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Balbharti Maharashtra State Board 12th Physics Important Questions Chapter 14 Dual Nature of Radiation and Matter Important Questions and Answers.

Maharashtra State Board 12th Physics Important Questions Chapter 14 Dual Nature of Radiation and Matter

Question 1.
Explain Planck’s idea of quantization of energy.
Answer:
Max Planck, in 1900, put forward the idea of quantization of energy to explain the blackbody radiation spectrum. He proposed that atoms behave as tiny oscillators and emit electromagnetic radiation, not continuously but as little packets of energy called quanta. He assumed that the energy associated with a quantum of radiation (now called a photon) is proportional to the frequency v of the oscillator. Thus, E = nhv, where n = 1, 2, 3, 4, … etc., and h is a universal constant, now called Planck’s constant. For n = 1, E = hv. A quantum of radiation is emitted when there is a transition from a higher quantized level of energy of an oscillator to lower quantized level.

[Note : Historically, various terms have been used to denote a particle of light; quantum of electromagnetic radiation ≡ photon ≡ packet of energy ≡ atom of energy ≡ quantum of radiation ≡ bundle of energy. Interaction between two charged particles involves exchange of photons. The photon has zero rest mass, no charge, unit spin and travels in free space at a speed of 2.99792458 × 10⁸ m/s exact by definition. There is no conservation law for photons, i.e., they can be produced / absorbed.]

Question 2.
What was Hertz’s observation regarding emission of electrons from a metal surface?
Answer:
During his experiments on electromagnetic waves in 1887, Heinrich Rudolph Hertz (1857-94), Ger-man physicist, noticed that electric sparks occurred more readily when one of the electrodes of his spark-gap transmitter was exposed to ultraviolet radiation. This discovery was called the Hertz effect and is now known as the photoelectric effect.

Although Hertz did not follow up his discovery, others quickly established that the cause of the sparking ease was due to emission of negatively charged particles from the electrode irradiated. These particles were identified as electrons after the discovery of the electron in 1897.

Question 3.
Draw a neat labelled diagram to illustrate photoelectric effect.
Answer:

[Note : Positive metal ions and atoms are not shown in the figure.]

Question 4.
What were the investigations of Hallwachs and Lenard regarding photoelectric effect?
Answer:
Wilhelm Hallwachs (1859-1922), German physicist, found that a metal plate irradiated with ultraviolet radiation lost its charge more rapidly when the plate is negatively charged than when it is neutral or positive.

Investigations of photoelectric effect by Phillipp Lenard (1862-1947), German physicist, showed that

1. electron emission occurs only with radiations below a critical wavelength, i.e., above a critical frequency.
2. kinetic energy of the emitted electrons increases as wavelength decreases i.e., frequency increases but is independent of the intensity of radiation which determines the rate of emission of electrons (the number of electrons emitted per unit time).

Question 5.
What is a photosensitive surface?
Answer:
The surface which emits electrons when illuminated by electromagnetic radiation of appropriate frequency is called photosensitive surface.
[Note : The material that exhibits photoelectric effect is called photosensitive material.]

Question 6.
Why are alkali metals most suitable as photo-sensitive surfaces?
Answer:
The alkali metals e.g., caesium, potassium and sodium emit photoelectrons even when visible radiation (light) is incident on them. Hence, they are most suitable as photosensitive surfaces.

Question 7.
With a neat diagram, describe the apparatus to study the characteristics of photoelectric effect.
Answer:
Apparatus : A photoelectric cell G consists of the emitting electrode E (emitter) of the material being studied and the collecting electrode C (collector). The electrodes are sealed in an evacuated glass envelope provided with quartz window W that allows the passage of UV radiation and visible light. Monochromatic light of variable frequency from a suitable source S (such as a carbon arc) passes through a pair of polarizers P (permitting a change in the intensity of radiation) and falls on the emitter.

The electric circuit, as shown in below figure, allows the collector potential to be varied from positive through zero to negative with respect to the emitter, and permits the measurement of potential difference and current between the electrodes. When the collector is made negative, the voltmeter is connected in reverse.

[ Note : The radiation coming out of a filter is not truly monochromatic, it lies in the wavelength range between λ and λ + ∆λ that depends on the source and the filter. ]

Question 8.
In the experiment to study photoelectric effect, describe the effects of the frequency and intensity of the incident radiation on the photoelectric current, for a given emitter material and potential difference across the photoelectric cell.
Answer:
The emitter of the photoelectric cell is irradiated with monochromatic light whose frequency and intensity can be varied continuously and measured. Initially, the collector is made positive with respect to the emitter, so that photoelectrons ejected move quickly from the emitter to the collector. About 10 V is sufficient to do this. The photoelectric current as a function of intensity and frequency of incident radiation is studied.

(1) Effect of frequency : Keeping the light intensity and the accelerating potential difference V constant, the frequency of the incident radiation is varied from that of far-UV to red. It is found that for every material (usually, a metal) irradiated there is a limiting frequency below which no photoelectrons are emitted irrespective of the intensity of the radiation. This frequency, v₀, called the threshold frequency or cut-off frequency, is a characteristic of the material irradiated.

The graph of photoelectric current against frequency is shown in below figure; A and B represent two different metals. The photoelectric current is not the same in the two cases, because the intensity of light is different for different frequencies.

(2) Effect of intensity : With an emitter of a given material, the light intensity is varied by keeping the frequency v (≥ v₀) of the light and the accelerating potential difference V constant. It is found that the rate of electron emission, as indicated by the photoelectric current, is proportional to the light intensity. The graph of photoelectric current against light intensity is a straight line through (0, 0), below figure; if we vary either the frequency of the light or the material irradiated, only the slope of the line changes. No electrons are emitted in the absence of incident radiation.

[Note : The dark current, i.e., the current observed in the absence of light, is extremely low. Hence, it is ignored.]

Question 9.
In the experiment to study photoelectric effect, describe the variation of the photoelectric current as a function of the potential difference across the photoelectric cell, for incident radiation of (1) a given frequency above the threshold but different intensities (2) a given intensity but different frequencies above the threshold.
Answer:
(1) The potential difference (p.d.) across the photo-electric cell is varied keeping both the frequency v (≥ threshold frequency v₀) and the intensity of the light constant. Starting with the collector at about 10 V positive, we reduce this potential to zero and then run it negative.

When the p.d. across the tube is 10 V or more, all the emitted electrons are accelerated and travel across the tube, constituting the saturation current for a given light intensity; an increase in the potential of the collector does not cause an increase in current. As the collector potential is reduced from positive values through zero to negative values, the tube current reduces because of the applied retarding potential. In this case, some electrons stop and turn back before they can reach the collector. Eventually the potential difference is large enough to stop the current completely. This is called the stopping potential or cut-off potential VQ. The product of the stopping potential and electronic charge, V₀e, is equal to the maximum kinetic energy that an electron can have at the time of emission.
V₀e = KE_max \(\frac{1}{2}\)v²_max

In above figure, I₁ and I₂ are two intensities of the incident radiation for the same frequency v ( > v₀); I₂ = 2I₀. Doubling the intensity of light doubles the current at each potential, as in I₂, but V₀ is independent of I.

(2) The above experiment is repeated with different light frequencies for a given emitter material and light intensity. It is found that the stopping potential increases linearly with the frequency in below figure. Therefore, when photoejection occurs for frequencies above v₀, the maximum kinetic energy of the photoelectrons increases linearly with the frequency of the radiation.

[Note : It was shown by Hughes that the stopping potential depends linearly on the frequency of incident radiation. Lawrence and Beams established that the time interval between arrival of a photon on a metal surface and emission of an electron is less than 3 × 10^-9 sc.]

Question 10.
What is the effect of the intensity of incident radiation on the stopping potential in photo-electric emission?
Answer:
V₀ is independent of intensity.

Question 11.
In the experiment to study photoelectric effect, discuss the effect and significance of extremely weak radiation of frequency greater than the threshold frequency for the emitter material.
Answer:
The emitter of the photoelectric cell is irradiated with monochromatic light whose frequency is greater than the threshold frequency for the emitter material. The collector is kept at 10 V positive with respect to the emitter, so that photoelectrons ejected move quickly from the emitter to the collector.

The light is made extremely dim (i.e., the intensity is extremely weak). In this case, the number of ‘ photoelectrons emitted per unit time is very small (and special techniques are required to detect them); but, however few, they are emitted almost instantaneously and with the same maximum kinetic energy as for bright light of the same frequency.

According to the wave theory of light, wave trains of pulsating electromagnetic field spread out from the source. Dim light corresponds to waves of small amplitudes and small energy. If dim light spreads over a surface, conservation of energy requires that the electrons must store energy over long periods of time, which can be several hours, before gathering enough energy to become free of the metal. The fact that photoelectrons appear immediately, within about 10^-9 s, can be explained only by assuming that the light energy is not spread over the surface uniformly as required by the wave theory, but falls on the surface in concentrated bundles.

Question 12.
Define (1) threshold frequency (2) threshold wavelength (3) stopping potential.
Answer:
(1) The threshold frequency for a given metal surface is the characteristic minimum frequency of the incident radiation below which no photoelectrons are emitted from that metal surface (whatever may be the intensity of incident light).

(2) The threshold wavelength for a given metal surface is the characteristic maximum wavelength of the incident radiation above which no photoelectrons are emitted from that metal surface (whatever may be the intensity of incident light).

(3) The stopping potential is the value of the retarding potential difference that is just sufficient to stop the most energetic photoelectrons emitted from reaching the collector so that the photoelectric current in a photocell reduces to zero.

[Note : The threshold wavelength λ₀ = c/v₀, where c is the speed of light in free space and v₀ is the threshold frequency for the metal.]

Question 13.
State the characteristics of photoelectric effect.
Answer:
Characteristics of photoelectric effect:
(1) For every metal surface there is a limiting frequency of incident radiation below which no photoelectrons are emitted from that metal surface. This frequency, called the threshold frequency, is characteristic of the metal irradiated.

(2) The time rate of emission of photoelectrons in-creases in direct proportion to the intensity, of incident radiation.

(3) The photoelectrons have different speeds at the time of emission ranging from zero to a certain maximum value, which is characteristic for a given metal for a given frequency of the incident radiation. The maximum kinetic energy of the photoelectrons at the time of emission is independent of the intensity but increases linearly with the frequency of the incident radiation.

(4) For incident radiation of frequency greater than or equal to the threshold frequency for a given metal surface, photoelectric emission from the surface is almost instantaneous, even under extremely weak irradiation.

Question 14.
Can we get photoemission with an intense beam of radio waves ? Is photoemission possible at all frequencies ?
Answer:
The frequency of the incident radiation and not its intensity is the criterion for photoelectric effect. The lowest frequency of electromagnetic waves that can cause photoemission is about 4.6 × 10¹⁴ Hz (for the alkali metal caesium). Since radio waves have frequencies 1 GHz or lower, they cannot cause photoemission.

Only alkali metals are photosensitive to visible light; other metals are photosensitive only to far ultraviolet radiations.

Question 15.
Explain how wave theory of light fails to explain the characteristics of photoelectric effect.
OR
Explain the failure of wave theory of light to account for the observations from experiments on photoelectric effect.
Answer:
According to the wave theory of light, electromagnetic waves carry the energy stored in oscillating electric and magnetic fields. When enough energy is absorbed by an electron in a substance, it should be liberated as a photoelectron. Frequency of light does not come into picture in this case. Hence, there should not be any threshold frequency for emission of electrons. But it is found that there exists threshold frequency and it depends on the metal.

Experimentally, the maximum kinetic energy of photoelectrons increases linearly with the frequency of light. This cannot be accounted by the wave theory of light.

If a source of light is weak or far away from a metal surface, emission of an electron will not be almost instantaneous. The electron may have to wait for several hours/days for absorption of enough energy from the incident light as by the wave theory of light, energy is spread over the wavefront. But experimentally, for an appropriate frequency of incident light, photoelectric effect is almost instantaneous.

Only one observation, photoelectric current ∝ intensity of incident light can be accounted by the wave theory of light.

Question 16.
Give Einstein’s explanation of the photoelectric effect.
Answer:
Max Planck put forward the quantum theory in 1900 to explain blackbody spectrum. In the theory, he proposed that the electromagnetic radiation emitted by the body consists of discrete concentrated bundles of energy, each equal to hv, where h is a universal constant (now called Planck’s constant) and v is the frequency of the radiation.

Einstein put forth (1905) that these energy quanta, called light quanta/later called photons, interact with matter much like a particle. When a photon collides with an electron in an atom, the electron absorbs whole of the photon energy hv in a single collision or nothing. The electron uses this energy (1) to liberate itself from the atom, (2) to overcome the potential energy barrier at the surface thus liberating itself from the metal, and (3) retains the remaining part as its kinetic energy. Different electrons need different energies in the first two processes. There are some electrons which use minimum energy in the two processes, and hence come out of the metal with maximum kinetic energy. The minimum energy required, in the form of electromagnetic radiation, to free an electron from a metal is called the photoelectric work function Φ of that metal. Thus, for the most energetic photoelectrons at the time of emission,
maximum kinetic energy of the electron = photon energy – photoelectric work function
∴ \(\frac{1}{2}\)\(m v_{\max }^{2}\) = hv – Φ ∴ hv = Φ + \(\frac{1}{2}\)\(m v_{\max }^{2}\)
The above equation is called Einstein’s photo-electric equation.

Light interacts with matter as concentrated bundles of energy rather than energy spread over a Huygens type wavefront. Even under weak irradiation, an electron absorbs a photon’s energy in a single collision. But the rate of incident photons in dim light being less, the chances of such absorption diminish and consequently the photoelectric current diminishes. However, a photoelectron is emitted as soon as a photon is absorbed.

[Note : Albert Einstein (1879-1955), German-Swiss- US theoretical physicist, gave his photoelectric equation in 1905. In the period 1912-1916, Robert Andrews Millikan (1868 -1953), US physicist, was the first to obtain the precise experimental data from which the straight-line graphs, like the one shown in Fig. 14.6, were plotted for various metals. Einstein’s theoretically predicted equation-clearly having the right form for a straight-line graph-was thus verified.]

Question 17.
Define photoelectric work function of a metal.
Answer:
The photoelectric work function of a metal is defined as the minimum photon energy that ejects an electron from the metal.
It is equal to hv₀, where h is Planck’s constant and v0 is the threshold frequency for the metal.

Question 18.
Write Einstein’s photoelectric equation and explain its various tends. How does the equation explain the various features of the photoelectric effect?
Answer:
Einstein’s photoelectric equation :
hv = Φ + \(\frac{1}{2}\)\(m v_{\max }^{2}\) ………… (1)
where h ≡ Planck’s constant, v ≡ frequency of the electromagnetic radiation, hv ≡ energy of the photon incident on a metal surface, Φ ≡ photo-electric work function, i.e., the minimum energy of light quantum required to liberate an electron from the metal surface, v_max and – \(\frac{1}{2}\)\(m v_{\max }^{2}\) ≡ the maximum speed and maximum kinetic energy of the photoelectrons at the time of emission. Φ = hv₀, where v₀ is the threshold frequency for the metal.

Explanation of the characteristics of photoelectric effect:
(1) From the above equation we find that for photoejection, hv ≥ Φ. That is, hv_min = hv₀ must be equal to Φ. Hence, photoelectric effect is observed only if hv ≥ hv₀, i.e., v ≥ v₀. This shows the existence of a threshold frequency v₀ for which photoelectrons are just liberated from a metal surface (with zero kinetic energy). Since different metals differ in electronic configuration, the work function hv₀ and, therefore, frequency v₀ are different and characteristic of different metals.

(2) In this particle model of light,’ intensity of incident radiation’ stands for the number of photons incident on a metal per unit surface area per unit time. As the number of photons incident on a metal per unit surface area per unit time increases, there is a greater likelihood of a photon being absorbed by any electron. Therefore, the time rate of photoejection and hence photoelectric current increases linearly with the intensity of the incident radiation (v ≥ v₀).

(3) From Eq. (1), \(\frac{1}{2}\)\(m v_{\max }^{2}\), = hv – Φ= h(v – v₀)
This shows that the maximum kinetic energy in-creases linearly with the frequency v of the incident photon (v ≥ v₀) and does not depend on the time rate at which photons are incident on a metal surface.

(4) As the incident energy is concentrated in the form of a photon, and not spread over a wavefront, it is expected that an electron is emitted from the metal surface as soon as a photon (v ≥ v₀) is absorbed. This is in agreement with the experimental observation.

[ Note : The frequency v that appears in the formula E = hv is the frequency of the oscillating electric field / magnetic field in the electromagnetic wave. ]

Question 19.
Obtain the dimensions of Planck’s constant.
Answer:
The energy of a photon of frequency v is E = hv, where h is the Planck’s, constant.
∴ [h] = \(\frac{[E]}{[v]}=\frac{\mathrm{ML}^{2} \mathrm{~T}^{-2}}{\mathrm{~T}^{-1}}\) = ML²T^-1

Question 20.
Is the kinetic energy of all photoelectrons the same when emitted from a certain metal ? Explain.
Answer:
No. Explanation : Depending upon the position and state of an electron in a metal when it absorbs an incident photon, a photoelectron can have kinetic energy ranging from 0 to a certain maximum value equal to the photon energy minus the work function of the metal. Hence, the emitted photoelectrons have this range of kinetic energies.

Question 21.
In photoelectric effect, what does the stopping potential depend upon ?
Answer:
In photoelectric effect, the stopping potential depends upon the energy of the incident photon and the work function for the metal irradiated (or upon the frequency/wavelength of the incident radiation and the threshold frequency/wavelength for the metal irradiated).

Question 22.
What does the maximum kinetic energy (or the maximum speed) of a photoelectron depend on?
Answer:
The maximum kinetic energy (or the maximum speed) of a photoelectron depends upon the energy of the incident photon and the work function for the metal irradiated (or upon the frequency /wavelength of the incident radiation and the threshold frequency/wavelength for the metal irradiated).

Question 23.
In photoelectric effect, if a graph of stopping potential versus frequency of the incident radiation is plotted, what does the intercept on the frequency axis (v corresponding to V_o = 0) represent?
Answer:
The intercept on the frequency axis (v corresponding to V_o = 0) represents the threshold frequency for the metal.

Question 24.
State the equation that relates the threshold wavelength (λ_o), the wavelength of incident radiation (λ) and the maximum speed of a photo-electron (v_max).
Answer:
\(\frac{h c}{\lambda}=\frac{h c}{\lambda_{0}}+\frac{1}{2} m v_{\max }^{2}\) is the required equation, where h is Planck’s constant, c is the speed of light in vacuum (free space) and m is the mass of the electron.

Question 25.
What is the energy of a photon (quantum of radiation) of frequency 6 × 10¹⁴ Hz?
[h = 6.63 × 10^-34 J∙s]
Answer:
hv = (6.63 × 10^-34)(6 × 10¹⁴)
= 3.978 × 10^-19 J is the energy of the photon.

Question 26.
If the total energy of a radiation of frequency 10¹⁴ Hz is 6.63 J, calculate the number of photons in the radiation.
Answer:
E = nhv, where hv is the energy of a photon in a radiation of frequency v and n is the number of photons in the radiation.
∴ n = \(\frac{E}{h v}=\frac{6.63}{\left(6.63 \times 10^{-34}\right)\left(10^{14}\right)}\) = 10²⁰

Question 27.
If in a photoelectric experiment, the stopping potential is 1.5 volts, what is the maximum kinetic energy of a photoelectron ? [e = 1.6 × 10^-19 C]
Answer:
\(\frac{1}{2}\)\(m v_{\max }^{2}\) = V_se = ( 1.5)(1.6 × 10^-19)
= 2.4 × 10^-19 J is the required kinetic energy.

Question 28.
What is the photoelectric work function for a metal if the threshold wavelength for the metal is 3.315 × 10^-7 m?
[h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s]
Answer:
Photoelectric work function for the metal =

Question 29.
Explain the utilization of energy absorbed by an electron in a metal during its collision with a photon.
Answer:
When a photon collides with an atomic electron inside an emitter metal, the electron absorbs whole of the photon energy in a single shot or nothing. The electron uses the absorbed energy (1) to liberate itself from the atom, (2) to overcome the potential energy barrier at the surface thus liberating itself from the metal, and (3) retains the remaining part as its kinetic energy.

30. Solve the following :
(h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s, e = 1.6 × 10^-19 J, m (electron) = 9.1 × 10^-31 kg)

Question 1.
Find the energy of a photon if
(i) the frequency of radiation is 100 MHz
(ii) the wavelength of radiation is 10000 Å.
Solution:
Data : h = 6.63 × 10^-34 J∙s, v = 100 MHz = 100 × 10⁶ Hz, λ = 10000 Å = 10⁶ m, c = 3 × 10⁸ m/s
(i) The energy of a photon, E = hv
= (6.63 × 10^-34)(100 × 10⁶) = 6.63 × 10^-26 J

(ii) The energy of a photon, E = \(\frac{h c}{\lambda}\)
= \(\frac{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{10^{-6}}\) = 1.989 × 10^-19 J

Question 2.
A monochromatic source emits light of wavelength 6000 Å. If the power of the source is 10 W, find the number of photons emitted by it per second assuming that 1% of electric energy is converted into light.
Solution:
Data : λ = 6000 Å = 6 × 10^-7 m ,h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s, electric energy converted into light per second = \(\frac{1}{100}\) × 10W = 0.1J/s

Question 3.
Radiation of intensity 4 × 10^-5 W/m² is incident uniformly on a metal surface with work function 2.4 eV and area 1 cm². Assume that the radius of a metal atom is 2.4 Å and photoelectrons are ejected only from the surface of the metal. On the basis of the wave theory of light, how long will it take for an electron to be ejected from the metal surface ? (Assume one free electron/metal atom.)
Solution:
Data : Power/area = 4 × 10^-5 W/m², Φ = 2.4 eV = 2.4 × 1.6 × 10^-19 J = 3.84 × 10^-19 J, A = 1 cm² = 10^-4 m², r = 2.4 Å = 2.4 × 10^-10 m
Number of metal atoms on the surface = \(\frac{A}{\pi r^{2}}\)

For a single free electron, radiant energy incident per unit time

Ignoring reflection/scattering of light, time needed to absorb energy equal to 3.84 × 10^-19 J is
\(\frac{3.84 \times 10^{-19} \mathrm{~J}}{7.24 \times 10^{-24} \mathrm{~J} / \mathrm{s}}\) = 5.304 × 10⁴ s = 53040s
= 14 hours 44 minutes.

Question 4.
The energy of a photon is 2 eV. Find its frequency and wavelength.
Solution:
Data : E = 2 eV = 2 × 1.6 × 10^-19 J, c = 3 × 10⁸ m/s, h = 6.63 × 10^-34 J∙s
(i) Frequency,

= 6.217 × 10^-7 = 6.217 × 10^-7 × 10¹⁰ Å
= 6217 Å = 621.7 nm

Question 5.
Find the wave number of a photon having an energy of 2.072 eV. [Given : e, c, h]
Solution:
Data : e = 1.6 × 10^-19 C, c = 3 × 10⁸ m / s, h = 6.63 × 10^-34 J∙s,
E = 2.072 eV = 2.072 × 1.6 × 10^-19 J
E = hv = \(\frac{h c}{\lambda}\)
Wave number, \(\frac{1}{\lambda}=\frac{E}{h c}\)
= \(\frac{2.072 \times 1.6 \times 10^{-19}}{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}\) = 1.666 × 10⁶ m^-1

Question 6.
Calculate the energy of a photon, in joule and eV, in a light of wavelength 5000 Å.
Solution :
Data : λ = 5000 Å = 5000 × 10^-10 m = 5 × 10^-7 m, c = 3 × 10⁸ m/s, h = 6.63 × 10^-34 J∙s
Energy of a photon, E = hv = \(\frac{h c}{\lambda}\)

Question 7.
The photoelectric work function for a metal surface is 2.3 eV. 1f the light of wavelength 6800 Å is incident on the surface of the metal, find the threshold frequency and the incident frequency. Will there be an emission of photoelectrons or not? [Given : c, h]
Solution:
Data: c = 3 × 10⁸ m/s, h = 6.63 × 10^-34 J.s, Φ = 2.3 eV = 2.3 × 1.6 × 10^-19 J, λ = 6800 Å = 6.8 × 10^-7 m
(i) Threshold frequency (v₀) : Φ = hv₀
∴ v₀ = \(\frac{\phi}{h}=\frac{2.3 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}}\) = 5.550 × 10¹⁴ Hz

(ii) Incident frequency (v) : c = vλ
∴ v = \(\frac{c}{\lambda}=\frac{3 \times 10^{8}}{6.8 \times 10^{-7}}\) = 4.412 × 10¹⁴ Hz

(iii) Thus, v <v₀
As the frequency of the incident orange light is less than the threshold frequency there will be no emission of photoelectrons.

Question 8.
If the work function of a metal is 3 eV, calculate the threshold wavelength of that metal. [Given : c, h, 1 eV = 1.6 × 10^-19 J]
Solution:
Data : Φ = 3 eV, c = 3 × 10⁸ m/s, h = 6.63 × 10^-34 J.s, 1 eV = 1.6 × 10^-19 J
∴ Φ = 3 × 1.6 × 10^-19 J

Question 9.
The photoelectric work function of copper is 4.7 eV. What are the threshold frequency and wavelength for photoemission from a copper surface? [1 eV = 1.6 × 10^-19 J]
Solution :
Data : Φ = 4.7 eV, 1 eV = 1.6 × 10^-19 J, c = 3 × 10⁸ m/s, h = 6.63 × 10^-34 J.s

Question 10.
The work functions for potassium and caesium are 2.25 eV and 2.14 eV respectively. Will the photoelectric effect occur for either of these elements
(i) with incident light of wavelength 5650 Å
(ii) with light of wavelength 5180 Å?
Solution:
Data : Φ (potassium) = 2.25 eV,
Φ (caesium) = 2.14 eV, λ₁ = 5650 Å = 5.650 × 10^-7 m, λ₂ = 5180 Å = 5.180 × 10^-7 m, h = 6.63 × 10^-34 J.s, c = 3 × 10⁸ m/s
Φ (potassium) = 2.25 eV
= 2.25 × 1.6 × 10^-19 J =3.6 × 10^-19 J
Φ (caesium) = 2.14 eV = 2.14 × 1.6 × 10^-19 J
= 3.424 × 10^-19 J
Photon energy, E = \(\frac{h c}{\lambda}\)

(i) For λ₁ = 5650 Å
E₁ = \(\frac{h c}{\lambda_{1}}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{5.650 \times 10^{-7}}\)
= 3.52 × 10^-19 J
This is greater than Φ (caesium), but less than Φ (potassium). Hence, photoelectric effect will occur in case of caesium, but not in case of potassium.

(ii) For λ₂ = 5180 Å
E₂ = \(\frac{h c}{\lambda_{2}}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{5.180 \times 10^{-7}}\)
= 3.84 × 10^-19 J
This is greater than 0 for potassium and for caesium. Hence, photoelectric effect will occur in both the cases.

Question 11.
Photoemission just occurs from a lead surface when radiation of wavelength 3000 Å is incident on it. Find the maximum kinetic energy of the photoelectrons when the surface is irradiated by UV radiation of wavelength 2500 Å.
Solution:
Data : λ₀ = 3000 Å = 3 × 10^-7 m, λ = 2500 Å = 2.5 × 10^-7 m, h = 6.63 × 10^-34 J∙s,c = 3 × 10⁸ m/s
According to Einstein’s photoelectric equation, the maximum kinetic energy of the photoelectrons

Question 12.
The photoelectric work function for a metal is 4.2 eV. If the stopping potential is 3 V, find the threshold wavelength and the maximum kinetic energy of emitted electrons. [Given : c, h, e]
Solution :
Data : e = 1.6 × 10^-19 C, c = 3 × 10⁸ m/s, Φ = 4.2 eV = 4.2 × 1.6 × 10^-19 J = 6.72 × 10^-19 J, h = 6.63 × 10^-34 J∙s, v₀ = 3 V
(i) Threshold wavelength.

= 2.960 × 10^-7 m or 2960 Å

(ii) Maximum kinetic energy of emitted electrons,
KE_max = eV₀ = (1.6 × 10^-19)(3) = 4.8 × 10^-19 J

Question 13.
Radiation of wavelength 2 × 10^-7 m is incident on the cathode of a photocell. The current in the photocell is reduced to zero by a stopping potential of 2 V. Find the threshold wavelength for the cathode.
Solution:
Data : λ = 2 × 10^-19 m, V₀ = 2 V, e = 1.6 × 10^-19 C, h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s
According to Einstein’s photoelectric equation,

∴ The threshold wavelength, λ₀ = 2.948 × 10^-7 m

Question 14.
The photoelectric current in a photoelectric cell can be reduced to zero by a stopping potential of 1.8 volts. Monochromatic light of wavelength 2200 Å is incident on the cathode. Find the maximum kinetic energy of the photoelectrons in joule. [Charge on the electron = 1.6 × 10^-19 C]
Solution:
Data: V₀ = 1.8 V, e = 1.6 × 10^-7 C .
The maximum kinetic energy of the photoelectrons,
KE_max = eV₀
= (1.6 × 10^-19) (1.8) = 2.88 × 10^-19 J

Question 15.
The photoelectric work function of a metal is 3 eV. Find the maximum kinetic energy and maximum speed of photoelectrons when radiation of wavelength 4000 Å is incident on the metal surface.
Solution:
Data : Φ = 3 eV = 3 × 1.6 × 10^-19 J, c = 3 × 10⁸ m / s, λ = 4000 Å = 4000 × 10^-10 m, h = 6.63 × 10^-34 J∙s, m = 9.1 × 10^-31 kg
(i) According to Einstein’s photoelectric equation, the maximum kinetic energy of photoelectrons,

= 1.725 × 10^-20 J

= 1.947 × 10⁵ m/s

Question 16.
The work function of tungsten is 4.50 eV. Calculate the speed of the fastest electron ejected from tungsten surface when light whose photon energy is 5.80eV shines on the surface.
Solution:
Data : Φ = 4.50 eV = 4.50 × 1.6 × 10^-19 J = 7.2 × 10^-19 J,
hv = 5.80eV = 5.80 × 1.6 × 10^-19 J = 9.28 × 10^-19 J,
m = 9.1 × 10^-31 kg

This is the speed of the fastest electron ejected.

Question 17.
If the work function for a certain metal is 1.8 eV,
(i) what is the stopping potential for electrons ejected from the metal when light of 4000 Å shines on the metal
(ii) what is the maximum speed of the ejected electrons?
Solution:
Data: Φ = 1.8eV, λ = 4000 Å = 4 × 10^-7 m,
h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s,
m = 9.1 × 10^-31 kg, e = 1.6 × 10^-19 C

This is the maximum speed of the ejected electrons.

Question 18.
The work function of caesium is 2.14 eV. Find
(i) the threshold frequency for caesium
(ii) the wavelength of the incident light if photocurrent is brought to zero by a stopping potential of 0.60 V.
Solution:
Data : Φ = 2.14eV = 2.14 × 1.6 × 10^-19 J = 3.424 × 10^-19 J, V_O = 0.60 V,h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s, e = 1.6 × 10^-19C
(i) Φ = hv₀
∴ The threshold frequency, v₀ = \(\frac{\phi}{h}\)
= \(\frac{3.424 \times 10^{-19}}{6.63 \times 10^{-34}}\) = 5.164 × 10¹⁴ Hz

(ii) V_Oe = 0.6 × 1.6 × 10^-19 = 0.96 × 10^-19 J
\(\frac{h c}{\lambda}\) – Φ = V_Oe ∴ \(\frac{h c}{\lambda}\) = Φ + V_Oe
∴ λ = \(\frac{h c}{\phi+V_{\mathrm{O}} e}\)

This is the required wavelength of the incident light.

Question 19.
The threshold wavelength for photoemission from silver is 3800 Å. Calculate the maximum kinetic energy in eV of photoelectrons emitted when ultraviolet radiation of wavelength 2600 Å falls on it. Also calculate the corresponding stop-ping potential. [1 eV = 1.6 × 10^-19 J]
Solution:
Data : λ₀ = 3800 A = 3.8 × 10^-7 m,
λ = 2600 Å = 2.6 × 10^-7 m, c = 3 × 10⁸ m/s, h = 6.63 × 10^-34 J∙s, 1 eV = 1.6 × 10^-19 J
(i) According to Einstein’s photoelectric equation, the maximum kinetic energy of photoelectrons emitted,

Question 20.
When a surface is irradiated with light of wavelength 4950 Å, a photocurrent appears. The current vanishes if a retarding potential greater than 0.6 V is applied across the phototube. When a different source of light is used, it is found that the critical retarding potential is 1.1 V. Find the work function of the emitting surface and the wavelength of light from the second source.
Solution :
Data : λ₁ = 4.95 × 10^-7 m, V_O1 = 0.6 V, V_O2 = 1.1 V, h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s, e = 1.6 × 10^-19 C

∴ The wavelength of light from the second source,

Question 21.
The work function for the surface of aluminium is 4.2 eV. What potential difference will be required to stop the most energetic electrons emitted by light of wavelength 2000 Å? What should be the wavelength of the incident light for which the stopping potential is zero?
Solution:
Data: Φ = 4.2eV, λ₁ = 2 × 10^-7 m,V_O2 = 0,
h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s, e = 1.6 × 10^-19 C

Question 22.
Radiation of wavelength 3000 Å falls on a metal surface having work function 2.3 eV. Calculate the maximum speed of ejected electrons.
Solution:
Data : λ = 3000 Å = 3 × 10^-7 m, h = 6.63 × 10^-34 J∙s, Φ = 2.3 eV = 2.3 × 1.6 × 10^-19 J, c = 3 × 10⁸ m/s, m = 9.1 × 10^-31 kg

Question 23.
If the total energy of radiation of frequency 10¹⁴ Hz is 6.63 J, calculate the number of photons in the radiation. [Planck’s constant = 6.63 × 10^-34 J∙s]
Solution:
Data: v = 10¹⁴ Hz, h = 6.63 × 10^-34 J∙s
The energy of a photon in a radiation of frequency v is hv.
∴ E = nhv,
so that the number of photons in the radiation is
n = \(\frac{E}{h v}\)
= \(\frac{6.63}{\left(6.63 \times 10^{-34}\right)\left(10^{14}\right)}\) = 10²⁰

Question 31.
Explain wave-particle duality of electromagnetic radiation.
Answer:
A particle is an object with a definite position in space at a given instant and having mass (or momentum), while a wave is a periodically repeated pattern in space and time, generally described by its velocity of propagation, wavelength and amplitude. It is a characteristic of a wave that it is not localized, i.e., it is spread over a region. Thus, these two concepts are contradictory and classical physics treats particles and waves as separate.

Under suitable circumstances, light and all other types of electromagnetic radiation exhibit typical ‘ wave phenomena like polarization, interference and diffraction. On the other hand, radiation exhibits a particle-like nature when it interacts with matter, as in the photoelectric effect and the Compton effect (scattering of X-rays by electrons in matter). It is emitted or absorbed only in terms of quanta of energy. This is the concept of photon : a particle with energy E = hv, where v is the frequency of the radiation and Planck’s constant h connects v and E, respectively the wave and particle aspects.

We see, therefore, that radiation exhibits a dual character. The synthesis of these two contradictory descriptions is called wave-particle duality of electromagnetic radiation.
[Notes : (1) Arthur Holly Compton (1892-1962), US physicist, discovered the effect, now known as the Compton effect, in 1923. (2) Planck’s constant h is also called the elementary quantum of action. Like e and c, it is one of the fundamental constants of nature.]

Question 32.
What is Compton effect? State the formula for the Compton shift and obtain its maximum value.
Answer:
When a high energy X-ray photon or γ-ray photon is scattered by an electron that is (almost) free, the photon loses energy and the electron gains energy shown in figure. This effect was discovered by A.H. Compton in 1923. It is now known as the Compton effect. This effect exhibits particle nature of electro-magnetic radiation.

If X is the wavelength of the incident photon, λ is the wavelength of the scattered photon, θ is the angle through which the photon is scattered, m₀ is the rest mass of an electron, c is the speed of light in free space and h is Plank’s constant, then, the wavelength shift, called the Compton shift is given by

[Note : In a collision between a low energy photon and a high energy electron, scattering results in loss in the energy of the electron and gain in the energy of the photon. This effect is known as the inverse Compton effect.]

Question 33.
What is the implication of Einstein’s interpretation of the photoelectric effect?
OR
What is the significance of the photoelectric effect?
Answer:
The phenomena of interference and polarization exhibit the wave nature of light, and James Clerk Maxwell (1831 – 79), British physicist, had established by 1865 that light is, and propagates as, an electromagnetic wave.

In his interpretation of the photoelectric effect in 1905, Einstein proposed that electromagnetic radiation behaves as a series of small packets or quanta of energy, later called photons. If the frequency of radiation is v, each photon has energy hv and momentum hv/c, where c is the speed of light in free space. Einstein’s photoelectric equation was verified experimentally by Robert Andrews Millikan (1868-1953), US physicist, in 1916.

A very strong additional evidence in support of the quantum theory of radiation was the discovery (in 1923) and explanation of the inelastic scattering of X-rays or γ-rays by electrons in matter by Arthur Holly Compton (1892-1962), US physicist. This inelastic scattering in which a photon transfers part of its energy to an electron is known as the Compton effect. It is similar to the Raman effect. The Compton effect shows particle nature of electro-magnetic radiation.

Since energy and momentum are considered in classical physics as characteristic properties of particles, the photoelectric effect and Compton effect exhibit the particle nature of radiation. But, to describe the photon energy, the quantum theory needs the frequency of the radiation, which is necessarily an attribute associated with a wave in classical physics. Thus, radiation exhibits the dual, seemingly contradictory, characters of particle and wave. In an experiment, we need to use only one of the descriptions, not both at the same time.

[Note : The momentum p and energy £ of a photon are related by the equation, p = E/c, where c is the speed of light in free space.]

Question 34.
Give a brief summary of the quantum theory of radiation.
OR
What is the photon picture of electromagnetic radiation?
Answer:
Quantum theory of radiation (The photon picture of electromagnetic radiation) :
(1) In its interaction with matter, electromagnetic radiation behaves as particles or quanta of energy. A quantum of energy is called a photon.

(2) If the frequency of radiation is v, irrespective of the intensity of radiation, each photon has energy hv and momentum hv/c, where c is the speed of light in free space.

(3) Intensity of radiation corresponds to the number of photons incident per unit time per unit surface area.

(4) Photons are electrically neutral and have zero rest mass.

(5) A photonRarticle collision (such as a photon-electron collision) obeys (he principles of conservation of energy arid momentum. However, in such a collision, an incident photon may be absorbed and/or a new photon may be created, so that the number of photons may not be conserved. For example, a γ-ray photon of energy greater than 1.02 MeV can produce an electron-positron pair in the presence of a heavy nucleus such as lead. In this case, the photon disappears and two particles (electron and positron) are produced. The total energy and momentum are conserved.

[Note: Photons have unit spin. Photons are influenced by gravitational field. A gravitational field can change the path and/or frequency/wavelength of a photon. Even after more than a century of its introduction, the concept of photon is not fully understood.]

Question 35.
What is the momentum of a photon of energy 3 × 10^-19 J? [c = 3 × 10⁸ m/s]
Answer:
Momentum of a photon = \(\frac{E}{c}=\frac{3 \times 10^{-19}}{3 \times 10^{8}}\)
= 10^-27 kg∙m/s .

Question 36.
What is the momentum of a photon of wave length 3.315 × 10^-7 m? [h = 6.63 × 10^-34J∙s]
Answer:
Momentum of a photon \(\frac{h}{\lambda}=\frac{6.63 \times 10^{-34}}{3.315 \times 10^{-7}}\)
= 2 × 10^-27 kg∙m/s .

37. Solve the following :

Question 1.
Find the momentum of a photon if the wavelength of the radiation is 6630 Å.
Solution:
Data : λ = 6.63 × 10^-7 m, h = 6.63 × 10^-34 J∙s
Energy of a photon, E = hv
c = λv
The momentum of a photon,

Question 2.
Find the momentum of a photon of energy 3 eV.
Solution :
Data : e = 1.602 × 10^-19 C,
E = 3 eV = 3 × 1.6 × 10^-19 J, c = 3 × 10⁸ m/s
The momentum of the photon,
p = \(\frac{E}{c}=\frac{3 \times 1.6 \times 10^{-19}}{3 \times 10^{8}}\) = 1.6 × 10^-27 kg∙m/s

Question 3.
Find the energy of a photon with momentum 2 × 10^-27 kg∙m/s.
Solution :
Data : p = 2 × 10^-27 kg∙m/s
The energy of the photon,
E = pc = (2 × 10^-27)(3 × 10⁸) = 6 × 10^-19 J
= \(\frac{6 \times 10^{-19}}{1.6 \times 10^{-19}}\) eV = 3.75 eV

Question 38.
What is a photocell or photoelectric cell?
Describe its construction and working with a neat labelled diagram.
Answer:
A photocell or photoelectric cell is a device in which light energy is converted into electrical energy by photoelectric effect.

Construction : One form of the photoelectric cell shown in figure consists of a highly evacuated or gas-filled glass tube, an emitter (cathode) and a collector (anode). The light enters through a quartz window W and falls on the semicylindrical cathode C coated with a photosensitive metal. The anode is in the form of a straight wire of platinum or nickel, coaxial with the cathode.

If the cell is required to respond to the visible part of the spectrum, the cathode is coated with potassium or rubidium and the quartz window is replaced by glass. If the UV radiation only is to be used, cadmium is used as the sensitive surface. The cell is either highly evacuated (for accurate photometry) or filled with an inert gas at low pressure (if a larger current is desired).

Working : A photocell is connected in series with a battery and a variable resistance. The collector is kept at a positive potential with respect to the emitter. When UV radiation or visible light of frequency greater than the threshold frequency for the emitter surface is incident on the emitter, the ejected photoelectrons are focused by the cylindrical emitter (cathode) towards the collector (anode).

The photoelectrons collected by the collector constitute a photocurrent which may be measured by a microammeter in series with the photocell, as in an exposure meter or lux meter. Otherwise, the photocurrent is used to operate a relay circuit as in an alarm, or to drive the coils of a speaker as in reading an optical sound track in a cine film. The photocurrent becomes zero when the incident light is cut off.

Question 39.
State any four applications of a photoelectric cell.
OR
Explain any two applications of photoelectric effect.
Answer:
Applications of a photoelectric cell :
(1) In an exposure meter used for photography: A photographic film must be exposed to correct amount of light which, for a given film speed and lens aperture, depends on the exposure time. An exposure meter consists of a photocell, battery and microammeter connected in series. When the meter is directed towards an object, light reflected by the object enters the photocell and the photocurrent is directly proportional to the intensity of this light.
Usually, the microammeter scale is calibrated to read the exposure time directly.

(2) As a lux meter : A lux meter is used to measure the illumination and is similar in working to an exposure meter, except that the scale is calibrated to read the illumination in lux.

(3) In a burglar alarm as a ‘normally closed’ light- activated switch : It consists of a photocell, battery, relay system and a small directed light source. The radiation from the source falls on the photocell. If the light beam is interrupted by an intruder, the photoelectric current stops. This activates the relay system which sets off an alarm.

(4) In an optical reader of sound track in a cine film : The sound track of a cine film is recorded on one side of the positive film that is run in a cinema hall. The track consists of a dark wavy patch modulated by the recorded sound. Light from the projector lamp also passes through the sound track and falls on a photocell behind. The photocurrent is proportional to the transmitted light intensity and changes according to the recorded sound wave. The photocurrent is amplified and is used to drive the loudspeaker.

(5) A photocell can be used to switch on or off street lights.

Question 40.
Name any two instruments in which photo-electric effect is used.
Answer:
Exposure meter used in photography and lux meter.

Question 41.
State the de Broglie hypothesis and the de Broglie equation.
Answer:
De Broglie hypothesis : Louis de Broglie (1892-1987), French physicist, proposed (in 1924) that the wave-particle duality may not be unique to light but a universal characteristic of nature, so that a particle of matter in motion also has a wave or periodicity associated with it which becomes evident when the magnitude of Planck’s constant h cannot be ignored.

De Broglie equation : A particle of mass m moving with a speed v should under suitable experimental conditions exhibit the characteristics of a wave of wavelength
λ = \(\frac{h}{m v}=\frac{h}{p}\)
where p = mv = momentum of the particle.
The relation λ = h/p is called the de Broglie equation, and the wavelength λ associated with a particle momentum is called its de Broglie wavelength. The corresponding waves are termed as matter waves or de Broglie waves or Schrodinger waves.

[Note: This hypothesis was revolutionary at that time and accepted by others because Einstein supported it. Erwin Schrodinger (1887-1961), Austrian physicist, formulated a wave equation for matter waves.]

Question 42.
Explain the concept of de Broglie waves or matter waves.
Answer:
According to de Broglie, a particle of mass m moving with a speed v should, under suitable experimental conditions, exhibit the characteristics of a wave of wavelength
λ = \(\frac{h}{m v}=\frac{h}{p}\) … ………. (1)
where p = mv ≡ momentum of the particle and h is Planck’s constant.

This dual character of matter contained in Eq. (1) is usually referred to as the wave nature of matter or matter waves. They are a set of waves that represent the behaviour of particles under appropriate conditions. It does not, however, mean that the particles themselves are oscillating in space.

Interpretation of matter waves by Max Born (1882-1970), German bom British physicist, is that they are waves of probability, since the square of their amplitude at a given point is linked to the likelihood of finding the particle there. Hence, the wavelength λ may be regarded as a measure of the degree to which the energy is localized. If λ is exceedingly small, the energy is very localized and the particle character of the object is dominant. On the other hand, if λ is very large, the energy is distributed over a large volume; under these circumstances, the wave behaviour is dominant.

The wave nature of material particles such as the electron, neutron and helium atom has been established experimentally beyond doubt.

Question 43.
Derive an expression for the de Broglie wavelength associated with an electron accelerated from rest through a potential difference V. Consider the nonrelativistic case.
Answer:
Consider an electron accelerated from rest through a potential difference V. Let v be the final speed of the electron. We consider the nonrelativistic case, v << c, where c is the speed of light in free space. The kinetic energy acquired by the electron is
\(\frac{1}{2}\) mv² = \(\frac{1}{2m}\) (mv)² = eV ………. (1)
where e and m are the electronic charge and mass (nonrelativistic).

Therefore, the electron momentum,
p = mv = \(\sqrt{2 m e V}\) ………… (2)
The de Broglie wavelength associated with the electron is
λ = \(\frac{h}{p}\) ………….. (3)
where h is Planck’s constant.
From Eqs. (2) and (3),
λ = \(\frac{h}{\sqrt{2 m e V}}\)
Equation (4) gives the required expression.
[Note : Substituting the values of h = 6.63 × 10^-34 J-s, m = 9.1 × 10^-31 kg, e = 1.6 × 10^-19 C in Eq. (4), we obtain λ = \(\sqrt{150 / V}\) × 10^-10 m = \(\sqrt{150 / V}\) Å = \(12.25 / \sqrt{V}\) Å, where V is in volt. Therefore, electrons accelerated from rest through 150 volts have a de Broglie wavelength of 1 Å. This corresponds to the X-ray region of the electromagnetic spectrum.]

Question 44.
Derive an expression for the de Broglie wavelength.
Answer:
For the particle-like aspects of electromagnetic radiation, we consider radiation to consist of particles whose motion is governed by the wave propagation properties of certain associated waves.

To determine the wavelength of such waves, consider a beam of electromagnetic radiation of frequency v whose quanta have energy E.
E = hv
where h is the Planck constant.
For a quantum of radiation of momentum p, by Einstein’s theory,
E = pc
where c is the speed of propagation of the radiation in free space.
∴ pc = hv
∴ p\(\frac{c}{v}\) = h
The wavelength X of the associated wave governing the motion of the quanta is given by the relation
λ = c/v.
∴ pλ = h ∴ λ = \(\frac{h}{p}\)
V ’
This is the required expression.

Question 45.
What is the de Broglie wavelength associated with a particle having momentum 10^-26 kg∙m/s? [h = 6.63 × 10^-34 J∙s]
Answer:
The de Broglie wavelength associated with the particle,
λ = \(\frac{h}{p}=\frac{6.63 \times 10^{-34}}{10^{-26}}\) = 6.63 × 10^-8 m

Question 46.
With a neat labelled diagram, describe the Davisson and Germer experiment in support of the concept of matter waves.
Answer:
Davisson and Germer experiment (1927) :
The experimental arrangement, as shown in below figure, consists of an electron gun, a crystal holder and an electron detector enclosed in a vacuum chamber. In the electron gun, electrons emitted by a heated metallic filament (cathode) are accelerated by a potential difference V between the cathode and the anode, and emerge through a small hole in the anode. The electron gun directs a narrow collimated beam of electrons at a nickel crystal. Scattered electrons are detected by a movable detector.

The angle Φ between the incident and scattered beams is the scattering angle. Polar graphs of the number of scattered electrons as a function of angle Φ are plotted for different values of the accelerating voltage.

It is found that the electrons are scattered at a certain angle more than at others. Also, the number of scattered electrons in this direction is maximum for a certain kinetic energy of the incident electrons.

The detector registered a maximum at a scattering angle Φ = 50° for V = 54 V from figure. This electron diffraction can be understood only on the basis of de Broglie’s matter wave model. The de Broglie wavelength of the electrons accelerated from rest through a p.d. of 54 V is λ = \(\sqrt{150 / 54}\) Å = 1.67 Å
The wavelength calculated from the diffraction effect is 1.65 Å, nearly 1.67 Å.
[ Note : Clinton Joseph Davisson (1881 -1958), US physicist. Lester Halbert Germer (1896-1971), US physicist.]

47. Solve the following

Question 1.
(2) Find the momentum of the electron having de Brogue wavelength of 0.5 Å.
Solution:
Data: λ = 0.5Å = 5 × 10^-11 m, h = 6.63 × 10^-34 J∙s
The momentum of the electron, .
p = \(\frac{h}{\lambda}=\frac{6.63 \times 10^{-34}}{5 \times 10^{-11}}\) = 1.326 × 10^-23 kg∙m/s

Question 2.
A cracker of mass M at rest explodes in two parts of masses m₁ and m₂ with non-zero velocities. Find the ratio of the de Broglie wavelengths of the two particles.
Solution:
The cracker has zero momentum before explosion. By the principle of conservation of momentum, after the explosion,

Question 3.
Calculate the de Brogue wavelength of a proton if it is moving with the speed of 2 × 10⁵ m/s. [m^p = 1.673 × 10^-27 kg]
Solution:
Data: m^p = 1673 × 10^-27 kg, v = 2 × 10⁵ m/s, h = 6.63 × 10^-34 J∙s
De Brogue wavelength, λ = \(\frac{h}{p}=\frac{h}{m v}\)
∴ λ = \(\frac{6.63 \times 10^{-34}}{\left(1.673 \times 10^{-27}\right)\left(2 \times 10^{5}\right)}\)
= 1.981 × 10^-12 m

Question 4.
Calculate the de Brogue wavelength of an electron moving with \(\frac{1}{300}\) of the speed of light in vacuum. [Take m (electron) = 9.11 × 10^-28 g]
Solution:

Question 5.
Find the de Broglie wavelength of a dust particle of radius 1 μm and density 2.5 g/cm³ drifting at 2.2 m/s. (Take π = 3.14)
Solution:
Data : r = 1 μm = 10^-6 m, h = 6.63 × 10^-34 J∙s, ρ = 2.5 g/cm³ = 2.5 × 10³ kg/m³, v = 2.2 m/s,

Question 6.
Find the de Broglie wavelength associated with a car (mass = 1000 kg) moving at 20 m/s.
Solution:
Data : m = 1000 kg, v = 20 m/s,h = 6.63 × 10^-34 J∙s
The de Broglie wavelength,
λ = \(\frac{h}{p}=\frac{h}{m v}=\frac{6.63 \times 10^{-34}}{(1000)(20)}\) = 3.315 × 10^-38 m

Question 7.
What is the de Broglie wavelength of an electron accelerated from rest through 25000 volts ?
Solution:
Data: V = 25 × 10³ V, e = 1.6 × 10^-19 C, m_e = 9.11 × 10^-31 kg, h = 6.63 × 10^-34 J∙s
Kinetic energy of the electron,
E = eV
=(1.6 × 10^-19)(25 × 10³ V)
=4 × 10^-15 j
The momentum of the electron,

I Note : Here, the kinetic energy of the electron, 25 keV, is far less than the electron’s rest mass energy (m0c²) which is about 0.51 MeV. Hence, it is a nonrelativistic case.]

Question 8.
Find the de Broglie wavelength of a proton accelerated from rest by a potential difference of 50 V. [m_p = 1.673 × 10^-27 kg]
Solution:
Data : m_p = 1.673 × 10^-27 kg, h = 6.63 × 10^-34 J∙s, KE = 50 eV = 50 × 1.6 × 10^-19 J = 8 × 10^-18 J
The kinetic energy of the proton,

= 4.053 × 10^-12 m = 0.04053 A

Question 9.
A moving electron and a photon have the same de Brogue wavelength. Show that the electron possesses more energy than that carried by the photon.
Solution:
The de Brogue wavelength, λ = \(\frac{h}{p}\)
If an electron and a photon have the same de Brogue wavelength, they must have the same momentum, p.
For the photon, E_p = hv = \(\frac{h c}{\lambda}=\left(\frac{h}{\lambda}\right) c\) = pc … (1)
For the electron, mass m = \(\frac{m_{0}}{\sqrt{1-v^{2} / c^{2}}}\)
where m₀ is the rest mass of the electron and y is its speed.
∴ \(m^{2}\left(\frac{c^{2}-v^{2}}{c^{2}}\right)=m_{0}^{2}\)
∴ m²c⁴ – m²v²c² = \(m_{0}^{2} c^{4}\)
∴ (m²c⁴ = (m₀c²)² + p²c² (where p = mv)
∴ \(E_{\mathrm{e}}^{2}\) = (m₀c²)² + p²c²
where E_e = mc² = m₀c² + K is the total energy of the electron, m₀c² being he rest mass energy and K, the kinetic energy.
∴ E_e = \(\sqrt{\left(m_{0} c^{2}\right)^{2}+p^{2} c^{2}}\) …………. (2)
From Eqs. (1) and (2), we have E_e > E_p.
[Note : The result m = \(\frac{m_{0}}{\sqrt{1-v^{2} / c^{2}}}\) was obtained by Einstein in 1905.]

Multiple Choice Questions

Question 1.
The energy of a photon of wavelength λ is

Answer:
(D) \(\frac{h c}{\lambda}\)

Question 2.
The number of photoelectrons emitted
(A) varies inversely with the frequency of radiation
(B) varies directly with the frequency of radiation
(C) varies inversely with the intensity of radiation
(D) varies directly with the intensity of radiation.
Answer:
(D) varies directly with the intensity of radiation.

Question 3.
A metal emits no electrons if the incident light energy falls below certain threshold. For photo-emission, you would decrease
(A) the intensity of light
(B) the frequency of light
(C) the wavelength of light
(D) the collector potential.
Answer:
(C) the wavelength of light

Question 4.
When light of wavelength 5000 Å falls on a metal surface whose photoelectric work function is 1.9 eV, the kinetic energy of the most energetic photoelectrons is
(A) 0.59 eV
(B) 1.39 eV
(C)1.59eV
(D)2.59eV.
Answer:
(A) 0.59 eV

Question 5.
The threshold wavelengths for photoemission of two metals A and B are 300 nm and 600 nm, respectively. The ratio Φ_A/Φ_B of their photoelectric work functions is
(A) \(\frac{1}{4}\)
(B) \(\frac{1}{2}\)
(C) 2
(D) 4.
Answer:
(C) 2

Question 6.
The photoelectric threshold wavelength of a certain metal is 3315 Å. Its work function is
(A) 6 × 10^-19 J
(B) 7.286 × 10^-19 J
(C) 9 × 10^-19 J
(D) 9.945 × 10^-19 J.
Answer:
(A) 6 × 10^-19 J

Question 7.
The photoelectric work function of a certain metal is 2.5 eV. If the metal is separately irradiated with photons of energy 3 eV and 4.5 eV, the ratio of the respective stopping potentials is
(A) 1
(B) \(\frac{2}{3}\)
(C) \(\frac{1}{4}\)
(D) \(\frac{1}{5}\)
Answer:
(B) \(\frac{2}{3}\)

Question 8.
Sodium and copper have photoelectric work functions 2.3 eV and 4.7 eV, respectively. The ratio λ_0Na/λ_oCu 0f the threshold wavelengths for photoemission is about
(A) 1 : 4
(B) 1 : 2
(C) 2 : 1
(D) 4 : 1.
Answer:
(C) 2 : 1

Question 9.
When light of wavelength A falls on the cathode of a photocell, the kinetic energy of the most energetic photoelectrons emitted is £. If light of wavelength λ/2 is used, what can be said about the new value E’?
(A) E’ = E/2
(B) E’ = E
(C) E’ = 2E
(D) E’ > 2E.
Answer:
(D) E’ > 2E.

Question 10.
Electrons are ejected from a metallic surface when light with a wavelength of 6250 Å is used. If light of wavelength 4500 Å is used instead,
(A) there may not be any photoemission
(B) the photoelectric current will increase
(C) the stopping potential will increase
(D) the stopping potential will decrease.
Answer:
(C) the stopping potential will increase

Question 11.
UV radiation of energy 6.2 eV falls on molybdenum surface whose photoelectric work function is 4.2 eV. The kinetic energy of the fastest photoelectrons is
(A) 3.2 × 10^-19 J
(B) 3.52 × 10^-19 J
(C) 6.72 × 10^-19 J
(D) 9.92 × 10^-19 J.
Answer:
(A) 3.2 × 10^-19 J

Question 12.
In a photocell, increasing the intensity of light increases
(A) the stopping potential
(B) the photoelectric current
(C) the energy of the incident photons
(D) the maximum kinetic energy of the photo-electrons.
Answer:
(B) the photoelectric current

Question 13.
In a photocell, doubling the intensity of the incident light (v > v₀) doubles the
(A) stopping potential
(B) threshold frequency
(C) saturation current
(D) threshold wavelength.
Answer:
(C) saturation current

Question 14.
In the usual notation, the momentum of a photon is
(A) hvc
(B) \(\frac{h v}{c}\)
(C) \(\frac{h \lambda}{c}\)
(D) hλc.
Answer:
(B) \(\frac{h v}{c}\)

Question 15.
The momentum of a photon with λ = 3315 Å is
(A) 2 × 10^-27 kg∙m/s
(B) 5 × 10^-27 kg∙m/s
(C) 2 × 10^-41 kg∙m/s
(D) 5 × 10^-41 kg∙m/s.
Answer:
(A) 2 × 10^-27 kg∙m/s

Question 16.
Let p and E denote the linear momentum and energy of emitted photon, respectively. If the wavelength of incident radiation is increased,
(A) both p and E decrease
(B) p increases and E decreases
(C) p decreases and E increases
(D) both p and E decrease.
Answer:
(C) p decreases and E increases

Question 17.
When radiations of wavelength λ₁ and λ₂ are incident on a certain photosensitive material, the energies of electron ejected are E₁ and E₂ respectively, such that E₁ > E₂. Then, Planck’s constant h is [c = speed of light]

Answer:
(C) \(\frac{\left(E_{1}-E_{2}\right) \lambda_{1} \cdot \lambda_{2}}{c\left(\lambda_{2}-\lambda_{1}\right)}\)

Question 18.
If the frequency of incident light falling on a photosensitive material is doubled, then the kinetic energy of the emitted photoelectron will be
(A) the same as its initial value
(B) two times its initial value
(C) more than two times its initial value
(D) less than two times its initial value.
Answer:
(C) more than two times its initial value

Question 19.
The kinetic energy of emitted photoelectrons is independent of
(A) the frequency of incident radiation
(B) the intensity of incident radiation
(C) the wavelength of incident radiation
(D) the collector plate potential.
Answer:
(B) the intensity of incident radiation

Question 20.
In a photon-electron collision
(A) only total energy is conserved
(B) only total momentum is conserved
(C) both total energy and total momentum are conserved
(D) both total momentum and total energy are not conserved.
Answer:
(C) both total energy and total momentum are conserved

Question 21.
The de Broglie equation for the wavelength of matter waves is

Answer:
(A) λ = \(\frac{h}{p}\)

Question 22.
The momentum associated with a photon is given by
(A) hv
(B) \(\frac{h v}{c}\)
(C) hE
(D) hλ
Answer:
(B) \(\frac{h v}{c}\)

Question 23.
The momentum of a photon of de Broglie wave-length 5000 Å is [h = 6.63 × 10^-34 J∙s]
(A) 1.326 × 10^-28 kg∙m/s
(B) 7.54 × 10^-28 kg∙m/s
(C) 1.326 × 10^-27 kg∙m/s
(D) 7.54 × 10^-27 kg∙m/s.
Answer:
(C) 1.326 × 10^-27 kg∙m/s

Question 24.
The de Broglie wavelength of a 100-m sprinter of mass 66 kg running at a speed of 10 m/s is about
[h = 6.63 × 10^-34 J∙s]
(A) 10^-34 m
(B) 10^-33 m
(C) 10^-32 m
(D) 10^-31 m.
Answer:
(C) 10^-32 m

Question 25.
Which of the following particles moving with the same speed has the longest de Broglie wavelength?
(A) Proton
(B) Neutron
(C) α-particle
(D) β-particle
Answer:
(D) β-particle

Question 26.
If p and E are respectively the momentum and energy of a photon, the speed of the photon is given by
(A) p∙E
(B) E/p
(C) (E/p)²
(D) \(\sqrt{E / p}\)
Answer:
(B) E/p

Question 27.
If the kinetic energy of a free electron is doubled, its de Broglie wavelength
(A) decreases by a factor of 2
(B) increases by a factor of 2
(C) decreases by a factor of \(\sqrt {2}\)
(D) increases by a factor of \(\sqrt {2}\).
Answer:
(C) decreases by a factor of \(\sqrt {2}\)

Question 28.
The de Broglie wavelength of an a-particle accelerated from rest through a potential difference V is λ. In order to have the same de Broglie wavelength, a proton must be accelerated from rest through a potential difference of .
(A) V
(B) 2V
(C) 4V
(D) 8V.
Answer:
(D) 8V.

Question 29.
If a photon has the same wavelength as the de Broglie wavelength of an electron, they have the same
(A) velocity
(B) energy
(C) momentum
(D) angular momentum.
Answer:
(C) momentum

Question 35.
The de Broglie wavelength of a grain of sand, of mass 1 mg, blown by a wind at the speed of 20 m/s is [h = 6.63 × 10^-34 J∙s]
(A) 33.15 × 10^-36m
(B) 33.15 × 10^-33 m
(C) 33.15 × 10^-30 m
(D) 33.15 × 10^-30 m.
Answer:
(C) 33.15 × 10^-30 m

Balbharti Maharashtra State Board 12th Physics Important Questions Chapter 13 AC Circuits Important Questions and Answers.

Maharashtra State Board 12th Physics Important Questions Chapter 13 AC Circuits

Question 1.
Write an expression for an alternating emf that varies sinusoidally with time. Show graphically variation of emf with time.
Answer:
An alternating emf that varies sinusoidally with time is given by e = e₀ sin ωt, where e₀ is the maximum value of the emf, called the peak value, and co is the angular frequency of the emf.
ω = 2πf = \(\frac{2 \pi}{T}\), where f is the frequency of the emf, expressed in Hz, and T is the periodic time of the emf, expressed in second.

Using these data, we can plot e versus t

Question 2.
An alternating emf is given by e = 2.20 sin ωt (in volt). What will be its value at time t = \(\frac{T}{12}\)?
Answer:
e = 220 sin[latex]\frac{2 \pi}{T}\left(\frac{T}{12}\right)[/latex]= 220 sin(\(\frac{\pi}{6}\))
= 220 \(\left(\frac{1}{2}\right)\) = 110 v.

Question 3.
What is the average or mean value of an alternating emf? Obtain the expression for it. (2 marks)
Answer:
The average or mean value of an alternating emf is defined as its average value over half cycle (because the average value over one cycle is zero) and is given as

Question 4.
If the peak value of an alternating emf is 10 V, what is its mean value over half cycle?
Answer:
e_av = 0.6365 e₀ = 0.6365(10) = 6.365 V
Note: In general, when e = e₀ sin ωt, the correspond ing current is j = i sin (ωt + α), where α is the phase difference between emf e and current j. ¿z may be positive or negative or zero.

i₀ is the peak value of the current and i_av (over half cycle)
= \(\frac{2}{\pi}\) i₀ = 0.6365 i₀].

Question 5.
What is the rms value of an alternating current? Find the relation between the rms value and peak value of an alternating current that varies sinusoidaily with time.
Answer:
The root mean square (rms) value of an alternating current i is, by definition,
i_rms = \(\left[\frac{\int_{0}^{I} i^{2} d t}{T}\right]^{\frac{1}{2}}\), where T is the periodic time, i.e., time for one cycle.

[Note: i_rms is also called the effective value or virtual value of the alternating current. In one cycle, the heat produced in a resistor by i = i₀ sin ωt is the same as that produced by a direct current (dc) equal to i_rms]

Question 6.
What is the relation between i,, (over half cycle) and i_rms?
Answer:
i_av (over half cycle) = \(\frac{2}{\pi}\) i₀, and i_rms = \(\frac{i_{0}}{\sqrt{2}}\)
∴ i_av (over half cycle) = \(\left(\frac{2}{\pi}\right)\left(\sqrt{2} i_{\mathrm{rms}}\right)=\frac{2 \sqrt{2}}{\pi} i_{\mathrm{rms}}\)

Question 7.
If i_rms = 3.142 A, what is i_av (over half cycle)?
Answer:
i_av (over half cycle) = \(\frac{2 \sqrt{2}}{\pi}\) i_rms
= \(\frac{(2)(1.414)}{3.142}\)(3.142) = 2.828 A
[Note: i_av (over half cycle) < i_rms]

Question 8.
For e = e₀ sin ωt, what is
(i) e_av (over half cycle)
(ii) r_rms
Answer:
For e = e₀ sin ωt, e_av (over half cycle) = \(\frac{2}{\pi}\) e₀ and e_rms = \(\frac{e_{0}}{\sqrt{2}}\)

9. Solve the following:
Question 1.
An alternating emf is given by e = 220 sin 314.2 t (in volt). Find its
(i) peak value
(ii) rms value
(iii) average value over half cycle
(iv) frequency
(iv) period
(vi) value at \(\frac{T}{4}\) .
Solution:
Data: e = 220 sin314.2t (in volt), t = \(\frac{T}{4}\)
(i) Comparing the given equation with e = e₀ sin ωt, we get, peak value, e₀ = 220V.

(ii) e_rms = e₀/\(\sqrt{2}\) = 155.6 V

(iii) e_av (over half cycle) = \(\frac{2}{\pi}\)e₀ = \(\frac{2(220)}{3.142}\) = 140V

(iv) ω = 2πf= 314.2 ∴ The frequency,
f = \(\frac{\omega}{2 \pi}=\frac{314.2}{2(3.142)}\) = 50 Hz

(v) The period, T = \(=\frac{1}{f}=\frac{1}{50}\) = 0.02 same

(vi) e = 220 sin(\(\frac{2 \pi}{T} \cdot \frac{T}{4}\)) = 220 sin \(\frac{\pi}{2}\) = 220 v

Question 2.
The peak value of AC through a resistor of 10 Ω is 10 mA. What is the voltage across the resistor at time
Solution:

This is the required voltage.

Question 3.
The peak value of AC through a resistor of 100 Ω is 2A If the frequency of AC is 50Hz, find the heat produced in the resistor in one cycle.
Solution:
Data: R = 100 Ω, i₀ = 2A, f = 50 Hz
H = \(\frac{R i_{0}^{2}}{2 f}=\frac{100(2)^{2}}{2(50)}\) = 4 J
This is the required quantity.

Question 10.
What is a phasor?
Answer:
A phasor is a rotating vector that represents a quantity varying sinusoidally with time.

Question 11.
What is a phasor diagram ? Illustrate it with an example.
Answer:
A diagram that represents a phasor is called phasor diagram. Consider an alternating emf e = e₀ sin ωt. The phasor representing it is inclined to the horizontal axis at an angle cot and rotates in an anticlockwise direction as shown in below figure.

The length (OP) of the arrow \(\overrightarrow{\mathrm{OP}}\) represents the peak value (maximum value), e₀, of the emf.
For e = e₀ sin ωt, the projection of \(\overrightarrow{\mathrm{OP}}\) on the y-axis gives the instantaneous value of the emf.

In above figure, OR = e₀ sin ωt.
For e = ₀ sin ωt, the projection of \(\overrightarrow{\mathrm{OP}}\) on the x-axis gives the instantaneous value of the emf.
In above figure, OQ = e₀ sin ωt.
Phasor diagrams are useful in adding harmonically varying quantities.

Question 12.
An alternating emf e = e₀ sin ωt is applied to a resistor of resistance R. Write the expression for the current through the resistor. Show the variation of emf and current with ωt. Draw a phasor diagram to show emf and current.
Answer:
Below figure shows an alternating emf e = e₀ sin ωt applied to a resistor of resistance R.

e₀ is the peak value and co is the angular frequency of the emf. The instantaneous current through the resistor is i = i₀ sin ωt, where i₀ is the peak value of the current.
Here, i and e are always in phase.
For ωt = 0, sin ωt = 0,e = 0,i = 0;
for ωt = π/2, sin ωt = 1, e = e₀, i = i₀;
for ωt = π, sin ωt = 0, e = 0, i = 0;
for ωt = 3π/2, sin ωt = -1, e= – e₀, i= -i₀;
for ωt = 2π, sin ωt = 0, e = 0, i = 0.
Below figure shows variation of e and i with cot.

Below figure shows phasors of e and i

Variation of e and i with time t for a purely resistive AC circuit

Question 13.
If the peak value of the alternating emf applied to a resistor of 100Ω is 100 V, what is the rms current through the resistor?
Answer:
The rms current through the resistor,
i_rms = \(\frac{i_{0}}{\sqrt{2}}=\frac{e_{0}}{R \sqrt{2}}=\frac{100}{100 \sqrt{2}}\) = 0.7071 A

Question 14.
An alternating emf e = e₀ sin ωt is applied to a pure inductor of inductance L. Show variation of the emf and current with ωt.
Answer:
Here, e = e₀ sin ωt and i = i₀ sin (ωt – π/2), where i₀ = e₀/ωL.

[Note : A pure inductor ≡ an ideal inductor.]

Question 15.
Draw a Phasor diagram showing e and i in the case of a purely inductive circuit.
Answer:
In this case, e = e₀ sin ωt and i = i₀ sin (ωt – \(\frac{\pi}{2}\)),
where i₀ = \(\frac{e_{0}}{\omega L}\) and L is the inductance of the inductor. In this case, the current j lags behind the emf e by a phase angle of \(\frac{\pi}{2}\) rad.

Question 16.
Explain the term inductive reactance. Show graphically variation of inductive reactance with the frequency of the applied alternating emf.
Answer:
When an alternating emf e = e₀ sin ωt is applied to a pure inductor of inductance L, the current in the
circuit is i = i₀ sin (ωt – \(\frac{\pi}{2}\)), where i₀ = \(\frac{\pi}{2}\), where i₀ = \(\frac{e_{0}}{\omega L}\) In the case of a pure resistor of resistance R, i = i₀ sin ωt for e = e₀ sin ωt, and i₀ = \(\frac{e_{0}}{R}\)

Comparison of Eqs. i₀ = \(\frac{e_{0}}{\omega L}\) and i₀ = \(\frac{e_{0}}{R}\) shows that ωL is the resistance offered by the inductor to the applied alternating emf. It is called the reactance. It increases linearly with the frequency because ωL = 2πfL. This is illustrated in the following figure. ωL is denoted by X_L.

[Note : Reactance has the same dimensions and unit as resistance.]

Question 17.
What is the reactance of a pure inductor with inductance 10H if the frequency of the applied alternating emf is 50 Hz?
Answer:
The reactance of the inductor,
X_L = ωL = 2πfL = 2(3.142)(50)(10) = 3142 Ω
[Note : In a DC circuit, f = 0 ∴ X_L = 2πfL = 0.]

Question 18.
How does a pure inductor behave when the frequency of the applied alternating emf is
(i) very high
(ii) very low?
Answer:
Inductive reactance = 2πfL.
(i) If the frequency (f) of the applied emf is very high, the inductive reactance (for reasonable value of inductance L) will be very high. Hence, the current through the inductor will be very low (for reasonable value of peak emf). Hence, it will practically block AC.

(ii) For very low f, 2πfL is low and hence the inductor will behave as a good conductor.

Question 19.
The capacitance of an ideal capacitor is 2 μF. What is its reactance if the frequency of the applied alternating emf is 1000 Hz?
Answer:
The reactance of the capacitor =

Question 20.
How does a pure (an ideal) capacitor behave when the frequency of the applied alternating emf is very low?
Answer:
Capacitive reactance = \(\frac{1}{2 \pi f C}\)
If the frequency (f) of the applied emf is very low, the capacitive reactance (for reasonable value of capacitance C) will be very high and hence the current through the circuit will be very low (for reasonable value of peak emf).

Question 21.
What will be the current through an ideal capacitor if it is connected across a 2 V battery ?
Answer:
In a DC circuit, the frequency (f) of the applied emf is zero.
∴ Capacitive reactance, \(\frac{1}{2 \pi f C}\) = ∞
∴ The current through the capacitor will be zero.
(Note : The capacitor blocks DC and acts as an open circuit while it passes AC of high frequency.]

Question 22.
An alternating emf is applied to an LR circuit. Assuming the expression for the current, obtain the expressions for the applied emf and the effective resistance of the circuit. Assume the inductor and resistor to be ideal. Draw the phasor diagram showing the emf and current.
Answer:
Below figure shows a source of alternating emf (e), key K, ideal inductor of inductance L and ideal resistor of resistance R connected to form a closed series circuit. Ignoring the resistance of the source andthekey,wehave,e = Ri + L\(\frac{d i}{d t}\) …………… (1)
where Ri is the potential difference across R and L\(\frac{d i}{d t}\) is the potential difference across L.

where e₀ = Zi₀ is the peak value of the applied emf.
Z = \(\frac{e_{0}}{i_{0}}=\sqrt{R^{2}+\omega^{2} L^{2}}\) is the effective resistance of the circuit. It is called the impedance. Here, the emf leads the current by phase angle Φ.

Question 23.
(a) What is the impedance of an LR circuit if R = 40 Ω and X_L = 30 Ω ?
(b) What is the peak current if the peak emf is 10 V, R = 0 and X_L = 30 Ω?
Ans.
(a) The impedance, Z = \(\sqrt{R^{2}+X_{\mathrm{L}}^{2}}\)
= \(\sqrt{1600+900}=\sqrt{2500}\) = 50 Ω.
(b) i₀ = \(\frac{e_{0}}{X_{\mathrm{L}}}=\frac{10}{30}=\frac{1}{3}\) A = 0.3333 A.

Question 24.
An alternating emf is applied to a CR circuit. Obtain an expression for the phase difference between the emf and the current. Also obtain the expression for the effective resistance of the cir-cuit. Assume the capacitor and resistor to be ideal. Draw the phasor diagram showing the emf and current.
Answer:
Below figure shows a source of alternating emf (e), key K, ideal capacitor of capacitance C and ideal resistor of resistance R to form a closed series circuit. Ignoring the resistance of the source and the key, we have,

where C is the time independent constant of integration which must be zero as j oscillates about zero when e oscillates about zero.
∴ e = R i₀ sin ωt – \(\frac{i_{0}}{\omega C}\) cos ωt
Let Z = \(\sqrt{R^{2}+\frac{1}{\omega^{2} C^{2}}}\), R = Z cos and \(\frac{1}{\omega C}\) = Z sin Φ
∴ e = i₀Z (cos Φ sin ωt – sin Φ cos ωt)
= Zi₀ (sin ωt cos Φ – cos ωt sin Φ)
= Zi₀ sin (ωt – Φ) = e₀ sin (ωt – Φ), where e₀ = Zi₀ is the peak emf. Here, the emf lags behind the current by phase angle Φ.

Question 25.
(a) What is the impedance of a CR circuit if R = 30 Ω and X_C = 40 Ω?
(b) What is the peak current if the peak emf is 10 V, R = 0 and X_C = 40 Ω ?
Ans.
(a) The impedance, Z = \(\sqrt{R^{2}+X_{\mathrm{C}}^{2}}=\sqrt{900+1600}\)
= \(\sqrt{2500}\) = 50
(b) The peak current i₀ = \(\frac{e_{0}}{X_{C}}=\frac{10}{40}\) = 0.25 A.

Question 26.
What is meant by the term impedance? State the formula for it in the case of an LCR series circuit.
Answer:
In an AC circuit containing resistance and inductance and / or capacitance, the effective resistance offered by the circuit to the flow of current is called impedance. It is denoted by Z.
For an LCR series circuit,
Z = \(\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}\) where
ω = 2πf is the angular frequency and f is the frequency of AC.
[Note: Here, in the absence of a capacitor.
Z = \(\sqrt{R^{2}+\omega^{2} L^{2}}\), and in the absence of an inductor,
Z = \(\sqrt{R^{2}+\frac{1}{\omega^{2} C^{2}}}\)].

Question 27.
Draw the impedance triangle for a series LCR AC circuit and write the expressions for the im-pedance and the phase difference between the emf and the current.
Answer:

28. Solve the following :
Question 1.
An alternating emf e = 40 sin (120 πt) (in volt) is applied across a 100 Ω resistor. Calculate the rms current through the resistor and the frequency of the applied emf.
Solution:
Data : e = 40 sin (120 πt) V, R = 100 Ω
The equation of a sinusoidally alternating emf is e = e₀ sin ωt
where e₀ is the peak value of the emf.
Comparing the given expression with this, we get, e₀ = 40 V
∴ The rms current,

Comparing e = 40 sin (120 πt) with
e = e₀ sin ωt, we get,
ω = 2πf= 120 π
∴ f = 60 Hz
This is the frequency of the applied emf.

Question 2.
In problem (1) above, what is the period of the AC?
Solution:
The period of the AC,
T = \(\frac{1}{f}=\frac{1}{60}\) s ≈ 0.01667 s

Question 3.
An alternating emf of frequency 50 Hz is applied a series combination of an inductor (L = 2 H) and a resistor (R = 100 Ω). What is the impedance of the circuit?
Solution:
Data : f = 50 Hz, L = 0.2 H, R = 100 Ω
The inductive reactance, X_L = 2πfL
= 2(3.142)(50)(0.2) = 62.84 Ω
The impedance of the circuit, Z = \(\sqrt{R^{2}+X_{\mathrm{L}}^{2}}\)
= \(\sqrt{(100)^{2}+(62.84)^{2}}=\sqrt{10000+3949}=\sqrt{13949}\)
= 118.1 Ω

Question 4.
An alternating emf is applied to a series combination of an inductor and a resistor (R = 100 Ω). If the impedance of the circuit is 100\(\sqrt {2}\) Ω, what is the phase difference between the emf and the current?
Solution:

This is the phase difference between the emf and the current.

Question 5.
When 100 V dc is applied across a coil, a current of 1 A flows through it. When 100 V ac of frequency 50 Hz is applied to the same coil, only 0.5 A current flows through it. Calculate the resistance, impedance and self-inductance of the coil.
Solution:
Data : V_dc = 100 V, I_dc = 1 A, V_rms = 100 V,
f = 50 Hz, I_rms = 0.5 A
(i) The resistance of the coil,
R = \(\frac{V_{\mathrm{dc}}}{I_{\mathrm{dc}}}=\frac{100}{1}\) = 100 Ω

(ii) The impedance of the coil,
Z = \(\frac{V_{\mathrm{rms}}}{I_{\mathrm{rms}}}=\frac{100}{0.5}\) = 200 Ω
Z² = R² + X²_L

Question 6.
A 20 µF capacitor is connected in series with a 25 Ω resistor and a source of alternating emf, 240 V (peak)/50 Hz. Calculate the capacitive reactance, circuit impedance and the maximum current in the circuit.
Solution:
Data : C = 20 µF = 20 × 10^-6 F, k = 25 Ω, e₀ = 240 V, f = 50 Hz
(i) Capacitive reactance,

Question 7.
A 25 µF capacitor, 0.1 H inductor and 25 Ω resistor are connected in series with an ac source of emf e = 220 sin 314t volt. What is the expression for the instantaneous value of the current?
Solution:
Data : C = 25 µF = 25 × 10^-6 F, L = 0.1 H,
R = 25 Ω, e = 220 sin 314t volt
The equation of a sinusoidally alternating emf is e = e₀ sin ωt, where e₀ is the peak emf. Comparing the given expression with this, we get,
e₀ = 220 V, ω = 314 rad/s
∴ Inductive reactance,
X_L = ωL = 314 × 0.1 = 31.4 Ω and capacitive reactance,
X_C = \(\frac{1}{\omega C}=\frac{1}{314 \times 25 \times 10^{-6}}\) = 127.4 Ω
∴ The reactance of the circuit,
|X_L – X_C| = 96 Ω (capacitive, ∵ X_C > X_L)

∴ Φ = – 75°24′
i. e., the applied emf lags behind the current by 75°24′.
The instantaneous value of the current is i = i₀ sin (ωt + Φ)
∴ i = 1.569 sin (314 f + 75°24′) ampere

Question 8.
An alternating emf of peak value 110 V and frequency 50 Hz is connected across an LCR series circuit with R = 100 Ω, L = 10 mH and C = 25 µF. Calculate the inductive reactance, capacitive reactance and impedance of the circuit.
Solution:
Data : e₀ = 110 V, f = 50 Hz, R = 100 Ω,
L = 10 mH = 10 × 10^-3 H, C = 25 µF = 25 × 10^-6 F
(i) Inductive reactance,
X_L = ωL = 2πfL
= 2 × 3.142 × 50 × 10 × 10^-3 = 3.142 Ω

(ii) Capacitive reactance,

Question 29.
An alternating emf with rms value 100 V is applied to a pure resistor of resistance 100 Ω. What is the power consumed over one cycle ?
Answer:
The power consumed over one cycle = e_rms i_rms
= e_rms \(\left(\frac{e_{\mathrm{rms}}}{R}\right)\) = (100) \(\left(\frac{100}{100}\right)\) = 100 W.

Question 30.
An alternating emf is applied to a pure resistor of 400 Ω. If the power consumed over one cycle is 100 W, what is the rms current through the resistor?
Answer:
P_av = R (i_rms)²

Question 31.
An alternating emf with e_rms = 100 V is applied to a series LR circuit with R = 100 Ω and Z = 200 Ω What is the average power consumed over one cycle?
Answer:
The average power consumed over one cycle

Question 32.
An alternating emf with e_rms = 60 V is applied to a series CR circuit with R = 100 \(\sqrt {3}\) Ω and capacitive reactance 100 V 3Q. What is the average power consumed over one cycle ?
Answer:
The average power consumed over one cycle

Question 33.
State the expression for the average power consumed over one cycle in the case of a series LCR AC circuit. What happens if the circuit is purely
(i) resistive
(ii) inductive
(iii) capacitive?
Answer:
Average power consumed over one cycle in the case of a series LCR AC circuit,

Question 34.
In the case of a series LCR AC circuit, what is the power factor if
(i) the resistance is far greater than the reactance
(ii) the resistance is far less than the reactance?
Answer:
Power factor, cos Φ = \(\frac{R}{\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}}\)
(i) For R >> (X_L – X_C), cos Φ ≅ 1
(ii) For R >> (X_L – X_e), cos Φ ≅ zero.

35. Solve the following.
Question 1.
An alternating emf e = 200 sin ωt (in volt) is connected to a 1000 Ω resistor. Calculate the rms current through the resistor and the average power dissipated in it in one cycle.
Solution:
Data: e = 200 sin ωt V, R = 1000 Ω
The equation of a sinusoidally alternating emf is e = e₀ sin ωt, where e₀ is the peak value of the emf.
Comparing the given expression with this, we get
∴ Peak current, i₀ = \(\frac{e_{0}}{R}=\frac{200}{1000}\) = 0.2 A
∴ rms current, i_rms = \(\frac{i_{0}}{\sqrt{2}}=\frac{0.2}{\sqrt{2}}\) = 0.1414 A
The average power dissipated in the resistor in one cycle,
P_av = e_rms i_rms = \(\frac{e_{0} i_{0}}{2}=\frac{200 \times 0.2}{2}\) = 20 W

Question 2.
A circuit has a resistance and a reactance, each equal to 100 Ω Find its power factor. If the rms value of the applied voltage is 200 V, what is the average power consumed by the circuit?
Solution:
Data : R = 100 Ω, X = 100 Ω, V_rms = 200 V

∴ The average power, P = e_rms i_rms cos Φ
= 200 × 1.415 × 0.7071 = 200 W

Question 3.
A dc ammeter and an ac hot-wire ammeter are connected to a circuit in series. When a direct current is passed through the circuit, the dc ammeter shows 6A. When a pure alternating current is passed, the ac ammeter shows 8 A. What will be the reading of each ammeter if the direct and alternating currents pass simultaneously through the circuit?
Solution:
Data: i_dc = 6 A, i_rms(ac) = 8A
A dc ammeter measures the average value of a current passing through it. Since the average value of an alternating current over one cycle is zero, when the direct and alternating currents are siniultaneously passed, the dc ammeter will read 6 A which is the dc part.

An ac hot-wire ammeter measures the effective value of a current using the heating effect of an electric current. When the direct and alternating currents are simultaneously passed through the ac ammeter, the average power dissipated is
P_av = i²_dcR + i²_rms = i²_eff R
where R is the resistance of the heating element of the ac ammeter.
∴ i_eff = \(\sqrt{i_{\mathrm{dc}}^{2}+i_{\mathrm{rms}}^{2}}\)
= \(\sqrt{(6)^{2}+(8)^{2}}\) = 10 A
Thus, the ac ammeter will read 10 A.

Question 4.
An alternating emf e = 100 sin [2π(1000) t] (in volt) is applied to a series LCR circuit with resistance 300 Ω, inductance 0.1 H and capacitance 1 µF. Find the power factor and the average power consumed over one cycle.
Solution:
Data: e = 100 sin[2π (1000)t] (in volt), R = 300 Ω L = 0.1H, C = 1 µF = 1 × 10^-6 F
Comparing e = e₀ sin 2πft with the given equation,
we get e₀ = 100 V, f = 1000 Hz
∴ X_L = 2πfL = 2(3.142)(1000)(0.1) = 628.4 Ω

= 4.835 W

Question 5.
An ac circuit with a 10 Ω resistor, 0.1 H inductor and 50 µF capacitor is connected across a 200 V/50 Hz supply. Compute
(i) the power factor
(ii) the average power dissipated in the circuit.
Solution:
Data : R = 10 Ω, L = 0.1 H, e_rms = 200 V, C = 50 µF = 50 × 10^-6 F, f = 50 Hz
(i) X_L = ωL = (2πf)L
= 2 × 3.142 × 50 × 0.1
= 31.42 Ω


[Note: An alternating emf is usually specified by giving its rms value.]

Question 36.
How are oscillations produced using an inductor and a capacitor?
Answer:
Consider a charged capacitor of capacitance C, with an initial charge q₀, connected to an ideal inductor of inductance L through a key K. We assume that the circuit does not include any resistance or a source of emf. At first, the energy stored in the electric field in the dielectric medium between the plates of the capacitor is U_E = \(\frac{1}{2} \frac{q_{o}^{2}}{C^{\prime}}\), while the energy stored in the magnetic field in the inductor is zero.

When the key is closed, the capacitor begins to discharge through the inductor and there is a clockwise current in the circuit, as shown in below figure (a). Let q and i are the instantaneous values of charge on the capacitor and current in the circuit, respectively.

As q decreases, i increases : i = – dq/dt. Thus, the energy U_B = \(\frac{1}{2}\) Li² stored in the magnetic field of the inductor increases from zero. Since the circuit is free of resistance, energy is not dissipated in the form of heat, so that the decrease in the energy stored in the capacitor appears as the increase in energy stored in the inductor. As the current reaches its maximum value i(y the capacitor is fully discharged and all the energy is stored in the inductor, from figure (b).

Although q = 0 at this instant, dq/dt is nonzero. The current in the inductor then continues to transfer charge from the top plate of the capacitor to its bottom plate, as in from figure (c). The electric field in the capacitor builds up again, but now in the opposite sense, as energy flows back into it from the inductor. Eventually, all the energy of the magnetic field of the inductor is transferred back into the electric field of the capacitor, which is now fully charged, from figure (d).

The capacitor then begins to discharge with an anticlockwise current until the energy is completely back with the inductor. The magnetic field in the inductor is in the opposite sense and becomes maximum when the current reaches its maximum minimum value – i₀. Subsequently, the current in the inductor charges the capacitor once again until the capacitor is fully charged and back to its original condition.

In the absence of an energy dissipative resistance (ideal condition), this cycle continues indefinitely. When the magnitude of the current is maximum, the energy is stored completely in the magnetic field. When the energy is stored entirely in the electric field, the current is zero. The current varies sinusoidally with time between i₀ and – i₀. The frequency of this electrical oscillation in the LC circuit is determined by the values of L and C.

[Notes : (1) Electrical oscillations in an LC circuit are analogous to the oscillations of an ideal mechanical oscillator. An LC circuit with resistance is analogous to a damped mechanical oscillator, while one with a source of alternating emf is analogous to a forced mechanical oscillator. (2) With suitable choices of L and C, it is possible to obtain frequencies ranging from 10 Hz to 10 GHz. (3) In practice, LC oscillations are damped because an inductor has some resistance (R) and hence Joule heat (izRt) is developed in it. The amplitude of oscillations goes on decreasing with time and becomes zero eventually. Also, part of energy stored in the inductor and capacitor is radiated in the form of electromagnetic waves. Working of radio and TV transmitters is based on such radiation.]

Question 37.
Explain electrical resonance in an LCR series circuit. Deduce the expression for the resonant frequency of the circuit.
Answer:
Suppose a sinusoidally alternating emf e, of peak value e₀ and frequency f, is applied to a circuit containing an inductor of inductance L, a resistor of resistance R and a capacitor of capacitance C, all in series, from figure (a) The inductive reactance, X_L, and the capacitive reactance, X_C, are
X_L = ωL and X_C = \(\frac{1}{\omega C}\)
where ω = 2πf.
The rms values i_rms and e_rms of current and emf are proportional to one another.
i_rms = \(\frac{e_{\mathrm{rms}}}{\mathrm{Z}}\)
where Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\) = the impedance of the circuit.

The impedance Z drops to a minimum at the frequency f_r for which the inductive and capacitive reactances are equal (and opposite, in a phasor diagram); i.e., when

At this frequency, Z = R and the phase angle Φ = 0, i.e., the combination behaves like a pure
resistance, and the current and emf are in phase. If R is small, the loss is small. Then, the current may be very large. At any other frequency, the impedance is greater than R. If a mixture of frequencies is applied to the circuit, the current only builds up to a large value for frequencies near the one to which the circuit is ‘tuned’, as given by Eq. (5). The resonance curve, from figure (b), shows the variation of the rms current with frequency. This is an example of electrical resonance. Equations (3) or (4) give the resonance condition and f_ris called the resonant frequency of the LCR series circuit.

At the resonant frequency, the potential differences across the capacitor and inductor are equal in magnitude but in exact antiphase; the current is in quadrature, i.e., 900 out of phase with them. The energy stored in the electric field of the capacitor changes periodically as the square of the potential difference across it; while the energy stored in the magnetic field of the inductor changes periodically as the square of the current. At moments when the potential difference across the capacitor is a maximum and the current through the inductor zero, there is then a maximum of energy stored in the electric field of the capacitor. At moments the potential difference across the capacitor is zero and the current through the inductor a maximum, there is then a maximum of energy stored in the magnetic field of the inductor.

At resonance, the total energy stored in the L-C system is constant, and is simply passed back and forth between the electric and magnetic fields. When the resonant current is first building up, this energy is drawn from the ac supply. After that, the supply only needs to make up the energy lost as heat in the resistor.

Question 38.
State the characteristics of a series LCR AC resonance circuit.
Answer:
Characteristics of a series LCR AC resonance circuit:

1. Resonance occurs when inductive reactance (X_L = 2πfL) equals capacitive reactance j (X_C = \(\frac{1}{2 \pi f C}\)). Resonant frequency, f_r = \(\frac{1}{2 \pi \sqrt{L C}}\).
2. Impedance is minimum and the circuit is purely resistive.
3. Current is maximum.
4. Frequencies, other than the resonant frequency (f_r) are rejected. Only f_r is accepted. Hence, it is called the acceptor circuit.

Question 39.
In LCR series circuit, what is the condition for current resonance ?
Answer:
In LCR series circuit, the condition for current resonance is ωL = \(\frac{1}{\omega C}\) or f = \(\frac{1}{2 \pi \sqrt{L C}},\), where L is the inductance, C is the capacitance and / is the frequency of the applied alternating emf.

Question 40.
In LCR series circuit, what is the
(i) reactance and
(ii) impedance at current resonance?
Answer:
In LCR series circuit, at current resonance,

1. reactance is zero and
2. impedance equals resistance R.

Question 41.
A series LCR circuit has resistance 5 Ω and reactance, for a certain frequency, is 10\(\sqrt {2}\) Ω, what is the impedance of the circuit?
Answer:
Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}=\sqrt{(5)^{2}+(10 \sqrt{2})^{2}}\)
= \(\sqrt{25+200}=\sqrt{225}\) = 15 Ω is the impedance of the circuit.

Question 42.
In LCR series circuit, what is the
(i) power factor and
(ii) phase difference between the emf and current, at resonance.
Answer:
At resonance,

1. the power factor is 1 and
2. the phase difference between the emf and current is zero.

Question 43.
What is an acceptor circuit ? State its use.
Answer:
An acceptor circuit is a series LCR resonant circuit used in communications and broadcasting to selec-tively pass a current for a signal of only the desired frequency.

The resonance curve of a series LCR resonant circuit with a small resistance exhibits a very sharp peak at a certain frequency called the resonant frequency f_r. For an alternating signal of this frequency, the impedance of the circuit is minimum, equal to R, and the current is maximum. That is, the circuit has a selective property as it prefers to pass a signal of frequency f_r and reject those of other frequencies.

Use : An acceptor circuit is used in a radio or television receiver to accept the signal of a desired broadcasting station or channel from all the signals that arrive concurrently at its antenna. Tuning a receiver means adjusting the acceptor circuit to be resonant at a desired frequency.

Question 44.
Explain electrical resonance in an LC parallel circuit. Deduce the expression for the resonant frequency of the circuit.
Answer:
Consider a capacitor of capacitance C, and an inductor of large self-inductance L and negligible resistance, connected in parallel across a source of sinusoidally alternating emf from below figure. Let the instantaneous value of the applied emf be
e = e₀ sin ωt

Let i_L and i_C be the instantaneous currents through the inductor and capacitor respectively.
As the current in the inductor lags behind the emf in phase by π/2 radian,
i_L = \(\frac{e_{0}}{X_{\mathrm{L}}} \sin \left(\omega t-\frac{\pi}{2}\right)=-\frac{e_{0}}{X_{\mathrm{L}}} \cos \omega t\)
where X_L is the inductive reactance.
As the current in the capacitor leads the emf by a phase angle of π/2 radian,
i_C = \(\frac{e_{0}}{X_{C}}\) sin (ωt + π/2) = \(\frac{e_{0}}{X_{C}}\) cos ωt
where X_C is the capacitive reactance.
The instantaneous current drawn from the source is
i = i_L + i_C = e₀ \(\left(\frac{1}{X_{\mathrm{C}}}-\frac{1}{X_{\mathrm{L}}}\right)\) cos ωt
If X_L = X_C, i = 0. Thus, no current is drawn from the source if X_L = X_C. In such a case, alternating current goes on circulating in the LC loop, though no current is supplied by the source. This condition is called parallel resonance and the frequency of ac at which it occurs is called the resonant frequency (f_r).
The condition for resonance is
X_L = X_C

In practice, every inductor possesses some resistance and hence even at resonance, some current is drawn from the source. Also, the resonant frequency is different from that for zero resistence.

The resonance curve shows the variation of current (i) and impedance with the frequency of the ac supply, from figure (b). At resonance the current supplied by the source is minimum and the impedance of the circuit is maximum.

Question 45.
State the characteristics of a parallel LC AC resonance circuit.
Answer:
Characteristics of a parallel LC AC resonance circuit:

1. Resonance occurs when inductive reactance (X_L = 2πfL) equals capacitive reactance (X_C = \(\frac{1}{2 \pi f C}\))
   Resonant frequency, f_r = \(\frac{1}{2 \pi \sqrt{L C}}\)
2. Impedance is maximum.
3. Current is minimum.
4. The circuit rejects f_r but allows the current to flow for other frequencies. Hence, it is called a rejector circuit.

Question 46.
What is a rejector circuit? State its use.
Answer:
A rejector circuit is a parallel LC resonant circuit used in communications and broadcasting as well as filter circuits to selectively reject a signal of a certain frequency.

The resonance curve of a parallel resonant circuit with a finite resistance of its inductor windings exhibits a sharp minimum at a certain frequency called the resonant frequency f_r. For an alternating signal of this frequency, the impedance of the circuit is maximum and the current is minimum. That is, the circuit has a selective property to reject a signal of frequency f_r while passing those of other frequencies.

Use : A rejector circuit is used at the output stage of a radiowave transmitter.

Question 47.
Distinguish between an acceptor circuit and a rejector circuit. (Any two points)
Answer:

Acceptor circuit	Rejector circuit
1. An acceptor circuit is a 1. series LCR resonant circuit.	1. A rejector circuit is a parallel LC resonant circuit.
2. For such a circuit with a 2. small resistance, the reson­ance curve has a sharp peak at the resonant frequency, i.e., at this frequency, the impedance is minimum so that the current is maxi­mum.	2. With a small resistance of its inductor windings, the res­onance curve has a sharp minimum at the resonant frequency, i.e., at this fre­quency, the impedance is maximum so that the cur­rent is minimum.
3. It selectively passes a signal 3. of frequency equal to the resonant frequency.	3. It selectively rejects a signal of frequency equal to the resonant frequency.

Question 48.
In an LC parallel circuit, under what condition, does the impedance become maximum?
Answer:
In an LC parallel circuit, the Impedance becomes maximum when ωL = \(\frac{1}{\omega C}\) or f = \(\frac{1}{2 \pi \sqrt{L C}^{\prime}}\) where f is the frequency 0f the applied alternating emf, L is the inductance and C is the capacitance.

Question 49.
Explain the terme sharpness of resonance and Q factor (quality factor).
Answer:
In a series LCR Ac circuit, the amplitude of the current, i.e., the peak value of the current, is
i₀ = \(\frac{e_{0}}{\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}}\)
If the angular frequency, n changed. at resonance.
ω_rL = \(\frac{1}{\omega_{\mathrm{r}} C}\) giving ω_r = \(\frac{1}{\sqrt{L C}}\)
For ω different from ω_r, the amplitude of i is less than the maximum value of i₀. which is \(\frac{e_{0}}{R}\).

Contider the value of ω for which i₀ = \(\frac{\left(i_{0}\right)_{\max }}{\sqrt{2}}\)
= \(\frac{e_{0}}{R \sqrt{2}}\) that the power dissipated by the circuit is half the maximum power. This ω is called the half power angular frequency. There are two such values of ω on either side of ω_r as shown in below figure.

circuit. \(\frac{\omega_{\mathrm{r}}}{2 \Delta \omega}\) is a measure of the sharpness of resonance If It is high, resonance is sharp; if it is low, resonance is not sharp.

The sharpness of resonance Is measured by a coefficient called the quality or Q fader of the cicuit.

The Q factor of a series LCR resonant circuit is defined as the ratio of the resonant angular frequency to the diference in two angular frequencies taken on both sides of the angular resonant ‘frequency such that at each angular frequency the current amplitude becomes \(\frac{1}{\sqrt{2}}\) times the value at resonant frequency.
∴ Q = \(\frac{\omega_{\mathrm{r}}}{\omega_{2}-\omega_{1}}=\frac{\omega_{\mathrm{r}}}{2 \Delta \omega}=\frac{\text { resonant frequency }}{\text { bandwidth }}\)

Q-factor is a dimensionless quantity. The larger the Q-factor, the smaller is the bandwidth i.e., the sharper is the peak in the current It means the series resonant circuit is more selective in this case. from figure shows that the lower angular frequency side of the resonance curve is dominated by the capacitive reactance, the higher angular frequency side is dominated by the inductive reactance and resonance occurs ¡n the middle. This follows from the formulae, X_L = ωL and X_C = \(\frac{1}{\omega C}\). The higher the ω, the greater ¡s X_L and smaller is X_C. At ω = ω_r, X_L = X_C.

Question 50.
What Is the natural frequency of LC circuit with inductance 1H and capacitance µF?
Answer:

Question 51.
What is a choke coil? What is it used for? Explain.
Answer:
A choke coil is an inductor of high inductance. It consists of a large number of turns of thick insulated copper wire wound closely over a soft iron laminated cure- Average power consumed by it over one cycle is P_av = r_rms i_rms cos Φ, where the power factor cos Φ = \(\frac{R}{\sqrt{R^{2}+\omega^{2} L^{2}}}\)

For ωL >> R. cos Φ is very low implying power consumption is reduced. The energy loss due to hysteresis in iron core is reduced by using a soft Iron core.

In an AC circuit a choke coil is used instead of a resistor to reduce power consumption In case of a pure resistor P_av is high as it is e_rms i_rms.

Question 52.
What is the approximate value of the power factor of a choke coil with R = 10 Ω and reactance = 100Ω ?
Answer:

53. Solve the following 
Question 1.
A coil of resistance S D and self-inductance 0.2 H is connected in series with a variable capacitor across a 30 V_(rms) 50 Hz supply. At what capacitance will resonance occur? Find the corresponding current.
Solution:
Data: R = 5 Ω. L = 0.2 H, e_rms = 30 V. f = 50 Hz
Let C be the capacitance of the capacitor at resonance.
(i) At resonance,

Question 2.
An ac circuit consists of a resistor of 5 0 and an inductor of 10 mH connected In series with a 50 V
(peak)/50 Hz supply. What capacitance should be connected in series with the circuit to obtain maximum current? What will be the maximum current?
Solution:
Data: R = 50 Ω, L = 10 mH = 10 × 10^-3 H, e₀ = 50 V, f = 50 Hz
(i) Maximum current is obtained at resonance.
The condition for resonance is

(ii) At resonance, Z = R
∴ Maximum current,
i₀ = \(\frac{e_{0}}{Z}=\frac{e_{0}}{R}=\frac{50}{5}\) = 10 A

Question 3.
An LCR series combination has R = 10 Ω, L = 1 mH and C = 2 µF. Determine (i) the resonant frequency (ii) the current in the circuit (iii) voltages across L and C, when an alternating voltage of rms value 10 mV operating at the resonant frequency is applied to the series combination.
Solution:
Data : R = 10 Ω, L = 1 mH = 10^-3 H, C = 2 × 10^-6 F, e_rms = 10 mV = 10^-2 V

Question 4.
In a parallel resonant circuit, the inductance of the coil is 3 mH and resonant frequency is 1000 kHz. What is the capacitance of the capacitor in the circuit?
Solution:
Data : L = 3 mH = 3 × 10^-3 Hz, f_r = 1000 kHz = 1000 × 10³ = 10⁶ Hz

= 8.441 × 10^-12 F or 8.441 pF

Question 5.
An ac circuit consists of an inductor of inductance 125 mH connected in parallel with a capacitor of capacity 50 µF. Determine the resonant frequency.
Solution :
Data : L = 125 mH = 0.125 H, C = 50 µF = 50 × 10^-6 F
Resonant frequency,

= 63.65 Hz

Question 6.
An ac voltage of rms value 1V is applied to a parallel combination of inductor L = 10mH and capacitor C = 4 µF. Calculate the resonant frequency and the current through each branch at resonance.
Solution:
Data : e_rms = 1 V, L = 10 mH = 10^-2H, C = 4 µF = 4 × 10^-6 F
(i) Resonant frequency,

= 795.7 Hz

(ii) At resonance, the currents through the inductor and capacitor are in exact antiphase but equal in magnitude : i_L = i_C.
∴ i_C = \(\frac{e_{\mathrm{rms}}}{X_{\mathrm{C}}}\) = (2πf_rC) e_rms
= (2 × 3.142 × 795.7 × 4 × 10^-6)(1) = 0.02A

Multiple Choice Questions

Question 1.
The motor of an electric fan has a self inductance of 10 H, and is connected to a 50-Hz ac supply in series with a capacitor. If maximum power transfer occurs when X_L = X_C, the capacitance of the capacitor is
(A) 0.5 µF
(B) 1 µF
(C) 10 µF
(D) 100 µF.
Answer:
(B) 1 µF

Question 2.
The reactance of a coil is 157 Ω. On connecting the coil across a source of frequency 100 Hz, the current lags behind the emf by 45°. The inductance of the coil is
(A) 0.25 H
(B) 0.5 H
(C) 4 H
(D) 314 H.
Answer:
(A) 0.25 H

Question 3.
In a series LCR circuit, the power factor at resonance is
(A) zero
(B) \(\frac{1}{2}\)
(C) \(\frac{1}{\sqrt{2}}\)
(D) 1.
Answer:
(D) 1.

Question 4.
The current in an LC circuit at resonance is called
(A) the displacement current
(B) the idle current
(C) the wattless current
(D) the apparent current.
Answer:
(C) the wattless current

Question 5.
In a series LCR circuit at resonance, the applied emf and current are
(A) out of phase
(B) in phase
(C) differ in phase by \(\frac{\pi}{4}\) radian
(D) differ in phase by \(\frac{\pi}{2}\) radian.
Answer:
(B) in phase

Question 6.
In a series LCR circuit, R = 3 Ω, X_L = 8 Ω and X_C = 4 Ω. The impedance of the circuit is
(A) 3 Ω
(B) 7 Ω
(C) 5 Ω
(D) 25 Ω
Answer:
(C) 5 Ω

Question 7.
A sinusoidal emf of peak value 150\(\sqrt {2}\) V is applied to a series LCR circuit in which R = 3 Ω and Z = 5 Ω. The rms current in the circuit is
(A) 30 A
(B) 30\(\sqrt {2}\) A
(C) 50 A
(D) 50\(\sqrt {2}\) A.
Answer:
(A) 30 A

Question 8.
In a series LCR circuit, R = 3 Ω, Z = 5 Ω, i_rms = 40 A and power factor = 0.6. The average power dissipated in the circuit is
(A) 2880 W
(B) 4800 W
(C) 8000 W
(D) 9600 W.
Answer:
(A) 2880 W

Question 9.
A parallel LC resonant circuit is used as
(A) a filter circuit
(B) a tuning circuit in a television receiver
(C) a transformer
(D) a rectifier.
Answer:
(A) a filter circuit

Question 10.
A senes LCR resonant circuit is used as
(A) a potential divider
(B) a tuning circuit in a television receiver
(C) a source of wattless current
(D) a radiowave trasmitter.
Answer:
(B) a tuning circuit in a television receiver

Question 11.
If AC voltage is applied to a pure capacitor. then voltage acrose the capacitor .
(A) leads the current by phase angle (\(\frac{\pi}{2}\)) rad
(B) leads the current by phase angle π rad
(C) lags behind the current by phase angle (\(\frac{\pi}{2}\)) rad
(D) lags behind the current by phase angle π rad.
Answer:
(C) lags behind the current by phase angle (\(\frac{\pi}{2}\)) rad

Question 12.
In a series LCR circuit at resonance, the phase difference between the current and emf of the source is
(A) π rad
(B) \(\frac{\pi}{2}\) rad
(C) \(\frac{\pi}{4}\) rad
(D) zero rad.
Answer:
(D) zero rad.

Question 13.
For e = e₀ sin ωt, (average) over one cycle is

Answer:
(D) \(\frac{2}{\pi} e_{0}\)

Question 14.
For i = i₀ sin ωt. i_rms/i_av is

Answer:
(A) \(\frac{\pi}{2 \sqrt{2}}\)

Question 15.
If i = 10sin(314t) [in ampere). i_av =
(A) 6.365 A
(B) 10/\(\sqrt{2}\) A
(C) 10/π A
(D) 5A.
Answer:
(A) 6.365 A

Question 16.
If e = 10 sin(400t) [in volt]. e_rms =
(A) \(\frac{10}{\pi}\) V
(B) \(\frac{10 \sqrt{2}}{\pi}\) V
(C) 5V
(D) 7.07V
Answer:
(D) 7.07V

Question 17.
In a purely resistive circuit, the heat produced by a sinusoidally varying AC over a complete cycle is given by H =

Answer:
(C) \(R\left(i_{\mathrm{rms}}\right)^{2} \cdot \frac{2 \pi}{\omega}\)

Question 18.
In a purely inductive AC circuit, i₀ =
(A) \(\frac{e_{0}}{L}\)
(B) \(\frac{e_{0}}{\omega L}\)
(C) \(\frac{e_{0}}{f L}\)
(D) ωLe₀.
Answer:
(B) \(\frac{e_{0}}{\omega L}\)

Question 19.
In a purely capacitive AC circuit, i₀ =
(A) e₀/C
(B) ωCe₀
(C) e₀/ωC
(D),fCe₀.
Answer:
(B) ωCe₀

Question 20.
The impedance of a series LCR circuit is
(A) R + (X_L – X_C)
(B) R + (X_C – X_L)
(C) \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\)
(D) \(\sqrt{R^{2}+X_{\mathrm{L}}^{2}-X_{\mathrm{C}}^{2}}\)
Answer:
(C) \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}\)

Question 21.
In a purely inductive circuit, P_av =

Answer:
(C) Zero

Question 22.
In a series LCR AC circuit, power factor is

Answer:
(D) \(\frac{R}{Z}\)

Question 23.
The Q factor of an LCR series resonant circuit is
(A) resonant frequency/bandwidth
(B) bandwidth / resonant frequency
(C) ω_r/(ω₁ + ω₂)
(D) (ω₁ + ω₂)/ ω_r
Answer:
(A) resonant frequency/bandwidth

Question 24.
The power factor for a choke coil is

Answer:
(A) \(\frac{R}{\sqrt{R^{2}+\omega^{2} L^{2}}}\)

Question 25.
The power factor for a purely resistive AC circuit is
(A) 0.5
(B) 1
(C) \(\frac{1}{\pi}\)
(D) \(\frac{\pi}{2}\)
Answer:
(B) 1

Balbharti Maharashtra State Board 12th Physics Important Questions Chapter 12 Electromagnetic Induction Important Questions and Answers.

Maharashtra State Board 12th Physics Important Questions Chapter 12 Electromagnetic Induction

Question 1.
Describe Faraday’s magnet and coil experiment. What conclusion can be drawn from the experiment?
Answer:
Faraday’s magnet and coil experiment:

1. The terminals of a copper coil of several turns are connected to a sensitive galvanometer.
   
2. A bar magnet is moved swiftly towards the coil with its N-pole facing the coil. As long as the magnet is in motion, the galvanometer shows a deflection [from figure (a)].
3. If the magnet is now moved swiftly away from the coil, again the galvanometer shows a deflection, but now in the opposite direction.
4. The galvanometer shows a deflection when the experiment is repeated with the S-pole of the magnet facing the coil [from figure (b)]. However, the effect of bringing the S-pole towards the coil is the same as that of taking the N-pole away from the coil and vice versa.
5. The same results are obtained when the magnet is held still and the coil is moved towards or away from the magnet.

Conclusion : A current is induced in an electric circuit whenever the magnetic flux linked with the circuit keeps on changing as a result of relative motion of a magnet and the circuit.

Question 2.
Describe Faraday’s coil-coil experiment. What conclusion can be drawn from the experiment?
Answer:
Faraday’s coil-coil experiment:
(1) A copper coil P of several turns is connected in series to a rheostat, a tap key and a battery. The terminals of another copper coil Q of several turns are connected to a sensitive galvanometer. The coils are placed close to each other such that when a current is passed through coil P by closing the key K, the magnetic flux through P is linked with coil Q.

(2) On closing the key K, the rise of current in coil P changes the flux linked with the coil Q nearby as shown by a momentary deflection (throw) of the galvanometer G, from below figure. A similar deflection in the same direction is seen if the key closed and either coil is moved swiftly towards the other.

(3) On releasing the tap key, the current in the coil P does not reduce to zero instantaneously. With the decreasing flux through its turns, and a consequent decrease in the flux linked with coil Q, there is an opposite throw of the galvanometer. A similar deflection in the same direction is seen if the key is kept closed and either coil is moved swiftly away from the other.

Conclusion : A current is induced in an electric circuit whenever the magnetic flux linked with the circuit keeps on changing, either as a result of changing current in a nearby circuit or due to relative motion between them.

Question 3.
Will an induced current be always produced in a coil whenever there is a change of magnetic flux linked with it ?
Answer:
Yes, provided the coil is in a closed circuit.

Question 4.
What is the basis of Lena’s law of electromagnetic Induction?
Answer:
Law of conservation of energy is the basis of Lenz’s law of electromagnetic inductIon.

Question 5.
Express Faraday-Lena’s law of electromagnetic induction in an equation form.
Answer:
Suppose dΦ_m Is the change in the magnetic flux through a coil or circuit in time dt. Then, by
Faraday’s second law of electromagnetic induction, the magnitude of the einf Induced is
e ∝ \(\frac{d \Phi_{\mathrm{m}}}{d t}\) or e = k\(\frac{d \Phi_{\mathrm{m}}}{d t}\)
where dΦ_m/dt is the rate of change of magnetic flux
linked with the coil and k is a constant of proportionality. The Sl units of e (the volt) and dΦ_m df (the weber per second) are so selected that the constant of proportionality, k, becomes unity. Combining Faraday’s law and Lents law of electromagnetic induction, the induced emf
e = – \(\frac{d \Phi_{\mathrm{m}}}{d t}\)
where the minus sign is Included to indicate the polarity of the induced emf as given by Lents law. This polarity simply determines the direction of the induced current in a dosed loop. If a coil has N tightly wound loops, the induced emf will be N times greater than for a single loop, so that
e = – N \(\frac{d \Phi_{\mathrm{m}}}{d t}\)
where \(\frac{d \Phi_{\mathrm{m}}}{d t}\) is the rate of change of magnetic flux through one loop.

Question 6.
State the causes of induced current and explain them on the basis of Lena’s law.
Answer:
According to Lena’s law, the direction of the induced emf or current is such as to oppose the change that produces it. The change that induces a current may be
(i) the motion of a conductor in a magnetic field or
(ii) the change of the magnetic flux through a stationary circuit.
In the first case, the direction of induced emf in the moving conductor Is such that the direction of the side-thrust exerted on the conductor by the magnetic field is opposite in direction to its motion. The motion of the conductor is, therefore, opposed.

In the second case, the induced current sets up a magnetic field of its own which within the area bounded by the circuit is (a) opposite to the original magnetic field if this field is increasing, but (b) is in the same direction as the original field, if the field is decreasing. Thus, it is the change in magnetic flux through the circuit (not the flux itself) which is opposed by the induced current.

Question 7.
In one version of Faraday’s coil-coil experiment, the two coils are wound on the same iron ring as shown, where closing and opening the switch induces a current in the other coil. How do the multiple-loop coils and iron ring enhance the observation of induced emf?

Answer:
The magnetic flux through a coil is directly proportional to the number of turns a coil has. Hence, with multiloop coils in Faraday’s coil-coil experiment, the induced emf is directly proportional to N. Also, the permeability of iron being many orders of magnitude greater than air, the magnetic field lines of the primary coil P are confined to the iron ring and almost all the flux is linked with the secondary coil S. Thus, increased flux and better flux linkage enhances the magnitude of the induced emf.

Question 8.
A circular conducting loop in a uniform magnetic field is stretched to an elongated ellipse as shown below. The magnetic field points into the page. Will an emf be induced in the loop? If so, state why and give the direction of the induced current.

Answer:
Looking in the direction of the magnetic field, there will be an induced current in the clockwise sense.

For the same perimeter, the area of a circle is greater than that of an ellipse. Hence, stretching the loop reduces the inward flux through its plane. To oppose this decreasing flux, a current is induced in the clockwise sense so that the field due to the induced current is into the plane of the diagram.

Question 9.
A bar magnet is dropped vertically through a thick copper ring as shown. What is the direction of the force exerted by the coil on the magnet? Explain.

Answer:
The magnetic flux through the loop increases when the magnet approaches the loop, and decreases after the magnet has passed through. The induced current in the loop opposes the cause producing the change in flux which, in this case, is the falling magnet. Therefore, the motion of the magnet’ is opposed, first with a repulsion and then with an attraction. The force, in both cases, is upward in the + z-direction.

The magnetic dipole moment of the falling magnet is directed up. Therefore, looking down the z-axis, the induced current is clockwise when the magnet is approaching the loop, so that the magnetic moment of the loop points down; subsequently, as the magnet recedes, the induced current is anticlockwise.

Question 10.
Briefly explain the jumping ring experiment.
Answer:
Elihu Thompson’s jumping ring experiment is an outstanding demonstration of Faraday’s laws and Lenz’s law of electromagnetic induction. The apparatus consists of a cylindrical laminated iron- cored solenoid. A conducting non-magnetic ring, usually copper or aluminium, is placed over the extended vertical core of the solenoid. When an alternating current is passed through the solenoid, the ring is thrown off high into the air.

Due to ac, the magnetic field of the solenoid changes continuously. This induces eddy current in the ring. By Lenz’s law, the magnetic field produced by the induced eddy current in the ring opposes the changing magnetic field of the solenoid. Consequently, the two magnetic fields repel each other, making the ring jump.

The iron core increases the magnetic field of the solenoid. Often, the ring is cooled with liquid nitrogen. The colder the ring, the less is its resistance and greater the eddy current in it. More current means a greater magnetic field and even higher jumps.

Question 11.
Explain what you understand by magnetic flux.
Answer:
The total number of magnetic lines of force passing normally through a given area in a magnetic field, is called the magnetic flux through that area.

Consider a very small area dA in a uniform magnetic field of induction \(\vec{B}\). The area dA can be represented by a vector \(\overrightarrow{d A}\) perpendicular to it.

[Note : The area vector is perpendicular to the sur-face, so it can point either up and to the right as shown or down and to the left. Although either choice is acceptable, choosing the direction that is closest to the magnetic field is convenient and usually the one we choose.]

Question 12.
How do you find the magnetic flux through a finite area A ?
Answer:
Consider a small area element \(\overrightarrow{d A}\) of a finite area A bounded by contour C, from below figure. Suppose this area is situated in a magnetic field \(\vec{B}\).

In general, the magnetic field may not be uniform over the area A. Then, the magnetic flux through the area element is dΦ_m = \(\vec{B} \cdot \overrightarrow{d A}\) = B (dA) cos θ
where θ is the angle between \(\vec{B}\) and \(\overrightarrow{d A}\), so that the flux through the area A is
Φ_m = \(\int d \Phi_{\mathrm{m}}=\int_{A} \vec{B} \cdot \overrightarrow{d A}=\int_{A}\) B(dA)cos θ
The integration is over the entire area A. \(\vec{B}\) can be taken out of the integral if and. only if \(\vec{B}\) is the same everywhere over A, in which case,
Φ_m = \(\int_{A}\) B (dA) cos θ = B cos θ \(\int_{A}\) dA = BA cos θ
where \(\int_{A}\) dA is just the total area A.

Question 13.
State an expression for the magnetic flux through a loop of finite area A inside a uniform magnetic field \(\vec{B}\). Hence discuss Faraday’s second law, given that the magnetic flux varies with time.
Answer:
Consider a conducting loop of finite area A, situated in a uniform magnetic field \(\vec{B}\). We choose the direction of the area vector \(\vec{A}\) that is closest to the magnetic field. For the area vector in below figure, the fingers of the right hand must be turned in the sense of the arrow on the contour of the loop.

Since \(\vec{B}\) is the same everywhere over A, the flux through the area A is
Φ_m = BA cos θ
where θ is the angle between \(\vec{B}\) and \(\vec{A}\).
Faraday’s discovery was that the rate of change of flux dΦ_m/ dt is related to the work done on taking a unit positive charge around the contour in the reverse direction. This work done is just the induced emf. Accordingly we express Faraday’s second law of electromagnetic induction as
|e| = \(\frac{d \Phi_{\mathrm{m}}}{d t}=\frac{d}{d t}\) (BA cos θ)
If B, A and θ are all constants in time, no emf is induced in the loop. An emf will be induced if at least one of these parameters changes with time. B and A may change in magnitude; the loop may turn, thereby changing θ.

Question 14.
When is the magnetic flux through an area element (i) maximum (ii) zero? Explain.
Answer:
When an area element dA is placed in a magnetic field \(\vec{B}\), the magnetic flux through the element is
dΦ_m = B(dA) cos θ …………. (1)
where 8 is the angle between \(\vec{B}\) and the area vector \(\overrightarrow{d A}\).
(i) The maximum value of cos θ = 1 when θ = 0. Thus, from Eq. (1), the magnetic flux is maximum, dΦ_m = B(dA), when the magnetic induction is in the direction of the area vector.
(ii) The minimum value of cos θ = 0 when θ = 90°. Then, the magnetic flux is minimum, dΦ_m = 0, when the magnetic induction is perpendicular to the area vector.

Question 15.
State the SI units and dimensions of
(i) magnetic induction
(ii) magnetic flux.
Answer:
(i) Magnetic induction, B :
SI unit : the tesla (T) : 1 T = 1 Wb / m²
Dimensions: [B] = [MT^-2I^-1].

(ii) Magnetic flux, Φ_m:
SI unit : the weber (Wb)
Dimensions : [Φ_m] = [B][A]
= [MT^-2I^-1][L²] = [ML²T^-2I^-1]

Question 16.
State the relation between the SI units volt and weber.
Answer:
1 volt = 1 weber per second (1 V = 1 Wb/s).

Question 17.
Explain how Lenz’s law is incorporated into Faraday’s second law of electromagnetic induction by introducing a minus sign.
Answer:
Consider a conducting loop of area A in a uniform external magnetic field \(\vec{B}\) with its plane perpendicular to the field, i.e., its area vector \(\vec{A}\) is parallel to \(\vec{B}\) , from below figure. We choose the x-axis along \(\vec{B}\), so that \(\vec{B}=B \hat{i}\) and \(\vec{A}=A \hat{i}\).

Suppose the magnitude of the magnetic induction increases with time. Then, \(\vec{A}\) remaining constant, the induced emf by Faraday-Lenz’s second law of electromagnetic induction is
e = \(-\frac{d \Phi_{\mathrm{m}}}{d t}=-\frac{d}{d t}(B A)=-A \frac{d B}{d t}\) ………….. (1)
Since we have assumed that B is increasing with time, dB / dt is a positive quantity. Also, A = |\(\vec{A}\)| is positive by definition. Hence, the right hand side of Eq. (1) is a negative quantity.

The right hand rule for area vector fixes the positive sense of circulation around the loop as the clockwise sense. Then, by Lenz’s law the induced current in the loop is in the anticlockwise sense. The sense of the induced emf is the same as the sense of the current it drives. With the clockwise sense fixed as positive, the anticlockwise sense of the induced current is negative. Hence, the sense of e is also negative. That is, the left hand side of Eq. (1) is indeed a negative quantity. Thus, introducing a minus sign in Faraday’s second law incorporates Ienz’s law into Faraday’s law.

18. Solve the following
Question 1.
A coil of effective area 25 m² is placed in a field-free region. Subsequently, a uniform magnetic field that rises uniformly from zero to 1.25 T in 0.15 s is applied perpendicular to the plane of the coil. What is the magnitude of the emf induced in the coil?
Solution:
Data : NA = 25 m², B_f = 1.25 T, B_i = 0, A t = 0.15 s
Initial magnetic flux, Φ_i = 0 (∵ B_i = 0)
Final magnetic flux, Φ_f = NAB_f
e = –\(\frac{d \Phi}{d t}=-\frac{\left(\Phi_{\mathrm{f}}-\Phi_{\mathrm{i}}\right)}{d t}\)

Question 2.
A rectangular coil of length 0.5 m and breadth 0.4 m has resistance of 5 Ω. The coil is placed in a magnetic field of induction 0.05 T and its direction is perpendicular to the plane of the coil. If the magnetic induction is uniformly reduced to zero in 5 milliseconds, find the emf and current induced in the coil.
Solution:
Data : l =0.5 m, b = 0.4 m, R = 5Ω, B = 0.05 T, B_f = 0, dt = 5 × 10^-3 s
Area of the coil, A = lb = 0.5 × 0.4 = 0.2 m²
Initial magnetic flux, Φ_i = AB_i
= 0.02 × 0.05 = 0.01 Wb
Final magnetic flux, Φ_f = 0 (∵ B_f = 0)

Question 3.
A square wire loop with sides 0.5 m is placed with its plane perpendicular to a magnetic field. The resistance of the loop is 5 Ω. Find at what rate the magnetic induction should be changed so that a current of 0.1 A is induced in the loop.
Solution:
Data : l = 0.5 m, R = 5 Ω, I = 0.1 A
A = l² = 0.5 × 0.5 = 0.25 m²
The magnitude of the induced emf,
|e| = \(\frac{d \Phi}{d t}=\frac{d}{d t}\) (BA) = A \(\frac{d B}{d t}\)
since the area (A) of the coil is constant. The induced current, I = \(\frac{|e|}{R}=\frac{A}{R} \frac{d B}{d t}\)
∴ The time rate of change of magnetic induction,
\(\frac{d B}{d t}=\frac{I R}{A}=\frac{0.1 \times 5}{0.25}\) = 2 T/s

Question 4.
The magnetic flux through a loop of resistance 0.1 Ω is varying according to the relation Φ = 6t² + 7t + 1, where Φ is in mihiweber and t is in second. What is the emf induced in the loop at t = 1 s and the magnitude of the current?
Solution:
Data: R = 0.1 Ω, Φ_m = 6t² + 7t + 1 mWb, t = 1 s
(i) The induced emf, |e| = \(\frac{d \Phi_{\mathrm{m}}}{d t}\) = \(\frac{d}{d t}\)(6t² + 7t + 1)
= (12t + 7) mV
= 12(1) + 7 = 19 mV

(ii) The magnitude of the current = \(\frac{|e|}{R}\)
= \(\frac{19 \mathrm{mV}}{0.1 \Omega}\) = 190 mA

Question 5.
A wire 88 cm long is bent into a circular loop and kept with its plane perpendicular to a magnetic field of induction 2.5 Wb/m². Within 0.5 second, the coil is changed to a square and the magnetic induction is increased by 0.5 Wb/m². Calculate the emf induced in the wire.
Solution:
Data: l = 88 cm, B_i = 2.5 Wb/m², B_f = 3 Wb/m², ∆t = 0.5 s
For the circular loop, l = 2πr
∴ r = \(\frac{l}{2 \pi}=\frac{88}{2 \times(22 / 7)}\) = 14 cm = 0.14 m
Area of the circular loop, A_i = πr²
= \(\frac{22}{7}\) (0.14)² = 0.0616 m²
Initial magnetic flux, Φ_i = A_iB_i
= 0.0616 × 2.5 = 0.154 Wb
For the square loop, length of each side
= \(\frac{88}{4}\) cm = 22 cm = 0.22 m 4
Area of the square loop, A_f = (0.22)²
= 0.0484 m²
∴ Final magnetic flux, Φ_f = A_fB_f
= 0.0484 × 3 = 0.1452 Wb
Induced emf, e = – \(\frac{\Phi_{\mathrm{f}}-\Phi_{1}}{\Delta t}=\frac{\Phi_{1}-\Phi_{\mathrm{f}}}{\Delta t}\)
∴ e = \(\frac{0.154-0.1452}{0.5}\) = 8.8 × 10^-3 × 2
= 1.76 × 10^-2 V

Question 6.
A 1000 turn, 20 cm diameter coil is rotated in the Earth’s magnetic field of strength 5 × 10^-5 T. The plane of the coil was initially perpendicular t0 the Earth’s field and is rotated to be parallel to the field in 10 ms? Find the average emf induced.
Solution:
Data: N = 1000, d = 0.2 m, B = 5 × 10^-5 T,
∆t = 10 ms = 10^-2 s
Radius of coil, r = d/2 = 10^-1 m
Induced emf, e = -N \(\frac{\Delta \Phi_{\mathrm{m}}}{\Delta t}=-N \frac{\Phi_{\mathrm{f}}-\Phi_{1}}{\Delta t}\)
Initial area, A_i = πr² and initial flux,
NΦ_i = NBA_i NB (πr²)
Final flux, Φ_f = 0, since the plane of the coil is parallel to the field lines.

Question 7.
A television loop antenna has diameter of 11 cm. The magnetic field of the TV signal is uniform, normal to the plane of the loop and changing at the rate of 0.16 T/s. What is the magnitude of the emf induced in the antenna?
Solution:

Question 8.
The magnetic field through a wire loop, of radius 12 cm and resistance 8.5 Ω, changes with time as shown in the graph below. The magnetic field is uniform and perpendicular to the plane of the loop. Calculate the emf induced in the loop as a function of time. Hence, find the induced emf in the time interval (a) t = 0 to t = 2 s (b) t = 2 s to t = 4s (c) t = 4s to t = 6s.

Solution :
Data : r = 0.12 m, R = 8.5 Ω

This is the emf induced in the loop as a function of time.
\(\frac{d B}{d t}\) is the slope of the B-t graph

Question 19.
What is motional emf?
Answer:
An emf induced in a conductor or circuit moving in a magnetic field is called motional emf.

Question 20.
Determine the motional emf induced in a straight conductor moving in a uniform magnetic field with constant velocity.
Answer:
Consider a straight wire AB resting on a pair of conducting rails separated by a distance l lying wholly in a plane perpendicular to a uniform magnetic field \(\vec{B}\). \(\vec{B}\) points into the page and the rails are stationary relative to the field and are connected to a stationary resistor R.

Suppose an external agent moves the rod to the right with a constant speed v, perpendicular to its length and to \(\vec{B}\). As the rod moves through a distance dx = vdt in time dt, the area of the loop ABCD increases by dA = ldx = lv dt.

Therefore, in time dt, the increase in the magnetic flux through the loop,
dΦ_m = BdA = Blvdt
By Faraday’s law of electromagnetic induction, the magnitude of the induced emf
e = \(\frac{d \Phi_{\mathrm{m}}}{d t}=\frac{B l v d t}{d t}\) = Blv

Question 21.
Determine the motional emf induced in a straight conductor moving in a uniform magnetic field with constant velocity on the basis of Lorentz force.
Answer:
Consider a straight rod or wire PQ of length l, lying wholly in a plane perpendicular to a uniform magnetic field of induction B , as shown in below figure; \(\vec{B}\) points into the page.

Suppose an external agent moves the wire to the right with a constant velocity \(\vec{v}\) perpendicular to its length and to \(\vec{B}\). The free electrons in the wire experience a Lorentz force \(\vec{F}\) ( = q\(\vec{v}\) × \(\vec{B}\)).

According to the right-hand rule for cross products, the Lorentz force on negatively charged electrons is downward. The Lorentz force \(\vec{F}\) moves the free electrons in the wire from P to Q so that P becomes positive with respect to Q. Thus, there will be a separation of the charges to the two ends of the wire until an electric field builds up to oppose further motion of the charges.

In moving the electrons a distance l along the wire, the work done by the Lorentz force is
W = Fl = (qvB sin θ) l = qvBl
since the angle between \(\vec{v}\) and \(\vec{B}\), θ = 90°. Since electrical work done per unit charge is emf, the induced emf in the wire is
e = \(\frac{W}{q}\) = vB l
Alternatively, the electric field due to the separation of charges is \(\vec{F} / q=\vec{v} \times \vec{B}\). Since \(\vec{v}\) is perpendicular to B, the magnitude of the field = vB.
Electric field = \(\frac{\text { p.d. }(e) \text { between } \mathrm{P} \text { and } \mathrm{Q}}{\text { distance } \mathrm{PQ}(l)}\)
Therefore, the p.d. or emf induced in the wire PQ is e = v B l

Question 22.
Determine the motional emf induced in a straight conductor rotating in a uniform magnetic field with constant angular velocity.
Answer:
Suppose a rod of length l is rotated anticlockwise, around an axis through one end and perpendicular to its length, in a plane perpendicular to a uniform magnetic field of induction \(\vec{B}\), as shown in below figure; \(\vec{B}\) points into the page. Let the constant angular speed of the rod be ω.

Consider an infinitesimal length element dr at a distance r from the rotation axis. In one rotation, the area traced by the element is dA = 2πrdr. Therefore, the time rate at which the element traces out the area is
\(\frac{d A}{d t}\) = frequency of rotation × dA = f dA
where f = \(\frac{\omega}{2 \pi}\) is the frequency of rotation.
∴ \(\frac{d A}{d t}=\frac{\omega}{2 \pi}\) (2πrdr) = ωr dr
Therefore, the magnitude of the induced emf in the element is
|de| = \(\frac{d \Phi_{\mathrm{m}}}{d t}=B \frac{d A}{d t}\) = B ωr dr
Since the emfs in all the elements of the rod will be in series, the total emf induced across the ends of the rotating rod is
|e| = \(\int d e=\int_{0}^{l} B \omega r d r=B \omega \int_{0}^{l} r d r=B \omega \frac{l^{2}}{2}\)
For anticlockwise rotation in B pointing into the page, the pivot point O\(\vec{B}\) is at a higher potential.

[Note : To understand the polarity of the emf across the ends of the rod, imagine that the rod slides along a wire that forms a circular arc MPN of radius /, as shown below. Assume that the resistor R furnishes all of the resistance in the closed loop. As 9 increases, so does the inward flux through the loop due to \(\vec{B}\).

To counteract this increase, the magnetic field due to the induced current must be directed out of the page in the region enclosed by the loop. Therefore, the current in the loop POMP circulates anticlockwise with the motional emf directed from P to O.]

23. Solve the following
Question 1.
A straight metal wire slides to the right at a constant 5 m/s along a pair of parallel metallic rails 25 cm apart. A 10 Ω resistor connects the rails on the left end. The entire setup lies wholly inside a uniform magnetic field of strength 0.5 T, directed into the page. Find the magnitude and direction of the induced current in the circuit.
Solution:
Data : v = 5 m/s, l = 0.25 m, R = 10 Ω, B = 0.5T
The induced current,
i = \(\frac{e}{R}=\frac{B l v}{R}=\frac{(0.5)(0.25)(5)}{10}\) = 0.0625 A
Since the magnetic flux into the page through the | closed conducting loop increases, the induced current in the loop must be anticlockwise. Alternatively, Fleming’s right hand rule gives the direction of induced current in the moving wire from bottom to top.

Question 2.
A straight conductor (rod) of length 0.3 m is rotated about one end at a constant 6280 rad/s in a plane normal to a uniform magnetic field of induction 5 × 10^-5 T. Calculate the emf induced between its ends.
Solution:
Data : l = 0.3 m, ω = 6280 rad/s, B = 5 × 10^-5 T In one rotation, the rod traces out a circle of radius l, i.e., an area, A = πl². Therefore, the time rate at which the rod traces out the area is

Question 3.
A metal rod 1/\(\sqrt{\pi}\) m long rotates about one of its ends in a plane perpendicular to a magnetic field of induction 4 × 10^-3 T. Calculate the number of revolutions made by the rod per second if the emf induced between the ends of the rod is 16 m V.
Solution :
Data : r = l = \(\frac{1}{\sqrt{\pi}}\) m, B = 4 × 10^-3 T, |e| = 16 mV = 16 × 10^-3 V
In one rotation, the rod traces out a circle of radius Z, i.e., an area, A = πl²
Therefore, the time rate at which the rod traces out the area is

Question 4.
A cycle wheel with 10 spokes, each of length 0. 5 m, is moved at a speed of 18 km/h in a plane normal to the Earth’s magnetic induction of 3.6 × 10^-5 T. Calculate the emf induced between
(i) the axle and the rim of the cycle wheel
(ii) ends of a single spoke and ten spokes.
Solution:
Data : r = l = 0.5 m, v = 18 km/h = \(\frac{18000}{3600}\) = 5 m/s,

Since the spokes have common ends (the axle and wheel rim), they are connected in parallel. Hence,
the emf induced between the end of a single spoke and the other common end of ten spokes is also 4.5 × 10^-5 V.

Since the total emf of this parallel combination of identical emfs e is equal to a single emf e, the emf induced between the axle and wheel rim is equal to 4.5 × 10^-5 V.

Question 24.
Briefly describe with necessary diagrams the experimental setup to investigate the phenomenon of electromagnetic induction for a magnet swinging through a coil.
Answer:
Apparatus: A permanent magnet is mounted at the centre of the arc of a semicircular aluminium frame of radius 50 cm. The whole frame is pivoted at its centre and can oscillate freely in its plane, from figure (a). Movable weights m₁ and m₂ on the radial arms of the frame can be symmetrically positioned to adjust the period of oscillation from about 1.5s to 3s. The magnet can freely pass through a copper coil of about 10000 turns.

When the magnet swings through and out of the coil, the magnetic flux through the coil changes, inducing an emf. The amplitude of the swing can be read from the graduations on the arc. Since the induced emf will be small, it may be measured by connecting the terminals of the coils to a CRO (cathode-ray oscilloscope, or they may be connected to a 100 pF capacitor through a diode, from figure (b), and the voltage across the capacitor is measured. The resistor in series with the diode helps to adjust the capacitor charging time ( = RC).

[Note : Real-time graphs can be captured using a datalogger connected to a computer. The datalogger uses rotary motion, voltage and magnetic field sensors to measure the angle, the induced voltage and the magnetic flux, respectively.]

Question 25.
In the experiment to investigate the phenomenon of electromagnetic induction for a magnet swinging through a coil, relate the graphical representations (flux-time and voltage-time) with the motion of the magnet.
Answer:
In the demonstration of a magnet swinging through a coil, a voltage is induced in the coil as the magnet swings through it. For the discussion, we assume the length of the magnet to be smaller (about half) than the length of the coil and the North pole of the magnet swings into the coil from the left. (The polarity of the induced voltage pulse depends on the polarity of the magnet.)

We take the magnetic flux linked with the coil to be nearly zero when the magnet is high up away from the coil. As the magnet moves through it the coil and recedes, the magnetic field through the coil increases to its maximum and then decreases. There is a substantial magnetic field at the coil only when it is very near the magnet. Moreover, the speed of the magnet is maximum when it is at the centre of the coil, since it is then at the mean position of its oscillation. Thus the magnetic field changes quite slowly when the magnet is far away and rapidly as it approaches the coil, from figure (a).

The flux through the coil increases as the north pole approaches the left end of the coil, and reaches a maximum when the magnet is exactly midway in the coil, as shown by the portion be in from figure (a). By Lenz’s law, the induced emf will produce a leftward flux that will seek to oppose the increasing magnetic flux of the magnet through the coil.

The interval cd, when the flux is maximum but remains constant and induced emf is zero, corresponds to the situation where the magnet is wholly inside the coil.

Once the magnet swings past the centre of the coil, the flux through the coil starts to decrease-the interval de. To reinforce the decreasing flux of the magnet through the coil, a rightward flux is now induced, thereby flipping the polarity of the induced emf.

If we use a coil that is shorter than the magnet, the time interval cd for which the induced emf remains zero would have been shorter. The times f₁ and f₂ in from figure (a) are the points of inflection of the curve, and in from figure (b) are obviously the minimum and maximum of the induced emf, respectively. The sequence of two pulses, one negative and one positive, occurs during just half a cycle. On the return swing of the magnet, they are repeated in the same order.

Question 26.
In the experiment to investigate the phenomenon of electromagnetic induction for a magnet swinging through a coil, show that the peak induced emf is directly proportional to the speed of the magnet (or show that the peak induced emf is directly proportional to the angular amplitude and inversely proportional to the time period).
Answer:
In the experiment, a magnet is swung through a coil in a radius R. The angular position θ of the magnet is measured from the vertical, the mean position of the swing. The angular amplitude is θ₀.

The kinetic energy of the system is \(\frac{1}{2}\) Iω² and the potential energy (relative to the lowest position of the magnet) is MgR(1 – cos θ), where M is mass of the system. Conservation of energy gives, for small θ,

as required. The rate of change of flux through the coil is essentially proportional to the velocity of the magnet as it passes through the coil. By choosing different amplitudes of oscillation of the magnet, we can alter this velocity.

Question 27.
What is an ac generator? State the principle of an ac generator.
Answer:
An electric generator or dynamo converts mechanical energy into electric energy, just the opposite of what an electric motor does.

Principle : An AC generator works on electro-magnetic induction : When a coil of wire rotates between two poles of a permanent magnet such that the magnetic flux through the coil changes periodically with time due to a change in the angle between the area vector and the magnetic field, an alternating emf is induced in the coil causing a current to pass when the circuit is closed.

Question 28.
Briefly describe the construction of a simple ac generator. Obtain an expression for the emf induced in a coil rotating with a uniform angular velocity in a uniform magnetic field. Show graphically the variation of the emf with time (t). OR Describe the construction of a simple ac generator and explain its working.
Answer:
Construction : A simplified diagram of an ac generator is shown in below figure 12.18. It consists of many loops of wire wound on an armature that can rotate in a magnetic field. When the armature is turned by some mechanical means, an emf is generated in the rotating coil.

Consider the coil to have N turns, each of area A, and rotated with a constant angular speed ω – about an axis in the plane of the coil and perpendicular to a uniform magnetic field \(\vec{B}\), as shown in the figure. The frequency of rotation of the coil is f = ω / 2π.

Working : The angle 9 between the magnetic field \(\vec{B}\) and the area of the coil \(\vec{A}\) at any instant t is θ = ωt (assuming θ = 0° at t = 0). At this position, the magnetic flux through the coil is
Φ_m = \(N \vec{B} \cdot \vec{A}\) = NBA cos θ = NBA cos ωt

∴ e = e₀ sin ωt, where e₀ = NBAω.
Therefore the induced emf varies as sin cot and is called sinusoidally alternating emf. In one rotation of the coil, sin cot varies between +1 and – 1 and hence the induced emf varies between +e₀ and -e₀. The maximum value e₀ of an alternating emf is called the peak value or amplitude of the emf.

The sinusoidal variation of emf with time t is shown in above figure. The emf changes direction at the end of every half rotation of the coil. The frequency of the alternating emf is equal to the frequency/of rotation of the coil. The period of the alternating emf is T = \(\frac{1}{f}\)

Imagine looking at the coil of the ac generator from the slip rings along the rotation axis in Fig. 12.18. The magnetic flux, rate of change of flux and sign of the induced emf are shown in the table below for the different orientations of the coil as in below figure.

Coil orientation	Flux Φm	dΦm/dt	Induced emf
1	Positive maximum	Momentarily zero (constant flux)	Zero
2	Positive	Decreasing (negative)	Positive
3	Zero	Decreasing (negative)	Positive
4	Negative	Decreasing (negative)	Positive
5	Negative maximum	Momentarily zero (constant flux)	Zero
6	Negative	Increasing (positive)	Negative
7	Zero	Increasing (positive)	Negative
8	Positive	Increasing (positive)	Negative
9	Return to positive maximum	Momentarily zero (constant flux)	Zero

Question 29.
How does a dc generator differ from an ac generator?
Answer:
A dc generator is much like an ac generator, except that the slip rings at the ouput are replaced by a split-ring commutator, just as in a dc motor.

The output of a dc generator is a pulsating dc as shown in Fig. 12.22. For a smoother output, a capacitor filter is connected in parallel with the output (see below figure for reference).

Question 30
Explain back emf in a motor.
Answer:
A generator converts mechanical energy into electrical energy, whereas a motor converts electrical energy into mechanical energy. Also, motors and generators have the same construction. When the coil of a motor is rotated by the input emf, the changing magnetic flux through the coil induces an emf, consistent with Faraday’s law of induction. A motor thus acts as a generator whenever its coil rotates. According to Lenz’s law, this induced emf opposes any change, so that the input emf that powers the motor is opposed by the motor’s self-generated emf. This self-generated emf is called a back emf because it opposes the change producing it.

Question 31.
A motor draws more current when it starts than when it runs at its full (i.e., operating) speed. Explain.
OR
When a pump or refrigerator (or other large motor) starts up, lights in the same circuit dim briefly.
Answer:
The back emf is effectively the generator output of a motor, and is proportional to the angular velocity co of the motor. Hence, when the motor is first turned on, the back emf is zero and the coil receives the full input voltage. Thus, the motor draws maximum current when it is first turned on. As the motor speeds up, the back emf grows, always opposing the driving emf, and reduces the voltage across the coil and the amount of current it draws. This explains why a motor draws more current when it first comes on, than when it runs at its normal operating speed.

The effect is noticeable when a high power motor, like that of a pump, refrigerator or washing machine is first turned on. The large initial current causes the voltage at the outlets in the same circuit to drop. Due to the IR drop produced in feeder lines by the large current drawn by the motor, lights in the same circuit dim briefly.

[Note : A motor is designed to run at a certain speed for a given applied voltage. A mechanical overload on the motor slows it down appreciably. If the rotation speed is reduced, the back emf will not be as high as designed for and the current will increase. At too low speed, the large current can even burn its coil. On the other hand, if there is no mechanical load on the motor, its angular velocity will increase until the back emf is nearly equal to the driving emf. Then, the motor uses only enough energy to overcome friction.]

Question 32.
What is back torque in a generator?
Answer:
In an electric generator, the mechanical rotation of the armature induces an emf in its coil. This is the output emf of the generator. Under no-load condition, there is no current although the output emf exists, and it takes little effort to rotate the armature.

However, when a load current is drawn, the situation is similar to a current-carrying coil in an external magnetic field. Then, a torque is exerted, and this torque opposes the rotation. This is called back torque or counter torque.

Because of the back torque, the external agent has to apply a greater torque to keep the generator running. The greater the load current, the greater is the back torque.

33. Solve the following 
Question 1.
An ac generator spinning at a rate of 750 rev/min produces a maximum emf of 45 V. At what angular speed does this generator produce a maximum emf of 102 V ?
Solution:
Data : e₁ = 45 V, f₁ = 750 rpm, e₂ = 102 V
e = NABω = NAB(2πf) ∴ e ∝ f
∴\(\frac{e_{2}}{e_{1}}=\frac{f_{2}}{f_{1}}\)
∴ f₂ = \(\frac{e_{2}}{e_{1}}\) × f₁ = \(\frac{102}{45}\) × 750 = 1700 rpm
This is the required frequency of the generator coil.

Question 2.
An ac generator has a coil of 250 turns rotating at 60 Hz in a magnetic field of \(\frac{0.6}{\pi}\) T. What must be the area of each turn of the coil to produce a maximum emf of 180 V ?
Solution:
Data : N = 250, f = 60 Hz, B = \(\frac{0.6}{\pi}\) T
e₀ = NABω = NAB (2πf)
∴ A = \(\frac{e_{0}}{N B 2 \pi f}=\frac{180}{(250)(0.6 / \pi)(2 \pi \times 60)}=\frac{18}{25 \times 72}\)
= 10^-2 m²
This must be the area of each turn of the coil.

Question 3.
A dynamo attached to a bicycle has a 200 turn coil, each of area 0.10 m². The coil rotates half a revolution per second and is placed in a uniform magnetic field of 0.02 T. Find the maximum voltage generated in the coil.
Solution:
Data : N = 200, A = 0.1 m², f = 0.5 Hz, B = 0.02T
e₀ = NABω = NAB (2πf)
Therefore, the maximum voltage generated,
e₀ = (200)(0.1)(0.02)(2 × 3.142 × 0.5) = 1.26 V

Question 4.
A motor has a coil resistance of 5 Ω. If it draws 8.2 A when running at full speed and connected to a 220 V line, how large is the back emf ?
Solution:
Data : R = 5 Ω, I = 8.2 A, e_appIied = 220 V
e_appIied – e_back =IR = 0
∴ e_back = _appIied – IR = 220 – (8.2)(5)
= 220 – 42 = 178 V

Question 5.
The back emf in a motor is 100 V when operating . at 2500 rpm. What would be the back emf at 1800 rpm? Assume the magnetic field remains unchanged.
Solution:
Data : e₁ = 100 V, f₁ = 2500 rpm, f₂ = 1800 rpm
The back emf is proportional to the angular speed.
∴ \(\frac{e_{2}}{e_{1}}=\frac{f_{2}}{f_{1}}\)
∴ e₂ = \(\frac{f_{2}}{f_{1}}\) × e₁ = \(\frac{1800}{2500}\) × 100 = 72V
This is the back emf at lower speed.

Question 6.
The armature windings of a dc motor have a resistance of 10 Ω. The motor is connected to a 220 V line, and when the motor reaches full speed at normal load, the back emf is 160 V. Calculate
(a) the current when the motor is just starting up
(b) the current at full speed,
(c) What will be the current if the load causes it to run at half speed ?
Solution:
Data : R = 10 Ω, e_appIied = 220 V, e_back = 160 V,
f₂ = f₁/2 .
e_appIied – e_back – IR = 0
(a) At start up, back emf is zero.
∴ I_start = \(\frac{e_{\text {applied }}}{R}=\frac{220}{10}\) = 22 A

(b) At full speed,
I_normal = \(\frac{e_{\text {applied }}-e_{\text {back }}}{R}=\frac{220-160}{10}=\frac{60}{10}\) = 6 A

(c) Back emf is proprtional to rotational speed. Thus, if the motion is running at half the speed, back emf is half the original value, i.e., 80 V. Therefore, at half speed,
I₂ = \(\frac{e_{\text {applied }}-e_{2}}{R}=\frac{220-80}{10}=\frac{140}{10}\) = 14 A

Question 34.
Find an expression for the power expended in pulling a conducting loop out of a magnetic field.
Answer:
When an external agent produces a relative motion between a conducting loop and an external magnetic field, a magnetic force resists the motion, requiring the applied force to do positive work. The work done is transferred to the material of the loop as thermal energy because of the electrical resistance of the material to the current that is induced by the motion.

Proof : Consider a rectangular wire loop ABCD of width l, with its plane perpendicular to a uniform magnetic field of induction \(\vec{B}\). The loop is being pulled out of the magnetic field at a constant speed v, as shown in below figure (a).

At any instant, let x be the length of the part of the loop in the magnetic field. As the loop moves to the right through a distance dx = vdt in time dt, the area of the loop inside the field changes by dA = ldx = lvdt. And, the change in the magnetic flux dΦ_m through the loop is
dΦ_m = BdA = Blvdt ………….. (1)
Then, the time rate of change of magnetic flux is
\(\frac{d \Phi_{\mathrm{m}}}{d t}=\frac{B l v d t}{d t}\) = B l v ……………. (2)
By Faraday’s second law, the magnitude of the induced emf is
|e| = \(\frac{d \Phi_{\mathrm{m}}}{d t}\) = B l v ………….. (3)

Due to the motion of the loop, the tree electrons (charge, e) in the wire inside the field experience Lorentz force \(e \vec{v} \times \vec{B}\). In the wire PQ this force moves the Free electrons 1mm P to Q making them travel in the anticlockwise sense around the 1oop. Therefore, the induced conventional current I is in the clockwise sense, as shown.

From figure (b) shows the equivalent circuit of the loop, where the induced emf e is a distributed emf and R is the total resistance of the loop.
∴ I = \(\frac{|e|}{R}=\frac{B l v}{R}\) …………… (4)
Now, a straight current carrying conductor of length L in a magnetic held experiences a torce
\(\vec{F}=I \vec{L} \times \vec{B}\)
whose direction can be found using Fleming’s Left hand rule.

Accordingly, forces \(\vec{F}_{2}\) and \(\vec{F}_{3}\) on wires AH and CD, respectively, are equal in magnitude (= Ix8), opposite in direction and have the same line of action- Hence, they balance each other. There is no torce on the wire BC as it hes outside the field.

The force \(\vec{F}_{1}\) on the wire AD has magnitude F₁ = IlB and Is directed towards the left. To move the loop with constant velocity \(\vec{v}\), an external force \(\vec{F}=-\vec{F}_{1}\) must be applied. Therefore, in magnitude,

Because B, l and R are constants a force of constant magnitude F is required to move the loop at constant speed v.

Thus, the power or the rate of doing work by the external agent is
P = \(\vec{F} \cdot \vec{v}\) = Fv = \(\frac{B^{2} l^{2} v^{2}}{R}\) ………….. (5)

Question 35.
Why and where are eddy currents undesirable ? How are they minimized ?
Answer:
Eddy currents result in generation of heat (energy loss) in the cores of transformers, motors, induction coils, etc.

To minimize the eddy currents, instead of a solid metal block, cores are made of thin insulated metal strips or laminae.

Question 36.
If a magnet is dropped through a long thick- walled vertical copper tube, it attains a constant velocity after some time. Explain.
Answer:
Every thin transverse section of a thick-walled vertical copper tube is an annular disc. The downward motion of the magnet causes increased magnetic flux through such conducting discs. By Lenz’s . law, the induced or eddy current around the discs produces a magnetic field of its own to oppose the change in flux due to the magnet’s motion.

Initially, as the magnet falls under gravity, its speed increases. But, quickly the vertically upward force on the magnet due to the induced current becomes equal in magnitude to the gravitational force on the magnet and the net force on the magnet becomes zero. The subsequent motion of the magnet is at this constant terminal speed.

Question 37.
Describe in brief an experiment to demonstrate that eddy currents oppose the cause producing them.
Answer:
Apparatus : A strong electromagnet; two thick copper discs (4″ dia, \(\frac{1}{4}\)” thick), each attached to a rod about 30″ long. One of the discs has several vertical slots, about 80 % of the way up. The pendulums can be suspended from a lab stand by a pivot mount and made to oscillate between closely-spaced pole pieces of the electromagnet.

Experiment: When the electromagnet is not turned on, both the pendulums swing freely with some damping due to air resistance. When the electromagnet is turned on, the slotted pendulum still swings, although a little more damped, but the solid pendulum practically stops dead between the pole pieces of the magnet immediately.

Conclusion : As the pendulums enter or exit the magnetic field, the changing magnetic flux sets up eddy currents in the discs. The sense of the eddy currents is so as to produce a torque that opposes the rotation of the discs about their pivot. This opposing torque produces a breaking action, damping the oscillations.

In the case of the solid disc, the continuous volume of the disc offers large unbroken path to the swirling electrons. Thus, the eddy current builds up to a large magnitude. The thicker the disc, the larger is the eddy current and, consequently, the larger the damping.

In the case of the slotted disc, the vertical slots do not allow large eddy current and, consequently, the damping is small.

Question 38.
A solid conducting plate swings like a pendulum about a pivot into a region of uniform magnetic field, as shown in the diagram. As it enters and leaves the field, show and explain the directions of the eddy current induced in the plate and the force on the plate.

Answer:
Figure shows the eddy currents in the conducting plate as it enters and leaves the magnetic field. In both cases, it experiences a force \(\vec{F}\) opposing its motion. As the plate enters from the left, the magnetic flux through the plate increases. This sets up an eddy current in the anticlockwise direction, as shown. Since only the right-hand side of the current loop is inside the field, by Fleming’s right hand rule (FRH rule), an unopposed force acts on it to the left. There is no eddy current once the plate is completely inside the uniform field. When the plate leaves the field on the right, the decreasing flux causes an eddy current in the clockwise direction. The damping magnetic force on the current is to the left, further slowing the motion.

The eddy current in the plate results in mechanical energy being dissipated as thermal energy. Each time the plate enters and leaves the field, a part of its mechanical energy is transformed into thermal energy. After a few swings, the mechanical energy becomes zero and the motion comes to a stop with the warmed-up plate hanging vertically.

39. Solve the following 
Question 1.
A metal rod of resistance of 15 Ω is moved to the right at a constant 60 cm/s along two parallel conducting rails-25 cm apart and shorted at one end. A magnetic field of magnitude 0.35 T points into the page, (a) What are the induced emf and current in the rod? (b) At what rate is thermal energy generated?
Solution:
Data: R = 15Ω, v = 0.6 m/s, l = 0.25m, B = 0.35T
(a) Induced emf, e = Blv = (0.35)(0.25)(0.6)
= 0.0525 V = 52.5 mV
The current in the rod, I = \(\frac{e}{\mathrm{R}}=\frac{52.5}{15}\) = 3 5 mA

(b) Power dissipated, P = eI = 0.0525 × 3.5 × 10^-4
= 0.184 mW

Question 2.
A conducting rod 10 cm long is being pulled along horizontal, frictionless conducting rails at a con-stant 5 m/s. The rails are shorted at one end with a metal strip. There is a uniform magnetic field of strength 1.2 T out of the page in the region in which the rod moves. If the resistance of the rod is 0.5 Ω, what is the power of the external agent pulling the rod? Assume that the resistance of the rails is negligibly small.
Solution:
Data: l = 0.1 m, B = 1.2T, v = 5 m/s. R = 0.5 Ω
Power, P = \(\frac{(B l v)^{2}}{R}=\frac{(1.2 \times 0.1 \times 5)^{2}}{0.5}\) = 0.72 W

Question 40.
Explain the concept of self induction.
Answer:
Consider an isolated coil or circuit in which there is a current I. The current produces a magnetic flux linked with the coil.

The magnetic flux linked with the coil can be changed by varying the current in the coil itself, e.g., by breaking and closing the circuit. This produces a self-induced emf in the coil, called a back emf because it opposes the change producing it. It sets up an induced current in the coil itself in the same direction as the original current opposing its decrease when the key K is suddenly opened. When the key K is closed, the induced current is opposite to the conventional current, opposing its increase.

When the current through a coil changes continuously, e.g., by a time-varying applied emf, the magnetic flux linked with the coil also goes on changing.

The production of induced emf in a coil, due to the changes of current in the same coil, is called self induction.

Question 41.
Explain and define the self inductance of a coil.
OR
Define the coefficient of self induction.
Answer:
When the current through a coil goes on changing, the magnetic flux linked with the coil also goes on changing. The magnetic flux (NΦ_m) linked with the coil at any instant is directly proportional to the current (I) through the coil at that instant.
NΦ_m ∝ I
∴ NΦ_m = LI
where L is a constant, dependent on the geometry of the coil, called the self inductance or the coefficient, of self induction of the coil.
The self-induced emf in the coil is

Definition : The self inductance or the coefficient of self induction of a coil is defined as the emf induced in the coil per unit time rate of change of current in the same coil. OR (using L = NΦ_m/I), the self inductance of a coil is the ratio of magnetic flux linked with the coil to the current in it.

Question 42.
State and define the SI unit of self inductance. Give its dimensions.
OR
Write the SI unit and dimensions of the coefficient of self induction.
Answer:
The SI unit of self inductance or coefficient of self induction or inductance as it is commonly called is called the henry (H).

The self-inductance of a coil is 1 henry, if an emf of 1 volt is induced in the coil when the current through the same coil changes at the rate of 1 ampere per second.

The dimensions of self inductance or coefficient of self induction are [ML²T^-2I^-2].
1 henry = 1 H = 1 V/A.s = 1 T.m²/A

[ Note : The unit henry is named in honour of Joseph Henry (1797-1878) US physicist.]

Question 43.
What is an inductor?
Answer:
An inductor is a coil of wire with significant self inductance. If the coil is wound on a nonmagnetic cylinder or former, such as ceramic or plastic, it is called an air-core inductor; its circuit symbol is . If the cod is wound on a magnetic former. such as laminated iron or ferrite. it Is called an iron core inductor; its circuit symbol is .

Question 44.
Current passes through a coil shown from left to right. In which direction is th induced emf. if the current is (a) increasing with time (b) decreasing in time?

Answer:
From Lent’s law, the induced emf must oppose the diange in the magnetic flux. (a) When the current mcreases to the right, so is the magnetic flux. To oppose the increasing flux to the tight. the induced emi Is to the left. i.e.. the point A is at a positive potential relative to point B.

(b) When the current to the right is decreasing the induced emf acts to boost up the flux to the right and points to the tight, so that the point A is at a negative potential relative to point B.

Question 45.
Derive an expression for the energy stored in the magnetic field of an inductor.
OR
Derive an expression for the electrical work done in establishing a steady current in a coil of self inductance L.
Answer:
Consider an inductor of sell inductance L connected in a circuit When the circuit is dosed, the current in the circuit increases and so does the magnetic flux linked with the coiL At any instant the magnitude of the induced emf is
e = L \(\frac{d i}{d t}\)
The power consumed in the inductor is
P = ei = L \(\frac{d i}{d t}\) ∙ i
[Alternatively, the work done in moving a charge dq against this emf e is
dw = edq = L \(\frac{d i}{d t}\) ∙ dq = Li ∙ di (∵ \(\frac{d q}{d t}\) = i)
This work done is stored in the magnetic field of the inductor. dw = du.]

The total energy stored In the magnetic field when the current increases from 0 to I In a time interval from 0 to t can be determined by integrating this expression :
U_m = \(\int_{0}^{t} P d t=\int_{0}^{I} L i d i=L \int_{0}^{I} i d i=\frac{1}{2} L I^{2}\)
which is the required expression for the stored magnetic energy.
[Note: Compare this with the electric energy stored in a capacitor, U_e = \(\frac{1}{2}\)CV²]

Question 46.
State the expression for the energy stored in’the magnetic field of an inductor. Hence, define its self inductance.
Answer:
When a steady current is passed through an inductor of self inductance L the energy stored in the
magnetic field of the inductor is U_m = \(\frac{1}{2}\)Li²]. Therefore, for unit current, L = 2U_m

Hence, we may define the self inductance of a coil as numerically equal to twice the energy stored in its magnetic field for unit current through the inductor.

Question 47.
What is the role of an inductor in an ac circuit ?
Answer:
As a circuit element, an inductor slows down changes in the current in the circuit. Thus, it provides an electrical inertia and is said to act as a ballast. In a non-inductive coil (L ≅ 0), electrical energy is converted into heat due to ohmic resistance of the coil (Joule heating). On the other hand, an inductive coil or an inductor stores part of the energy in the magnetic field of its coils when the current through it is increasing; this energy is released when the current is decreasing. Thus, an inductor limits an alternating current more efficiently than a non-inductive coil or a pure resistor.

Question 48.
State the expressions for the effective or equivalent inductance of a combination of a number of inductors connected (a) in series (b) in parallel. Assume that their mutual inductance can be ignored.
Answer:
We assume that the inductors are so far apart that their mutual inductance is negligible.
(a) For a series combination of a number of inductors, L₁, L₂, L₃, …, the equivalent inductance is
L_series = L₁ + L₂ + L₃+ ……..

(b) For a parallel combination of a number of inductors, L₁, L₂, L₃, …, the equivalent inductance is
\(\frac{1}{L_{\text {parallel }}}=\frac{1}{L_{1}}+\frac{1}{L_{2}}+\frac{1}{L_{3}}+\ldots\)

Question 49.
Obtain an expression for the self inductance of a solenoid.
Answer:
Consider a long air-cored solenoid of length Z, diameter d and N turns of wire. We assume that the length of the solenoid is much greater than its diameter so that the magnetic field inside the solenoid may considered to be uniform, that is, end effects in the solenoid can be ignored. With a steady current I in the solenoid, the magnetic field within the solenoid is
B = µ₀nI ………….. (1)
where n = N/l is the number of turns per unit length. So the magnetic flux through one turn is
Φ_m = BA = µ₀nIA ……….. (2)
Hence, the self inductance of the solenoid,
L = \(\frac{N \Phi_{\mathrm{m}}}{I}\) =(nl)µ₀nA = µ₀n²lA = µ₀n² V ………….. (3)
= µ₀n²l\(\frac{\pi d^{2}}{4}\) …………. (4)
where V = lA is the interior volume of the solenoid. Equation (3) or (4) gives the required expression.

[Note: It is evident thatthe self inductance of a long solenoid depends only on its physical properties – such as the number of turns of wire per unit length and the volume, and not on the magnetic field or the current. This is true for inductors in general.] .

Question 50.
State the expression for the self inductance of a solenoid. Hence show that the SI unit of magnetic permeability is the henry per metre.
Answer:
The self inductance of an air-cored long solenoid of volume V and number of turns per unit length n is L = µ₀n²V. Since [n²] = [L^-2], n²V has the dimension of length. The SI unit of the L being the henry, the SI unit of magnetic permeability (µ₀) is the henry per metre (H / m). .
µ₀ = 4π × 10^-7 H/m = 4π × 10^-7 T∙m/A

Question 51.
Derive an expression for the self inductance of a narrow air-cored toroid of circular cross section.
Answer:
Consider a narrow air-cored toroid of circular cross section of radius r, central radius R and number of turns N. So that, assuming r << R, the magnetic field in the toroidal cavity is considered to be uniform, equal to
B = \(\frac{\mu_{0} N I}{2 \pi R}\) = µ₀nI ………….. (1)
where n = \(\frac{N}{2 \pi R}\) is the number of turns of the wire 2nR per unit length. The area of cross section, A = πr².
The magnetic flux through one turn is
Φ_m = BA = µ₀nIA ………… (2)
Hence, the self inductance of the toroid,
L = \(\frac{N \Phi_{\mathrm{m}}}{I}\) = (2πRn) µ₀nA = µ₀2πRn²A = µ₀n²V …………… (3)
= \(\frac{\mu_{0} N^{2} r^{2}}{2 R}\) ………….. (4)
where V = 2πRA is the volume of the toroidal cavity. Equation (3) or (4) gives the required expression.

Question 52.
Obtain an expression for the energy density of a magnetic field.
Answer:
Consider a short length ¡ near the middle of a long, tightly wound solenoid, of cross-sectional area A, number of turns per unit length n and carrying a steady current I. For such a solenoid, the magnetic field is approximately uniform everywhere inside and zero outside. So, the magnetic energy U_m stored by this length l of the solenoid lies entirely within the volume Al.

The magnetic field inside the solenoid is
B = µ₀nI …………… (1)
and if L be the inductance of length l of the solenoid,
L = µ₀ n²lA …………… (2)
The stored magnetic energy,
U_m = \(\frac{1}{2}\)LI² …………. (3)
and the energy density of the magnetic field (energy per unit volume) is

Equation (6) gives the magnetic energy density in vacuum at any point in a magnetic field of induction B, irrespective of how the field is produced.

[Note : Compare Eq.(6) with the electric energy density in vacuum at any point in an electric field of intensity
e, u_e = \(\frac{1}{2}\) ε₀e². Both u_e and u_m are proportional to the square of the appropriate field magnitude.]

Question 53.
Determine the magnetic energy stored per unit length of a coaxial cable, represented by two coaxial cylindrical shells of radii a (inner) and b (outer), and carrying a current I. Hence derive an expression for the self inductance of the coaxial cable of length l.
Answer:
Figure (a) shows a coaxial cable represented by two hollow, concentric cylindrical conductors along which there is electric current in opposite directions. The magnetic field between the conductors can be found by applying Ampere’s law to the dashed path of radius r{a < r < b) in figure (a). Because of the cylindrical symmetry, B is constant along the path, and
\(\oint \vec{B} \cdot \overrightarrow{d l}\) = B (2πr) = u₀I
∴ B = \(\frac{\mu_{0} I}{2 \pi r}\) ……………… (1)

A similar application of Ampere’s law for r > b and r < a, shows that B = 0 in both the regions. Therefore, all the magnetic energy is stored between the two conductors of the cable.
The energy density of the magnetic field is
u_m = \(\frac{B^{2}}{2 \mu_{0}}\) …………….. (2)
Therefore, substituting for B from Eq. (1) into Eq. (2), the magnetic energy stored in a cylindrical shell of radius r, thickness dr and length l is
dU_m = u_mdV = u_m(2πr ∙ dr ∙ l)

Equating the right hand sides of Eqs. (4) and (6),

54. Solve the following
Question 1.
A coil of self inductance 5 H is connected in series with a switch and a battery. After the switch is closed, the steady state value of the current is 5 A. The switch is then suddenly opened, causing the current to drop to zero in 0.2 s. Find the emf developed across the inductor (coil) as the switch is opened.
Solution:
Data : L = 5 H, I_i = 5 A, I_f = 0, ∆t = 0.2 s
The rate of change of current,
\(\frac{d I}{d t}=\frac{I_{\mathrm{f}}-I_{\mathrm{i}}}{\Delta t}=\frac{0-5}{0.2}\) = – 25 A/s
∴ The induced emf,
e = -L \(\frac{d I}{d t}\) = -5(-25) = 125 V

Question 2.
A toroidal coil has an inductance of 47 mH. Find the maximum self-induced emf in the coil when the current in it is reversed from 15 A to -15 A in 0.01 s.
Solution:
Data : L = 4.7 × 10^-2 H, I_i = 15A, I_i = -15 A,
∆f = 0.01 s
The rate of change of current,
\(\frac{d I}{d t}=\frac{I_{\mathrm{f}}-I_{\mathrm{i}}}{\Delta t}=\frac{(-15)-15}{0.01}\) = – 3000 A/s
∴ The maximum self-induced emf,
e = – L \(\frac{d I}{d t}\) (4.7 × 10^-2) (- 3000) = 141 V

Question 3.
An emf of 2 V is induced in a closely-wound coil of 50 turns when the current through it increases uniformly from O to 5 A in 0.1 s. (a) What is the self inductance of the coil? (b) What is the flux through each turn of the coil for a steady current at 5A?
Solution:
Data : e = 2 V, N = 50, I_i = 0, I_f = 5A, ∆t = 0.1 s
(a) The rate of change of current

This is the flux through each turn.

Question 4.
At the instant the current through a coil is 0.2 A, the energy stored in its magnetic field is 6 mJ. What is the self indudance of the coil ?
Solution:
Data: I = 0.2A, U_m = 6 × 10^-3 J
U_m = \(\frac{1}{2}\) LI²
Therefore, self inductance of the coil is

Question 5.
A coil of self inductance 3 H and resistance 100 Ω carries a steady current of 2 A. (a) What is the energy stored in the magnetic field of the coil? (b) What is the energy per second dissipated in the resistance of the coil ?
Solution:
Data : L = 3 H, R = 100 Ω, I = 2 A
(a) Magnetic energy stored,
U_m = \(\frac{1}{2}\) LI² = \(\frac{1}{2}\) (3) (2)² = 6 J

(b) Power dissipated in the resistance of the coil,
P = I²R = (2)²(100) = 400 W

Question 6.
A 10 H inductor carries a current of 25 A. Flow much ice at 0 °C could be melted by the energy stored in the magnetic field of the inductor ? [Latent heat of fusion of ice, L_f = 335 J/g]
Solution:
Data : L = 10 H, Z = 25 A, L_f = 335 J/g
Magnetic energy stored,
U_m = \(\frac{1}{2}\) LI² = \(\frac{1}{2}\) (10) (25)² = 3125 J
Heat energy required to melt ice at 0 °C of mass m,
H = mL_f
Equating H with U_m,
m = \(\frac{U_{\mathrm{m}}}{L_{\mathrm{f}}}=\frac{3125}{335}\) = 9.328 g
Therefore, 9.328 g of ice could be melted by the energy stored.

Question 7.
A solenoid 40 cm long has a cross-sectional area of 0.9 cm2 and is tightly wound with wire of diameter 1 mm. Calculate the self inductance of the solenoid.
Solution:
Data : D = 1 mm, l = 40 cm = 0.4 m, A = 0.9 cm² = 9 × 10^-5 m², I_i = 10 A, I_f = 0, ∆t = 0.1 s,
μ₀ = 4π × 10^-7 H/m
The number of turns per unit length,
n = \(\frac{1}{1 \mathrm{~mm}}\) = 1 mm^-1 = 10³ m^-1
Self inductance of the solenoid,
L = μ₀n²lA = (4π × 10^-7)(10³)²(0.4)(9 × 10^-5)
= 16 × 9 × 3.142 × 10^-7 = 4.524 × 10^-5 H

Question 8.
A solenoid of 1000 turns is wound with wire of diameter 0.1 cm and has a self inductance of 2.4 π × 10^-5 H. Find (a) the cross-sectional area of the solenoid (b) the magnetic flux through one turn of the solenoid when a current of 3 A flows through it.
Solution:
Data: N = 1000, D = 0.1 cm, L = 2.4π × 10^-5 H,
I = 3A, μ₀ = 4π × 10^-7 H/m
The number of turns per unit length.
n = \(\frac{1}{1 \mathrm{~mm}}\) = 1 mm^-1 = 10³ m^-1
and the length of the solenoid,
l = ND = 1000 × 0.1 = 100 cm = 1 m
L = μ₀n²lA

(a) The area of cross section,
A = \(\frac{L}{\mu_{0} n^{2} l}=\frac{2.4 \pi \times 10^{-5}}{\left(4 \pi \times 10^{-7}\right)\left(10^{3}\right)^{2}(1)}=\frac{24 \pi}{4 \pi} \times 10^{-5}\)
= 6 × 10^-5 m²

(b) Magnetic flux through one turn,
Φ_m = BA = (μ₀nI)A
= (4π × 10^-7)(10³)(3)(6 × 10^-5)
= 72π × 10^-9 Wb

Question 9.
A toroid of circular cross section of radius 0.05 m has 2000 windings and a self inductance of 0.04 H. What is (a) the current through the windings when the energy in its magnetic field is 2 × 10^-6 J (b) the central radius of the toroid ?
Solution:
Data : r = 0.05 m, N = 2000, L = 0.04 H,
U_m = 2 × 10^-6 J, μ₀ = 4π × 10^-7 H/m
(a) U_m = \(\frac{1}{2}\) LI²
Therefore, the current in the windings,

Question 10.
A coaxial cable, whose outer radius is five times its inner radius, is carrying a current of 1.5 A. What is the magnetic field energy stored in a 2 m length of the cable ?
Solution:
Data : b/a = 5, I = 1.5A, l = 2m, \(\frac{\mu_{0}}{4 \pi}\) = 10^-7 H/m
The total magnetic energy in a given length of a current-carrying coaxial cable,
U_m = \(\left(\frac{\mu_{0}}{4 \pi}\right) I^{2} l \log _{e} \frac{b}{a}\)
Therefore, the required magnetic energy is
U_m = (10^-7)(1.5)²(2)log_e5
= 4.5 × 10⁷ × 2.303 × log₁₀5
= 4.5 × 10^-7 × 2.303 × 0.6990 = 7.24 × 10^-7 J

Question 55.
Explain the concept/phenomenon of mutual induction.
OR
Explain and define mutual inductance of a coil with respect to another coil.
OR
Define the coefficient of mutual induction.
Answer:

The production of induced emf in a coil due to the change of current in the same coil is called self induction.

In above figure (a), a current I₁ in coil 1 sets up a magnetic flux Φ₂₁ through one turn of a neighbouring coil 2, magnetically linking the two coils. Then, the flux through the N₂ turns of coil 2, i.e., the flux linkage of coil 2, is N₂Φ₂₁.
N₂Φ₂₁ ∝ I₁
∴ N₂Φ₂₁ = M₂₁I₁ …………. (1)
where the constant of proportionality, M₂₁, is called the coefficient of mutual induction of coil 2 with respect to coil 1. If the current I₁ in coil 1 changes with time, the varying flux linkage induces an emf e₂ in coil 2.
e₂ = – \(\frac{d}{d t}\) (N₂Φ₂₁) = – M₂₁ \(\frac{d I_{1}}{d t}\) …………. (2)
Similarly, if we interchange the roles of the two coils and set up a current I₂ in coil 2 [from figure (b)], Then, the flux linkage of N1 turns of coil 1 is N₁Φ₁₂ and
N₁Φ₁₂ = M₁₂I₂ ………… (3)
where M₁₂ is the coefficient of mutual induction of coil 1 with respect to coil 2. And, for a varying current I₂(t), the induced emf in coil 1 is

We define mutual inductance using Eq. (5) or Eq. (6).

The mutual inductance or the coefficient of mutual induction of two magnetically linked coils is equal to the flux linkage of one coil per unit current in the neighbouring coil.
OR
The mutual inductance or the coefficient of mutual induction of two magnetically linked coils is numerically equal to the emf induced in one coil (secondary) per unit time rate of change of current in the neighbouring coil (primary).

Question 56.
State and define the SI unit of mutual inductance. Give its dimensions.
Answer:
The SI unit of mutual inductance is called the henry (H).

The mutual inductance of a coil (secondary) with respect to a magnetically linked neighbouring coil (primary) is one henry if an emf of 1 volt is induced in the secondary coil when the current in the primary coil changes at the rate of 1 ampere per second.

The dimensions of mutual inductance are [ML²T^-2I^-2] (the same as those of self inductance).

Question 57.
Two coils A and B have mutual inductance 2 × 10^-2 H. If the current in the coil A is 5 sin (10πt) ampere, find the maximum emf induced in the coil B.
Ans;
The emf induced in the coil B,
|e_B| = M \(\frac{d I_{\mathrm{A}}}{d t}\)
=(2 × 10^-2)[5 cos (10πt)] × 10π
∴ |e_B|_max = π volts.

Question 58.
A long solenoid, of radius R, has n turns per unit length. An insulated coil C of IV turns is wound over it as shown. Show that the mutual inductance for the coil-solenoid combination is given by M = μ₀πR²nN.

Answer:
We assume the solenoid to be ideal and that all the flux from the solenoid passes through the outer coil C. For a steady current Is through the solenoid, the uniform magnetic field inside the solenoid is
B = μ₀nI_s ……………… (1)
Then, the magnetic flux through each turn of the coil due to the current in the solenoid is
Φ_CS = BA = (μ₀nI_s)(πR²) ………….. (2)
Thus, their mutual inductance is
M = \(\frac{N \Phi_{\mathrm{CS}}}{I_{\mathrm{S}}}\) = μ₀πR²nN ………….. (3)
Equation (2) is true as long as the magnetic field of the solenoid is entirely contained within the cross section of the coil C. Hence, M does not depend on the shape, size, or possible lack of close packing of the coil.

Question 59.
A solenoid of N₁ turns has length l₁ and radius R₁, and a second smaller solenoid of N₂ turns has length l₂ and radius R₂. The smaller solenoid is placed coaxially and completely inside the larger solenoid. What is their mutual inductance ?
Answer:
Assuming the larger solenoid to be ideal, the magnetic field within it may be considered uniform, so the flux through the small solenoid due to the larger solenoid is also uniform. Assuming a current I₁ in the larger solenoid, the magnitude of the magnetic field at points within the small solenoid due to the larger one is
B₁ = μ₀\(\frac{N_{1}}{l_{1}}\) I₁
Then, the flux Φ₂₁ through each turn of the small coil is
Φ₂₁ = B₁A₂
where is A₂ = πR²₂, the area enclosed by the turn. Thus, the flux linkage in the small solenoid with its N₂ turns is
N₂Φ₂₁ = N₂B₁A₂
Thus, their mutual inductance is
M = \(\frac{N_{2} \Phi_{21}}{I_{1}}=N_{2}\left(\mu_{0} \frac{N_{1}}{l_{1}}\right)\left(\pi R_{2}^{2}\right)=\mu_{0} \pi \frac{N_{1} N_{2}}{l_{1}} R_{2}^{2}\)
which is the required expression.

Question 60.
What is meant by coefficient of magnetic coupling?
Answer:
For two inductively coupled coils, the fraction of the magnetic flux produced by the current in one coil (primary) that is linked with the other coil (secondary) is called the coefficient of magnetic coupling between the two coils.

The coupling coefficient K shows how good the coupling between the two coils is; 0 ≤ K ≤ 1. In the ideal case when all the flux of the primary passes through the secondary, K=l. For coils which are not coupled, K = 0. Two coils are tightly coupled if K > 0.5 and loosely coupled if K < 0.5.

[ Note ; For iron-core coupled circuits, the value of K may be as high as 0.99, for air-core coupled circuits, K varies between 0.4 to 0.8. ]

Question 61.
State the factors which magnetic coupling coefficient of two coils depends on.
Answer:
The coefficient of magnetic coupling between two coils depends on

1. the permeability of the core on which the coils are wound
2. the distance between the coils
3. the angle between the coil axes.

Question 62.
When is the magnetic coupling coefficient of two coils (i) maximum (ii) minimum?
Answer:
The coefficient of magnetic coupling between two coils is

1. maximum when the coils are wound on the same ferrite (iron) core such that the flux linkage is maximum,
2. minimum for air-cored coils with the coil axes perpendicular.

Question 63.
Show that the mutual inductance for a pair of inductively coupled coils/circuits of self inductances L₁ and L₂ is given by M = K\(\sqrt{L_{1} L_{2}}\), where K is the coupling coefficient.
Answer:
Consider a pair of inductively coupled coils having N₁ and N₂ turns, shown in figure

A current l₁(t) sets up a flux N₁Φ₁(t) in coil 1 and induces a current l₂(t) and flux N₂Φ₂(t) in coil 2. Then, the self inductances of the coils are

Alternate method :
Consider a pair of inductively coupled coils shown in above figure.We assume that I₁(t), I₂(t) are zero at t = 0. as also the magnetic energy of the system.
The induced emfs are

The net energy Input to the system shown in figure at time t is given by

If one current enters a dot marked terminal while the other leaves a dot marked terminal, Eq. (2) becomes
W(t) = \(\frac{1}{2}\) L₁(I₁)² + \(\frac{1}{2}\) L₂(I₁)² – MI₁I₂ …………. (3)
The net electrical energy input to the system is non-negative, W(t) ≥ 0. We rearrange Eq.(3) as

The first term in the parenthesis on the right hand side of Eq. (4) is positive for all values of I₁ and I₂ Thus, for the second term also to be non-negative,

where the coupling coefficient K is a non-negtive number, 0 ≤ K ≤ 1, and is independent of the reference directions of the currents in the coils.

Question 64.
What is a transformer?
State the principle of working of a transformer.
Answer:
A transformer is an electrical device which uses mutual induction to transform electrical power at one alternating voltage into electrical power at another alternating voltage (usually different), without change of frequency of the voltage.

Principle : A transformer works on the principle that a changing current through one coil creates a changing magnetic flux through an adjacent coil which in turn induces an emf and a current in the second coil.

Question 65.
What are step-up and step-down transformers?
Answer:

1. Step-up transformer : It increases the amplitude of the alternating emf, i.e., it changes a low voltage alternating emf into a high voltage alternating emf with a lower current.
2. Step-down transformer : It decreases the amplitude of the alternating emf, i.e., it changes a high voltage alternating emf into a low voltage alternating emf with a higher current.

Question 66.
Describe the construction and working of a transformer with a neat labelled diagram.
Answer:
Construction : A transformer consists of two coils, primary and secondary, wound on two arms of a rectangular frame called the core.
(1) Primary coil : It consists of an insulated copper wire wound on one arm of the core. Input voltage is applied at the ends of this coil.

In a step-up transformer, thick copper wire is used for primary coil. In a step-down transformer, thin copper wire is used for primary coil.

(2) Secondary coil : It consists of an insulated copper wire wound on the other arm of the core. The output voltage is obtained at the ends of this coil.

In a step-up transformer, thin copper wire is used for secondary coil. In a step-down transformer, thick copper wire is used for secondary coil.

(3) Core : It consists of thin rectangular frames of soft iron stacked together, but insulated from each other. A core prepared by stacking thin sheets rather than using a single thick sheet helps reduce eddy currents.

Working : When the terminals of the primary coil are connected to a source of an alternating emf (input voltage), there is an alternating current through it. The alternating current produces a time varying magnetic field in the core of the transformer. The magnetic flux associated with the secondary coil thus varies periodically with time according to the current in the primary coil. Therefore, an alternating emf (output voltage) is induced in the secondary coil.

Question 67.
Derive the relationship \(\frac{V_{\mathrm{P}}}{V_{\mathrm{S}}}=\frac{I_{\mathrm{S}}}{I_{\mathrm{P}}}\) for a transformer.
Answer:
An alternating emf VP from an ac source is applied across the primary coil of a transformer. This sets up an alternating current IP in the primary circuit and also produces an alternating magnetic flux through the primary coil such that
V_P = -N_P \(\frac{d \Phi_{\mathrm{P}}}{d t}\),
where N_P is the number of turns of the primary coil and Φ_P is the magnetic flux through each turn.

Assuming an ideal transformer (i.e., there is no leakage of magnetic flux), the same magnetic flux links both the primary and the secondary coils,
i. e., Φ_P = Φ_S
As a result, the alternating emf induced in the secondary coil,
V_S = = N_S \(\frac{d \Phi_{\mathrm{S}}}{d t}\) = – N_S \(\frac{d \Phi_{\mathrm{P}}}{d t}\)

where N_S is the number of turns of the secondary coil. If the secondary circuit is completed by a resistance R, the secondary current is I_S = V_S/R, assuming the resistance of the coil to be far less than R. Ignoring power losses, the power delivered to the primary coil equals that taken out of the secondary coil, so V_PI_P = V_SI_S.
∴ \(\frac{V_{\mathrm{P}}}{V_{\mathrm{S}}}=\frac{I_{\mathrm{S}}}{I_{\mathrm{P}}}\)
which is the required expression.

Question 68.
Derive the relation \(\frac{V_{\mathrm{S}}}{V_{\mathrm{P}}}=\frac{N_{\mathrm{S}}}{N_{\mathrm{P}}}\) for a transformer. Hence, explain a step-up and a step-down trans-former. Also, show that \(\frac{I_{P}}{I_{\mathrm{S}}}=\frac{N_{\mathrm{S}}}{N_{\mathrm{P}}}\)
OR
Derive expressions for the emf and current for a transformer in terms of the turns ratio.
Answer:
An alternating emf VP from an ac source is applied across the primary coil of a transformer, shown in figure.

This sets up an alternating current fP in the primary circuit and also produces an alternating magnetic flux through the primary coil such that
V_P = -N_P \(\frac{d \Phi_{\mathrm{P}}}{d t}\) ………….. (1)
where N_P is the number of turns of the primary coil and Φ_P is the magnetic flux through each turn.

Assuming an ideal transformer (i.e., there is no leakage of magnetic flux), the same magnetic flux links both the primary and the secondary coils, i.e., Φ_P = Φ_S.
As a result, the alternating emf induced in the secondary coil,
V_S = – N_S \(\frac{d \Phi_{\mathrm{S}}}{d t}\) = – N_S \(\frac{d \Phi_{\mathrm{P}}}{d t}\) ……………… (2)
where N_S is the number of turns of the secondary coil.
From Eqs. (1) and (2),
\(\frac{V_{\mathrm{S}}}{V_{\mathrm{P}}}=\frac{N_{\mathrm{S}}}{N_{\mathrm{P}}}\) or V_S = V_P \(\frac{N_{\mathrm{S}}}{N_{\mathrm{P}}}\) …………… (3)

Case (1) i If N_S > N_P, V_S > V_P. Then, the trans-former is called a step-up transformer.
Case (2) : If N_S < N_P, V_S < V_P. Then the transformer is called a step-down transformer.

Ignoring power losses, the power delivered to the primary coil equals that taken out of the secondary coil, so that V_PI_P = V_SI_S …………. (4)
From Eqs. (3) and (4),
\(\frac{I_{\mathrm{P}}}{I_{\mathrm{S}}}=\frac{V_{\mathrm{S}}}{V_{\mathrm{P}}}=\frac{N_{\mathrm{S}}}{N_{\mathrm{P}}}\)

Question 69.
What is the turns ratio of a transformer? What can you say about its value for a (1) step-up transformer (2) step-down transformer?
Answer:
The ratio of the number of turns in the secondary coil (N_S) to that in the primary coil (N_P) is called the turns ratio of a transformer.
The turns ratio \(\frac{N_{\mathrm{S}}}{N_{\mathrm{P}}}\) > 1 for a step-up transformer.
The turns ratio \(\frac{N_{\mathrm{S}}}{N_{\mathrm{P}}}\) < 1 for a step-down transformer.

Question 70.
State any two factors on which the maximum value of the alternating emf induced in the secondary coil of a transformer depends.
Answer:
The maximum value of the alternating emf induced in the secondary coil of a transformer depends on

1. the ratio of the number of turns of the secondary coil to that of the primary coil
2. the maximum value of the alternating emf applied to the primary coil
3. the core of the transformer.

Question 71.
The primary coil of a transformer has 100 turns and the secondary coil has 200 turns. If the peak value of the alternating emf applied to the primary coil is 100 V, what is the peak value of the alternating emf obtained across the secondary coil?
Answer:

Question 72.
Distinguish between a step-up and a step-down transformers. (Any two points)
Answer:

Step-up transformer	Step-down transformer
1. The output voltage is more than the input voltage.	1. The output voltage is less than the input voltage.
2. The number of turns of the secondary coil is more than that of the primary coil.	2. The number of turns of the secondary coil is less than that of the primary coil.
3. The output current is less than the input current.	3. The output current is more than that of the input current.
4. The primary coil is made of thicker copper wire than the secondary coil.	4. The secondary coil is made of thicker copper wire than the primary coil.

72. Solve the following
Question 1.
When a current changes from 4 A to 12 A in 0.5 s in the primary coil, an induced emf of 50 mV is generated in the secondary coil. What is the mutual inductance between the two coils ? What will be the emf induced in the secondary, if the current in the primary changes from 3 A to 9 A in 0.02 s ?
Solution:
Data : I_i1 =4 A, I_f1 = 12 A, ∆t₁ = 0.5 s, ∆t₂ = 0.02 s

Question 2.
A plane coil of lo turns is tightly wound around a solenoid of diameter 2 cm having 400 turns per centimeter. The relative permeability of the core is 800. Calculate the mutual inductance.
Solution:
Data: N = 10, R = 1 cm = 10^-2 m,
n = 400 cm^-1 = 4 × 10⁴ m^-1, k = 800,
μ₀ = 4π × 10⁴ H/m
Mutual inductance,
M = kμ₀πR²nN
=(800)(4π × 10^-7)[π × (10²)²](4 × 10⁴)(10)
= 0.1264 H

Question 3.
Two coils of 100 turns and 200 turns have self inductances 25 mH and 40 mH, respectively. Their mutual inductance is 3 mH. If a 6 mA current in the first coil is changing at the rate of 4 A/s, calculate (a) 2 that links the first coil (b) self induced emf in the first coil (c) Φ₂₁ that links the second coil (d) mutually induced emf in the second coil.
Solution:
Data : N₁ = 100, N₂ = 200, L₁ = 25 mH, L₂ = 40 mH,
I₁ = 6 mA, dI₁ /dt = 4 A/s
(a) The flux per unit turn in coil 1,
Φ₂₁ = \( \frac{L_{1} I_{1}}{N_{1}}=\frac{\left(25 \times 10^{-3}\right)\left(6 \times 10^{-3}\right)}{100}\)
= 1.5 × 10^-6 Wb =1.5 μ Wb

(b) The magnitude of the self induced emf in coil 1 is
L₁ = \(\frac{d I_{1}}{d t}\) = (25 × 10^-3)(4) = 0.1 V

(c) The flux per unit turn in coil 2,
Φ₂₁ = \(\frac{M I_{1}}{N_{2}}=\frac{\left(3 \times 10^{-3}\right)\left(6 \times 10^{-3}\right)}{200}\)
= 90 × 10^-9 Wb = 90 nWb

(d) The mutually induced emf in coil 2 is
e₂₁ = M \(\frac{d I_{1}}{d t}\) = (3 × 10^-3)(4) = 12 × 10^-3 V
= 12 mV

Question 4.
The coefficient of mutual induction between primary and secondary coils is 2 H. Calculate the induced emf if a current of 4A is cut off in 2.5 × 10^-4 second.
Solution:
Data : M = 2 H, dI = – 4 A, dt = 2.5 × 10^-4 s
The induced emf, e = – M \(\frac{d I}{d t}=-\frac{2 \times(-4)}{2.5 \times 10^{-4}}\)
= \(\frac{8}{2.5}\) × 10⁴ = 3.2 × 10⁴ V

Question 5.
A current of 10 A in the primary of a transformer is reduced to zero at the uniform rate in 0.1 second. If the mutual inductance be 3 H, what is the emf induced in the secondary and change in the magnetic flux per turn in the secondary if it has 50 turns?
Solution:

This gives the change in the magnetic flux per turn in the secondary.

Question 6.
The primary and secondary coils of a transformer, assumed to be ideal, have 20 and 300 turns of wire, respectively. If the primary voltage is VP = 10 sincot (in volt), what is the maximum voltage in the secondary coil?
Solution:
Data : N_P = 20, N_S = 300, V_P = 10 sin ωt V
V_S = \(\frac{N_{\mathrm{S}}}{N_{\mathrm{P}}}\) V_P
= \(\frac{300}{20}\) × 10 sin ωt
= 150 sin ωt V
This is of the form V₀ sin ωt, where V₀ is the peak (or maximum) voltage.
∴ The maximum voltage in the secondary coil is 150 V.

Question 7.
A transformer converts 200 V ac to 50 V ac. The secondary has 50 turns and the load across it draws 300 mA current. Calculate (i) the number of turns in the primary (ii) the power consumed.
Solution:
Data: V_P = 200 V, V_S = 50 V, N_S = 50, I_S = 300mA = 0.3 A
(i) \(\frac{N_{\mathrm{P}}}{N_{\mathrm{S}}}=\frac{V_{\mathrm{P}}}{V_{\mathrm{S}}}\)
∴ The number of turns in the primary,
N_P = N_S\(\frac{V_{\mathrm{P}}}{V_{\mathrm{S}}}\)
= 50 × \(\frac{200}{50}\) = 200

(ii) Power consumed = V_SI_S = 50 × 0.3 = 15 W

Question 8.
A resistance of 3 Ω is connected to the secondary coil of 60 turns of an ideal transformer. Calculate the current (peak value) in the resistor if the primary has 1200 turns and is connected to 240 V (peak) ac supply. Assume that all the magnetic flux in the primary coil passes through the secondary coil and that there are no other losses.
Solution:
Data : R = 3 Ω, N_S = 60, N_P = 1200, V_P = 240 V
V_S = \(\frac{N_{\mathrm{S}}}{N_{\mathrm{P}}}\) × V_P
= \(\frac{60}{1200}\) × 240 = 12 V (peak)
∴ The peak value of the current in the resistor in the transformer secondary coil is
I_S = \(\frac{V_{\mathrm{S}}}{R}=\frac{12}{3}\) = 4 A

Question 9.
The primary of a transformer has 40 turns and works on 100 V and 100 W. Find the number of turns in the secondary to step up the voltage to 400 V. Also calculate the current in the secondary and primary.
Solution :
Data : N_P = 40, V_P = 100 V, P_P = 100 W, V_S = 400 V

This gives the number of turns in the secondary coil.

(ii) Assuming P_S = P_S = 100 W,
V_SI_S = 100 W
∴ I_S = \(\frac{100}{V_{\mathrm{S}}}=\frac{100}{400}\) = 0.25 A
This gives the current in the secondary coil.

(iii) V_P . I_P = P_P ∴ I_P = \(\frac{P_{\mathrm{P}}}{V_{\mathrm{P}}}=\frac{100}{100}\) = 1 A
This gives the current in the primary coil.

Question 10.
A transformer converts 400 volt ac to 100 volt ac The secondary of the transformer has 50 turns and the load across it draws a current of 600 mA. What is the current in the primary, the power consumed and the number of turns in the primary?
Solution:
Data : V_P = 400 V. V_S = 100 V, N_S = 50, I_S = 0.6 A
Assuming no power loss. P_PV_P = I_SV_S
∴ The current in the primary,

Question 11.
A step down transformer works on 220 V a mains. What is the efficiency of the transformer when a bulb of 100 Wf20 V is connected to the a mains and the current in the primary is 0.5 A ?
Solution:
Data: V_P = 220V, V_S = 20V, P_S = 100W, I_P = 0.5 A
The Input power. P_P = I_PV_P = (0.5)220) = 110 W
The output power, P_S = 100 W
∴ The efficiency of the transformer
= \(\frac{\text { output power }}{\text { input power }}=\frac{100}{110}\) = 0.9091 or 90.91%

Multiple Choice Questions

Question 1.
A circular loop is placed in a uniform magnetic field. The total number of magnetic field lines passing normally through the plane of the coil is called
(A) the displacement current
(B) the eddy current
(C) the self inductance
(D) the magnetic flux
Answer:
(D) the magnetic flux

Question 2.
According to Lenz’s law, the direction of the induced current in a closed conducting loop is such that the induced magnetic field attempts to
(A) maintain the original magnetic flux through the loop
(B) maximize the magnetic flux through the loop
(C) maintain the magnetic flux through the loop to zero
(D) minimize the magnetic flux through the loop.
Answer:
(A) maintain the original magnetic flux through the loop

Question 3.
A metallic conductor AB moves across a magnetic field as shown in the following figure.

Which of the following statements is correct?
(A) The free electrons experience a magnetic force and move to the lower part of the conductor.
(B) The free electrons experience a magnetic force and move to the upper part of the conductor.
(C) The positive and negative charges experience a magnetic force and move, respectively, to the upper and lower parts of the conductor.
(D) The moving conductor gives rise to an emf but there is no separation of charges as they are bound in the solid structure.
Answer:
(A) The free electrons experience a magnetic force and move to the lower part of the conductor.

Question 4.
A bar magnet moves vertically down, approaching a circular conducting loop in the x-y plane. The direction of the induced current in the loop (looking down the z-axis) is

(A) anticlockwise
(B) clockwise
(C) alternating
(D) along negative z-axis.
Answer:
(A) anticlockwise

Question 5.
A moving conductor AB of length 1 makes a sliding electrical contacts at its ends with two parallel conducting rails. The rails are joined at the left edge (CD) by a resistance R to form a complete circuit.

The rate at which the magnetic flux through the area bounded by the circuit changes is
(A) Bv
(B) Bl/v
(C) Bvl
(D) Bv/l.
Answer:
(C) Bvl

Question 6.
A metre gauge train is heading north with speed 54 km/h in the Earth’s magnetic field 3 × 10^-4 T. The emf induced across the axle joining the wheels is
(A) 0.45 mV
(B) 4.5 mV
(C) 45 mV
(D) 450 mV.
Answer:
(B) 4.5 mV

Question 7.
A conducting rod of length l rotates about one of its ends in a uniform magnetic field \(\vec{B}\) with a constant angular speed ω. If the plane of rotation is perpendicular to \(\vec{B}\), the emf induced between the ends of the rod is
(A) \(\frac{1}{2}\)Bωl2
(B) πl²Bω
(C) Bωl²
(D) 2Bωl².
Answer:
(A) \(\frac{1}{2}\)Bωl2

Question 8.
A circular conducting loop of area 100 cm2 and resistance 3 Ω is placed in a magnetic field with its plane perpendicular to the field. If the field is spatially uniform but varies with time t (in second) as B(f) = 1.5 cos ωt tesla, the peak value of the current is
(A) 3 mA
(B) 5ω mA
(C) 300ω mA
(D) 500 mA.
Answer:
(B) 5ω mA

Question 9.
In a simple rectangular-loop ac generator, the time rate of change of magnetic flux is a maximum when
(A) the induced emf has a minimum value
(B) the plane of the coil is parallel to the magnetic field
(C) the plane of the coil is perpendicular to the magnetic field
(D) the emf varies sinusoidally with time.
Answer:
(B) the plane of the coil is parallel to the magnetic field

Question 10.
A simple generator has a 300 loop square coil of side 20 cm turning in a field of 0.7 T. How fast must it turn to produce a peak output of 210 V ?
(A) 25 rps
(B) 4 rps
(C) 2.5 rps
(D) 0.4 rps
Answer:
(B) 4 rps

Question 11.
A rectangular loop generator of 100 turns, each of area 1000 cm², rotates in a uniform field of 0.02 π tesla with an angular velocity of 60 π rad/s. The maximum value of \(\frac{d \Phi_{\mathrm{m}}}{d t}\) is
(A) 12π V
(B) 12π² Wb
(C) 6π² V
(D) 12π² V.
Answer:
(D) 12π² V.

Question 12.
A 250 loop circular coil of area 16π² cm² rotates at 100 rev/s in a uniform magnetic field of 0.5 T. The rms voltage output of the generator is nearly
(A) 200\(\sqrt {2}\) V
(B) 20\(\sqrt {2}\) V
(C) 400 V
(D) 2\(\sqrt {2}\) MV.
Answer:
(A) 200\(\sqrt {2}\) V

Question 13.
Two tightly wound solenoids have the same length and circular cross-sectional area, but the wire of solenoid 1 is half as thick as solenoid 2. The ratio of their inductances is
(A) \(\frac{1}{4}\)
(B) \(\frac{1}{2}\)
(C) 2
(D) 4
Answer:
(D) 4

Question 14.
The wire of a tightly wound solenoid is unwound and used to make another tightly wound solenoid of twice the diameter. The inductance changes by a factor of
(A) 4
(B) 2
(C) \(\frac{1}{2}\)
(D) \(\frac{1}{4}\)
Answer:
(B) 2

Question 15.
The back emf of a dc motor is 108 V when it is connected to a 120 V line and reaches full speed against its normal load. What will be its back emf if a change in load causes the motor to run at half speed ?
(A) 66 V
(B) 12 V
(C) 60 V
(D) 54 V
Answer:
(D) 54 V

Question 16.
A single rectangular loop of wire, of dimensions 0.8 m × 0.4 m and resistance 0.2 Ω, is in a region of uniform magnetic field of 0.5 T in a plane perpendicular to the field. It is pulled along its length at a constant velocity of 5 m/s. Once one of its shorter side is just outside the field, the force required to pull the loop out of the field is

(A) 0.2 N
(B) 0.5 N
(C) 1 N
(D) 2 N.
Answer:
(C) 1 N

Question 17.
A pivoted bar with slots falls through a magnetic field. The bar falls the quickest if it is made of [Assume identical plate and slot dimensions. Ignore air resistance.]

(A) copper
(B) a ferromagnetic
(C) aluminium
(D) plastic
Answer:
(D) plastic

Question 18.
Eddy currents are also called
(A) Maxwell currents
(B) Faraday currents
(C) displacement currents
(D) Foucault currents
Answer:
(D) Foucault currents

Question 19.
At a given instant the current and self-induced emf (e) in an inductor are directed as shown. If e = 60 V,
which of the following is true?

(A) The current is increasing at 2 A/s. 12 H
(B) The current is decreasing at 5 A/s.
(C) The current is increasing at 5 A/s.
(D) The current is decreasing at 6 A/s.
Answer:
(C) The current is increasing at 5 A/s.

Question 20.
A metal ring is placed in a region of uniform magnetic field such that the plane of the ring is perpendicular to the direction of the field. The field strength is increasing at a constant rate. Which of the following graphs best shows the variation with time t of the induced current I in the ring ?

Answer:
(C)

Question 21.
At a given instant, the current through a 60 mH inductor is 50 mA and increasing at 100 mA/ s. The energy stored at that instant is
(A) 150 µJ
(B) 75 µJ
(C) 0.6 mJ
(D) 0.3 mJ
Answer:
(B) 75 µJ

Question 22.
The magnetic field within an air-cored solenoid is 0.8 T. If the solenoid is 40 cm long and 2 cm in diameter, the energy stored in its magnetic field is
(A) 32 J
(B) 3.2 J
(C) 6.4 kJ
(D) 64 kJ
Answer:
(A) 32 J

Question 23.
The adjacent graph shows the E induced emf against time of a coil rotated in a uniform magnetic field at a certain frequency. 0;
If the frequency of rotation is reduced to one half of its initial value, which one of the following graphs correctly shows the new variation of the induced emf with time.

[All the graphs are drawn to the same scale.]

Answer:
(A)

Question 24.
A transformer has 320 turns primary coil and 120 turns secondary coil. Which of the following statements is true?
(A) It changes current by a factor of \(\frac{8}{3}\) and is a step-up transformer.
(B) It is a step-down transformer and changes current by a factor of \(\frac{8}{3}\).
(C) It changes current by a factor of \(\frac{8}{3}\) and is a step-up transformer.
(D) It is a step-down transformer and changes current by a factor of \(\frac{8}{3}\).
Answer:
(B) It is a step-down transformer and changes current by a factor of \(\frac{8}{3}\).

Question 25.
Input power at 11000 V is fed to a step-down transformer which has 4000 turns in its primary winding. In order to get output power at 220 V, the number of turns in the secondary must be
(A) 20
(B) 80
(C) 400
(D) 800.
Answer:
(B) 80

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) \(\frac{S}{S+G}\)