mica, glass.

Question 53.
Define electric polarization in dielectrics.
Answer:
Definition The electric polarization at every point within a dielectric is defined as the electric dipole moment per unit volume. It has the direction of the external electric field.

Question 54.
What is electric susceptibility ?
Ans. In a linearly isotropic dielectric placed in a uniform electric field, the electric polarization \(\vec{P}\) is directly proportional to the electric field intensity \(\vec{E}\) inside the dielectric.
∴ \(\vec{P}\) = χ_eε₀\(\vec{E}\)
where the proportionality constant χ_e, a positive quantity, is called the electric susceptibility of the dielectric.
[Note : χ_e is dimensionless. SI units of P and E are C/m² and N/C.]

Question 55.
Explain the reduction of electric field inside a polarized dielectric.
OR
Explain the behaviour of a dielectric slab which is subjected to an external electric field.
Answer:
Consider a rectangular slab of a linear isotropic dielectric placed in a uniform external electric field.

In case of a nonpolar dielectric, the applied field slightly separates the centres of negative and positive charge in a molecule inducing an electric dipole moment in the direction of the field. In most cases, this separation is a very small fraction of a molecular diameter. In case of a polar dielectric, the permanent dipole moments of its molecules are partially aligned with the field. In either case, the dielectric is said to become polarized.

In a uniformly polarized dielectric, the charges of adjacent interior dipoles add to zero. But due to the unbalanced positive ends of dipoles at one face of the slab, bound positive charge appears on that exterior surface. Similarly, bound negative charge appears on the opposite exterior surface of the slab. These bound surface charges are called polarization charges. There is, however, no excess charge in any volume element within the slab and the slab as a whole remains electrically neutral.

The external electric field \(\overrightarrow{E_{0}}\) polarizes the dielectric, with a net polarization \(\vec{P}\) parallel to \(\overrightarrow{E_{0}}\). Within the dielectric, the induced field \(\vec{E}_{\mathrm{p}}=-\vec{P} / \varepsilon_{0}\) due to the polarization charges is opposite to the applied field, from figure.
\(\vec{E}=\vec{E}_{0}+\vec{E}_{\mathrm{p}}=\vec{E}_{0}-\vec{P} / \varepsilon_{0}\)
i.e., E = E₀ – E_p in magnitude.
Assuming the dielectric to be isotropic and linear,
\(\vec{P}\) = χ_eε₀ \(\vec{E}\)
where χ_e is the electric susceptibility of the dielectric.
∴ \(\overrightarrow{E_{0}}\) = (1 + χ_e)\(\vec{E}\) = k\(\vec{E}\)
where k = 1 + χ_e is the dielectric constant. This implies E = E₀/k, thus less than E₀. Thus, the effect of a dielectric material is always to decrease the electric field below the applied electric field.

[Note : The charges + Q and – Q on the plates of a capacitor are said to be free charges because they can move when the potential difference between the plates is changed.]

Question 56.
How does the electric field inside a dielectric decrease when it is placed in an external electric field?
Answer:
Suppose a rectangular slab of dielectric is placed in an electric field \(\overrightarrow{E_{0}}\), with two of its parallel sides perpendicular to the field. The dielectric becomes polarized. Polarization charges appear on the external surfaces of these two parallel sides such that within the dielectric the field due to the polarization charges is opposite to \(\overrightarrow{E_{0}}\). Thus, the magnitude of the net electric field \(\vec{E}\) within the dielectric is less than \(\left|\overrightarrow{E_{0}}\right| \cdot E=\frac{E_{0}}{k}\), where k is the relative permittivity (dielectric constant) of the dielectric.

Question 57.
Is vacuum the best dielectric, meaning is its dielectric strength infinite ?
Answer:
In principle, an absence of particles mean no breakdown. However, practical vacuum still has large number of particles of residual gases. In practice, only pressures lower than 10^-2 mbar can be considered to provide a real dielectric insulation. In such vacuum, free electrical charges under a sufficiently high force can produce ionization and breakdown of its insulation properties. Thus, the dielectric strength of vacuum is only about 20 kV/mm, better than air but lower than most solid dielectrics.

Question 58.
What is a capacitor ?
Explain capacitance of a capacitor.
Answer:
A capacitor is a device used to store electrical charge and electrical energy. It consists of at least two electrical conductors, called as capacitor plates, separated by a distance. The space between the plates may simply be air or, usually, filled with a dielectric.

Consider a capacitor having two conducting plates close to, but not touching, one another shown in figure. Imagine that each plate is neutral, so the potential difference between the plates is zero. If a small positive charge q is transferred from one plate

to the other, the second plate acquires a charge + q while that on the first plate is — q and a small potential difference appears between the plates. As the amount of charge on each plate increases, so does the potential difference between the plates. The potential difference AV between the plates is directly proportional to the magnitude of charge Q on each plate : Q ∝ ∆V.
∴ Q = C∆V
The constant of proportionality C is called the capacitance. The capacitance depends only on the geometry of the plates and the type of dielectric between the plates.

When the terminals of a battery are connected to the plates of an initially uncharged capacitor, the battery potential V moves a small amount of charge of magnitude Q from the plate at the higher potential to the other plate. The capacitor remains neutral overall, but with charges + Q and – Q on opposite plates. The capacitance C is then the ratio of the maximum charge Q that can be stored in the capacitor to the applied voltage V across its plates or, in other words, capacitance is the largest amount of charge per unit potential difference that can be stored on the device.

[Note : A capacitor is represented in circuit diagrams by the symbol, ]

Question 59.
The graph shows the potential difference ∆V between the plates of a capacitor versus the charge Q on its plates. Use the graph to find (1) the capacitance of the capacitor (2) the magnitude of the excess charge on its plates.

Solution:
C = \(\frac{Q}{\Delta V}=\frac{1}{\text { slope }}=\frac{1.5 \mathrm{nC}}{30 \mathrm{~V}}\) = 0.05 nF = 50 pF
From the graph, for Q = 0.5 nC, ∆V = 10 V.

Question 60.
Define the capacitance of a capacitor. State and define the SI unit of capacitance.
Answer:
(1) Definition : The capacitance (capacity) of a capacitor is defined as the ratio of the charge on either conductor to the potential difference between the two conductors forming the capacitor.

(2) The SI unit of capacitance is the farad.
Definition : The capacitance of a capacitor is said to be one farad if a charge of one coulomb is required to increase the potential difference between the two conductors forming the capacitor by one volt.
1 farad = 1 coulomb/volt; 1 F = 1 C/V
[Note : Capacitors in common electronic circuits are in microfarad (10^-6 F), nanofarad (10^-9 F) or picofarad (10^-12 F).]

Question 61.
State and explain the principle of a capacitor.
Answer:
Principle of a capacitor : Any conductor can be used to store charges, however, its capacity can be increased by keeping a grounded conductor near it.

Consider a metal plate A whose potential is raised to V by depositing a charge + Q on it, so that its capacity is C = Q/V. Now, if an uncharged metal plate B is brought close to plate A, then negative bound charge – Q will be induced on the surface of B near A and positive free charge + Q on the other side of B, from figure (a).

If plate B is grounded, i.e., connected to the Earth, the free charge on it will escape to the Earth, from figure (b). The bound charge (- Q) thus remaining on B will lower the potential of A, as if superimposing a potential – V₁ on the potential V of plate A. The resultant potential of A will become V – V₁ and its capacity will be Q/(V – V₁).

C’ = \(\frac{Q}{V-V_{1}}\)
Keeping plate B very close to A, V – V₁ can be made very small, so that the capacity of the combination can become very much greater than the capacity of conductor A alone. C’ >> C

Question 62.
What are the different types of capacitors? Describe in brief.
Answer:
The three main types of capacitors depending on their shape are (1) parallel-plate capacitor (2) spherical capacitor (3) cylindrical capacitor.
(1) Parallel-plate capacitor: It consists of two parallel metal plates, separated by a small gap [from figure (a)] of air or filled with a dielectric. The charge to be stored is given to one plate (A) while the other plate (B) is earthed.

(2) Spherical capacitor: It consists of two concentric spherical conductors, separated by a small gap of air or filled with a dielectric [from figure (b)]. The charge to be stored is given to the inner sphere (A), while the outer sphere (B) is earthed.

(3) Cylindrical capacitor : It consists of two coaxial, cylindrical conductors separated by a small gap of air or filled with a dielectric [from figure (c)]. The charge to be stored is given to the inner cylinder (A), while the outer cylinder (B) is earthed.

Depending on the dielectric used, the capacitors of different types are (1) mica capacitor (2) air capacitor (3) paper capacitor (4) electrolytic capacitor, etc.

Question 63.
Derive an expression for the effective or equivalent capacitance (capacity) of a combination of a number of capacitors connected in series.
OR
Derive an expression for the effective capacitance of three capacitors connected in series.
Answer:
In the series arrangement of capacitors, the capacitors are connected end to end and a cell is connected across the combination of the capacitors as shown in figure.

Let C₁, C₂, C₃ be the capacitances of the three capacitors connected in series and Q, the charge on each capacitor. Let V₁, V₂, V₃ be the potential differences across the capacitors.

Now, charge = capacitance x potential difference
∴ Q = C₁V₁ = C₂V₂ = C₃V₃
∴ V₁ = \(\frac{Q}{C_{1}}\), V₂ = \(\frac{Q}{C_{2}}\) and V₃ = \(\frac{Q}{C_{3}}\)
If Vis the potential difference across the combination and C is the equivalent or effective capacitance of the combination, we have,
C = \(\frac{Q}{V}\) ∴ V = \(\frac{Q}{V}\)

In general, if n capacitors of capacitances C₁, C₂, C₃, …, C, are connected in series, the equivalent capacitance (C) of the combination is given by
\(\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}+\ldots+\frac{1}{C_{n}}\)

Question 64.
Derive an expression for the effective or equivalent capacitance (capacity) of a combination of a
number of capacitors connected in parallel.
OR
Derive an expression for the effective or equivalent capacitance of three capacitors connected in parallel.
Answer:
In the parallel arrangement of capacitors, the capacitors are connected between two common points and a cell is connected across the combination of the capacitors as shown in below figure. Thus, the potential difference (V ) across each capacitor is the same, Let C₁, C₂, C₃ be the capacitances of the three capacitors connected in parallel and C, the equivalent or effective capacitance of the combination. The charge Q supplied by the cell is distributed as Q₁, Q₂ and Q₃ on the capacitors.

∴ Q₁ = C₁V, Q₂ = C₂V, Q₃ = C₃V and Q = CV
Since, Q = Q₁ + Q₂ + Q₃
CV = C₁V + C₂V + C₃V
∴ C = C₁ + C₂ + C₃
In general, if n capacitors are connected in parallel,
C = C₁ + C₂ + ……. + C_n .

65. Solve the following

Question 1.
Three capacitors have capacities 2 µF, 4 µF and 8 µF. Find the equivalent capacity when they are connected in (a) series (b) parallel.
Solution:
Data : C₁ = 2 µF, C₂ = 4 µF, C₃ = 8 µF
(a) Series arrangement:
\(\frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=\frac{7}{8}\)
The equivalent capacity,
∴ C_s = \(\frac{8}{7}\) = 1.143 µF

(b) Parallel arrangement:
The equivalent capacity is
C_p = C₁ + C₂ + C₃ = 2 + 4 + 8 = 14 µF

Question 2.
The equivalent capacitance of two capacitors is 6 µF when they are connected in series and 25 µF when they are connected in parallel. Find the capacitance of each capacitor.
Solution:
Data : C_s = 6 µF, C_p = 25 µF
Let C₁ and C₂ be the capacitances of the two capacitors respectively.
In parallel combination,
C_p = C₁ + C₂ = 25
In series combination,
\(\frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{C_{1}+C_{2}}{C_{1} C_{2}}\)
∴ C_s = \(\frac{C_{1} C_{2}}{C_{1}+C_{2}}\)
∴ 6 = \(\frac{C_{1} C_{2}}{25}\)
∴ C₁C₂ = 150
∴ C₁(25 – C₁) = 150
∴ 25C₁ – C₁² = 150
∴ C₁² – 25C₁ + 150 = 0
∴ (C₁ – 15) (C₁ – 10) = 0
∴ C₁ = 15 µF or 10 µF
∴ C₂ = 25 – C₁ = 10 µF or 15 µF
∴ The capacitances of the capacitors are 15 µF and 10 µF.

Question 3.
With four capacitors of the same capacity, when three of them are connected in parallel and the remaining one in connected in series with this combination, the resultant capacity is 3.75 µF. Find the capacity of each capacitor.
Solution:
Data : C_eff = 3.75 µF
Let the capacity of each of the four capacitors be C. The equivalent capacity of three of them in parallel is
C_p = C + C + C = 3C
The equivalent capacity of the series combination of C_p and the fourth capacitor is
C_eff = \(\frac{C_{\mathrm{p}} C}{C_{\mathrm{P}}+C}=\frac{(3 C) C}{3 C+C}=\frac{3}{4} C\)
∴ By the data, \(\frac{3}{4}\) C = 3.75 µF
∴ C = \(\frac{4}{3}\) × 3.75 = 5 µF
∴ The capacity of each capacitor = 5 µF.

Question 4.
Two capacitors of capacities C₁ and C₂ are joined in series and this combination is joined in parallel with a capacitor of capacity C₃. Show that the capacity of the system is C = \(\frac{C_{1}\left(C_{2}+C_{3}\right)+C_{2}
Solution:
The equivalent capacitance of the series combination of C₁ and C₂ is
C_s = [latex]\frac{C_{1} C_{2}}{C_{1}+C_{2}}\)
The equivalent capacitance of the parallel combination of C_s and C₃ is

Question 5.
Capacitors of capacities 5 µF and 10 µF are connected in parallel in a circuit with a cell of emf 2 V. What should be the capacity of the capacitor to be connected in series with the parallel combination of the capacitors to get 1 µC charge on the combination?
Solution:
Data : C₁ = 5 µF, C₂ = 10 µF, V = 2 V, Q = 1 µC
The effective capacity of the parallel combination of C₁ and C₂ is
C_p = C₁ + C₂ = 5 + 10 = 15 µF

Let C be the capacity of the capacitor to be connected in series with C_p to get the charge of 1 coulomb on the combination. The equivalent capacity (C_s) of the series combination is given by

Question 6.
Three capacitors of capacities 8 µF, 8 µF and 4 µF are connected in series and a potential difference of 120 V is maintained across the combination. Calculate the charge on the capacitor of capacity 4 µF. Also calculate the potential difference across it.
Solution:
Data : C₁ = 8 µF, C₂ = 8 µF, C₃ = 4 µF, V = 120 V
Let C_s = equivalent capacity of the series combination of the capacitors
∴ \(\frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}=\frac{1}{8}+\frac{1}{8}+\frac{1}{4}\)
= \(\frac{1+1+2}{8}=\frac{4}{8}=\frac{1}{2}\)
∴ C_s = 2 µF = 2 × 10^-6 F
In series combination, the charge on each capacitor is the same. It is given by Q = C_sV
∴ Q = 2 × 10^-6 × 120
= 2.4 × 10^-4 coulomb
V₃ = \(\frac{Q}{C_{3}}=\frac{2.4 \times 10^{-4}}{4 \times 10^{-6}}=\frac{240}{4}\) = 60 V (∵ C₃ = 4 µF = 4 × 10^-6 F)
The charge on the 4 µF capacitor is 2.4 × 10^-4 C and the potential difference across it is 60 V.

Question 7.
A 100 V battery is connected across the combination of capacitors of capacities 4 µF and 8 µF in parallel and then in series. Calculate the charge on each capacitor in parallel and in series combination.
Solution:
Data : V = 100 V, C₁ = 4 × 10^-6 F, C₂ = 8 × 10^-6 F
(i) Parallel combination :
Q₁ = C₁V = (4 × 10^-6)(100) = 4 × 10^-4 C
Q₂ = C₂V = (8 × 10^-6)(100) = 8 × 10^-4 C

(ii) Series combination :

Question 8.
Three capacitors are connected as shown in the figure below.

Calculate the effective capacitance between A and B.
Solution:
Data : C₁ = 2 µF, C₂ = 3 µF, C₃ = 4 µF
The resultant capacitance C_s of C₁ and C₂ in series is given by
\(\frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
∴ C_s = \(\frac{6}{5}\) = 1.2 µF
The effective capacitance between A and B is due to the parallel combination of C_s and C₃.
C_p = C_s + C₃ = 1.2 + 4 = 5.2 µF

Question 9.
Four capacitors are of the same capacitance.
(a) If three of them are connected in parallel and the remaining one is connected in series with this combination, the resultant capacitance is 3.75 µF. Find the capacitance of each capacitor.
(b) When three of them are connected in series and the remaining one is connected in parallel with this combination, find the resultant capacitance of the combination.
Solution:
(a) Let C be the capacitance of each capacitor.

Between points A and B, three capacitors, each of capacitance C, are connected in parallel from figure. Their effective capacitance (C_p) is given by
C_p = C + C + C = 3C
When a capacitor of capacitance C is connected in series with C_p, their resultant capacitance (C_s) is given by
\(\frac{1}{C_{s}}=\frac{1}{C_{p}}+\frac{1}{C}=\frac{1}{3 C}+\frac{1}{C}=\frac{4}{3 C}\)
∴ C_s = \(\frac{3 C}{4}\)
Given : C_s = 3.75 µF
∴ \(\frac{3 C}{4}\) = 3.75
∴ C = \(\frac{4 \times 3.75}{3}=\frac{15}{3}\) = 5 µF

(b) The arrangement is shown in below figure.

The effective capacitance (C_s) of the series combination of three capacitors, each of capacitance C, is given by
\(\frac{1}{C_{s}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}=\frac{3}{C}\)
∴ C_s = \(\frac{C}{3}\)
When another capacitor of capacitance C is connected in parallel with the combination, the result ant capacitance is
C_p = C_s + C = \(\frac{C}{3}\) + c
= \(\frac{4 C}{3}=\frac{4 \times 5}{3}=\frac{20}{3}\)
= 6.667 µF

Question 10.
In below figure, C₁ = 10 µF, C₂ = 30 µF, C₃ = 20 µF, C₄ = 40 µF. Find the capacitance between the points A and B when (i) the key K is closed (ii) the key K is open.

Solution:
(i) Key K closed :
The parallel combination of C₁, and C₃ is in series with the parallel combination of C₂ and C₄. Let
C₅ = C₁ || C₃ and C₆ = C₂ || C₄.
∴ C₅ = C₁ + C₃ = 10 + 20 = 30 µF
and C₆ = C₂ + C₄ = 30 + 40 = 70 µF
The capacitance between A and B is the equivalent capacitance of C₅ and C₆ in series, i.e.,
C_AB = \(\frac{C_{5} C_{6}}{C_{5}+C_{6}}=\frac{30 \times 70}{30+70}=\frac{2100}{100}\) = 21 µF

(ii) Key K open :
The series combination of C₁ and C₂ is in parallel with the series combination of C₃ and C₄. Let C₇ and C₈ be the equivalent capacitances of the respective series combinations.

The capacitance between A and B is the equivalent capacitance of C₇ and C₈ in parallel, i.e.,
C_AB = C₇ + C₈ = \(\frac{15}{2}+\frac{40}{3}=\frac{45+80}{6}\)
= \(\frac{125}{6}\) = 20.83 µF

Question 11.
A network of four capacitors, 5 µF each, are connected to a 240 V supply. Determine the equivalent capacitance of the network and the charge on each capacitor.

Solution:
Data : C₁ = C₂ = C₃ = C₄ = 5 µF, V = 240 V

The equivalent capacitance C₅ of the series combination is given by
\(\frac{1}{\mathrm{C}_{5}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{3}{5}\)
∴ C₅ = \(\frac{5}{3}\) µF
The equivalent capacitance of the parallel combination of C₅ with C₄ is
C = C₅ + C₄ = \(\frac{5}{3}\) + 5 = \(\frac{20}{3}\) µF = 6.667 µF
The potential difference across each of the three capacitors in series = \(\frac{V}{3}=\frac{240}{3}\) = 80 V
and the charge on each of them is the same.
∴ Q₁ = Q₂ = Q₃ = C₁ × \(\frac{V}{3}\) = (5 × 10^-6)(80)
= 4 × 10^-4 C
The charge on the capacitor C₄ is
Q₄ = C₄V = (5 × 10^-6)(240) = 1.2 × 10^-3 C

Question 12.
Six capacitors of capacities 5 µF, 5 µF, 5 µF, 5 µF, 10 µF, and X µF are connected as shown in the network given below :

Find (a) the value of X if the network is balanced (b) the resultant capacitance between A and C.
Solution:
(a) The effective capacitance of the series combination in the arm DC is \(\frac{10 X}{10+X}\) µF
Using the balancing condition for the network,

(b) As the network is balanced, no charge flows through the arm BD.
Effective capacitance 5×5 25
C_ABC = \(\frac{5 \times 5}{5+5}=\frac{25}{10}\)= 2.5 µF (series combination)
Similarly,
C_ADC =\(\frac{5 \times 5}{5+5}=\frac{25}{10}\) = 2.5 µF (series combination)
Effective (resultant) capacitance between A and C .
= C_ABC + C_ADC (parallel combination)
= 2.5 + 2.5 = 5 µF

Question 66.
Derive an expression for the capacitance of a parallel-plate air/vacuum capacitor.
Answer:
Consider a parallel-plate capacitor, consisting of two parallel plates A and B separated by a distance d as shown in below figure. Let A be the area of each plate. Plate B is connected to the Earth. Suppose that the capacitor is connected to the terminals of a battery of potential difference V. The battery transfers a charge + Q to the insulated plate A. A charge – Q is induced on the near surface of the grounded plate B while the + Q charge on the far side of B flows to the ground.

1f the area A is very large and the distance between the plates is very small, the electric field in the region between the plates is almost uniform, except near the edges. The magnitude of the electric field E at a point between the plates and the potential difference V between the plates are related by E = V/d. Outside the capacitors, the electric fields due to the two charged plates cancel out:
\(\frac{\sigma}{2 \varepsilon_{0}}-\frac{\sigma}{2 \varepsilon_{0}}\) = 0.
But, E = \(\frac{\sigma}{\varepsilon_{0}}\), where σ is the surface charge density on the plates.
∴ \(\frac{\sigma}{\varepsilon_{0}}=\frac{V}{d}\) …………. (1)
Now, = \(\frac{Q}{A}\) ∴ E = \(\frac{Q}{\varepsilon_{0} A}=\frac{V}{d}\) …………… (2)
The capacity (capacitance) of a capacitor is, by definition, C = \(\frac{Q}{V}\)
∴ C = \(\frac{\varepsilon_{0} A}{d}\) …………… (3)
This gives the capacitance of a parallel-plate capacitor without a dielectric, i.e., an air or vacuum capacitor.

Question 67.
What is the electric field intensity in the region between the plates of a parallel-plate capacitor if the separation between the plates is 1 mm and the potential difference across the plates is 2V?
Answer:
E = \(\frac{V}{d}=\frac{2}{1 \times 10^{-3}}\) = 2000 N/C is the required electric field intensity.

Question 68.
The capacitance of a parallel-plate air capacitor is 12 µF. If the separation between its plates is doubled, what will be the new capacitance?
Answer:
\(\frac{C_{2}}{C_{1}}=\frac{d_{1}}{d_{2}}\)
∴ C₂ = \(\frac{d_{1}}{d_{2}}\) × C₁ = \(\frac{1}{2}\) × 12 = 6pF
is the new capacitance.

Question 69.
Explain the effect of a dielectric on the capacitance of a isolated charged parallel-plate capacitor.
Hence, show that if a dielectric of relative permit-tivity (dielectric constant) k completely fills the space between the plates, the capacitance increases by a factor k.
Answer:
Consider a parallel-plate capacitor without a dielectric, of plate area A, plate separation d and capacitance C₀, charged to a potential difference V₀ and then isolated.
Suppose the charges on its conducting plates are + Q and – Q, from figure (a). The surface density of free charge is

If A is very large and d is very small, the electric field in the region between the plates is almost uniform, except near the edges. The magnitude of the electric field intensity is
E₀ = \(\frac{V_{0}}{d}=\frac{\sigma}{\varepsilon_{0}}=\frac{Q}{\varepsilon_{0} A}\)
Without the dielectric, the capacitance of the parallel plate capacitor is, by definition, C₀ = \(\frac{Q}{V_{0}}=\frac{\varepsilon_{0} A}{d}\) ……………. (3)

Now, suppose a dielectric slab of permittivity e and thickness t (t < d) is introduced in the space between, and parallel to, the charged plates, from figure (b). A polarisation charge – Q_p appears on the exterior surface of the dielectric nearer to the positive plate while a polarisation charge + Q_p appears on its opposite face. Since the capacitor was isolated after charging, the free charge Q on the plates is the same as earlier. Within the dielectric, the induced field \(\vec{E}_{\mathrm{p}}\) due to the polarisation charges is opposite to the applied field. The net electric field within the dielectric \(\vec{E}\) is less than the applied field. In magnitude,
E = E₀ – E_p …………. (4)
By definition, the relative permittivity (dielectric constant) of the dielectric, k = \(\frac{\varepsilon}{\varepsilon_{0}}=\frac{E_{0}}{E}\) ………. (5)

Between the plates, the field within the dielectric of thickness t is E = E₀/k, and that in the region (d – t) is E₀. Therefore, the new potential difference between the plates is

Equations (7) and (8) give the capacitance of a capacitor with a dielectric.

Special case: If the dielectric completely fills the space between the plates, t = d. Therefore, from Eq. (7),
C = C₀ \(\frac{d}{\left(d-d+\frac{d}{k}\right)}\) = C₀ \(\frac{d}{d / k}\) = kC₀
Thus, the capacitance increases by the factor of k.

[Notes : (1) It‘is useful to check the other limits :
(i) As the thickness of the dielectric approaches zero, i.e., t → 0, we have C → ε₀A/ d = C₀, as expected for no dielectric.
(ii) If k = 1, we again have C = ε₀A/d = C₀, as expected for air or vacuum, capacitor where the dielectric is absent.
(2) The equation, V = \(\frac{V_{0}}{d}\) ( d – t + \(\frac{t}{k}\)), shows that V too is less than V0 for the same charge Q.
(3) Note that the configuration is equivalent to two capacitors of plate separations (d -1) and t connected in series, as shown in below figure. For capacitors connected in series, the equivalent capacitance is given by

Thus, if the dielectric slab is replaced by a conducting slab of the same thickness f, the right-hand plates in above figure will be shorted and the equivalent capacitance C = C₁ = \(\frac{\varepsilon_{0} A}{d-t}=\frac{d}{d-t}\) C₀. Therefore, C > C₀.

(4) If a battery supplying a potential difference V0 remains connected as a dielectric is inserted filling the space, the charge on the plates is increased by a factor k, the p.d. across the capacitor remaining constant Q = kQ₀/ where Q₀ is the charge on the plates without the dielectric. The capacitance becomes
C = \(\frac{Q}{V_{0}}=\frac{k Q_{0}}{V_{0}}\) = kC₀
which is the same as in Eq. (9) where the charge remained constant, but now the charge has increased.]

Question 70.
State the expression for the capacitance of a parallel-plate capacitor filled with a dielectric. Explain how its capacitance can be increased.
Answer:
The capacitance of a parallel-plate capacitor filled with a dielectric is C = \(\frac{A k \varepsilon_{0}}{d}\)

where A is the area of each plate, k is the relative permittivity (dielectric constant) of the medium between the plates, ε₀ is the permittivity of free space and d is the uniform plate separation.

The capacitance of a parallel-plate capacitor can be increased by

1. increasing the area of each plate
2. decreasing the distance between the two plates
3. filling the space between the two plates by a medium of greater relative permittivity.

Question 71.
What are the functions of a dielectric in a capacitor?
Answer:
A dielectric material between the plates of a capacitor

1. increases the capacitance of the capacitor
2. provides mechanical support to the plates
3. increases the maximum operating voltage, i.e., the maximum voltage to which the capacitor may be charged without breakdown of the insulating property of the medium between the plates.

Question 72.
A parallel-plate capacitor consists of n dielectric slabs of end-face areas A₁, A₂, ………, A_n and respective relative permittivities (dielectric constants) k₁, k₂, ….., k_n, in the space between the plates as shown in below figure. Find the equivalent capacitance of this arrangement. Hence, show that if the areas are equal, the capacitance is C = \(\frac{\varepsilon_{0} A}{n d} \sum_{j} k_{j}\).

Answer:
Consider a parallel-plate capacitor without a dielectric, of plate area A, plate separation d and capacitance C₀, charged to a potential difference V₀ and then isolated. Without a dielectric, the capacitance of the parallel-plate capacitor is, by definition
C₀ = \(\frac{Q}{V_{0}}=\frac{\varepsilon_{0} A}{d}\) …………… (1)
Now, suppose n dielectric slabs of end-face areas
A₁, A₁, …, A₃ and respective relative permittivities
(dielectric constants) k₁, k₂, ………, k_n fills the space between the plates as shown in above figure. The charge on the plates remaining the same, the reduced potential difference (V) between the plates is the same across each dielectric slab. Therefore, the capacitor is equivalent to a parallel combination of n capacitors of the same plate separation d but different plate areas and dielectric constants, as shown in below figure.

Then, C₁ = k₁0A₁/d, C₂ = k₂e0A₂/d, … and so on.
Therefore, their equivalent capacitance is
C = C₁ + C₂ + ………… + C_n
= \(\frac{\varepsilon_{0}}{d}\)(k₁A₁ + k₂A₂ + …….. + k_nA_n) ………… (2)
If the dielectric slabs have equal thickness such that their end-face areas are equal, i.e., A₁ = A₂ = …….. = A_n = A/n.
C = \(\frac{\varepsilon_{0} A}{n d}\)(k₁ + k₂ + ………… + k_n) = \(\frac{\varepsilon_{0} A}{n d}\) Σ_j K\(\frac{\varepsilon_{0} A}{n d}\)

Question 73.
A dielectric of relative permittivity (dielectric constant) k completely fills the space between the plates of a parallel-plate capacitor with a surface charge density σ. Show that the induced density of surface charge on the dielectric is σ_p = σ (1 – \(\frac{1}{k}\))
Answer:
Consider a parallel-plate capacitor without a dielectric, of plate area A, plate separation d and capacitance C₀, charged to a potential difference V and then isolated.

Suppose the charges on its conducting plates are +Q and -Q, from figure (a). The surface density of free charge is
σ = \(\frac{Q}{A}\) …………. (1)

When no dielectric is present, the electric field \(\overrightarrow{E_{0}}\) in the region between the plates can be found by applying Gauss’s law to the Gaussian surface :

When a dielectric is inserted as in figure (b), there is an induced charge – Q, on the surface, and . the net charge enclosed by the Gaussian surface is Q – Q_p. Then, by Gauss’s law,

The effect of the dielectric is to weaken the original field E₀ by a factor k.
Writing E = \(\frac{E_{0}}{k}\),

Question 74.
The capacitance of a parallel-plate air capacitor is 2 pF. If the air is replaced by a medium of dielectric constant 10. What will be its capacitance?
Answer:
\(\frac{C_{2}}{C_{1}}=\frac{k_{2}}{k_{1}}\)
C₂ = \(\frac{k_{2}}{k_{1}}\) × C₁ = \(\frac{10}{1}\) × 2 = 20 pF
is the required capacitance.

Question 75.
Explain the concept of displacement current.
Answer:
Consider an uncharged parallel-plate capacitor with capacitance C connected to a resistor R and a battery of voltage V, below figure. For a parallel-plate capacitor, of plate area A, separation d and filled with a dielectric of permittivity ¡, its capacitance is C = εA/d.

When the key is closed, charge begins to flow into the capacitor via the resistor. The initial voltage across the capacitor is zero and the current in the circuit now is called the charging current. As the capacitor charges up, the potential difference across its plates slowly increases. During the charging, the charge on the capacitor grows as Q(t) = Q(1 – e ^-t/T).

While a capacitor is being charged, there is no current through the capacitor but the electric field between the plates, and the electric flux through a Gaussian surface enclosing one of the plates, increases as the charge on its plates increases. The instantaneous charge on the capacitor is
Q(t) = (εA)E(f) [ε = kε₀]
∴ \(\frac{d}{d t}\) Q(t) = (εA)\(\frac{d}{d t}\)E(t) = ε\(\frac{d}{d t}\)Φ_e(t)
where Φ_e(t) = AE(t) is the electric flux through the surface. The rate of change of charge, \(\frac{d}{d t}\) Q(t), has the
dimension of current and is the conduction current i_c. The fictitious current arising from the changing electric flux is called the displacement current:
i_d = ε\(\frac{d}{d t}\)Φ_e(t)
That is, we imagine the changing flux through the dielectric is somehow equivalent to the conduction current in the circuit. This lets us generalize Kirchhoff’s junction rule : considering the top plate of the capacitor in the figure, when we include the displacement current, we see that the conduction current coming in from top and an equal displacement current coming out of the bottom. With this generalized meaning of the term “current”, we speak of current going through the capacitor.

Maxwell introduced the concept of displacement current to generalize Ampere’s law in his formulation of electromagnetic theory that led to the discovery of electromagnetic waves.

[Notes : (1) The lowercase i is used to denote instantaneous value of current. (2) The actual time taken for the charge on the capacitor to reach 63% of its maximum possible voltage is known as one time constant (T = RC). For all practical purposes, a capacitor is considered to be fully charged to Q after a time t = 5T. (3) In the pheno-menon of electromagnetic induction (Chapter 12), we see that a changing magnetic field gives rise to an induced electric field. Remarkably, it turns out that a changing electric field gives rise to a magnetic field – an example of the symmetry in nature. These two effects together explain the existence of all types of electromagnetic waves.]

Question 76.
Show that the energy of a charged capacitor is \(\frac{1}{2}\) CV². Also, express this in other forms.
OR
Derive an expression for the energy stored in a charged capacitor. Express it in different forms.
Answer:
To charge a capacitor, an external agent has to do work against the electrostatic forces due to the charges already present on the plates of the capacitor.

Let C be the capacitance of the capacitor. Let Q and V be the final charge and the potential difference respectively when the capacitor is charged. Let q be the charge on the capacitor at some stage during the charging and v, the corresponding potential difference between the plates. The work done by an external agent in bringing additional small charge dq from infinity and depositing it on the capacitor is
dW = potential difference × charge = v dq
But C = \(\frac{q}{v}\) ∴ v = \(\frac{q}{C}\)
∴ dW = \(\frac{q}{C}\) dq
The total work done in charging the capacitor is
W = \(\int d W=\int_{0}^{Q} \frac{q d q}{C}=\frac{1}{C}\left[\frac{q^{2}}{2}\right]_{0}^{Q}=\frac{1}{2} \frac{Q^{2}}{C}\)
Now, Q = CV
W = \(\frac{1}{2}\)CV² = \(\frac{1}{2}\)(CV)V = \(\frac{1}{2}\) QV
This work is stored in the form of potential energy, in the electric field in the medium between the plates of the capacitor.
∴ Energy of a charged capacitor
= \(\frac{1}{2} \frac{Q^{2}}{C}=\frac{1}{2} C V^{2}=\frac{1}{2} Q V\)

Question 77.
How does the energy stored in a charged capacitor change if the plates of the capacitor are moved farther apart
(i) after the battery is disconnected
(ii) the battery remaining connected?
Answer:
(i) If the plates of a charged capacitor are moved farther apart after the battery is disconnected, the energy stored increases by the amount of work done by the external agent in pulling the plates apart against the force of attraction between the opposite charges on the plates.

(ii) With the battery still connected, increasing the separation between the plates decreases the energy stored in the charged capacitor.

[Notes : (1) The charge on the capacitor does not change after the battery is disconnected. Because the electric field of a large plate is independent of the distance from the plate, the electric field \(\vec{E}\) between the plates also remains the same. Therefore, since E = V/d, the p.d. between the plates (V) changes in the same proportion as d. Since the energy stored,
U = \(\frac{1}{2} C V^{2}=\frac{1}{2}\left(\frac{A \varepsilon}{d}\right)(E d)^{2}=\frac{1}{2}(A \varepsilon D) E^{2}\)
U also changes in the same proportion as d. The addi-tional energy is transferred to the system from the work done by the external agent.

(2) With the battery still connected, the p.d. between the plates remains the same. Since the energy stored,
U = \(\frac{1}{2} C V^{2}=\frac{1}{2}\left(\frac{A \varepsilon}{d}\right) V^{2}\), U is inversely proportional to d. With decrease in capacitance, the charge on the plates decreases.]

Question 78.
When a capacitor is charged by a battery, is the energy stored in the capacitor the same as the energy supplied by the battery? Why?
Answer:
Yes. To charge a capacitor, an external agent (the battery) has to do work against the electrostatic force due to the charges already present on the plates of the capacitor. This work done is stored in the form of potential energy in the electric field in the medium between the plates of the capacitor.

Question 79.
If the charge on a capacitor is 2 µC and the potential difference across the conductors forming the capacitor is 100 V, what is the electrostatic energy stored in the capacitor?
Answer:
U = \(\frac{1}{2}\)QV = \(\frac{1}{2}\) × 2 × 10^-6 C × 100V = 10^-4 J
is the required energy.

Question 80.
The capacitance of a charged capacitor is C and the energy stored in it is U. What is the value of the charge on the capacitor ?
Answer:
Let Q be the charge on the capacitor. Then, the energy stored in it is
U = \(\frac{1}{2} \frac{Q^{2}}{C}\)
∴ Q = \(\sqrt{2 U C}\)

Question 81.
Which combination of four identical capacitors has the maximum capacitance? Which combination of these capacitors will store minimum energy when a constant p.d. is applied across it ?
Answer:
The maximum capacitance of four identical capacitors, each of capacitance C, is obtained for their parallel combination : C_p = 4C.

Their series combination has the minimum capacitance. The charge stored in their parallel combination is four times that in their series combination. For the same constant p.d. V, the energy stored in the parallel combination is \(\frac{1}{2}\) (4Q)V and that in the series combination is \(\frac{1}{2}\) QV. Thus, the series combination will store minimum energy.

82. Solve the following

Data : ε₀ = 8.85 × 10^-12 F/m]
Question 1.
A 3 mm gap between the square metal plates of area 0.25 m2 that form a parallel plate capacitor is partly filled, as in figure (b), by a dielectric slab of the same shape and area, of thickness 2 mm and relative permittivity 2.5. Ignoring edge effects, calculate : .
(i) the capacitance when the dielectric slab covers the full area of the plates
(ii) the capacitance when the dielectric slab is partly withdrawn a distance x(x = 0.2 m) in a direction parallel to an edge of the plates.
Solution:
Data : d = 3mm = 3 × 10^-3 m, A = l² = 0.25m², t = 2mm = 2 × 10^-3 m, k = 2.5, x = 0.3m, ε₀ = 8.85 × 10^-12 F/m

(ii) From below figure, the system is equivalent to a parallel combination of two capacitors : C, is air-filled with plate area A₁ = lx and C₂ has dielectric with plate area A₂ = l (l – x).

since l² = 0.25 m², Z = 0.5 m

Question 2.
A parallel-plate air capacitor has an area 2 × 10^-4 m² and the separation between the two plates is 1 mm. Find its capacitance.
Solution:
Data : A = 2 × 10^-4 m², d = 1 mm = 10^-3 m,
ε₀ = 8.85 × 10^-12 C²/N.m², k = 1 (for air)
The capacitance,
C = \(\frac{A k \varepsilon_{0}}{d}=\frac{\left(2 \times 10^{-4}\right)(1)\left(8.85 \times 10^{-12}\right)}{10^{-3}}\)
= 1.77 × 10^-12 F = 1.77 pF

Question 3.
A parallel-plate air capacitor has circular plates, each of diameter 20 cm, separated by a distance of 2 mm. The potential difference between the plates is maintained at 360 volts. Calculate its capacitance and charge. What is the intensity of the electric field between the plates of the capacitor? [k = 1]
Solution:
Data : The diameter of each plate is 20 cm. Hence, its radius is r = 10 cm = 0.1 m, d = 2 mm = 2 × 10^-3 m, V = 360 V, k = 1, ε₀ = 8.85 × 10^-12 F/m
(i) Capacitance :

Question 4.
A parallel-plate air capacitor has rectangular plates, each of area 20 cm² separated by a distance of 2 mm. The potential difference between the plates is 500 volts. Calculate (i) its capacitance (ii) the charge on each plate (iii) the electric field intensity between the two plates.
Solution:
Data : A = 20 cm² = 20 × 10^-4 m² = 2 × 10^-3 m², k = 1, V = 500V, d = 2 mm = 2 × 10^-3 m, ε₀ = 8.85 × 10^-12 F/m
(i) Capacitance :
C = \(\frac{A \varepsilon_{0} k}{d}=\frac{\left(2 \times 10^{-3}\right)\left(8.85 \times 10^{-12}\right)(1)}{2 \times 10^{-3}}\)
= 8.85 × 10^-12 F ( = 8.85 pF)

(ii) Charge :
Q = CV = (8.85 × 10^-12)(500)
= 4.425 × 10^-9 C (= 4.425 nC)

(iii) Intensity of the electric field :
E = \(\frac{V}{d}=\frac{500}{2 \times 10^{-3}}\) = 2.5 × 10⁵ V/m

Question 5.
A parallel-plate air capacitor has rectangular plates each of length 20 cm and breadth 10 cm. The separation between the plates is 1 mm.
(a) Calculate the potential difference between the plates if 1 µC charge is given to the capacitor.
(b) With the same charge of 1 µC, if the separation between the plates is doubled, what is the new potential difference?
(c) Calculate the electric field between the plates.
Solution:
Data : Q = 1 nC = 10^-9 C, l = 20 cm, b = 10 cm, d = 1 mm = 10^-3 m, ε₀ = 8.85 × 10^-12 F/m, k = 1 (air)
Area of the plates, A = lb = 20 × 10 = 200 cm² = 0.02 m²
(a) The capacitance of the capacitor,

V₂ = V₁ × \(\frac{d_{2}}{d_{1}}\) = 5.65 × 2 = 11.3 V (∵\(\frac{d_{2}}{d_{1}}\) = 2, by the data )

(c) The electric field between the plates,
E = \(\frac{V}{d}=\frac{5.65}{10^{-3}}\) = 5650 N/C

Question 6.
A parallel-plate air capacitor has a capacity (capacitance) of 20 µF. What will be its new capacity if
(i) the distance between the plates is doubled
(ii) a marble slab of dielectric constant 8 is introduced filling the entire space between the two plates?
Solution:
Data : C₁ = 20 µF, d₂ = 2d₁, k₁ = 1 (air), k₂ = 8 (marble)
C = \(\frac{A k \varepsilon_{0}}{d}\)
(i) With air as the dielectric,

This gives the new capacity (capacitance) on doubling the plate separation.

(ii) With the plate separation d = d₁,

This gives the new capacity (capacitance) with marble as the dielectric.

Question 7.
A parallel-plate air capacitor of plate separation 2 mm and capacitance 1 µF is charged to 200 V. A dielectric of relative permittivity 50 is now inserted so as to fill the space between the plates.
(i) Find the new value of capacitance.
(ii) Find the polarisation charge on one of the boundaries of the dielectric.
(iii) Find the magnitude of the polarisation of the dielectric.
(iv) What is the magnitude of the electric field inside the dielectric?
Solution:
Data : C₀ = 10^-6 F, d = 2 × 10^-3 m, k = 50, V = 200 V, ε₀ = 8.85 × 10^-12 F/m
(i) The new value of the capacitance is
C = kC₀ = 50 × 10^-6 F (or 50µF)

Question 8.
A parallel-plate capacitor consists of two identical metal plates. Two dielectric slabs having dielectric constants (relative permittivities) k₁ and k₂ are introduced in the space between the plates as shown in figures. Show that the capacity (capacitance) of the capacitor in figure (a) is given by C’ = \(\frac{A \varepsilon_{0}\left(k_{1}+k_{2}\right)}{2 d}\) and that in figure (a) is given by C” = \(\frac{2 A \varepsilon_{0}}{d} \cdot \frac{k_{1} k_{2}}{k_{1}+k_{2}}\)
Solution:
(i) The capacitor in figure(a) is equivalent to a parallel combination of two capacitors of plate separation d and plate area A/2, with C₁ filled with dielectric of relative permittivity k₁ and C₂ filled with dielectric of relative permittivity k₂, as shown in figure (b).

The equivalent capacitance of their parallel combination is
C’ = C₁ + C₂ = \(\frac{A \varepsilon_{0} k_{1}}{2 d}+\frac{A \varepsilon_{0} k_{2}}{2 d}=\frac{A \varepsilon_{0}\left(k_{1}+k_{2}\right)}{2 d}\) ……………. (1)

(ii) Suppose the capacitor in figure (a) is charged by connecting it to a battery. Let the potential of the positive plate a be V_a, the potential of the negative plate c be V_c, and the potential midway between the plates at the interface b of the dielectrics be V_b.

The electric field in the absence of any dielectric is
E₀ = \(\frac{Q}{\varepsilon_{0} A}\)
The electric field in the first dielectric, E₁ = \(\frac{E_{0}}{k_{1}}\)

Equations (1) and (2) give the required expressions.

INote : The capacitor in figure (a) is equivalent to a series combination of two capacitors of plate separation d/2 and plate area A, with C₃ filled with dielectric of relative permittivity k₁ and C₄ filled with dielectric of relative permittivity k₂, as shown in below figure.]

Question 9.
The energy stored in a charged capacitor of capa city 25 µF is 4J. Find the charge on its plate.
Solution:
Data: C = 25 pF = 25 × 10^-12 F, U = 4 J
U = \(\frac{1}{2} \frac{Q^{2}}{C}\)
∴ The charge, Q = \(\sqrt{2 U C}\)
= \(\sqrt{2 \times 4 \times 25 \times 10^{-12}}\)
= 1.414 × 10^-5 C (= 14.14 µC)

Question 10.
The electrostatic energy of 3.5 × 10^-4 J is stored in a capacitor at 700 V. What is the charge on the capacitor?
Solution:
Data: U = 3.5 × 10^-4 J, V = 700V
U = \(\frac{1}{2}\) QV
∴ Q = \(\frac{2 U}{V}=\frac{2 \times 3.5 \times 10^{-4}}{700}\) = 10^-6 C
This is the charge on the capacitor.

Question 11.
Two capacitors, each of capacity 5 µF, and a battery of emf 180 V are given to you. Which combination gives the maximum energy? What is its value? Also find the charge on each capacitor of that combination.
Solution:
Data : C₁ = C₂ = 5 µF = 5 × 10^-6 F, V = 180 V
(i) The effective capacity (C_p) of their parallel combination is
C_p = C₁ + C₁
= 5 × 10^-6 F + 5 × 10^-6 F
= 10 × 10^-6F

Series combination :

(ii) The energy of a charged capacitor is U = \(\frac{1}{2}\) CV²
For a given value of V, the combination of maximum capacity will give maximum energy.
As C_p > C_s, the parallel combination gives maximum energy.

(iii) The value of maximum energy is
U_max = \(\frac{1}{2}\) C_pV² = \(\frac{1}{2}\) × 10 × 10^-6 × (180)²
= 0.162 J

(iv) If Q₁ and Q₂ are the charges on C₁ and C₂,
Q₁ = C₁V= 5 × 10^-6 × 180 = 9 × 10^-4 C
Q₂ = C₂V = 5 × 10^-6 × 180 = 9 × 10^-4C

Question 83.
What is the Van de Graaff generator?
Answer:
The Van de Graaff generator is a high-voltage electrostatic generator that is used to produce very high potential differences of several million volts. It was constructed by US physicist Robert Van de Graaff in 1931. The high potential difference produced by the Van de Graaff generator is used to produce high-energy ion beams in a linear accelerator inside.

Question 84.
State the principle of working of the Van de Graaff generator. Describe its construction with a neat labelled diagram.
Answer:
Principle of working : The Van de Graaff generator works on the principles of corona or point discharge, that the charge on hollow conductor resides entirely on its outer surface and that the charge supplied to an insulated conductor increases its potential.

Construction : A hollow spherical conductor C is supported and insulated from the ground by a tower of ceramic insulators. A long, vertical, endless belt made of special insulating paper or fabric (rubberised silk) is continuously

driven by an electric motor from the ground up to the inside of the conductor. Near the ground, the belt passes close to a spraycomb A which is connected to a high-voltage source. The spraycomb consists of a set of sharp needle points. Another spraycomb collector B is connected inside conductor C at the top. The entire apparatus is usually enclosed in a pressurized vessel containing a gas such as nitrogen or Freon.

Housed inside the assembly is an evacuated tube through which positively charged particles may be accelerated from a source at the same potential as the conductor C to a target at the ground potential.

Question 85.
Describe the working and uses of the Van de Graaff generator with a neat labelled diagram.
Answer:
Working : A high-voltage source, consisting of a high-voltage transformer rectifier circuit, is used to apply a potential difference of several thousand volts (about 50 kV to 100 kV) between spraycomb A and the ground. As a result of the high potential to spraycomb A, a continuous electric discharge takes place between comb A and the nonconducting belt that sprays electric charges onto the belt by corona or point discharge.

The moving belt carries this charge upward and transfers it to the hollow conductor. Collector comb B y inside the hollow conductor removes the charge from the belt by point discharge. Then, the charge flows to the outer surface of the hollow conductor where it accumulates as the process continues. The potential difference between the conductor and the Earth cumulatively increases until the energy density of the electric field builds up to such a high value that the insulating property of the surrounding gas breaks down and a corona discharge takes place through the surrounding gas between the conductor and the ground. If C is the capacitance of – the system and |Q| is the magnitude of the charge transferred from the ground to the conductor, the potential difference V between them is given by V = |Q|/C.

If the hollow conductor is well insulated from the Earth, the accumulated charge |Q| can be large enough and V can build up to several million volts. Since V is limited by the breakdown voltage of the surrounding medium, the entire apparatus is usually enclosed in a pressurized vessel containing a gas such as nitrogen or Freon. This raises the breakdown voltage considerably.

Positively charged particles may be accelerated in the evacuated tube from a source at the same potential as the dome conductor toward a target at the ground potential.

Uses : Machines equipped with positive ion sources and arranged for positive ion acceleration are widely used for research in nuclear structure and nuclear reactions, and for production of radioisotopes. Machines equipped with electron sources are used for X-ray therapy, industrial radiography, food and drug sterilization as well as research.

[Note : in a Van de Graaff single-stage accelerator, an ion source is located inside the high-voltage terminal (the dome conductor). Since the electric field inside a charged conductor is zero, the source must be at the same potential as the terminal. The ions are accelerated to the target (at the ground potential) by repulsion. Early quan-titative data on nuclear properties and processes came from Van de Graaff positive ion accelerators, which had a positive ion (proton, deuteron, alpha particle) source and the high voltage terminal raised to a high positive potential.

Van de Graaff electron accelerators had also been very successful as sources for X-rays for radiotherapy and industrial radiography. For an electrostatic electron accel-erator, the terminal is charged to a high negative potential and a thermionic cathode replaces the ion source. Accel-erated electrons impinging on watercooled gold targets provided for the first time high energy X-rays (~2 MeV) for cancer therapy. A number of such X-ray generators were developed by Prof. J. G. Trump and R.J. Van de Graaff which were installed in hospitals and industries. These instruments were about 3 ft in diameter and 6 ft long, and could be swung into any position to direct the X-ray beam.]

Multiple Choice Questions

Question 1.
Electric Intensity due to a charged sphere at a point outside the sphere dectea.ses with
(A) an increase in the charge on the sphere
(B) an increase in the dielectric constant
(C) a decrease m the distance from the centre of the sphere
(D) a decrease in the square of the distance from the centre of the sphere.
Answer:
(B) an increase in the dielectric constant

Question 2.
A charged spherical conductor in a medium of permittivity e basa surface charge density σ. At an outside point, a distance r from the centre of the conductor, the electric field intensity is

Answer:
(A) \(\frac{\sigma}{\varepsilon}\)

Question 3.
The electric field intensity in free space at a distance r outside a charged conducting sphere of radius R, in terms of its surface charge density e is

Answer:
(A) \(\frac{\sigma}{\varepsilon_{0}}\left(\frac{R}{r}\right)^{2}\)

Question 4.
If the radius of a sphere is doubled without chianging the charge on it then the electric flux originating from the sphere is
(A) double
(B) half
(C) the same
(D) zero.
Answer:
(C) the same

Question 5.
The intensity of electric field at a point clone but outside a charged conducting cylinder is proportional to
[r is the distance of the point from the axis of the cylinder]
(A) \(\frac{1}{r}\)
(B) \(\frac{1}{r^{2}}\)
(C) \(\frac{1}{r^{3}}\)
(D) r.
Answer:
(A) \(\frac{1}{r}\)

Question 6.
In the diagram, \(\vec{E}\) is a uniform electric field. Point B is to the west of point A while point C is to the east and point D is to the south. Which of the following is correct?

(A)V_B > V_A > V_C
(B) V_B < V_A < V_C
(C) V_B = V_C
(D) V_D = V_A
Answer:
(C) V_B = V_C

Question 7.
The figure shows three capacitors C₁, C₂ and C₃. The dashed lines are equipotential surfaces within each capacitor. In which of the capacitors is the pd. between the two equipotentials ∆V = 50V?

(A) Only in C₁
(B) Only in C₁ and C₂
(C) Only in C₃
(D) In all three capacitors
Answer:
(D) In all three capacitors

Question 8.
The potential energy of a dipole iii a uniform electric held \(\vec{E}\) is minimum when the dipole moment is
(A) transverse to \(\vec{E}\)
(B) parallel to \(\vec{E}\)
(C) antiparallel to \(\vec{E}\)
(D) either parallel or antiparallel to \(\vec{E}\).
Answer:
(B) parallel to \(\vec{E}\)

Question 9.
Which of the following molecules is nonpoLar?
(A) N₂O
(B) H₂O
(C) HCl
(D) CO₂
Answer:
(D) CO₂

Question 10.
A paralel-plate capacitor is charged by connecting it to a battery. The battery is then disconnected and the distance between the plates is doubled. This doubles
(A) the electric field at each point
(B) the charge density on each conductor
(C) the potential difference between the conductors
(D) the stored energy.
Answer:
(D) the stored energy.

Question 11.
A dielectric of relative permittivity k completely fills the space between the plates of a parallel.plate capacitor. When the surface charge density on the plates is σ, the polarization of the dielectric is
(A) σ (k – \(\frac{1}{k}\))
(B) \(\frac{\sigma}{k}\)
(C) σ (1 – \(\frac{1}{k}\))
(D) σ (k – 1)
Answer:
(C) σ (1 – \(\frac{1}{k}\))

Question 12.
If at a certain stage during the charging of a capacitor of capacitance C. the charge and potential difference are q and e, the work dW required to transfer an additional amount of charge dq is
(A) vdq
(B) \(\frac{d q}{v}\)
(C) \(\frac{v d q}{C}\)
(D) \(\frac{q^{2}}{2 C}\)
Answer:
(A) vdq

Question 13.
The energy density in the region between the plates of a charged parallel-plate air capacitor is given by the expression
(A) \(\frac{1}{2}\)ε₀E²
(B) \(\frac{1}{2}\)ε₀E
(C) \(\frac{E^{2}}{2 \varepsilon_{0}}\)
(D) \(\frac{\sigma^{2}}{\varepsilon_{0}}\)
Answer:
(A) \(\frac{1}{2}\)ε₀E²

Question 14.
A 5 μF capacitor is charged to a p.d. of 10 V. If it is further charged, so that its p.d. increases to 20 V, the electric energy stored in it increases by
(A) 450 μJ
(B) 500 μJ
(C) 750 μJ
(D) 900 μJ
Answer:
(C) 750 μJ

Question 15.
A 5 pF capacitor is connected in series with a 10 μF capacitor and the combination is connected across a 9 V battery. The potential differences across the capacitors are in the ratio
(A) 4 : 1
(B) 3 : 1
(C) 2 : 1
(D) 1 : 1
Answer:
(C) 2 : 1

Question 16.
A 2 μF capacitor, charged to a p.d. of 200 V, is connected across an uncharged capacitor. If the common p.d. is 20 V, the capacitance of the second capacitor is
(A) 18 μF
(B) 20 μF
(C) 36 μF
(D) 40 μF
Answer:
(A) 18 μF

Question 17.
Three capacitors of capacitances 2 μF, 3 μF and 6 μF are connected in series. The equivalent capacitance of the combination is
(A) 0.5 μF
(B) 1 μF
(C) 1.1 μF
(D) 11 μF
Answer:
(B) 1 μF

Question 18.
Two capacitors each of capacitance 4 μF are connected in series, and a third capacitor of capacitance 4 μF is connected in parallel with the combination. Then, the equivalent capacitance of the arrangement is
(A) 12 μF
(B) 8 μF
(C) 6 μF
(D) 2.65 μF
Answer:
(C) 6 μF

Question 19.
A parallel-plate air capacitor of a plate area A and plate separation d has capacitance C₀. A dielectric of thickness d, width A/2 and relative permittivity k, is inserted between the plates as shown. The capacitance of the capacitor is

(A) 2(k + 1) C₀
(B) (k + 1 )C₀
(C) kC₀
(D) \(\frac{k+1}{2}\) C₀
Answer:
(D) \(\frac{k+1}{2}\) C₀

Question 20.
The resultant capacitance between the points A and B in the figure below is

(A) 1 μF
(B) 1.5 μF
(C) 2 μF
(D) 3 μF.
Answer:
(A) 1 μF

Question 21.
Two parallel plates, separated by a distance d, are kept at potential difference V volt. A charge q of mass m enters between the parallel plates with some velocity. The acceleration of the charged particle will be
(A) \(\frac{q V}{d m}\)
(B) \(\frac{d m}{q V}\)
(C) \(\frac{q m}{d V}\)
(D)\(\frac{d V}{q m}\)
Answer:
(A) \(\frac{q V}{d m}\)

Question 22.
The capacitance of an isolated spherical conductor of radius R in vacuum is
(A) not defined
(B) zero
(C) 4πε₀R
(D) infinite.
Answer:
(C) 4πε₀R

Question 23.
Three parallel plates, each of area A, form a capacitor. The separation between the first and second plates is d1 and that between the second and third is d₂. The gaps are completely filled with dielectrics of dielectric constant k₁ and k₂, respectively. The capacitance of this capacitor is

Answer:
(D) \(\frac{\varepsilon_{0} A k_{1} k_{2}}{k_{2} d_{1}+k_{1} d_{2}}\)

Question 24.
A parallel-plate capacitor, of plate area A and plate separation d, is filled with dielectrics of dielectric constants k₁, k₂ and k₃, as shown

Answer:
(C) \(\frac{\varepsilon_{0} A}{d}\left(\frac{k_{1}}{2}+\frac{k_{2} k_{3}}{k_{2}+k_{3}}\right)\)

Question 25.
With three 6 μF capacitors, which of the capacitance values are available to you ?
(A) 2 μF and 18 μF
(B) 2 μF, 9 μF, 12 μF and 18 μF
(C) 2 μF, 9 μF and 18 μF
(D) 2 μF, 6 μF, 9 μF, 12 μF and 18 μF
Answer:
(D) 2 μF, 6 μF, 9 μF, 12 μF and 18 μF

Question 26.
A capacitor of plate separation 0.02 mm is completely filled with a dielectric material of strength 20 kV/mm. The maximum voltage rating of the capacitor is
(A) 100 V
(B) 200 V
(C) 400 V
(D) 800 V
Answer:
(C) 400 V

Question 27.
Three capacitors C₁, C₂ and C₃ are connected to a battery of p.d. V as shown. Which of the following are the correct relations for the charges on the capacitors and the p.d.s across them ?

(A) Q₁ = Q₁ = Q₃ andV₁ = V₂ = V₃ = V
(B) Q₁ = Q₂ + Q₃ and V = V₁ + V₂ + V₃
(C) Q₁ = Q₂ + Q₃ and V= V₁ + V₂
(D) Q₂ – Q₃ and V₂ = V₃
Answer:
(C) Q₁ = Q₂ + Q₃ and V= V₁ + V₂

Question 28.
A copper plate of thickness b is inserted between the plates of a parallel-plate capacitor of plate separation d. If b = d/3, the capacitances before and after the insertion of the plate are in the ratio

(A) 2 : 3
(B) 3 : 2
(C) 1 : \(\sqrt {3}\)
(D) \(\sqrt {3}\) : 1.
Answer:
(A) 2 : 3

Question 29.
The energy stored in a charged capacitor is U. The capacitor is isolated and connected across the terminals of an identical uncharged capacitor. The energy stored in each capacitor is
(A) U
(B) 3U/4
(C) U/2
(D) U/4
Answer:
(D) U/4

Question 30.
An uncharged parallel-plate capacitor filled with a material of dielectric constant k is connected to a parallel-plate air capacitor of identical geometry charged to a potential V. At equilibrium, common potential difference across them is V’. The dielectric constant k is equal to V’-V

Answer:

Question 31.
Two parallel capacitors of capacitances C and 2C are connected in parallel and charged to a potential V. The battery is then disconnected and the space between the plates of the first capacitor is filled with a material of dielectric constant k. The potential difference across the capacitors is
(A) \(\frac{3 V}{k}\)
(B) \(\frac{3 V}{k+2}\)
(C) \(\frac{3 V}{3k+2}\)
(D) \(\frac{4 V}{2k+3}\)
Answer:

Balbharti Maharashtra State Board 12th Physics Important Questions Chapter 7 Wave Optics Important Questions and Answers.

Maharashtra State Board 12th Physics Important Questions Chapter 7 Wave Optics

Question 1.
Name the various theories of the nature of light.
Answer:
The various theories of the nature of light from the 17th century to modern times are

1. Descartes’ corpuscular theory of light (1637)
2. Newton’s corpuscular theory (1666)
3. Huygens’ wave theory of light propagation (1678), modified, verified, and put on a firm mathematical base by Young, Fraunhofer, Fresnel, and Kirchhoff (in the 1800s)
4. Maxwell’s electromagnetic theory (1865)
5. the light quantum, i.e., the photon model of the modern quantum theory by Planck (1900) and Einstein (1905).

Question 2.
State the postulates of Newton’s corpuscular theory of light.
Answer:
Sir Isaac Newton developed the corpuscular theory of light proposed by Rene’ Descartes (1596-1650), French philosopher and mathematician. The theory assumed that light consists of a stream of corpuscles emitted by a luminous source.
Postulates of Newton’s corpuscular theory of light:

1. Light corpuscles are minute, light and perfectly elastic particles.
2. A luminous source emits light corpuscles in all directions which then travel at high speed in straight lines in a given medium.
3. The constituent colours of white light are due to different sizes of the corpuscles.
4. The light corpuscles stimulate the sense of sight on their impact on the retina of the eye.
5. A reflective surface exerts a force of repulsion normal to the surface on the light corpuscles when they strike the surface.
6. A transparent medium exerts a force of attraction normal to the surface on the light corpuscles striking the surface. This force is different for different mediums.

Notes :

1. A consequence of the assumption (6) is that, according to the corpuscular theory, the speed of light in a denser medium is greater than that in air and has different values for different mediums.
2. It was known from earliest recorded times that when light is incident on the surface of glass or water, it is partly reflected and partly transmitted, simultaneously. To explain this, Newton postulated that the corpuscles must have fits of easy reflection and fits of easy transmission and must pass periodically from one state to the other.

Question 3.
State the drawbacks of Newton’s corpuscular theory of light.
Answer:
Drawbacks of Newton’s corpuscular theory of light:

1. The theory predicted that the speed of light in a
   denser medium should be greater than that in air. This was disproved when experiment showed that the speed of light in water is less than that in air (carried out in 1850 by French physicist Jean Bernard Leon Foucault). ,
2. The theory could not satisfactorily explain the phenomenon of polarization and the simultaneousness of reflection and refraction.
3. The corpuscular theory failed to explain the phenomena of diffraction and interference.
4. There was no basis for the hypothesis that the constituent colours of white light are due to different sized corpuscles.

Question 4.
What is a ray of light?
Answer:
A ray of light is the path along which light energy is transmitted from one point to another in an optical system.

Question 5.
What is meant by ray optics or geometrical optics?
Answer:
The formation of

1. shadows, and
2. images by mirrors and lenses can be explained by assuming that light propagates in a straight line in terms of rays. The study of optical phenomena under this assumption is called ray optics. It is also called geometrical optics as geometry is used in this study.

Question 6.
Give a brief account of Huygens’ wave theory of light. State its merits and demerits.
Answer:
Huygens’ wave theory of light [Christiaan Huygens (1629-95), Dutch physicist] :

1. Light emitted by a source propagates in the form of waves. Huygens’ original theory assumed them to be longitudinal waves.
2. In a homogeneous isotropic medium, light from a point source spreads by spherical waves.
3. It was presumed that a wave motion needed a medium for its propagation. Hence, the theory postulated a medium called luminiferous ether that exists everywhere, in vacuum as well as in transparent bodies. Ether had to be assigned some extraordinary properties, a high modulus of elasticity (to account for the high speed of light), zero density (so that it offers no resistance to planetary motion) and perfect transparency.
4. The different colours of light are due to different wavelengths.

Merits :

1. Huygens’ wave theory satisfactorily explains reflection and refraction as well as their simultaneity.
2. In explaining refraction, the theory concludes that the speed of light in a denser medium is less than that in a rarer medium, in agreement with experimental findings.
3. The theory was later used by Young in 1800-04, Fraunhofer and Fresnel in 1814 to satisfactorily explain interference, diffraction and rectilinear propagation of light. The phenomenon of polarization could also be explained considering the light waves to be transverse.

Demerits :

1. It was found much later that the hypothetical medium, luminiferous ether, has no experimental basis. Einstein discarded the idea of ether completely in 1905.
2. Phenomena like absorption and emission of light, photoelectric effect and Compton effect, cannot be explained on the basis of the wave theory.

[Note: To decide between the particle and wave theories of light, Dominique Francois Jean Arago (1786-1853), French physicist, suggested the measurement of the speed of light in air and water. The experiment was performed in 1850 by Leon Foucault (1819-68), French physicist, using Arago’s experimental equipment. He found that the speed of light in water is less than that in air.]

Question 7.
What is meant by wave optics?
Answer:
It is not possible to explain certain phenomena of light, such as interference, diffraction and polarization, with the help of ray optics (geometrical optics). The branch of optics which uses wave nature of light to explain these optical phenomena is called wave optics.

Question 8.
What was Maxwell’s concept of light?
Answer:
In 1865, James Clerk Maxwell developed a mathematical theory on the intimate relationship between electricity and magnetism. His theory predicted light to be a high-frequency transverse electromagnetic wave in ether. Electric and magnetic fields in the wave vary periodically in space and time at right angles to each other and to the direction of propagation of the wave.

The speed of the electromagnetic waves in a medium, as calculated on the basis of Maxwell’s theory, was experimentally found to be equal to the measured speed of light in that medium. Maxwell’s electromagnetic theory of light, with addition by others till 1896, could account for all the known phenomena regarding the propagation (or transmission) of light through space and through matter.

Question 9.
What is the photon model or quantum hypothesis of light ?
Answer:
To explain the interaction of light and matter (as in the emission or absorption of radiation), Max Planck, in 1900, and Einstein, in 1905, hypothesized light as concentrated or localized packets of energy. Such an energy packet is called a quantum of energy, which was given the name photon much later, in 1926, by Frithiof Wolfers and Gilbert Newton Lewis. For a radiation of frequency v, a quantum of energy is hv, where h is a universal constant, now called Planck’s constant.

[Note : ‘Localisation’ of energy in a region gives light its particle nature while frequency is a wave characteristic. The complementary properties of particle and wave of light quanta are reconciled as follows : light propagates as wave but interacts with matter as particle.]

Question 10.
Give a brief account of the wave nature of light.
Answer:

1. Light is a transverse, electromagnetic wave.
2. A light wave consists of oscillating electric and magnetic fields that are perpendicular to each other and also perpendicular to the direction of propagation of the wave.
3. Like all other electromagnetic waves, light waves do not require any material medium as they can travel even through vacuum.
4. In a material medium, the speed of light depends on the refractive index of the medium, which, in turn depends on the permeability and permittivity of the medium.

Question 11.
Define absolute refractive index of a medium.
Answer:
The absolute refractive index of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in the medium.
[Note : Absolute refractive index of a medium (n) =

The absolute refractive index of vacuum is 1 (by definition) and that of air is greater than 1, but very nearly equal to 1.]

Question 12.
State the characteristics of the electromagnetic waves.
Answer:

1. The electromagnetic waves are transverse in nature as they propagate by oscillating electric and magnetic fields that are perpendicular to each other and also perpendicular to the direction of propagation of the wave.
2. These waves do not require any material medium for their propagation, i.e., they can travel even through vacuum.
3. The wavelength of the electromagnetic waves ranges from very small (< 1 fm) to very large (> 1 km). The waves are classified in the order of increasing wavelength as γ-rays, X rays, ultraviolet, visible, infrared, microwave and radio waves.
4. In vacuum, the speed of electromagnetic waves does not depend on the frequency of the wave. But, in a material medium, it depends on the frequency. For a given frequency, the speed is different in different mediums.
   [Note : 1 fm (femtometre) = 10^-15 m]

Question 13.
Define and explain :
(a) a wave normal
(b) a ray of light.
Answer:
(a) Wavenormal: A wave normal at a point on a wavefront is defined as a line drawn perpendicular to the wavefront in the direction of propagation of the wavefront.

In a homogeneous isotropic medium, a wavefront moves parallel to itself. Thus, at any point in the medium, the direction in which the wavefront moves is always perpendicular to the wavefront at that point. This direction is given by the wave normal at that point.

(b) Ray of light: The direction in which light is propagated is called a ray of light.

This term (ray of light) is also used to mean a narrow beam of light waves. Only in a homogeneous isotropic medium is a ray of light the same as a wave normal. For spherical wavefronts spreading out from a point source, the rays are radially divergent. The rays corresponding to a plane’ wavefront form a parallel beam.

Question 14.
What is a cylindrical wavefront? Draw the corresponding diagram.
Answer:
Cylindrical wavefront: An extended linear source, such as an aperture in the form of a narrow slit, gives rise to cylindrical wavefronts. All the points equidistant from the source lie on the curved surface of a cylinder. Thus, the shape of the wavefront is cylindrical.

Question 15.
What is a plane wavefront? Draw the corresponding diagram.
Answer:
Plane wavefront: It may be treated as a part of a spherical or cylindrical wavefront at a very great distance from the source, such that the wavefront has a negligible curvature.

Question 16.
Draw a neat labelled diagram illustrating spherical wavefronts corresponding to a diverging beam of light.
Answer:

Note : Lenses can be used to obtain

1. a converging beam of light
2. a diverging beam of light.

Question 17.
State Huygens’ principle.
Answer:
Huygens’ principle : Every point on a wavefront acts as a secondary source of light and sends out secondary wavelets in all directions. The secondary wavelets travel with the speed of light in the medium. These wavelets are effective only in the forward direction and not in the backward direction. At any instant, the forward-going envelope or the surface of tangency to these wavelets gives the position of the new wavefront at that instant.

Question 18.
Explain the construction and propagation of a plane wavefront using Huygens’ principle.
Answer:
Huygens’ construction nf a plane wavefront: A plane wavefront may be treated as a part of a spherical or cylindrical wave at a very great distance from a point source or an extended source, such that the wavefront has a negligible curvature. Let A, B, C, D, …, be points on a plane wavefront in a homogeneous isotropic medium in which the speed of light, taken to be monochromatic, is v.

In a time, t = T, secondary wavelets with points A, B, C, D,…, as secondary sources travel a distance vT. To find the position of the wavefront after a time t = T, we draw spheres of radii vT with A, B, C,…, as centres. The envelope or the surface of tangency to these spheres is a plane A’B’C’. This plane, the new wavefront, is at a perpendicular distance vT from the original wavefront in the direction of propagation of the wave. Thus, in an isotropic medium, plane wavefronts are propagated as planes.

Question 19.
Explain the construction and propagation of a spherical wavefront using Huygens’ principle.
Answer:
Huygens’ construction of a spherical wavefront: Consider a point source of monochromatic light S in a homogeneous isotropic medium. The light waves travel with the same speed v in all directions. After time f, the wave will reach all the points which are at a distance vt from S. This is spherical wavefront XY. Let, A, B, C, …,, be points on this wavefront.

To find the new wavefront after time T, we draw spheres of radius vT with A, B, C,…, as centres. The envelope or the surface of tangency of these spheres is the surface A’B’C’. This is the new spherical wavefront X’Y’. Thus, in an isotropic medium, spherical wavefronts are propagated as concentric spheres.

Question 20.
Suppose a parallel beam of monochromatic light is incident normally at a boundary separating two media. Explain what happens to the wavelength and frequency of the light as it propagates from medium 1 to medium 2. What happens when the medium 1 is vacuum ?
Answer:
Consider a parallel beam of monochromatic light incident normally on interface PQ separating a rarer medium (medium 1) and a denser medium (medium 2).

The three successive wavefronts AB, CD and EF are separated by a distance λ₁, the wavelength of light in first medium. The corresponding three wavefronts after refraction, are A’B’, C’D’ and E’F’. Due to the denser medium, the speed of light reduces and hence the wavefronts cover a less distance than that covered in the same time in the first medium. Thus, the wavefronts are comparitively more closely spaced than in the first medium. This distance between successive wavefronts is λ₂, the wavelength of light in the second medium. Thus, λ₂ is less than λ₁,. To find the relation between λ₁ and λ₂, let us consider the wavefront AB reaching

PQ at time t = 0. The next wavefront CD, separated from AB by distance λ₁, will reach PQ at time t = T. Let v₁ and v₂ be the speeds of light in medium 1 and medium 2 respectively. T is the time during which light covers distance λ₁ in medium 1 and λ₂ in medium 2.

This shows how the wavelength of light changes in refraction.

If the first medium is vacuum where the wavelength of light is λ₀ and n is the absolute refractive index of medium 2, then
λ₂ = λ₀\(\left(\frac{v_{2}}{c}\right)\) = \(\frac{\lambda_{0}}{n}\) (as v₁ = c) …(3)
Now, speed = frequency × wavelength.
Hence, the ratio of the frequencies v₁ and v₂, of the wave in the two mediums can be written using
EQ. (2) as, \(\frac{v_{1}}{v_{2}}\) = \(\frac{v_{1} / \lambda_{1}}{v_{2} / \lambda_{2}}\) = 1 …(4)

Thus, the frequency of a wave remains unchanged while going from one medium to another. Thus, v₀ = v₁ = v₂, where v₀ is the frequency of light in vacuum.

Question 21.
The refractive indices of diamond and water with respect to air are 2.4 and 4/3 respectively. What is the refractive index of diamond with respect to water ?
Answer:

is the refractive index of diamond with respect to water.

Question 22.
What is the refractive index of water with respect to diamond ? For data, see Question 21. above.
Answer:

is the refractive index of water with respect to diamond.

Question 23.
What happens to the frequency, wavelength and speed of light as it passes from one medium to another?
Answer:
The wavelength and speed of light change, but the frequency remains the same.

Question 24.
The refractive index of water with respect to air, for light of wavelength Aa in air, is 4/3. What is the wavelength of the light in water?
Answer:

as the wavelength of the light in water.

Question 25.
If the frequency of certain light is 6 × 10¹⁴ Hz, what is its wavelength in free space ? [c = 3 × 10⁸ m/s]
Answer:

is the required wavelength.

Question 26.
If the wavelength of certain light in air is 5000 Å and that in a certain medium is 4000 Å, what is the refractive index of the medium with respect to air ?
Answer:

is the refractive index of the medium with respect to air.

Data : c = 3 × 10⁸ m/s

Question 27.
Solve the following :

Question 1.
If the refractive index of glass is 3/2 and that of water is 4/3 respectively, find the speed of light in glass and in water.
Solution:
Let n_g and n_w be the refractive indices of glass and water respectively. Also let v_a, v_g and v_w be the speeds of light in air, glass and water, respectively.

Question 2.
The refractive indices of water and diamond are 4/3 and 2.42 respectively. Find the speed of light in water and diamond.
Solution :

This is the speed of light in diamond.

Question 3.
The refractive indices of glass and water with respect to air are \(\frac{3}{2}\) and \(\frac{4}{3}\), respectively. Determine the refractive index of glass with respect to water.
Solution :


This is the refractive index of glass with respect to water.

Question 4.
A diamond (refractive index = 2.42) is dipped into a liquid of refractive index 1.4. Find the refractive index of diamond with respect to the liquid.
Solution :
Data : n_d = 2.42, n₁ = 1.4
The refractive index of diamond with respect to liquid,

Question 5.
The refractive index of glycerine is 1.46. What is the speed of light in glycerine ? [Speed of light in vacuum = 3 × 10⁸ m/s]
Solution :
Data : n_g = 1.46, c = 3 × 10⁸ m/s c
n_g = \(\frac{c}{v}\)
v = \(\frac{c}{n}\) = \(\frac{3 \times 10^{8}}{1.46}\) = 2.055 × 10⁸ m/s
This is the speed of light in glycerine.

Question 6.
The refractive indices of glycerine and diamond with respect to air are 1.46 and 2.42 respectively. Calculate the speed of light in glycerine and in diamond. From these calculate the refractive index of diamond with respect to glycerine.
Solution:
Let n_g and n_d be the refractive indices of glycerine and diamond respectively. Also, let v_a, v_g and v_d be the speeds of light in air, glycerine and diamond respectively.
Data : v_a = 3 × 10⁸ m/s, n_g = 1.46, n_d = 2.42
(i) Refractive index of glycerine with respect to air,

(ii) Refractive index of diamond with respect to air,
n_d = \(\frac{v_{\mathrm{a}}}{v_{\mathrm{d}}}\)

(iii) Refractive index of diamond with respect to glycerine,

Question 7.
The wavelengths of a certain light in air and in a medium are 4560 Å and 3648 Å, respectively. Compare the speed of light in air with its speed in the medium.
Solution:
Let v_a and v_m be the speeds of light in air and in the medium respectively and let λ_a and λ_m be the wavelengths of light in air and in the medium respectively. Let v be the frequency of light in air.
When light passes from one medium to another, its frequency remains unchanged.

Question 8.
Monochromatic light of wavelength 632.8 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected and (b) refracted light ? [Given : Refractive index of water = 1.33]
Solution:
Let v be the frequency of the light, and λ₁ and λ₂ the wavelengths of reflected and refracted light respectively.
Data : λ₁ = 632.8 nm = 632.8 × 10^-9 m,
c = 3 × 10⁸ m/s, _an_w = 1.33
(a) For reflected light:
When the wave travels in air, its speed v₁ = c

(b) For refracted light:
Frequency does not change while going from one medium to other.
∴ V = 4.741 × 10¹⁴ Hz

This is the speed of the light in water.

Question 9.
The refractive indices of water for red and violet colours are 1.325 and 1.334, respectively. Find the difference between the speeds of the rays of these two colours in water. [c = 3 × 10⁸ m/s]
Solution :
Data : n_r = 1.325, n_v = 1.334, c = 3 × 10⁸ m/s

Question 10.
If the difference in speeds of light in glass and water is 2.505 × 10⁷ m/s, find the speed of light in air. [Refractive index of glass = 1.5, refractive index of water = 1.333]
Solution :


This is the speed of light in air.

Question 11.
If the difference in speeds of light in glass and water is 0.25 × 10⁸ m/s, find the speed of light in air. [n_g = 1.5 and n_w = \(\frac{4}{3}\)]
Solution :

∴ The speed of light in air, c = 12 (n_w – v_g)
= 12 × 0.25 × 10⁸
= 3 × 10⁸ m/s

Question 12.
Red light of wavelength 6400 Å in air has wavelength 4000 Å in glass. If the wavelength of violet light in air is 4400 Å, what is its wavelength in glass? Assume that the glass has the same refractive index for red and violet colours.
Solution :

[ Note : This problem was asked in Board examination in October 2014. The assumption n_r(glass) = n_v(glass) is manifestly gross. No glass can have the same refractive index for red and violet colours.]

Question 13.
A ray of light passes from air to glass. If the angle of incidence is 74° and the angle of refraction is 37°, find the refractive index of the glass.
Solution :
Data : i = 74°, r = 37°
n = \(\frac{\sin i}{\sin r}\) = \(\frac{\sin 74^{\circ}}{\sin 37^{\circ}}\) = \(\frac{0.9613}{0.6018}\)
∴ n = 1.597
This is the refractive index of the glass.

Question 14.
A ray of light is incident on a glass slab making an angle of 30° with the surface. Calculate the angle of refraction in glass and the speed of light in glass. The refractive index of glass and speed of light in air are 1.5 and 3 × 10⁸ m/s, respectively. Solution :
Data : v_air = 3 × 10⁸ m/s; n = 1.5
The angle of incidence (i) is the angle made by the incident ray with the normal drawn to the refracting surface.
∴ i = 90° – 30° = 60°
(a) Angle of refraction (r) in glass :

(b) Speed of light (v_glass) in glass :

Question 15.
A ray of light is incident on a water surface of refractive index \(\frac{4}{3}\) making an angle of 40° with the surface. Find the angle of refraction.
Solution :

∴ The angle of refraction, r = sin^-1(0.5745) = 35°4′

Question 16.
A ray of light travelling in air is incident on a glass slab making an angle of 30° with the surface. Calculate the angle by which the refracted ray in glass is deviated from its original path and the speed of light in glass [Refractive index of glass = 1.5].
Solution:
Solve for the speed (ug) of light in glass and the angle of refraction (r) in glass as in Solved Problem (14). ug = 2 × 10⁸ m/s and r = 35°16′
The angle of incidence (z), i.e., the angle between the incident ray and the normal to the glass surface, is i = 90° – 30° = 60°.
Hence, the angle by which the refracted ray is deviated from the original path is δ = i – r = 60° – 35°16′ = 24°44′

Question 17.
The wavelength of blue light in air is 4500 A. What is its frequency? If the refractive index of glass for blue light is 1.55, what will be the wavelength of blue light in glass ?
Solution :
Data : λ_a = 4500 Å = 4.5 × 10^-7 m, n_g = 1.55, v_a = 3 × 10⁸ m/s
v_a = v_aλ_a

This is the wavelength of blue light in glass.

Question 18.
White light consists of wavelengths from 400 nm to 700 nm. What will be the wavelength range seen when white light is passed through glass of refractive index 1.55 ?
Solution:
Let λ₁ and λ₂ be the wavelengths of light in water for 400 nm and 700 nm (wavelengths in vacuum) respectively. Let λ_a be the wavelength of light in vacuum.

The wavelength range seen when white light is passed through the glass would be 258.06 nm to 451.61 nm.

Question 19.
Determine the change in wavelength of electro-magnetic radiation as it passes from air to glass, if the refractive index of glass with respect to air for the radiation under consideration is 1.5 and the frequency of the radiation is 3.5 × 10¹⁴ Hz. [Speed of the radiation in air (c) = 3 × 10⁸ m/s]
Solution:
Let v be the frequency of the electromagnetic radiation and c its speed in air. Let λ_a and λ_b be the wavelengths of the radiation in air and glass, respectively.

Question 20.
Determine the change in wavelength of light during its passage from air to glass, if the refractive index of glass with respect to air is 1.5 and the frequency of light is 5 × 10¹⁴ Hz. [Speed of light in air = (c) = 3 × 10⁸ m/s] Solution:
Let v be the frequency of the electromagnetic radiation and c its speed in air. Let λ_a and λ_b be the wavelengths of the radiation in air and glass, respectively.

λ_a = 6000 Å, λ_g = 4000 Å ∴ λ_a – λ_g = 2000 Å

Question 21.
A light beam of wavelength 6400 Å is incident normally on the surface of a glass slab of thickness 5 cm. Its wavelength in glass is 4000 A. The beam of light takes the same time to travel from the source to the surface as it takes to travel through the glass slab. Calculate the distance of the source from the surface.
Solution :
Data : λ_a = 6400 Å, s_g = thickness of the glass slab = 5 cm, λ_g = 4000 Å
The speeds of light in glass and air are, respectively, v_g = λ_gv and v_a = λ_av
where the frequency of the light v remains un-changed with the change of medium.
The time taken to travel through the glass slab,
t_g = \(\frac{\mathrm{Sg}}{v_{\mathrm{g}}}\) = \(\frac{S_{g}}{\lambda_{g} v}\)
The time taken to travel through air,
t_a = \(\frac{S_{a}}{v_{a}}\) = \(\frac{d}{\lambda_{\mathrm{a}} v}\)
where s_a = d is the distance of the glass surface from the source.
Since t_a = t_g
\(\frac{d}{\lambda_{\mathrm{a}} v}\) = \(\frac{s_{\mathrm{g}}}{\lambda_{\mathrm{g}} v}\)
∴ d = \(\frac{\lambda_{\mathrm{a}}}{\lambda_{\mathrm{g}}} \mathrm{s}_{\mathrm{g}}\)S_g = \(\frac{6400}{4000}\) × 5 = 1.6 × 5 = 8 cm

Question 22.
A parallel beam of monochromatic light is incident on a glass slab at an angle of incidence 60°. Find the ratio of the width of the beam in glass to that in air if the refractive index of glass is \(\frac{3}{2}\).
Data: i = 60, _an_g = 1.5
By Snell’s law, _an_g = \(\frac{\sin i}{\sin r}\) ∴ sin r = \(\frac{\sin i}{\mathrm{a} n_{\mathrm{g}}}\)
Solution:

Question 23.
The width of a plane incident wavefront is found to be doubled in a denser medium if it makes an angle of 70° with the surface. Calculate the refractive index for the denser medium.
Solution :

Data : i = 70°, \(\frac{\mathrm{CD}}{\mathrm{AB}}\) = 2
\(\frac{\cos r}{\cos i}\) = \(\frac{\mathrm{CD}}{\mathrm{AB}}\) = 2
∴ cos r = 2 cos 70° = 2 × 0.3420 = 0.6840
∴ r = cos^-10.6840 = 46°51′
The refractive index of the denser medium relative to the rarer medium
= \(\frac{\sin i}{\sin r}\) = \(\frac{\sin 70^{\circ}}{\sin 46^{\circ} 51^{\prime}}\) = \(\frac{0.9397}{0.7296}\) = 1.288

Question 24.
A ray of light travelling through air falls on the surface of a glass slab at an angle i. It is found that the angle between the reflected and the refracted rays is 90°. If the speed of light in glass is 2 × 10⁸ m/s, find the angle of incidence.
Solution :
Data : c = 3 × 10⁸ m/s, v_g = 2 × 10⁸ m/s, angle between the reflected ray and the refracted ray = 90°
n_g = \(\frac{c}{v_{g}}\) = \(\frac{3 \times 10^{8}}{2 \times 10^{8}}\) = 1.5
The angle between the reflected and refracted rays = (90° – i) + (90° – r) = 180° – (i + r) = 90° (by the data)
i + r = 90° ∴ r = 90° – i
∴ sin r = sin(90° – i) = cos i
n_g = \(\frac{\sin i}{\sin r}\) = \(\frac{\sin i}{\cos i}\) = tan i
∴ The angle of incidence,
i = tan^-1 ng = tan^-1 1.5 = 56°19′

Question 28.
What is meant by polarized light ? How does it differ from unpolarized light?
Answer:
According to the electromagnetic theory of light, a light wave consists of electric and magnetic fields vibrating at right angles to each other and to the direction of propagation of the wave. If the vibrations of the electric field \(\vec{E}\) in a light wave are confined to a single plane containing the direction of propagation of the wave so that its electric field is restricted along one particular direction at right angles to the direction of propagation of the wave, the wave is said to be plane-polarized or linearly polarized.

If the vibrations of \(\vec{E}\) in a light wave are in all directions perpendicular to the direction of propagation of the light wave, the light wave is said to be unpolarized. Ordinary light, e.g. that emitted by a bulb, is unpolarized.

According to Biot, unpolarized light may be considered as a superposition of many linearly polarized waves, with random orientations. Also, these component waves are noncoherent, that is, irregular in their phase relationships.

[Note : Ordinary light consists of wave trains, each coming from a separate atom in the source. A beam of ordinary light in a single direction consists of millions of such wave trains from the very large number of atoms in the source radiating in that direction. Hence, the vibrations of \(\vec{E}\) are in all transverse directions with equal probability. Thus, light from an ordinary source is.un-polarized.]

Question 29.
How are polarized light and unpolarized light represented in a ray diagram ?
OR
How will you distinguish between polarized and unpolarized light in a ray diagram?
Answer:
Linearly polarized light is represented in a ray diagram by double-headed arrows or short lines drawn perpendicular to the direction of propagation of light, as shown in below figure.

An unpolarized light beam is represented by both dots and arrows, as shown in above figure. The dots are the ‘end views’ of arrows that are oriented normal to the plane of the diagram.

[Note : The length of an arrow or line represents the amplitude of the electric field (\(\vec{E}\)) in the plane of the diagram and the direction indicates the polarization axis of the beam, i.e., the direction of vibration of \(\vec{E}\). According to Jean Biot (1774-1862), French physicist, unpolarized light may be considered as a superposition of many linearly polarized waves, with random orientations in planes perpendicular to the direction of propagation. In an analytical treatment, the electric vectors may be resolved along two mutually perpendicular directions so that the multiplicity of vectors in can be replaced by the two mutually perpendicular vectors as in, both perpendicular to the direction of

propagation of light. Sir David Brewster (1781-1868), British physicist, conceived the ordinary unpolarized light to consist of two perpendicular, polarized components of equal intensity.]

Question 30.
What is a polarizer?
Answer:
When a beam of unpolarized light is passed through certain types of materials (or devices), these materials (or devices) allow only those light waves to pass through which have their electric field along a particular direction. All the other waves with the electric field in other directions are blocked. A material (or a device) which exhibits this special property is called a polarizer.

Question 31.
Explain the terms :

1. polarizing axis of a polarizer
2. plane of vibration
3. plane of polarization.

Answer:

1. Polarizing axis of a polarizer : When unpolarized light is incident on a polarizer, the particular direction along which the electric field of the emergent wave is oriented is called the polarizing axis of the polarizer.
2. Plane of vibration : The plane of vibration of an electromagnetic wave is the plane of vibration of the electric field vector containing the direction of propagation of the wave. Experiment shows that it is the electric field vector E which produces the optical polarization effects.
3. Plane of polarization : The plane of polarization of an electromagnetic wave is defined as the plane perpendicular to the plane of vibration. It is the plane containing the magnetic field vector and the direction of propagation of the wave.
   

Note :
For vision, photography, action of light on electrons and many other observed effects of light, it is the electric vector that is more important than the magnetic one. Hence, nowadays the plane of polarization is taken to be the plane of vibration. Most authoritative; modern books on Optics do not, therefore, explicitly define the plane of vibration. The above convention is obsolete and redundant.

Question 32.
What is a Polaroid? Explain its construction.
Answer:
A Polaroid is a synthetic dichroic sheet polarizer packed with tiny dichroic crystals oriented parallel to each other such that the transmitted light is plane polarized.

Construction : The first large polarizing sheet filter was made by US inventor Edwin H. Land (1909-91). He used the microscopic needlelike crystals of iodoquinine sulphate (known as herapathite) made into a thick colloidal dispersion in nitrocellulose. This material was squeezed through a long narrow slit which forced the needles to orient parallel to one another. The material was then dried to form a solid plastic sheet.

[Note : The modern version of Polaroid is made from long-chain polymer, polyvinyl alcohol. The transmission axis of a Polaroid is the plane of vibration of a plane polarized light which passes through with minimum absorption.]

If the vibrations of the electric field \(\vec{E}\) in a light wave are confined to a single plane containing the direction of propagation of the wave so that its electric field is restricted along one particular direction at right angles to the direction of propagation of the wave, the wave is said to be plane-polarized or linearly polarized.

This phenomenon of restricting the vibrations of light, i.e., of the electric field vector in a particular direction, which is perpendicular to the direction of the propagation of the wave is called polarization of light.

Consider an unpolarized light wave travelling along the x-direction. Let c, v and λ be the speed, frequency and wavelength, respectively, of the wave. The magnitude of its electric field (\(\vec{E}\)) is,
E = E₀ sin (kx – ωt), where E₀ = E_max = amplitude of the wave, ω = 2πv = angular frequency of the wave and k = \(\frac{2 \pi}{\lambda}\) = magnitude of the wave vector or propagation vector.
The intensity of the wave is proportional to |E₀|². The direction of the electric field can be anywhere in the y-z plane. This wave is passed through two identical polarizers as shown.

When a wave with its electric field inclined at an angle φ to the axis of the first polarizer is passed through the polarizer, the component E₀ cos φ will pass through it. The other component E₀ sin φ which is perpendicular to it will be blocked.

Now, after passing through this polarizer, the intensity of this wave will be proportional to the square of its amplitude, i.e., proportional to |E₀ cos φ|².

The intensity of the plane-polarized wave emerging from the first polarizer can be obtained by averaging E₀ cos φ|² over all values of φ between 0 and 180°. The intensity of the wave will be
proportional to \(\frac{1}{2}\) |E₀|² as the average value of cos² φ over this range is \(\frac{1}{2}\). Thus the intensity of an unpolarized wave reduces by half after passing through a polarizer.

When the plane-polarized wave emerges from the first polarizer, let us assume that its electric field (\(\overrightarrow{E_{1}}\)) is along the y-direction. Thus, this electric field is,
\(\overrightarrow{E_{1}}\) = \(\hat{i}\)E₁₀ sin (kx – ωt) … (1)
where, E₁₀ is the amplitude of this polarized wave. The intensity of the polarized wave,
I₁ ∝|E₁₀|² …(2)

Now this wave passes through the second polarizer whose polarization axis (transmission axis) makes an angle θ with the y-direction. This allows only the component E₁₀ cos θ to pass through it. Thus, the amplitude of the wave which passes through the second polarizer is E₂₀ = E₁₀ cos θ and its intensity,
I₂ ∝|E₂₀|²
∴ I₂ ∝ | E₁₀|² cos² θ
∴ I₂ = I₁ cos² θ … (3)

Thus, when plane-polarized light of intensity I₁ is incident on the second identical polarizer, the intensity of light transmitted by the second polarizer varies as cos² θ, i.e., I₂ = I₁ cos² θ, where θ is the angle between the transmission axes of the two polarizers. This is known as Malus’ law.

[Note : Etienne Louis Malus (1775-1812), French military engineer and physicist, discovered in 1809 that light can be polarized by reflection. He was the first to use the word polarization, but his arguments were based on Newton’s corpuscular theory.]

Question 33.
Unpolarized light is passed through two polarizers. Under what condition is the intensity of the emergent light
(i) maximum
(ii) zero?
Answer:
If the angle θ between the axes of polarization of the two polarizers, is 0°, i.e., the polarization axes of the two polarizers are parallel, the intensity of emergent light is maximum. If θ = 90°, i.e., the polarization axes of the two polarizers are perpendicular, no light emerges from the second polarizer, thus the intensity of emergent light is minimum.
[Note: θ = 0° and 90° are known as parallel and cross settings of the two polarizers.]

Question 34.
With a neat labelled diagram, explain the use of a pair of polarizers to vary the intensity of light. What are crossed polarizers?
Answer:
Consider an unpolarized light beam incident perpendicularly on the first polarizing sheet, called the polarizer, whose transmission axis is vertical, say. The light emerging from this sheet is polarized vertically, and the transmitted electric field is \(\overrightarrow{E_{0}}\).

The polarized light beam then passes through a second polarizing sheet, called the analyser, which is placed parallel to the polarizer with its transmission axis at an angle θ to the transmission axis of the polarizer. The component of \(\overrightarrow{E_{0}}\) which is perpendicular to the axis of the analyser is completely absorbed, and the component parallel to that axis is E₀ cos θ.

Since intensity varies as the square of the amplitude, the transmitted intensity varies as I = I₀ cos² θ

where I₀ is the incident light intensity on the analyser. This expression (known as Malus’s law) shows that the transmitted intensity is maximum when the transmission axes are parallel and zero when the transmission axes are perpendicular to each other.

Crossed polarizers are a pair of polarizers with their transmission axes perpendicular to each other so that the transmitted light intensity is zero.

Question 35.
If the angle made by the axis of polarization of the second polarizer to that of the first polarizer is 60°, what can you say about the intensity of the light transmitted by the second polarizer?
Answer:
If I₁ is the intensity of the light incident on the second polarizer and I₂ is the intensity of the light transmitted by the second polarizer,
I₂ = I₁ cos² θ = I₁ cos² 60°
= I₁\(\left(\frac{1}{2}\right)^{2}\)
= \(\frac{\boldsymbol{I}_{1}}{4}\)

Question 36.
If 75% of incident light is transmitted by the second polarizer, what is the angle made by the transmission axis of the second polarizer to the transmission axis of the first polarizer?
Answer:

Question 37.
Explain the phenomenon of polarization of light by reflection.
Answer:
Consider a ray of unpolarized monochromatic light incident at an angle θ_B on a boundary between two transparent media as shown. Medium 1 is a rarer medium with refractive index n₁ and medium 2 is a denser medium with refractive index n₂. Part of incident light gets refracted and the rest

gets reflected. The degree of polarization of the reflected ray varies with the angle of incidence.

The electric field of the incident wave is in the plane perpendicular to the direction of propagation of incident light. This electric field can be resolved into a component parallel to the plane of the paper, shown by double arrows, and a component perpendicular to the plane of the paper shown by dots, both having equal magnitude. Generally, the reflected and refracted rays are partially polarized, i.e., the two components do not have equal magnitude.

In 1812, Sir David Brewster discovered that for a particular angle of incidence. θ_B, the reflected wave is completely plane-polarized with its electric field perpendicular to the plane of the paper while the refracted wave is partially polarized. This particular angle of incidence (θ_B) is called the Brewster angle.

For this angle of incidence, the refracted and reflected rays are perpendicular to each other. For angle of refraction θ_r,
θ_B + θ_r = 90° …… (1)
From Snell’s law of refraction,
∴ n₁ sin θ_B = n₂ sin θ_r … (2)
From Eqs. (1) and (2), we have,

This is called Brewster’s law.

Question 38.
Define polarizing angle. At the polarizing angle, what is the plane of polarization of the reflected ray ?
Answer:
Polarizing angle : The polarizing angle for an interface is the angle of incidence for a ray of unpolarized light at which the reflected ray is completely polarized.

At the polarizing angle, the reflected ray is completely plane polarized in the plane of incidence.

Question 39.
Give one example in which polarization by reflection is used.
Answer:
Polarization by reflection is used to cut out glare from nonmetallic surfaces. Special sunglasses are used for this purpose. Sunglasses fitted with Polaroids reduce the intensity of partially or completely polarized / reflected light incident on the eyes from reflecting surfaces.

Question 40.
State any four uses of a Polaroid.
Answer:

1. A Polaroid lens filter makes use of polarization by reflection. This filter is used in photography to reduce or eliminate glare from reflective nonmetallic surfaces like glass, rock faces, water and foliage. It can also deepen the colour of the skies.
2. Polaroid sunglasses reduce the transmitted intensity by a factor of one half and also reduce or eliminate glare from nonmetallic surfaces such as asphalt roadways and snow fields.
3. Polaroid filters are used in liquid crystal display (LCD) screens.
4. Polaroids are used to produce and show 3-D movies to give the viewer a perception of depth.

[Note :Three-dimensional movies are filmed from two slightly different camera locations and shown at the same time through two projectors fitted with polarizers having different transmission axes. The audience wear glasses which have two Polaroid filters whose transmission axes are the same as those on the projectors.]

Question 41.
Explain the phenomenon of polarization by scattering.
Answer:
When a beam of sunlight strikes air molecules or dust particles whose size is of the order of wavelength of light, the beam gets scattered. The scattered light observed in a direction perpendicular to the direction of incidence, is plane polarized. This phenomenon is called polarization by scattering.

As shown in above figure, a beam of an un-polarized light is incident along the Z-axis on a molecule. Light waves being transverse in nature, all the possible directions of vibration of electric field vector in the unpolarized light are confined to the XY plane. The light incident on the molecule is scattered by the electromagnetic field of the molecule.

When observed along the X-axis, only the vibrations of electric field vector which are parallel to Y-axis can be seen. In the similar manner, when observed along the Y-axis, only the vibrations of electric field vector which are parallel to the X-axis can be seen. Thus, the light scattered in a direction perpendicular to the incident light is plane polarized.

Question 42.
Solve the following.

Question 1.
The angle between the transmission axes of two polarizers is 45°. What will be the ratio of the intensities of the original light and the transmitted light after passing through the second polarizer?
Solution :
Data : θ = 45°
According to Malus’ law,
I₂ = I₂ cos² θ

= 2
This is the required ratio.

Question 2.
Two polarizers are so oriented that the maximum amount of light is transmitted. To what fraction of its maximum value is the intensity of the transmitted light reduced when the second polarizer is rotated through
(a) 300
(b) 600?
Solution:
Data : θ = 30°, 60°,
I₂ = I₂ cos²θ
where I₁ is the intensity of the incident light and I₂ is that of the transmitted light.
(i) For θ = 30°

I₂ = 0.75 I₁
= 75% of I₁

(ii) For θ = 60°
I₂ = I₁ (cos 60°)²
= I₁\(\left(\frac{1}{2}\right)^{2}\) = \(\frac{1}{4}\)I₁
= 0.25 I₁ = 25% of I₁

Question 3.
For a glass plate as a polarizer with refractive index 1.633, calculate the angle of incidence at which reflected light is completely polarized.
Solution :
Data : n = 1.633
tan θ_B = n = 1.633
∴ The polarizing angle,
θ_B = tan^-1 1.633 = 58°31′

Question 4.
Find the refractive index of glass if the angle of incidence at which the light reflected from the surface of the glass is completely polarized is 58°.
Solution :
Data : θ_B = 58°
n_g = tan θ_B = tan 58° .
= 1.6003
This is the refractive index of the glass.

Question 5.
The critical angle for a glass-air interface is sin^-1\(\frac{5}{8}\). A ray of unpolarized monochromatic light in air is incident on the glass. What is the polarizing angle?
Solution :
Data : θ_c = sin^-1 \(\frac{5}{8}\)
∴ sin θ_c = \(\frac{5}{8}\)
∴ n = \(\frac{1}{\sin \theta_{\mathrm{c}}}\) = \(\frac{8}{5}\) = 1.6
Also, n = tan θ_B
∴ θ_B = tan^-1 n
∴ The polarizing angle, θ_B = tan^-1 1.6 = 58°

Question 6.
The refractive index of a medium is \(\sqrt{3}\). What is the angle of refraction, if the unpolarized light is incident on it at the polarizing angle of the medium?
Solution :

This is the angle of refraction.

Question 7.
For a given medium, the polarizing angle is 60°. What is the critical angle for this medium ?
Solution :

This is the critical angle for the medium.

Question 8.
Unpolarized light is incident on a plane glass surface. What should be the angle of incidence so that the reflected and refracted rays are perpendicular to each other? [Given n = 1.5]
Solution :
Data : Refractive index, n = 1.5
The reflected and refracted rays will be perpendicular to each other, when the angle of incidence = the polarizing angle θ_B,
tan θ_B = n = 1.5
∴ θ_B = tan^-1 (1.5) = 56°19′

Question 9.
If a glass plate of refractive index 1.732 is to be used as a polarizer, what would be the
(i) polarizing angle and
(ii) angle of refraction?
Solution :
Data : n_g = 1.732
n_g = tan θ_B
0B = tan^-1 (1.732) = 60°
This is the polarizing angle,

∴ θ_r = sin^-1\(\left(\frac{0.8660}{1.732}\right)\)
= sin^-1 (0.5) = 30°
This is the angle of refraction.

Question 10.
For a certain unpolarized monochromatic light incident on glass and water, the polarizing angles are 59°32′ and 53°4′, respectively. What would be the polarizing angle for the light if it is incident from water on to the glass?
Solution:

Question 11.
The wavelengths of a certain blue light in air and in water are 4800 Å and 3600 Å, respectively. Find the corresponding Brewster angle.
Solution:

This is the Brewster angle for the given light incident on water surface.

Question 12.
A ray of light is incident on a glass slab at the polarizing angle of 58°. Calculate the change in the wavelength of light in glass.
Solution :


∴ The change in the wavelength of the light in glass, λ_a – λ_b = 0.37514λ_a, i.e., 37.51% of its wavelength in air.

Question 13.
For what angle of incidence will light incident on a bucket filled with liquid having refractive index 1.3 be completely polarized after reflection ?
Solution:
The reflected light will be completely polarized when the angle of incidence is equal to the Brewster’s angle which is given by θ_B = tan^-1 \(\frac{n_{2}}{n_{1}}\), where n₁ and n₂ are refractive indices of the first and the second medium respectively. In this case, n₁ = 1 and n₂ = 1.3.
Thus, the required angle of incidence = Brewster’s angle = tan^-1\(\frac{1.3}{1}\) = 52.26°

Question 43.
State and explain the principle of superposition of waves.
Answer:
Principle of superposition of waves : The dis-placement at a point due to the combined effect of a number of waves arriving simultaneously at the point is the vector sum of the displacements due to the individual waves arriving at the point.

Explanation : This is a general principle of linear systems applied to wave phenomena. When two or more wave trains arrive at a point simultaneously, they interpenetrate without disturbing each other. The resultant displacement at that point is the vector sum of the displacements due to the individual waves arriving at the point. The amplitude and phase angle of the resulting disturbance are functions of the individual amplitudes and phases.

Notes :

1. In the case of mechanical waves, e.g., sound, the displacement is that of a vibrating particle of the medium.
2. When we consider superposition of electromagnetic waves, light in this chapter, the term displacement refers to the electric field component of the electromagnetic wave.

Question 44.
Explain what you understand by interference of light.
Answer:
The phenomenon in which the superposition of two or more light waves produces a resultant disturbance of redistributed light intensity or energy is called the interference of light.

Light waves are transverse in nature. If two monochromatic light waves of the same frequency arrive in phase at a point, the crest of one wave coincides with the crest of the other and the trough of one wave coincides with the trough of the other. Therefore, the resultant amplitude and hence the resultant intensity of light at that point is maximum and the point is bright. This phenomenon is called constructive interference. If two light waves having the same amplitude are in opposite phase, the crest of one wave coincides with the trough of the other. Therefore, the resultant amplitude, and hence the intensity, at that point is minimum (zero) and the point is dark. This phenomenon is called destructive interference. If the amplitudes are unequal, the resultant amplitude is minimum, but not zero. At other points, the intensity of light lies between the maximum and zero.

Question 45.
Explain how the phenomenon of interference can be demonstrated in a ripple tank.
Answer:

1. ‘Two pins, a small distance d apart, are attached to the electrical vibrator or an electrically maintained tuning fork of a ripple tank. The pins are kept vertical with their tips in contact with the surface of water in the ripple tank.
2. When the vibrator is switched on, the two pins vibrate together in phase with the same frequency and the same amplitude. Their tips form the sources S₁ and S₂ of circular waves which spread outward along the water surface.
   
3. The waves from the two sources interfere constructively at points where they meet in phase. Thus, where the crest of a wave from S₁ is superposed on the crest of a wave from S₂ (such as point A), and where the trough of a wave from S₁ is superposed on the trough of a wave from S₂ (such as point B), the water molecules have maximum amplitude of vibration.
4. The waves from the two sources interfere destructively at points where they meet in opposite phase. Thus, where the crest of a wave from S₁ is superposed on the trough of a wave from S₂ (such as point C), and where the trough of a wave from S₁ is superposed on the crest of a wave from S₂ (such as point D), the water molecules have minimum amplitude of vibration.

Question 46.
What are coherent sources? Is it possible to observe interference pattern with light from any two different sources ? Why ?
Answer:
Coherent sources : Two sources of light are said to be coherent if the phase difference between the emitted waves remains constant.

It is not possible to observe interference pattern with light from any two different sources. This is because, no observable interference phenomenon occurs by superposing light from two different sources. This happens due to the fact that different sources emit waves of different frequencies. Even if the two sources emit light of the same frequency, the phase difference between the wave trains from them fluctuates randomly and rapidly, i.e., they are not coherent.

Consequently, the interference pattern will change randomly and rapidly, and steady interference pattern would not be observed.

Question 47.
State the conditions for constructive and destructive interference of light.
Answer:
(1) Constructive interference (brightness) : There is constructive interference at a point and the brightness or intensity is maximum there, if the two waves of light of the same frequency arrive at the point in phase, i.e., with a phase difference of zero or an integral multiple of 2π radians.
A phase difference of 2π radians corresponds to a path difference λ, where λ is the wavelength of light. Since

for constructive interference with maximum intensity of light, phase difference = 0, 2π, 4π, 6π, … rad
= n(2π) rad
or path difference = 0, λ, 2λ, 3λ, …, etc.
= nλ
where n = 0, 1, 2, 3, …, etc.

(2) Destructive interference (darkness) : There is destructive interference at a point and the point is the darkest, i.e., the intensity of light is minimum, if the two waves of light of the same frequency and intensity arrive at the point in opposite phase, i.e., with a phase difference of an odd-integral multiple of π radians. A phase difference 2π radians corresponds to a path difference λ, where λ is the wavelength of light.
∴ For destructive interference with minimum intensity of light, phase difference = π, 3π, 5π, … rad
= (2m – 1)π rad
or path difference = λ/2, 3λ/2, 5λ/2, …, etc.
= (2m – 1)\(\frac{\lambda}{2}\)
where m = 1, 2, 3, …, etc.

Question 48.
In Young’s double-slit experiment using light of wavelength 5000Å, what phase difference corresponds to the 11th dark fringe from the centre of the interference pattern?
Answer:
The required phase difference is (2m – 1)π rad = (2 × 11 – 1)π rad = 21π rad.

Question 49.
In Young’s double-slit experiment, if the path difference at a certain point on the screen is 2.999 λ, what can you say about the intensity of light at that point ?
Answer:
The intensity of light at that point will be close to the maximum intensity and the point will be nearly bright as the path difference = 2.999 λ \(\approx\) 3λ (integral multiple of λ).

Question 50.
In Young’s double-slit experiment, if the path difference at a certain point on the screen is 7.4999 λ, what can you say about the nature of the illumination at that point ?
Answer:
The point will be nearly dark as the path difference = 7.4999 λ \(\approx\) (8 – 0.5) λ, which is of the form
(m – \(\frac{1}{2}\))λ, where m = 1, 2, … …. 8

Question 51.
How is Young’s interference experiment performed using a single source of light?
Answer:
When a narrow slit is placed in front of an intense source of monochromatic light, cylindrical wave-fronts propagate from the slit. In Young’s experiment, two coherent sources are then obtained by wavefront splitting by placing a second screen with two narrow slits at a small distance from the first slit.

Question 52.
State any two points of importance of Young’s experiment to observe the interference of light.
Answer:
Importance of Young’s experiment observe the interference of light:

1. It was the first experiment (1800-04) in which the interference of light was observed.
2. This experiment showed that light is propagated in the form of waves.
3. From this experiment, the wavelength of monochromatic light can be determined.

Question 53.
What is the nature of the interference pattern obtained using white light?
Answer:
With white light, one gets a white central fringe at the point of zero path difference along with a few coloured fringes on both the sides, the colours soon fade off to white.

The central fringe is white because waves of all wavelengths constructively interfere here. For a path difference of \(\frac{1}{2}\)λ_violet, complete destructive interference occurs only for the violet colour; for waves of other wavelengths, there is only partial destructive interference. Consequently, we have a line devoid of violet colour and thus reddish in appearance. A point for which the path difference = \(\frac{1}{2}\)λ_red is similarly devoid of red colour, and appears violettish. Thus, following the white central fringe we have coloured fringes, from reddish to violettish. Beyond this, the fringes disappear because there are so many wavelengths in the visible region which constructively interfere that we observe practically uniform white illumination.

Question 54.
In Young’s double-slit experiment, the second minimum in the interference pattern is exactly in front of one slit. The distance between the two slits is d and that between the source and screen is D. What is the wavelength of the light used ?
Answer:
The distance of the mth minimum from the central fringe is,

This is the wavelength of the light used.

Question 55.
In Young’s double-slit experiment, a glass slide of refractive index n_g and thickness b is placed in front of one of the slits. What happens to the interference pattern and fringe width ? Derive an expression for the positions of the bright fringes in the interference pattern.
Answer:
Suppose, in Young’s double-slit experiment, a glass slide of refractive index n_g and thickness b is placed in front of the slit S₁.

If the experiment is set up in free space, this introduces an additional optical path length (n_g – 1)b in the path S₁P.
Then, the optical path difference to the point P from S₁ and S₂ is,
∆l = S₂P – [S₁P + (n_g – 1)b] = (S₂P – S₁P) – (n_g – 1)b = y\(\frac{d}{D}\) – (n_g – 1)b … (1)
where y = PO’, d is the distance between S₁ and S₂, and D is the distance of the screen from S₁ and S₂. Thus, point P will be bright (maximum intensity) if ∆l = nλ, where n = 0, 1, 2, … . The central bright fringe, corresponding to n = 0, will be obtained at a distance y₀ from O’ such that,
∆l = y_o\(\frac{d}{D}\) – (n_g – 1)b = 0
∴ y_o\(\frac{d}{D}\) = (n_g – 1)b ∴ y₀ = \(\frac{D}{d}\)(ng-l) b …(2)
Therefore, the central bright fringe and the interference pattern will shift up (towards P) by a distance y₀ given by EQ. (2).
The distance of the nth bright fringe from O’ towards P is,

The distance of the (n + l)th bright fringe from O’ towards P is

Therefore, the fringe width,
w = y_n+1 – y_n = \(\frac{\lambda D}{d}\) ….(5)
Thus, the fringe width remains unchanged.

Question 56.
What happens to the interference pattern when the phase difference between the two sources of light changes with time ?
Answer:
If the two sources do not maintain their phase relation during the time required for observation, the intensity of light at any point on the screen and consequently the interference pattern changes rapidly, and hence steady interference pattern is not observed.

Question 57.
Obtain expressions in terms of electric field, for the resultant amplitude and the intensity for the interference pattern produced by monochromatic light waves from two coherent sources.
Answer:
Consider a two-source interference pattern produced by superposition of monochromatic light waves of angular frequency ω, wavelength λ and constant phase difference, φ. At the centre of a bright fringe, there is constructive interference. Here, the amplitude of the resultant wave is double the amplitude of the wave incident on AB. Now, the intensity is proportional to the square of the amplitude of the wave. Hence, the resultant intensity is, I = 4I₀, where I₀ is the intensity of the incident wave. At the centre of a dark fringe, there is destructive interference. Here, the amplitude of the resultant wave and hence the intensity is zero. At other points the intensity is between 4I₀ and zero, depending on the phase difference.

Let the equations of the two waves coming from S₁ and S₂ at some point in the interference pattern be, E₁ = E₂ sin ωt and E₂ = E₀ sin (ωt + φ) respectively, where E₀ is the amplitude of the electric field vector.

By the principle of superposition of waves, the resultant electric field at that point is the algebraic sum, E = E₁ + E₂
∴ E = E₀ sin ωt + E₀ sin (ωt + φ)
= E₀ [sin ωt + sin (ωt + φ)]
= 2E₀ sin (ωt + φ/ 2) cos (φ/2)
The amplitude of the resultant wave is 2 E₀ cos (φ/2).
Therefore, the intensity at that point is I ∝ |2 E₀ cos (φ/2) |²
∴ I = 4I₀ cos² (φ/2)
as I₀ ∝ |E₀|².

Question 58.
Monochromatic light waves of amplitudes E₁₀ and E₂₀ and a constant phase difference φ produce an interference pattern. State an expression for the resultant amplitude at a point in the pattern. Hence, deduce the conditions for
(i) constructive interference with maximum intensity
(ii) destructive interference with minimum intensity. Also show that the ratio of the maximum and minimum intensities is
\(\frac{I_{\max }}{I_{\min }}\) = \(\left(\frac{\boldsymbol{E}_{10}+\boldsymbol{E}_{20}}{\boldsymbol{E}_{10}-\boldsymbol{E}_{20}}\right)^{2}\)
Answer:
Consider a two-source interference pattern produced by monochromatic light waves of angular frequency ω, wavelength λ, amplitudes E₁₀ and E₂₀ and a constant phase difference φ.

Let the individual electric fields along the same line due to the waves at some point in the interference pattern be
E₁ = E₁₀ sin ωt and E₂ = E₂₀ sin (ωt + φ)
By the principle of superposition of waves, the resultant electric field (displacement) at that point is the algebraic sum

Since, the intensity of a wave is proportional to the square of its amplitude, the resultant intensity at P,
I ∝ |R|²

Thus, the intensity depends on cos φ.
The condition for constructive interference with maximum intensity is cos φ is maximum, equal to 1, i.e., φ = 2nπ (n = 0, 1, 2, 3…) … (3)
The condition for destructive interference with minimum intensity is cos φ is minimum equal to

Question 59.
If the amplitude ratio (E₂₀/E₁₀) of two interfering coherent waves producing an interference pattern is 0.5, what is the ratio of the minimum intensity to maximum intensity?
Answer:

Question 60.
If the amplitude ratio (E₂₀/E₁₀) of two interfering coherent waves producing an interference pattern is 0.8, what is the ratio of the maximum intensity to the minimum intensity?
Answer:

Question 61.
Two slits in Young’s experiment have widths in the ratio 2 : 3. What is the ratio of the intensities of light waves coming from them?
Answer:
The required ratio is \(\frac{I_{1}}{I_{2}}\) = \(\frac{w_{1}}{w_{2}}\) = \(\frac{2}{3}\)

Question 62.
Monochromatic light waves of intensities I₁ and I₂, and a constant phase difference φ produce an interference pattern. State an expression for the resultant intensity at a point in the pattern. Hence deduce the expressions for the resultant intensity, maximum intensity and minimum intensity if I₁ = I₂ = I₀.
Answer:
Consider a two-source interference pattern produced by monochromatic light waves of intensities I₁ and I₂, and a constant phase difference φ. The resultant intensity at a point in the pattern is

At a point of constructive interference with maximum intensity, cos φ = 1.
∴ I_max = 2I₀(1 + 1) = 4I₀ …(3)
At point of destructive interference, with minimum intensity, cos φ = – 1.
∴ I_min = 2I₀(1 – 1) = 0… (4)
Notes :

(1) Since 1 + cos φ = 2 cos²φ, EQ. (2) above can be expressed as I = 4I₀ cos²\(\frac{\phi}{2}\) = I_max cos²\(\frac{\phi}{2}\).

(2) The average of cos² \(\frac{\phi}{2}\), averaged over one cycle, is \(\frac{1}{2}\).
Therefore, the average intensity of a bright and dark fringe in an interference pattern is I_av = 4I₀ × \(\frac{1}{2}\) = 2I₀. The graph of l versus φ is shown, in which the dotted line shows I_av.

In the absence of interference phenomenon, the intensity at a point due to two waves each of intensity I₀ is 2I₀, which is the same as the average intensity in an interference pattern. Thus, interference of light is consistent with the law of conservation of energy. Interference produces a redistribution of energy out of regions where it is destructive into the regions where it is constructive.

Question 63.
At a point in an interference pattern, the two interfering coherent waves of equal intensity I₀ have phase difference 60°. What will be the resultant intensity at that point ?
Answer:
Resultant intensity, I₀ = 2I₀ (1 + cos φ)
= 2I₀(1 + cos 60°) = 2I₀(1 + \(\frac{1}{2}\)) = 3I₀.

Question 64.
Solve the following :

Question 1.
Find the ratio of intensities at two points X and Y on a screen in Young’s double-slit experiment where waves from the slits S₁, and S₂ have path difference of 0 and \(\frac{\lambda}{4}\) respectively.
Solution :

Question 2.
Two coherent sources, whose intensity ratio is 81 : 1, produce interference fringes. Calculate the ratio of the intensities of maxima and minima in the fringe system.
Solution :
Data : I₁ : I₂ = 81 : 1
If E₁₀ and E₂₀ are the amplitudes of the interfering waves, the ratio of the maximum intensity to the minimum intensity in the fringe system is

Question 3.
In Young’s double-slit experiment, the ratio of the intensities at the maxima and minima in the interference pattern is 36 : 16. What is the ratio of the widths of the two slits?
Answer:

Question 4.
In Young’s double-slit experiment, the ratio of the intensities of the maxima and minima in an interference pattern is 36 : 9. What is the ratio of the intensities of the two interfering waves?
Answer:

∴ The ratio of the intensities of the two interfering waves is 9 : 1.

Question 5.
Two slits in Young’s double-slit experiment have widths in the ratio 81 : 1. What is the ratio of the amplitudes of light waves coming from them?
Solution:
Data : w₁ : w₂ = 81 : 1
Since the intensity of a wave is directly proportional to the square of its amplitude,
\(\frac{I_{1}}{I_{2}}\) = \(\left(\frac{E_{10}}{E_{20}}\right)\) …. (1)
Also, the intensity of a wave coming out of a slit is directly propotional to the slit width.
∴ \(\frac{I_{1}}{I_{2}}\) = \(\frac{w_{1}}{w_{2}}\) …. (2)
From Eqs. (1) and (2),

The ratio of the amplitudes of light waves from the slits is 9: 1.

Question 6.
Two monochromatic light waves of equal intensities produce an interference pattern. At a point in the pattern, the phase difference between the interfering waves is π/3 rad. Express the intensity at this point as a fraction of the maximum intensity in the pattern.
Solution :
Data : I₁ = I₂ = I₀, φ = π/3 rad
The resultant intensity at the point in the pattern is

Question 7.
In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where the path difference is λ is I. What is the intensity of light at a point where the path difference is λ/3 ?
Solution :
Data : ∆l₁ = λ, I₁ = I, ∆l₂ = λ/3
We assume that light waves coming out of the two slits are of equal intensity I₀.
Then, at a point in the interference pattern where the phase difference between the interfering waves is φ, the resultant intensity is,
I = 2I₀ (1 + cos φ)
Phase difference (φ) = \(\frac{2 \pi}{\lambda}\) × path difference (∆l)


∴ A The intensity of light at a point where the path difference is λ/3 is λ/4.

Question 8.
The optical path difference between identical waves from two coherent sources and arriving at a point is 87λ. What can you say about the resultant intensity at the point? If the path difference is 49.19 μm, calculate the wavelength of light.
Solution :
Data : Path difference = 87λ = 49.19 /rm.
(i) Path difference = 87λ = nλ, where n = 87. As the path difference is an integral multiple of the wavelength λ, the point where the waves interfere is a bright point with maximum intensity.

(ii) 87 λ = 49.19 μn = 49.19 × 10^-6 m .
∴ λ = \(\frac{49.19 \times 10^{-6}}{87}\) = 5.654 × 10^-7 m = 5654 Å

Question 9.
The optical path difference between identical waves from two coherent sources and arriving at a point is 172. What can you say about the resultant intensity at the point? If the path difference is 9.18 μm, calculate the wavelength of light.
Solution :
Data : Path difference = 17λ = 9.18 μm
(i) Path difference = 17λ = nλ, where n = 17
As the path difference is an integral multiple of the wavelength λ, the point where the waves interfere is a bright point with maximum intensity.

(ii) 17λ = 9.18 μm = 9.18 × 10^-6 m

Question 10.
At a point on the two-slit interference pattern obtained using a source of green light of wavelength 5500 Å, the path difference is 4.125 pm. Is the point at the centre of a bright or dark fringe ? Hence, find the order of the fringe.
Solution :
Data : Path difference, ∆l = 4.125 × 10^-6 λ = 5500 Å = 5.5 × 10^-7 m

As the path difference is an odd integral multiple of \(\frac{\lambda}{2}\) the point is at the centre of a dark fringe.
∴ p = 2m – 1 (m = 1, 2, 3…)
∴ 2m – 1 = 15 ∴ m = 8
∴ The order of the fringe is 8 (i.e., the point lies at the centre of the 8th dark fringe.)

Question 11.
The two slits in an interference experiment are illuminated by light of wavelength 5600 A. Deter-mine the path difference and phase difference between the waves arriving at the centre of the eighth dark fringe on the screen.
Solution :
Data : λ = 5600 A = 5.6 × 10^-7 m
Order of the dark fringe is 8, ∴ m = 8
For the fringe to be dark, path difference
= (2m – 1)\(\frac{\lambda}{2}\)
∴ Path difference, ∆l = (2 × 8 – 1)\(\frac{\lambda}{2}\) = 7.5λ
= 7.5 × 5.6 × 10^-7 m
= 4.2 × 10^-6 m
Phase difference, φ = \(\frac{2 \pi}{\lambda}\) × ∆l
= \(\frac{2 \pi}{5.6 \times 10^{-7}}\) × 4.2 × 10^-6 = 15π

Question 12.
In Young’s double-slit experiment, interference fringes are observed on a screen 1 m away from the two slits which are 2 mm apart. A point P on the screen is 1.8 mm from the central bright fringe,
(i) Find the path difference at P.
(ii) If the wavelength of the light used in 4800 Å, what can you say about the illumination at P?
Solution :
Data : D = 1 m, d = 2 mm = 2 × 10^-3 m, y = 1.8 mm = 1.8 × 10^-3 m, λ = 4800 Å = 4.8 × 10^-7 m

∴ Path difference = 15\(\frac{\lambda}{2}\)
As the path difference is an odd integral multiple of \(\frac{\lambda}{2}\), point P is a dark point with minimum intensity.

Question 13.
Two slits 1.25 mm apart are illuminated by light of wavelength 4500 Å. The screen is 1 m away from the plane of the slits. Find the separation between the 2nd bright fringe on one side and the 2nd bright fringe on the other side of the central maximum.
Solution :
Data : D = 1 m, d = 1.25 mm = 1.25 × 10^-3 m,
λ = 4500 Å = 4500 × 10^-10 m = 4.5 × 10^-7 m
Since, it is a second bright fringe, n = 2. If s is the distance between the 2nd bright fringe on one side and the 2nd bright fringe on the other side of the central maximum, then
s = 2y = 2nλ\(\frac{D}{d}\)

= 14.4 × 10^-4 m
W = 1.44 mm

Question 14.
A plane wavefront of light of wavelength 5000 Å is incident on two slits in a screen perpendicular to the direction of light rays. If the total separation of 10 bright fringes on a screen 2 m away is 4 mm, find the distance between the slits.
Solution:
Given : λ = 5000 Å = 5 × 10^-7 m,
D = 2 m and the total separation of 10 fringes = 4 mm = 4 × 10^-3 m

Question 15.
In Young’s double-slit experiment with slits of equal width, a point P on the screen is at a distance equal to one-fourth of the fringe width from the central maximum. If the intensity at the central maximum is I_c, find the intensity at P.
Solution :

Since, the slits have equal width, the intensities of the two interfering waves are equal, say I₀. Then the intensity at a point on the screen is
I = 4I₀ cos² \(\frac{\phi}{2}\)
At the central maximum, φ = 0.

The intensity at P is half that at the central maximum.

Question 16.
In a double-slit experiment, the optical path difference between the waves coming from two coherent sources at a point P on one side of the central bright band is 7.5 × 10^-6 m and that at a point Q on the other side of the central bright band is 1.8 × 10^-6 m. How many bright and dark bands are observed between points P and Q if the wavelength of light used is 6 × 10^-7 m ?
Solution :
Data : ∆l₁ = 7.5 × 10^-6 m, ∆l₂ = 1.8 × 10^-6 m λ = 6 × 10^-7 m
For point P : Let p\(\frac{\lambda}{2}\) = ∆l₁

The path difference ∆l₁ is an odd integral multiple of λ/2 : ∆l₁ = (2m – 1)\(\frac{\lambda}{2}\), where m is an integer,
∴ 2m – 1 = 25 ∴ m = 13
∴ Point P is at the centre of the 13th dark band.
For point Q :
Let q\(\frac{\lambda}{2}\) = ∆l₂

The path difference ∆l₂ is an even integral multiple of \(\frac{\lambda}{2}\) : ∆l₂ = (2n)\(\frac{\lambda}{2}\), where n is an integer
∴ 2n = 6 ∴ n = 3
∴ Point Q is at the centre of the 3rd bright band. Between points P and Q, excluding the respective bands at P and Q, the number of dark bands = 12 + 3 = 15 and the number of bright bands (including the central bright band) = 12 + 2 + 1 = 15

Question 17.
In Young’s double-slit experiment, light waves of wavelength 5.2 × 10^-7 m and 6.5 × 10^-7 m are used in turn keeping the same geometry. Compare the fringe widths in the two cases.
Solution :
Data : λ₁ = 5.2 × 10^-7 m, λ₂ = 6.5 × 10^-7 m
As, W = \(\frac{\lambda D}{d}\) and the geometry is the same, i.e., D and d remain the same,

Question 18.
Two coherent sources are 1.8 mm apart and the fringes are observed on a screen 80 cm away from them. It is found that with a certain source of light, the fourth bright fringe is situated at a distance of 1.08 mm from the central fringe. Calculate the wavelength of light.
Solution :
Data : d = 1.8 mm = 1.8 × 10^-3 m
D = 80 cm = 0.8 m
For fourth bright fringe, n = 4

This is the wavelength of light.

Question 19.
Green light of wavelength 5100 Å from a narrow slit is incident on a double-slit. If the overall separation of 10 fringes on a screen 200 cm away from it is 2 cm, find the slit-separation.
Solution :
Data : λ = 5100 Å = 5.1 × 10^-7 m,
W = \(\frac{2}{10}\) cm = 2 × 10^-3 m,
D = 200 cm = 2 m
W = \(\frac{\lambda D}{d}\)
∴ d = \(\frac{\lambda D}{W}\) = \(\frac{5.1 \times 10^{-7} \times 2}{2 \times 10^{-3}}\)
= 5.1 × 10^-4 m
This is the slit-separation.

Question 20.
Sodium light of wavelength 5.896 × 10^-7 m is passed through two pinholes 0.5 mm apart, and an interference pattern is formed on a screen kept parallel to the plane of the pinholes and 1.2 m from them. Find the distance between
(i) the second and the fifth bright fringes
(ii) the third and the seventh dark fringes on the same side of the central bright point.
Solution :
Data : λ = 5.896 × 10^-7 m, d = 0.5 mm= 0.5 × 10^-3 m, D = 1.2 m
(i) The distance of the nth bright fringe from the central bright point is

The distance between the second and the fifth bright fringes on the same side of the central bright point is 4.245 × 10^-3 m.

(ii) The distance of the mth dark fringe from the central bright point is

= 5.66 × 10^-3 m
The distance between the third and the seventh dark fringes on the same side of the central bright point is 5.66 × 10^-3 m.

Question 21.
In Young’s double-slit experiment, the two slits are 2 mm apart. The interference fringes for light of wavelength 6000Å are formed on a screen 80 cm away from them.
(i) How far is the second bright fringe from the central bright point?
(ii) How far is the second dark fringe from the central bright point?
Solution:
Data: D = 80 cm = 0.8 m,
d = 2 mm = 2 × 10^-3 m, λ = 6000Å = 6 × 10^-7 m
(i) For the second bright fringe from the central bright point, n = 2

This is the required distance.

(ii) For the second dark fringe from the central bright point, m = 2

This is the required distance.

Question 21.
In Young’s double-slit experiment, the distance between two consecutive bright fringes on a screen placed at 1.5 m from the two slits is 0.6 mm. What would be the fringe width, if the screen is brought towards the slits by 50 cm, keeping rest of the setting the same?
Solution:
Let λ be the wavelengths of light used and d the distance between the two sources (i.e., slits), if D is the distance between the sources and the screen, the fringe width is
w = \(\frac{\lambda D}{d}\)
For the same λ and d, W ∝ D.
∴ \(\frac{W_{2}}{W_{1}}\) = \(\frac{D_{2}}{D_{1}}\)
Data:
W₁ = 0.6 mm = 6 × 10^-4 m,
D₁ = 1.5 m, D₂ = 1 m

This is the required fringe width.

Question 22.
On passing light of wavelength 5000 Å through two pinholes 2 mm apart, an interference pattern is formed on a screen kept parallel to the plane of the pinholes and 100 cm from them. Find the distance between the fifth bright band on one side of the central bright band and the sixth dark band on the other side.
Solution :
Data : λ = 5000 Å = 5 × 10^-7 m, d = 2 mm = 2 × 10^-3 m, D = 100 cm = 1 m

∴ y_6d = (6 – \(\frac{1}{2}\)) × 2.5 × 10^-4m = 1.375 × 10^-3 m
∴ y_5b + y_6d = 1.25 × 10^-3m + 1.375 × 10^-3m
= 2.625 × 10^-3 m = 2.625 mm
This is the required distance.

Question 23.
Monochromatic light from a narrow slit illuminates two narrow slits 3 mm apart, producing an interference pattern with bright fringes 0.15 mm apart on a screen 75 cm away from the slits. Find the wavelength of the light. How will the fringe width be altered if
(a) the distance of the screen from the slits is doubled
(b) the separation between the slits is doubled ?
Solution :
Data : d = 3 mm = 3 × 10^-3 m,
W = 0.15 mm = 1.5 × 10^-4 m, D = 75 cm = 0.75 m

Question 24.
In Young’s double-slit experiment the slits are 2 mm apart and interference is observed on a screen placed at a distance of 100 cm from the slits. It is found that the ninth bright fringe is at a distance of 2.208 mm from the second dark fringe from the centre of the fringe pattern. Find the wavelength of light used.
Solution :
Data : d = 2 mm = 2 × 10^-3 m,

Question 25.
In Young’s double-slit experiment, the two slits separated by 4 mm are illuminated by light of wavelength 6400 Å. Interference fringes are obtained on a screen placed at a distance of 60 cm from the slits. Find the change in the fringe width if the separation between the slits is
(i) increased by 1 mm
(ii) decreased by 1 mm.
Solution :
Data : d = 4 mm = 4 × 10^-3 m, λ = 6.4 × 10^-7m, D = 0.6 m, d’ = 5 mm = 5 × 10^-3 m, d” = 3 mm = 3 × 10^-3m
Fringe width, W = \(\frac{\lambda D}{d}\) ∴ W ∝ \(\frac{1}{d}\)
(i) Since d’ > d, W’ < W, i.e., the fringe width decreases.
Decrease in the fringe width = W – W

(ii) Since d” < d, W” > W, i.e., the fringe width increases.
Increase in the fringe width = W” – W

Question 26.
In Young’s double-slit experiment, the separation between the slits is 3 mm and the distance between the slits and the screen is 1 m. If the wavelength of light used is 6000 Å, calculate the fringe width. What will be the change in the fringe width if the entire apparatus is immersed in a liquid of refractive index \(\frac{4}{3}\)?
Solution :
Data : d = 3 × 10^-3 m, D = 1 m, λ = 6 × 10^-7 m, n = \(\frac{4}{3}\)

Question 27.
In Young’s double-slit experiment using monochromatic light, the fringe pattern shifts by certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 microns is introduced in the path of one of the interfering waves. The mica sheet is then removed and the slits-to-screen distance is doubled. It is found that the distance between successive maxima now is the same as the observed fringe shift with the mica sheet. Calculate the wavelength of the monochromatic light used.
Solution :
Data : n_m = 1.6, b = 1.964 microns
= 1.964 × 10^-6 m,
D₂ = 2D₁, W₂ = y₀
The fringe shift with the mica sheet,
y₀ = \(\frac{D_{1}}{d}\)(n_m – 1)b
Subsequent to the removal of the mica sheet and doubling the slits-to-screen distance, the new fringe width is,

Question 28.
What must be the thickness of a thin film which, when kept near one of the slits shifts the central fringe by 5 mm for incident light of wavelength 5890 Å in Young’s double-slit interference experiment ? The refractive index of the material of the film is 1.1 and the distance between the slits is 0.5 mm.
Solution :
Data : λ = 5890 Å, nm = 1.1, the shift of the central bright fringe = 5 mm

Let t be the thickness of the film and P the point on the screen where the central fringe has shifted. Suppose the film is kept in front of slit S1. Due to the film, the optical path travelled by the light passing through it increases by (1.1 – 1)f = 0.1t. Thus, the optical paths between the two beams passing through the two slits are not equal at the midpoint of the screen but are equal at P, 5 mm away from the centre. At this point the distance travelled by light from the other slit S₂ to the screen is larger than that travelled by light from S₁ by 0.1t.

The difference in distances, S₂P – S₁P = yλ/d, where y is the distance along the screen = 5 mm = 5 × 10^-3m and d = 0.5 mm = 5 × 10^-4 m.

This has to be equal to the difference in optical paths introduced by the film.
Thus, 0.1t = 5 × 10^-3 × 5890 × 10^-10/5 × 10^-4.
∴ t = 5890 × 10^-8m = 5.89 × 10^-5 m = 0.0589 mm

Question 29.
In a biprism experiment, the eyepiece is placed at a distance of 1.2 metres from the source. The distance between the virtual sources was found to be 7.5 × 10^-4 m. Find the wavelength of light if the eyepiece is to be moved transversely through a distance of 1.888 cm for 20 fringes.
Solution :
Data : D = 1.2 m, d = 7.5 × 10^-4 m,
20 W = 1.888 cm = 1.888 × 10^-2 m

Question 30.
A biprism is placed 5 cm from the slit illuminated by sodium light of wavelength 5890 Å. The width of the fringes obtained on a screen 75 cm from the biprism is 9.424 × 10^-2 cm. What is the distance between the two coherent sources?
Solution :
Data : D = 5 cm + 75 cm = 80 cm = 0.8 m,
λ = 5890 Å = 5.890 × 10^-7 m,
W = 9.424 × 10^-2 cm = 9.424 × 10^-4 m
Fringe width, W = \(\frac{\lambda D}{d}\)
∴ The distance between the two coherent sources,

= 5 × 10^-4 m = 0.5 mm

Question 31.
In a biprism experiment, the distance of the 20th bright band from the centre of the interference pattern is 8 mm. Calculate the distance of the 30th bright band from the centre.
Solution :
Data : y₂₀ = 8 mm (bright band)
The distance of the nth bright band from the centre of the interference pattern,

The distance of the 30th bright band from the centre of the interference pattern is 12 mm.

Question 32.
In a biprism experiment, a source of light having wavelength 6500 Å is replaced by a source of light having wavelength 5500 Å. Calculate the change in the fringe width, if the screen is at a distance of 1 m from the sources which are 1 mm apart.
Solution :

The fringe width decreases by 1 × 10^-4 m = 0.1 mm.

Question 33.
In a biprism experiment, light of wavelength 5200 Å is used to get an interference pattern on the screen. The fringe width changes by 1.3 mm when the screen is moved towards the biprism by 50 cm. Find the distance between the two virtual images of the slit.
Solution :
Data : λ = 5200 Å = 5.2 × 10^-7 m,
W₁ – W₂ = 1.3 mm = 1.3 × 10^-3 m,
D₁ – D₂ = 50 cm = 0.5 m

This is the distance between the two virtual sources.

Question 34.
In a biprism experiment, the 10th dark band is observed at 2.09 mm from the central bright point on the screen with red light of wavelength 6400 Å. By how much will the fringe width change if blue light of wavelength 4800 A is used with the same setting?
Solution :

Question 35.
In a biprism experiment, the slit is illuminated by red light of wavelength 6400 Å, and the cross wire of the eyepiece is adjusted at the centre of the 3rd bright band. On using blue light, it is found that the 4th bright band is on the cross wire. Find the wavelength of blue light.
Solution :
Data : λ_r = 6400 Å, y₃ (red, bright)

This is the wavelength of blue light.

Question 36.
In a biprism experiment, the wavelength of red light used is 6000 Å and the nth bright band is obtained at a point P on the screen. Keeping the same setting, the source is replaced by a source of green light of wavelength 5000 Å and the (n + 1)th bright band of green light coincides with point P. Find n.
Solution :

Question 37.
In a biprism experiment, the slit is illuminated by light of wavelength 4800 Å. The distance between the slit and the biprism is 15 cm and that between the biprism and the eyepiece is 85 cm. If the distance between the virtual sources is 3 mm, determine the distance between the 4th bright band on one side and the 4th dark band on the other side of the central band.
Solution :
Data : λ = 4800 Å = 4.8 × 10^-7 m,
d = 3 mm = 3 × 10^-3 m,
D = distance between the slit and the biprism + distance between the biprism and the eyepiece = 15 + 85 = 100 cm = 1 m The distance of the nth bright band from the central band is

This is the required distance.

Question 38.
An isosceles prism of refracting angle 179° and refractive index 1.6 is used as a biprism by keeping it 10 cm away from a slit, the edge of the biprism being parallel to the slit. The slit is illuminated by a light of wavelength 600 nm and the screen is 90 cm away from the biprism. Calculate the location of the centre of the 10th dark band from the centre of the interference pattern and the path difference at this location.
Solution :

Question 39.
In a biprism experiment, the distance between the second and tenth dark bands on the same side of the central bright band is 0.12 cm, that between the slit and the biprism is 20 cm and that between the biprism and the eyepiece is 80 cm. If the slit images given by the lens in the two positions are 4.5 mm and 2 mm apart, find the wavelength of light used.
Solution :
The distance between the second and tenth dark bands on the same side of the central band is equal to 8 times the fringe width (W).
∴ 8 W = 0.12 cm (by the data)
∴ W = \(\frac{0.12}{8}\) cm = 0.015 cm = 0.015 × 10^-2 m
The distance (D) between the slit and the eyepiece is equal to the sum of the distance between the slit and the biprism and the distance between the biprism and the eyepiece.
∴ D = 20 + 80 = 100 cm = 1 m (by the data)
Also, d₁ = 4.5 mm and d₂ = 2 mm
∴ The distance (d) between the virtual images of the slit is
d = \(\sqrt{d_{1} d_{2}}\) = \(\) mm = 3 mm
= 3 × 10^-3 m
∴ The wavelength of light,

Question 40.
In a biprism experiment, the slit and the eyepiece. are 10 cm and 80 cm away from the biprism. When a convex lens was interposed at 30 cm from the slit, the separation of the two magnified images of the slit was found to be 4.5 mm. If the wavelength of the source is 4500 Å, calculate the fringe width.
Solution:
Data : d₁ = 4.5 mm = 4.5 × 10^-3 m,
λ = 4500 Å = 4.5 × 10^-7 m,
distance between the slit and the eyepiece (D) = distance between the slit and the biprism + distance between the biprism and the eyepiece = 10 cm + 80 cm = 90 cm = 0.9 m, u₁ = 30 cm = 0.3 m v₁ = D – u₁ = 0.9m – 0.3m = 0.6 m
Linear magnification of a lens,

Question 65.
Describe with a neat labelled ray diagram the Fraunhofer diffraction pattern due to a single slit. Obtain the expressions for the positions of the intensity minima and maxima. Also obtain the expression for the width of the central maximum.
Answer:
When a parallel beam of monochromatic light of wavelength λ illuminates a single slit of finite width a, we observe on a screen some distance from the slit, a broad pattern of alternate dark and bright fringes. The pattern consists of a central bright fringe, with successive dark and bright fringes of diminishing intensity on both sides. This is called ‘ the diffraction pattern of a single slit.

Consider a single slit illuminated with a parallel beam of monochromatic light perpendicular to the plane of the slit. The diffraction pattern is obtained on a screen at a distance D (» a) from the slit and at the focal plane of the convex lens.

We can imagine the single slit as being made up of a large number of Huygens’ sources evenly distributed over the width of the slit. Then the maxima and minima of the pattern arise from the interference of the various Huygens’ wavelets.
Now, imagine the single slit as made up of two adjacent slits, each of width a/2. Since, the incident plane wavefronts are parallel to the plane of the slit, all the Huygens sources at the slit will be in phase. They will therefore also in phase at the point P₀ on the screen, where P₀ is equidistant from all the Huygens sources. At P₀, then, we get the central maximum.

For the first minimum of intensity on the screen, the path difference between the waves from the Huygens sources A and O (or O and B) is λ/2, which is the condition for destructive interference. Suppose, the nodal line OP for the first minimum subtends an angle θ at the slit; θ is very small. With P as the centre and PA as radius, strike an arc intersecting PB at C. Since, D » a, the arc AC can be considered a straight line at right angles to PB. Then, ∆ ABC is a right-angled triangle similar to A OP₀P.
This means that, ∠BAC = θ
∴ BC = a sin θ
∴ Difference in path length,
BC = PB – PA = (PB – PO) + (PO – PA)

(∵ θ is very small and in radian)
The other nodal lines of intensity minima can be understood in a similar way. In general, then, for the with minimum (wi = +1, ±2, ±3, …).
θ_m = \(\frac{m \lambda}{a}\) (with minimum) … (2)
as θ_m is very small and in radian.
Between the successive minima, the intensity rises to secondary maxima when the path difference is an odd-integral multiple of \(\frac{\lambda}{2}\) :

(mth secondary maximum) … (3)
Width of the central maximum :
Equation (1) gives the angular half width of the central maximum. Therefore, the angular width of the central maximum is,
2θ = \(\frac{2 \lambda}{a}\) … (4)
From ∆OP₀P, P₀P = D tan θ \(\simeq\) D sin θ
(∵ θ is very small and in radian)
∴ y₁ = P₀P = \(\frac{D \lambda}{a}\) [from Eq. (1)] … (5)
This is the distance of the first minimum from the centre of the central maximum.
∴ Width of the central bright fringe :
W_c = 2y_1d = 2W = 2\(\left(\frac{\lambda D}{a}\right)\) …(6)

The central bright fringe is spread between the first dark fringes on either side. Thus, the width of the central bright fringe is the distance between the centres of the first dark fringe on either side.

If the lens is very close to the slit, D is very nearly equal to f, where f is the focal length of the lens. Then W_c = 2\(\left(\frac{\lambda D}{a}\right)\) = 2\(\left(\frac{\lambda f}{a}\right)\) … (7)

Question 66.
Represent graphically intensity distribution in
(a) Young’s double-slit interference
(b) single- slit diffraction and
(c) double-slit diffraction
Answer:

Question 67.
State the characteristics of a single-slit diffraction pattern.
Answer:
Characteristics of a single-slit diffraction pattern :

1. The image cast by a single-slit is not the expected purely geometrical image.
2. For a given wavelength, the width of the diffraction pattern is inversely proportional to the slit width.
3. For a given slit width a, the width of the diffraction pattern is proportional to the wavelength.
4. The intensities of the non-central, i.e., secondary, maxima are much less than the intensity of the central maximum.
5. The minima and the non-central maxima are of the same width, Dλ/a.
6. The width of the central maximum is 2Dλ/a. It is twice the width of the non-central maxima or minima.

Question 68.
Explain briefly the double-slit diffraction pattern.
Answer:
The double-slit diffraction pattern is determined by the diffraction patterns due to the individual slits, and by the interference between them.

We can see narrow interference fringes similar to those obtained in Young’s double-slit experiment. These fringes vary in brightness and the shape of their envelope as that of the single-slit diffraction pattern.

Question 69.
What should be the order of the size of an obstacle or aperture to produce diffraction of light?
Answer:
For pronounced diffraction, the size of an obstacle or aperture should be of the order of the wavelength of light or greater.
[Note : For diffraction from a single slit of width a with monochromatic light of wavelength λ, the condition for first minimum (dark fringe) is
sin θ₁ = \(\frac{\lambda}{a}\)
When a = λ, θ₁ = 90° and the central maximum spreads over 180°; then, while the diffraction is maximum, no fringe pattern is seen at all.

When a » X (say, a is of the order of a centimetre or more), θ₁ is so small that there is practically no diffraction and the illuminated region on the screen is almost as given by geometrical optics. However, diffraction pattern due to a straight-edge will always be seen at the edge of the illuminated region. Hence, for an observable fringe pattern due to a single slit, a should be of the order of X with a > λ..]

Question 70.
Solve the following.

Question 1.
Plane waves of light from a sodium lamp are incident on a slit of width 2 μm. A screen is located 2 m from the slit. Find the spacing between the first secondary maxima of two sodium lines as measured on the screen.
(Given : λ₁ = 5890 Å and ₂ = 5896 Å)
Solution:


This is the required spacing.

Question 2.
The diffraction pattern of a single slit of width 0. 5 cm is formed by a lens of focal length 40 cm. Calculate the distance between the first dark and the next bright fringe from the axis. The wavelength of light used is 4890 Å.
Solution :
Data : a = 0.5 cm = 0.5 × 10^-2 m,
D \(\simeq\) f = 40 cm = 0.4 m, λ = 4890 Å = 4.890 × 10^-7 m
The distance between the first dark fringe and the next bright fringe = \(\frac{\lambda D}{2 a}\)

Question 3.
Light of wavelength 6000 Å is incident on a slit of width 0.3 mm. A screen is placed parallel to the slit 2 m away from the slit. Find the position of the first dark fringe from the centre of the central maximum. Also, find the width of the central maximum.
Solution :
Data : λ = 6 × 10^-7 m, a = 0.3 mm = 3 × 10^-4 m,
D = 2 m
The distance y_m of the m th minimum from the centre of the central maximum is

The first dark fringe is 4 mm from the centre of the central fringe.
∴ Half-width of the central maximum = 4 mm
∴ The width of the central maximum = 2 × 4 mm = 8 mm

Question 4.
In a single-slit diffraction pattern, the distance between the first minimum on the right and the first minimum on the left is 5.2 mm. The screen on which the pattern is displayed is 80 cm from the slit and the wavelength of light is 5460 Å. Calculate the slit width.
Solution :
Data : y₁ (minimum, right) + y₁ (minimum, left)

Question 5.
In a single-slit diffraction pattern, the distance between the first minimum on the right and the first minimum on the left of the central maximum is 4 mm. The screen is 2 m from the slit and the wavelength of light used is 6000 Å. Calculate the width of the slit and the width of the central maximum.
Solution :
Data : y₁ (minimum, right) + y₁ (minimum, left)
= 4 mm = 4 × 10^-3 m, D = 2 m, λ = 6 × 10^-7 m

Question 6.
Determine the angular spread between the central maximum and first order maximum of the diffraction pattern due to a single slit of width 0.25 mm, when light of 6650 Å is incident on it normally.
Solution:
Data : λ = 6650 Å = 6650 × 10^-10m, a = 0.25 mm

This is the required angular spread.

Question 71.
Explain and define the resolving power of an optical instrument.
Answer:
Resolving power of an optical instrument:
The primary aim of using an optical instrument is to see fine details, whether observing a star system through a telescope or a living cell through a microscope. After passing through an optical system, light from two adjacent parts of the object should produce sharp, distinct (separate) images of those parts. The objective lens or mirror of a telescope or microscope acts like a circular aperture. The diffraction pattern of a circular aperture consists of a central bright spot (called the Airy disc and corresponds to the central maximum) and concentric dark and bright rings.

Light from two close objects or parts of an object after passing through the aperture of an optical system produces overlapping diffraction patterns that tend to obscure the image. If these diffraction patterns are so broad that their central maxima overlap substantially, it is difficult to decide if the intensity distribution is produced by two separate objects or by one.

The resolving power of an optical instrument, e.g., a telescope or microscope, is a measure of its ability to produce detectably separate images of objects that are close together.

Definition : The smallest linear or angular separation between two point objects which appear just resolved when viewed through an optical instrument is called the limit of resolution of the instrument and its reciprocal is called the resolving power of the instrument.

Question 72.
State and explain Rayleigh’s criterion for minimum resolution.
Answer:
Rayleigh’s criterion for minimum resolution : Two overlapping diffraction patterns due to two point sources are acceptably or just resolved if the centre of the central peak of one diffraction pattern is as far as the first minimum of the other pattern.

The ‘sharpness’ of the central maximum of a diffraction pattern is measured by the angular separation between the centre of the peak and the first minimum. It gives the limit of resolution.

Two overlapping diffraction patterns due to two point sources are not resolved if the angular separation between the central peaks is less than the limit of resolution. They are said to be just separate, or resolved, if the angular separation between the central peaks is equal to the limit of resolution. They are said to be well resolved if the angular separation between the central peaks is more than the limit of resolution.

Question 73.
Explain the Rayleigh criterion for the limit of resolution for
(i) two linear objects
(ii) a pair of point objects.
Answer:
(i) The Rayleigh criterion for the limit of resolution for two linear objects : Consider, two self luminous objects or slits separated by some distance. Let λ be the wavelength of the light and a the width of the slits. As per the Rayleigh criterion, the first minimum of the diffraction pattern of one of the sources should coincide with the central maximum of the other. Thus, it is at the just resolved condition.
The angular separation dθ (position) of the first principal minimum is,
dθ = \(\frac{\lambda}{a}\) …(1)
This angular separation between the two objects must be minimum as this minimum coincides with the central maximum of the other. This is called the limit of resolution of that instrument. It is written as,
limit of resolution, dθ = \(\frac{\lambda}{a}\)
Minimum separation between the two linear objects that are just resolved, at distance D from the instrument is, ni
y = D(dθ) = \(\frac{D \lambda}{a}\) …(2)
It is the distance of the first minimum from the centre.

(ii) The Rayleigh criterion for the limit of resolution for a pair of point objects : The objects to be viewed through a microscope are often of the point-size. The diffraction pattern of such objects consists of a central bright spot called the Airy disc and corresponds to the central maximum surrounded by concentric dark and bright rings called Airy rings.

If a red laser beam passes through a 90 µm pinhole aperture, the Airy disc and several orders (rings) of diffraction are as shown.

According to Lord Rayleigh, for such objects to be just resolved, the first dark ring of the diffraction pattern of the first object should be formed at the centre of the diffraction pattern of the second object and vice versa. Thus, the minimum separation between the images on the screen should be equal to the radius of the first dark ring.
This is applicable to the eye, microscope, telescope, etc.

Question 74.
Define and explain the resolving power of a microscope. State the expressions for the resolving power of
(i) a microscope with a pair of non-luminous objects
(ii) a microscope with self luminous point objects.
OR
What is meant by the limit of resolution and the resolving power of a microscope?
Answer:
Definition : The limit of resolution of a microscope is the least separation between two-closely spaced points on an object which are just resolved when viewed through the microscope.

Definition : The resolving power of a microscope is defined to be the reciprocal of its limit of resolution.

In a compound microscope, the objective lens forms a real, magnified image of an object placed just beyond the focal length of the lens. The objective has a short focal length (for greater magnification) and is held close to the object so that it gathers as much of the light scattered by the object as possible.

Let a be the least separation between two point objects O and O’ viewed through an objective AB of a compound microscope. The medium between the object and the objective has a refractive index n. The images of the objects O and O’ are I and I’ respectively. In this case, the angular separation between the objects, at the objective is 2α. D is the diameter of the objective AB.

According to the Rayleigh criterion, the first dark ring due to O’ should coincide with I and that of O should coincide with I’. The nature of illumination at a point on the screen is determined by the effective path difference at that point. Let us consider point I to be symmetric with respect to O. Paths of the extreme rays reaching I from O’ are O’AI and O’BI. The paths AI and BI are equal. Thus, the actual path difference is O’B – O’A.

The enlarged view of the region around O and O’ is as shown.
From this,
path difference = DO’ + O’C
= 2a sin α

(i) Microscope with a pair of non-luminous objects (dark objects):
In actual practice, the objects O and O’ viewed through a microscope are illuminated by the same source. Often the eyepiece of the microscope is filled with some transparent material of refractive index n. Then the wavelength of light in this material is
λ_n = \(\frac{\lambda}{n}\) where λ is the wavelength of light in air.

In such a set up the path difference at the first dark ring in λ_n. Thus, from eq (1),

The factor n sin a is called numerical aperture (NA). The resolving power of the microscope is
R = \(\frac{1}{a}\) = \(\frac{2 \mathrm{NA}}{\lambda}\) …. (3)

(ii) Microscope with self luminous point objects : Applying Abbe’s theory of Airy discs and rings to Fraunhofer diffraction due to a pair of self luminous point objects, the path difference between the extreme rays, at the first dark ring is 1.22 λ, thus, for the requirement of just resolution,

The resolving power of the microscope is,
R = \(\frac{1}{a}\) = \(\frac{\mathrm{NA}}{0.61 \lambda}\) …. (5)
When the value of a is minimum, the quality of resolution is high.
Notes:

1. If f and D are the focal length and the diameter of the microscope objective.
   sin i_max \(\simeq \frac{D / 2}{f}\) so that Eq. (3) can be written as resolving power = \(\frac{n D}{f \lambda}\)
2. Ernst Abbe(1840 -1905), German physicist and developer of optical instruments.

Question 75.
Explain why microscopes of high magnifying power have oil filled (oil-immersion) objectives.
Answer:
Higher angular magnification of a high magnifying power microscope is of little use if the finer details in a tiny object are obscured by diffraction effects. Hence, a microscope of high magnifying power must also have a high resolving power.
Resolving power ot a microscope = \(\frac{2 n \sin \alpha}{\lambda}\)
Where α ≡ the half angle of the angular separation between the objects, at the objective lens. n ≡ the refractive index of the medium between the object and the objective, λ ≡ the wavelength of the light used to illuminate the object.

The factor n sin α is called the numerical aperture of the objective and the resolving power increases with increase in the numerical aperture. To increase α the diameter of the objective would have to be increased. But this increase in aperture would degrade the image by decreasing the resolving power. Hence, in microscopes of high magnifying power, the object is immersed in oil that is in contact with the objective. Usually cedarwood oil having a refractive index 1.5 (close to that of the objective glass) is used. Closeness of the refractive indices also reduces loss of light by reflection at the objective lens.

Question 76.
On what factors does the resolving power of a microscope depend? How can it be increased?
Answer:
Resolving power of a microscope
= \(\frac{2 n \sin \alpha}{\lambda}\) = \(\frac{2 \mathrm{NA}}{\lambda}\)

where, α ≡ the half angle of the angular separation between the objects at the objective lens. n ≡ the refractive index of the medium between the object and the objective, λ ≡ the wavelength of the light used to illuminate the object, NA = n sin α = the numerical aperture of the objective.

Thus, the resolving power of a microscope depends directly on the NA and inversely on λ.
The resolving power is increased by,

1. increasing the numerical aperture using oil-immersion objective.
2. illuminating the object with smaller wavelength radiation. But our eyes are not very sensitive to the shorter wavelength blue end of the visible spectrum. Hence, ultraviolet radiation is used for illumination with quartz lenses, but then photographs must be taken to examine the image.

Question 77.
With a neat ray diagram, explain the resolving power of a telescope. On what factors does it depend ?
OR
What is meant by the angular limit of resolution and resolving power of a telescope?
Answer:
The resolving power of a telescope is defined as the reciprocal of the angular limit of resolution between two closely-spaced distant objects so that they are just resolved when seen through the telescope.

Consider two stars seen through a telescope. The diameter (D) of the objective lens or mirror corresponds to the diffracting aperture. For a distant point source, the first diffraction minimum is at an angle θ away from the centre such that

Dsin θ = 1.22 λ
where λ is the wavelength of light. The angle 9 is usually so small that we can substitute sin θ \(\approx\) θ (θ in radian). Thus, the Airy disc for each star will be spread out over an angular half-width θ = 1.22 λ/D about its geometrical image point. The radius of the Airy disc at the focal plane of the objective lens is r = fθ = 1.22fλ/D, where f is the focal length of the objective.

When observing two closely-spaced stars, the Rayleigh criterion for just resolving the images as that of two point sources (instead of one) is met when the centre of one Airy disc falls on the first minimum of the other pattern. Thus, the angular limit (or angular separation) of resolution is
θ = \(\frac{1.22 \lambda}{D}\) … (1)

and the linear separation between the images at the focal plane of the objective lens is
y = fθ … (2)
∴ Resolving power of a telescope.
R = \(\frac{1}{\theta}\) = \(\frac{D}{1.22 \lambda}\) … (3)

It depends

1. directly on the diameter of the objective lens or mirror,
2. inversely on the wavelength of the radiation.

Question 78.
How can the resolving power of a telescope be increased?
Answer:
The resolving power of an astronomical telescope depends directly on the diameter of the objective lens or mirror, and inversely on the wavelength of radiation. Hence, the resolving power can be increased by

1. using an objective lens/mirror of larger diameter
2. observing a celestial object at smaller wavelengths.

Question 79.
Define the resolving power of a telescope and state its formula. What are the advantages of using a large objective lens in an astronomical telescope:
Answer:
Definition: The resolving power of a telescope is defined as the reciprocal of the angular limit of resolution between two closely-spaced distant objects so that they are just resolved when seen through the telescope.
Formula : Resolving power of a telescope
R = \(\frac{1}{\theta}\) = \(\frac{D}{1.22 \lambda}\)
where θ ≡ the minimum angular separation of two closely-spaced celestial objects or the angular limit of resolution, D ≡ the diameter of the objective lens of the telescope, λ ≡ the wavelengh of light.

Advantages of a large objective lens in an astronomical telescope :

1. The resolving power is directly proportional to the diameter of the objective lens. Hence, a large objective lens results in a smaller Airy disc and a sharper image.
2. It collects more of the incident radiation from a distant object which results in a brighter image.

Question 80.
What is the resolving power of a telescope if the diameter of the objective of the telescope is 1.22 m and the wavelength of light is 5000 Å ?
Answer:
Resolving power of the telescope = \(\frac{D}{1.22 \lambda}\)
= \(\frac{1.22}{1.22 \times 5 \times 10^{-7}}\)
= \(\frac{10 \times 10^{6}}{5}\) = 2 × 10⁶ rad^-1

Question 81.
Briefly explain about radio telescope.
Answer:
A radio telescope is an astronomical instrument consisting of a radio receiver and an antenna system used to detect radio-frequency radiation emitted by extraterrestrial sources. The wavelength of radio-frequency radiation is very long. Hence, it is expressed in metre. In order to attain the resolution of an optical telescope, the radio telescope should be very large. A single disc of such large diameters is impracticable. In such cases arrays of antennae spread over several kilometres are used. Giant Metrewave Radio Telescope (GMRT) located at Narayangaon, Pune, Maharashtra uses such an array consisting of 30 dishes of diameter 45 m each. It spreads over 25 km. It is the largest distance between two of its antennae. The highest angular resolution achievable ranges from about 60 arcsec at the lowest frequency of 50 MHz to about 2 arcsec at 1.4 GHz.

Question 82.
Solve the following :

Question 1.
Light of wavelength 650 nm falls on the objective of a microscope having total angle of angular w separation as 60°. Calculate the resolving power of the microscope. [Refractive index of air = 1]
Solution :
Data : λ = 650 nm = 6.5 × 10^-7m, n = 1,
2α = 60°
∴ α = 30°
Resolving power of the microscope = \(\frac{2 n \sin \alpha}{\lambda}\)
= \(\frac{2 \times 1 \times \sin 30^{\circ}}{6.5 \times 10^{-7}}\) = \(\frac{10}{6.5}\) × 10⁶

Question 2.
The semi vertical angle of the cone of the rays incident on the objective of a microscope is 20°. If the wavelength of incident light is 6600 Å, calculate the smallest distance between two points which can be just resolved.
Solution :
Data : α = 20°, λ = 6600 Å = 6.6 × 10^-7 m, n = 1 (air) resolution between two closely-spaced distant objects so that they are just resolved when seen through the telescope.
Formula : Resolving power of a telescope
R = \(\frac{1}{\theta}\) = \(\frac{D}{1.22 \lambda}\)
where θ ≡ the minimum angular separation of two closely-spaced celestial objects or the angular limit of resolution, D ≡ the diameter of the objective lens of the telescope, λ ≡ the wavelengh of light.

Advantages of a large objective lens in an astronomical telescope :

1. The resolving power is directly proportional to the diameter of the objective lens. Hence, a large objective lens results in a smaller Airy disc and a sharper image.
2. It collects more of the incident radiation from a distant object which results in a brighter image.

Question 83.
What is the resolving power of a telescope if the diameter of the objective of the telescope is 1.22 m and the wavelength of light is 5000 Å ?.
Answer:
Resolving power of the telescope = \(\frac{D}{1.22 \lambda}\)

Question 84.
Briefly explain about radio telescope.
Answer:
A radio telescope is an astronomical instrument consisting of a radio receiver and an antenna system used to detect radio-frequency radiation emitted by extraterrestrial sources. The wavelength of radio-frequency radiation is very long. Hence, it is expressed in metre. In order to attain the resolution of an optical telescope, the radio telescope should be very large. A single disc of such large diameters is impracticable. In such cases arrays of antennae spread over several kilometres are used. Giant Metrewave Radio Telescope (GMRT) located at Narayangaon, Pune, Maharashtra uses such an array consisting of 30 dishes of diameter 45 m each. It spreads over 25 km. It is the largest distance between two of its antennae. The highest angular resolution achievable ranges from about 60 arcsec at the lowest frequency of 50 MHz to about 2 arcsec at 1.4 GHz.

Question 85.
Solve the following :

Question 1.
Light of wavelength 650 nm falls on the objective of a microscope having total angle of angular separation as 60°. Calculate the resolving power of the microscope. [Refractive index of air = 1]
Solution :
Data : λ = 650 nm = 6.5 × 10^-7m, n = 1, 2α = 60°
∴ α = 30°
Resolving power of the microscope = \(\frac{2 n \sin \alpha}{\lambda}\)

= 1.538 × 10⁶ m^-1

Question 2.
The semi vertical angle of the cone of the rays incident on the objective of a microscope is 20°. If the wavelength of incident light is 6600 Å, calculate the smallest distance between two points which can be just resolved.
Solution :
Data : α = 20°, λ = 6600 Å = 6.6 × 10^-7 m, n = 1 (air)
The numerical aperture, NA = n sin α
= 1 × sin 20° = 0.3420
The limit of resolution for an illuminated object,

Question 3.
What is the minimum distance between two objects which can be resolved by a microscope having the visual angle of 30° when light of wavelength 600 nm is used ?
Solution :

Question 4.
The two headlights of an approaching automobile are 1.22 m apart. At what maximum distance will eye resolve them? Assume a pupil diameter of 5.0 mm and λ = 5500 Å. Assume also that this distance is determined only by the diffraction effect at the circular aperture.
Solution:
Data : y = 1.22 m, diameter D = 5 × 10^-3 m,

This is the required distance.

Question 5.
What is the minimum angular separation between two stars if a telescope is used to observe them with an objective of aperture 20 cm? The wavelength of light is 5900 Å.
Solution :
Data : D = 20 cm = 0.2 m,
λ = 5900 Å = 5.9 × 10^-7 m

This is the required angular separation.

Question 6.
The diameter of the objective of a telescope is 10 cm. Find the resolving power of the telescope is 10 cm. Find the resolving power of the telescope if the wavelength of light is 5000 Å.
Solution :
Data : D = 10 cm = 0.1 m, λ = 5000 Å = 5 × 10^-7 m
The resolving power of the telescope D 0.1

Question 7.
A telescope has an objective of diameter 2.44 m. What is its angular resolution when it observes at 5500 A?
Solution :
Data : λ = 5500A = 5.5 × 10^-7 m,
D = 2.44 m
Angular resolution. ∆θ = 1.22λ/D, D being the diameter of the aperture.
∴ ∆θ = 1.22 × 5.5 × \(\frac{10^{-7}}{2.44}\)= 2.75 × 10^-7 rad
= 0.0567 arcsec

Question 8.
The minimum angular separation between two stars is 4 × 10^-6 rad when a telescope is used to observe them with an objective of aperture 16 cm. Find the wavelength of the light.
Solution :
Data : θ = 4 × 10^-6 rad, D = 16 cm = 0.16 m
θ = \(\frac{1.22 \lambda}{D}\)
∴ The wavelength of the light used,

Question 9.
Estimate the smallest angular separation of two stars which can be just resolved by the telescope having objective of diameter 25 cm. The mean wavelength of light is 555 nm.
Solution :
Data : λ = 555 nm 555 × 10^-9 m
D = 25 cm = 25 × 10^-2 m

= 2.708 × 10^-6 rad
This is the required angular separation.

Question 86.
Huygens’ wave theory could not explain
(A) interference
(B) reflection
(C) photoelectric effect
(D) refraction.
Answer:
(C) photoelectric effect

Question 87.
The wavefront originating from a point source of light at finite distance is a wavefront.
(A) circular
(B) spherical
(C) plane
(D) cylindrical
Answer:
(B) spherical

Question 88.
In an isotropic medium, the secondary wavelets centred on every point of a given wavefront are all
(A) spherical
(B) cylindrical
(C) oval
(D) of arbitrary shape.
Answer:
(A) spherical

Question 89.
Consider a medium through which light is propagating with a speed v. Given a wavefront, in order to determine the wavefront after a time interval ∆t, the secondary wavelets are drawn with
a radius.
(A) of unit length
(B) v ∆ t
(C) \(\frac{\Delta t}{v}\)
(D) \(\frac{v}{\Delta t}\)
Answer:
(B) v ∆ t

Question 90.
Two points, equidistant from a point source of light, are situated at diametrically opposite positions in an isotropic medium. The phase difference between the light waves passing through the two points is
(A) zero
(B) π/2 rad
(C) π rad
(D) finite, but not these.
Answer:
(A) zero

Question 91.
Wavenormals to spherical wavefronts can be
(A) only diverging
(B) only converging
(C) parallel to each other
(D) diverging or converging.
Answer:
(D) diverging or converging.

Question 92.
Huygens’ principle is used to
(A) obtain the new position of wavefront geometrically
(B) explain the principle of superposition of waves
(C) explain the phenomenon of interference
(D) explain the phenomenon of polarization.
Answer:
(A) obtain the new position of wavefront geometrically

Question 93.
When a ray of light enters into water from air,
(A) its wavelength decreases
(B) its wavelength increases
(C) its frequency increases
(D) its frequency decreases.
Answer:
(A) its wavelength decreases

Question 94.
A parallel beam of light travelling in water is incident obliquely on a glass surface. After refraction, its width
(A) decreases
(B) increases
(C) remains the same
(D) becomes zero.
Answer:
(B) increases

Question 95.
Light of a certain colour has 1800 waves to the millimetre in air. What is its frequency in water?
[n = \(\frac{4}{3}\) for water]
(A) 1.67 × 10⁶ Hz
(B) 4.05 × 10¹⁴ Hz
(C) 5.4 × 10¹⁴ Hz
(D) 7.2 × 10¹⁴ Hz.
Answer:
(C) 5.4 × 10¹⁴ Hz

Question 96.
A ray of light, in passing from vacuum into a medium of refractive index n, suffers a deviation d equal to half the angle of incidence. Then, the refractive index is
(A) sin δ
(B) 2 sin δ
(C) cos δ
(D) 2 cos δ.
Answer:
(D) 2 cos δ.

Question 97.
A ray of light passes from vacuum to a medium of refractive index n. The angle of incidence is found to be twice the angle of refraction. The angle of incidence is
(A) cos^-1 \(\left(\frac{n}{2}\right)\)
(B) cos^-1 (n)
(C) 2cos^-1 \(\left(\frac{n}{2}\right)\)
(D) 2sin^-1 \(\left(\frac{n}{2}\right)\).
Answer:
(C) 2cos^-1 \(\left(\frac{n}{2}\right)\)

Question 98.
If the polarizing angle for a given medium is 60°, then the refractive index of the medium is
(A) \(\frac{1}{\sqrt{3}}\)
(B) \(\frac{\sqrt{3}}{2}\)
(C) 1
(D) \(\sqrt{3}\)
Answer:
(D) \(\sqrt{3}\)

Question 99.
A narrow beam of light in air is incident on glass at an angle of incidence of 58°. If the reflected beam is completely plane polarized, the refractive index of the glass is
(A) 1.9
(B) 1.8
(C) 1.7
(D) 1.6.
Answer:
(D) 1.6.

Question 100.
Polarization of light CANNOT be produced by
(A) reflection
(B) double refraction
(C) dichroism
(D) diffraction.
Answer:
(D) diffraction.

Question 101.
A glass plate of refractive index 1.732 is to be used as a polarizer. Its polarizing angle is
(A) 30°
(B) 45°
(C) 60°
(D) 90°.
Answer:
(C) 60°

Question 102.
Light transmitted through a Polaroid P₁ has an intensity I and is incident on a crossed Polaroid P₂. The intensity of the light transmitted by P₂ is
(A) zero
(B) \(\frac{1}{2}\)I
(C) I
(D) 2I.
Answer:
(A) zero

Question 103.
At points of constructive interference with maximum intensity of two coherent monochromatic waves (wavelength λ), the path difference between them is
(A) zero or an integral multiple of λ
(B) zero or an integral multiple of λ/2
(C) zero or an even integral multiple of λ/2
(D) an odd integral multiple of λ/2.
Answer:
(A) zero or an integral multiple of λ

Question 104.
Two sources of light are said to be coherent if light from them have
(A) the same speed and the same phase
(B) the same phase and the same or nearly the same amplitude
(C) constant phase difference and nearly the same frequency
(D) zero, or some constant, phase difference.
Answer:
(D) zero, or some constant, phase difference.

Question 105.
In a two-source interference pattern, the phase difference between the waves reaching a dark point in radian is (m = 1, 2, 3, …)
(A) 0
(B) mπ
(C) (2m – 1)\(\frac{\pi}{2}\)
(D) (2m – 1)π
Answer:
(D) (2m – 1)π

Question 106.
For destructive interference, the phase difference (in radian) between the two waves should be
(A) 0, 2π, π, …
(B) 0, 2π, 4π, …
(C) π, 3π, 5π, …
(D) \(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\), \(\frac{5 \pi}{2}\), …..
Answer:
(C) π, 3π, 5π, …

Question 107.
For constructive interference, the phase difference (in radian) between the two waves should be
(A) 0, \(\frac{\pi}{2}\), π, …
(B) 0, 2π, 4π, …
(C) π, 3π, 5π, …
(D) \(\frac{\pi}{4}\), \(\frac{\pi}{2}\), \(\frac{3 \pi}{4}\), …..
Answer:
(B) 0, 2π, 4π, …

Question 108.
If λ is the wavelength of light used in Young’s double-slit experiment, the path difference for a phase difference of 11π rad is
(A) 23 λ
(B) 11 λ
(C) 11\(\frac{\lambda}{2}\)
(D) 23\(\frac{\lambda}{2}\)
Answer:
(C) 11\(\frac{\lambda}{2}\)

Question 109.
The fringe width in an interference pattern is W. The distance between the 6th dark fringe and the 4th bright fringe on the same side of the central bright fringe is
(A) 1.5 W
(B) 2 W
(C) 2.5 W
(D) 10.5 W.
Answer:
(A) 1.5 W

Question 110.
Which of the following graphs shows the variation of the fringe width with the frequency of light in a two-source interference pattern ?

Answer:

Question 111.
In an interference pattern using two coherent sources of light, the fringe width is
(A) directly proportional to the wavelength
(B) inversely proportional to the square of the wavelength
(C) inversely proportional to the wavelength
(D) directly proportional to the square of the wavelength.
Answer:
(A) directly proportional to the wavelength

Question 112.
Two slits, 2 mm apart, are placed 300 cm from a screen. When light of wavelength 6000 Å is used, the separation (in mm) between the successive bright lines of the interference pattern is
(A) 0.9
(B) 4.5
(C) 6
(D) 9.
Answer:
(A) 0.9

Question 113.
In two separate setups of Young’s double-slit experiment, the wavelengths of the lights used are in the ratio 1 : 2 while the separation between the slits are in the ratio 2 : 1. If the fringe widths are equal, the ratio of the distances between the slit and the screen is
(A) 1 : 4
(B) 1 : 2
(C) 2 : 1
(D) 4 : 1.
Answer:
(D) 4 : 1.

Question 114.
In Young’s double-slit experiment, the slit separation is reduced to half while the distance of the screen from the slits is increased by 50%. In terms of the initial fringe width, W, the new fringe width is,
(A) \(\frac{1}{4}\) W
(B) \(\frac{3}{4}\) W
(C) \(\frac{3}{2}\) W
(D) 3 W.
Answer:
(D) 3 W.

Question 115.
A pair of slits 1.5 mm apart is illuminated with monochromatic light of wavelength 5500 Å and the interference pattern is obtained on a screen 1.5 m from the slits. The least distance of a point from the central maximum where the intensity is minimum is
(A) 0.275 mm
(B) 0.55 mm
(C) 2.75 mm
(D) 5.5 mm.
Answer:
(A) 0.275 mm

Question 116.
In Young’s double-slit experiment, if a thin transparent sheet covers one of the slits, the optical path of the wave from that slit
(A) increases
(B) decreases
(C) changes by \(\frac{D}{d}\) (n_m – 1)t
(D) is not affected.
Answer:
(A) increases

Question 117.
In Young’s double-slit experiment, if a thin mica sheet of thickness t and refractive index n_m covers one of the slits, the optical path of the wave from that slit
(A) increases by (n_m – 1 )t
(B) decreases by (n_m – 1) t
(C) changes by \(\frac{D}{d}\) (n_m – 1) t
(D) is not affected.
Answer:
(A) increases by (n_m – 1 )t

Question 118.
In a two-slit intereference experiment, if a thin transparent sheet of thickness f and refractive index n_m covers both the slits, the optical path difference between the two interfering waves
(A) increases by (n_m – 1)t
(B) decreases by (n_m – 1)t
(C) changes by \(\frac{D}{d}\)(n_m – 1)t
(D) is not affected.
Answer:
(D) is not affected.

Question 119.
In a biprism experiment two interfering waves are produced by division of
(A) amplitude
(B) wavefront
(C) amplitude and wavefront
(D) neither wavefront nor amplitude.
Answer:
(B) wavefront

Question 120.
In Fresnel’s biprism experiment, with the eyepiece 1 m from the two coherent sources, the fringe width obtained is 0.4 mm. If just the eyepiece is moved towards the biprism by 25 cm, then the fringe width
(A) decreases by 0.01 mm
(B) decreases by 0.1 mm
(C) increases by 0.01 mm
(D) increases by 0.1 mm.
Answer:
(B) decreases by 0.1 mm

Question 121.
In a biprism experiment, keeping the experimental setup unchanged, the fringe width
(A) increases with increase in wavelength
(B) decreases with increase in wavelength
(C) increases with decrease in wavelength
(D) remains unchanged with change in wavelength.
Answer:
(A) increases with increase in wavelength

Question 122.
In finding the distance between the two coherent sources in Fresnel’s biprism experiment by the conjugate foci method, one uses
(A) a long focus convex lens that forms real images of the virtual sources
(B) a short focus concave lens that forms real images of the virtual sources
(C) a short focus convex lens that forms virtual images of the virtual sources
(D) a short focus convex lens that forms real images of the virtual sources.
Answer:
(D) a short focus convex lens that forms real images of the virtual sources.

Question 123.
Using a light of wavelength 4800 Å in Fresnel’s biprism experiment, 21 fringes are obtained in a given region. If light of wavelength 5600 Å is used, the number of fringes in the same region will be
(A) 14
(B) 18
(C) 21
(D) 24.
Answer:
(B) 18

Question 124.
To obtain pronounced diffraction with a single slit illuminated by light of wavelength λ, the slit width should be
(A) of the same order as λ
(B) considerably larger than λ
(C) considerably smaller than λ
(D) exactly equal to λ/2.
Answer:
(B) considerably larger than λ

Question 125.
In single-slit diffraction, which of the following are equal ?
(A) Widths of all bright and dark fringes
(B) Intensities of non-central bright fringes
(C) Widths of non-central bright fringes
(D) Both widths and intensities of noncentral bright fringes.
Answer:
(C) Widths of non-central bright fringes

Question 126.
In single slit diffraction (at a narrow slit of width a), the intensity of the central maximum is
(A) independent of a
(B) proportional to a
(C) proportional to a²
(D) inversely proportional to a.
Answer:
(D) inversely proportional to a.

Question 127.
For a single slit of width a, the diffraction pattern minima are located at angles θ_m, where m is a positive, non-zero integer. Which of the following expressions is most correct ?
(A) a sin θ_m = mλ
(B) a sin θ_m = \(\frac{m \lambda}{2}\)
(C) a sin θ_m = (2m + 1)\(\frac{\lambda}{2}\)
(D) a sin θ_m = (2m – 1)\(\frac{\lambda}{2}\)
Answer:
(A) a sin θ_m = mλ

Question 128.
In a diffraction pattern due to a single slit of width a with incident light of wavelength λ, at an angle of diffraction θ, the condition for the first minimum is
(A) λ sin θ = a
(B) a cos θ = λ
(C) a sin θ = λ
(D) λ cos θ = a.
Answer:
(C) a sin θ = λ

Question 129.
The fringes produced in a diffraction pattern are of
(A) equal width with the same intensity
(B) unequal width with varying intensity
(C) equal intensity
(D) equal width with varying intensity.
Answer:
(B) unequal width with varying intensity

Question 130.
For a single slit of width a, the first diffraction maximum with light of wavelength λ subtends an angle θ such that sin θ is equal to
(A) \(\frac{\lambda}{2 a}\)
(B) \(\frac{\lambda}{a}\)
(C) \(\frac{1.5 \lambda}{a}\)
(D) \(\frac{2 \lambda}{a} .\)
Answer:
(C) \(\frac{1.5 \lambda}{a}\)

Question 131.
Fraunhofer diffraction pattern of a parallel beam of light (wavelength λ) passing through a narrow slit (width a) is observed on a screen using a convex lens (focal length f). The angular half-width of the central fringe is
(A) \(\frac{2 \lambda f}{a}\)
(B) \(\frac{\lambda f}{a}\)
(C) \(\frac{2 \lambda}{a}\)
(D) \(\frac{\lambda}{a}\)
Answer:
(D) \(\frac{\lambda}{a}\)

Question 132.
A parallel beam of light (λ = 5000 Å) is incident normally on a narrow slit of width 0.2 mm. The angular separation between the first two minima is
(A) 2.5 × 10^-3 degree
(B) 2.5 × 10^-3 rad
(C) 5 × 10^-3 degree
(D) 5 × 10^-3 rad.
Answer:
(B) 2.5 × 10^-3 rad

Question 133.
A parallel beam of light (λ = 5000 Å) is incident normally on a narrow slit of width 0.2 mm. The Fraunhofer diffraction pattern is observed on a screen placed at the focal plane of a convex lens (f = 20 cm). The first two minima are separated by
(A) 0.005 cm
(B) 0.05 cm
(C) 2.5 mm
(D) 5 mm.
Answer:
(B) 0.05 cm

Question 134.
A plane wave of wavelength 5500 Å is incident normally on a slit of width 2 × 10^-2 cm. The width of the central maximum on a screen 50 cm away is
(A) 2.50 × 10^-3 cm
(B) 2.75 × 10^-3 cm
(C) 2.75 × 10^-3 m
(D) 5.50 × 10^-3 m.
Answer:
(D) 5.50 × 10^-3 m.

Question 135.
If NA is the numerical aperture of a microscope objective, then its resolving power with an illumination of wavelength λ is
(A) \(\frac{0.5 \lambda}{\mathrm{NA}}\)
(B) \(\frac{0.61 \lambda}{\mathrm{NA}}\)
(C) \(\frac{1.22 \mathrm{NA}}{\lambda}\)
(D) \(\frac{2 \mathrm{NA}}{\lambda}\)
Answer:
(A) \(\frac{0.5 \lambda}{\mathrm{NA}}\)

Question 136.
A microscope with numerical aperture 0.122 is used with light of wavelength 6000 A. The limit of resolution is
(A) 3.33 × 10⁶ m
(B) 3.33 mm
(C) 3 × 10^-6 m
(D) 3 × 10^-7 m.
Answer:
(C) 3 × 10^-6 m

Question 137.
The objective lens of a telescope has a diameter D. The angular limit of resolution of the telescope for light of wavelength λ is
(A) \(\frac{D}{1.22 \lambda}\)
(B) \(\frac{1.22 \lambda}{D}\)
(C) \(\frac{D}{0.61 \lambda}\)
(D) \(\frac{0.61 \lambda}{D} .\)
Answer:
(B) \(\frac{1.22 \lambda}{D}\)

Question 138.
Two closely-spaced distant stars are just resolved when seen through a telescope with an objective lens of diameter D and focal length f. The separation between their images is given by.
(A) \(\frac{D}{1.22 \lambda f}\)
(B) \(\frac{f \mathcal{D}}{1.22 \lambda}\)
(C) \(\frac{1.22 \lambda f}{D}\)
(D) \(\frac{1.22 D \lambda}{f}\)
Answer:
(C) \(\frac{1.22 \lambda f}{D}\)

Question 139.
High magnifying power microscopes have oil-immersion objectives
(A) to increase the fringe width
(B) to increase the numerical aperture of the objective
(C) to decrease the wavelength of light
(D) because oil does not damage the observed sample.
Answer:
(B) to increase the numerical aperture of the objective

Question 140.
If the numerical aperture of a microscope is increased, then its
(A) resolving power decreases
(B) limit of resolution decreases
(C) resolving power remains constant
(D) limit of resolution increases
Answer:
(B) limit of resolution decreases

Question 141.
The numerical aperture of the objective of a microscope is 0.12. The limit of resolution, when light of wavelength 6000 A is used to view an object, is
(A) 0.25 × 10^-7 m
(B) 2.5 × 10^-7m
(C) 25 × 10^-7 m
(D) 250 × 10^-7 m.
Answer:
(C) 25 × 10^-7 m

Question 142.
The resolving power of a refracting telescope is increased by
(A) using oil-immersion objective
(B) increasing the diameter D of the objective lens
(C) resorting to short-wavelength radiation
(D) increasing D and using smaller λ.
Answer:
(D) increasing D and using smaller λ.

Question 143.
The resolving power of a telescope of aperture 100 cm, for light of wavelength 5.5 × 10^-7 m, is
(A) 0.149 × 10⁷
(B) 1.49 × 10⁷
(C) 14.9 × 10⁷
(D) 149 × 10⁷
Answer:
(A) 0.149 × 10⁷

Question 144.
The resolving power of a telescope depends upon the
(A) length of the telescope
(B) focal length of the objective
(C) diameter of the objective
(D) focal length of the eyepiece.
Answer:
(C) diameter of the objective

Question 145.
If a is the aperture of a telescope and 2 is the wavelength of light then the resolving power of the telescope is
(A) \(\frac{\lambda}{1.22 a}\)
(B) \(\frac{1.22 a}{\lambda}\)
(C) \(\frac{1.22 \lambda}{a}\)
(D) \(\frac{a}{1.22 \lambda}\)
Answer:
(D) \(\frac{a}{1.22 \lambda}\)

Question 146.
Using a monochromatic light of wavelength 2 in Young’s double-slit experiment, the eleventh dark fringe is obtained on the screen for a phase difference of
(A) \(\frac{11}{2}\) π rad
(B) \(\frac{21}{2}\) π rad
(C) 13 π rad
(D) 21 π rad
Answer:
(D) 21 π rad