# NCERT Solutions Class 12 Physics Chapter 6

## NCERT Solutions Class 12 Physics Chapter 6 Electromagnetic Induction

The NCERT Solutions Class 12 Physics Chapter 6 are beneficial for students preparing for Class 12 exams. To score good marks, one must prepare well for exams, and these solution sets help students understand the topic more easily. First, students understand the concepts by referring to the NCERT book. Then, after reading the theoretical topics from the textbook, students may use NCERT Solutions Class 12 Physics Chapter 6 to practise regularly.

Class 12 Physics Chapter 6 NCERT Solutions is a comprehensive guide for Class 12 students. All solutions are explained in simple language. Questions from NCERT textbooks, previous year question papers, practise tests and sample papers are all included in NCERT Solutions Class 12 Physics Chapter 6.

### Key Topics Covered In NCERT Solutions Class 12 Physics Chapter 6

Following are the topics covered in NCERT Solutions Class 12 Physics Chapter 6

Section Number
Section Name
6.1
Introduction
6.2
The Experiments Of Faraday And Henry
6.3
Magnetic Flux
6.4
6.5
Lenz’S Law And Conservation Of Energy
6.6
Motional Electromotive Force
6.7
Energy Consideration: A Quantitative Study
6.8
Eddy Currents
6.9
Inductance
6.9.1
Mutual Inductance
6.9.2
Self-Inductance
6.10
AC Generator.

Students may click on the respective topics to access the detailed Chapter 6 Physics Class 12 NCERT Solutions.

The various topics covered under NCERT Solutions Class 12 Physics Chapter 6 are explained in brief below.

6.1 Introduction
Electromagnetic or magnetic induction is inducing an electromotive force throughout an electrical conductor in a converting magnetic field. Michael Faraday is usually credited for discovering induction in 1831, and James Clerk Maxwell mathematically defined it as Faraday’s law of induction.

First Experiment:
In the initial test of Faraday and Henry, a coil was connected to a galvanometer. Then a bar magnet was kept closer to the coil. It was observed that the galvanometer showcased deflection because the bar magnet shifted. The same element became accomplished with the South Pole.
In this test of Faraday and Henry, the shift and deflection passed off only while the magnet was in movement and no longer deflected while it was stationary. The deflection point is small or huge, relying on the movement’s rate.
From Faraday and Henry’s experiment, a relative movement among the coil and magnet was ensured to produce current inside the coil.

Second Experiment:
In this experiment, the bar magnet of the circuit was changed into every other coil that had current generated inside it,connected to a battery. The current coil connected to the battery produces a constant current. The second coil that was the first coil suggests deflection inside the galvanometer pointer, suggesting the presence of current.
Here, the degree of deflection relied on the movement of the secondary coil in the direction of the first coil. The magnitude additionally depends upon the rate with which it moves. This suggests how the second case has similarities to the first.

Third Experiment:
Faraday concluded from the above experiments that the relative movement among the magnet and the coil resulted in the current technology inside the first coil. However, any other test with the aid of Faraday confirmed that relative movement among the coils was now no longer important for the first current to be generated.
He used stationary coils on this test. One was connected to the galvanometer and the opposite to a battery through a push-button. The galvanometer inside the different coil deflected because the button was pressed, displaying the presence of current in that coil. Furthermore, the deflection inside the pointer was only temporary.
However, the first and third experiment of Faraday and Henry showcases that the relative movement isn’t essential to supply current.

6.3 Magnetic Flux
Magnetic flux is the total magnetic field that passes via a given region. It is a beneficial device for assisting in describing the magnetic force results on something occupying a given region.
=B A Cosθ
Here, =magnetic flux
B=magnetic field
A=area
θ=angle between a perpendicular vector to the area and the magnetic field.

First law:
Whenever a conductor is placed in different magnetic fields, an electromotive force is induced. If the conductor circuit is closed, a current is induced, which is referred to as induced current.

Second law:
The induced emf in a coil is the same as the rate of change of flux linkage.
=-Nt

6.5 Conservation of Energy and Lenz law
Lenz law states that the induced current constantly tends to oppose the purpose which produces it. So to do work in opposition to opposing forces, we ought to put in more effort. This more-work results in periodic alternations in magnetic flux; subsequently, a greater current is induced.

6.6 Motional Electromotive Force
An emf brought on through the conductor’s movement throughout the magnetic field is a motional electromotive force. The equation is given through E = -vLB. This equation is so long as the velocity, field, and length are perpendicular. The minus sign is related to Lenz’s law.

6.7 Energy Consideration: A Quantitative Study

Suppose there may be a rectangular conductor. From the given figure, it can be stated that the sides of the rectangular conductor are PQ, RS, QR, and SP. Now on this rectangular conductor, the three sides are fixed, whilst one in each side PQ is readily free.
Let r be that movable resistance of the conductor. So the resistance of the remaining alternative sides of the rectangular conductor is the resistance of sides RS, SP, and QR may be very small compared to this movable resistance. If we change the flux in a consistent magnetic field, an emf is induced. i.e E =- dΦ/dt
If there may be induced emf E and a movable resistance r inside the conductor, we can say that I = Blv. As the magnetic field is present, there can also be a force F acting, as F = ILB. This force is directed outwards inside the direction contrary to the velocity of the rod, given through F = B²l²vR
Power = force × velocity = B²l²v²R
Now, the work completed is mechanical, and this mechanical energy is dissipated as Joule heat. This is given as PJ = I²R = B²l²v²R. Further, the mechanical energy converts into electric energy and ultimately into thermal energy. From Faraday’s law, we’ve found out that |E| =ΔΦBΔt
So we get, |E| = IR = ΔQΔtR
Hence we get ΔQ= ΔΦBR

6.8 Eddy Currents
Eddy currents (additionally referred to as Foucault’s currents) are loops of electrical current brought about inside conductors using a converting magnetic field inside the conductor in step with Faraday’s law of induction. Eddy currents flow in closed loops inside conductors, in planes perpendicular to the magnetic field.

6.9 Inductance
Induction is a property of an electric-powered circuit with the aid of using which an electromotive force is brought about in it using a version of current both inside the circuit itself or in a neighbouring circuit.

6.9.1 Mutual Inductance
A degree of mutual induction among magnetically related circuits is given because of the ratio of the electromotive force to the rate of alternation of current generating it.

6.9.2 Self-Inductance
It is the property of the current-carrying coil that resists or opposes the alternation of current flowing via it. This occurs precisely because of the self-induced emf produced inside the coil itself.

6.10 AC Generator
It is a device that converts mechanical power into electric power, generated as an alternating current sinusoidal output waveform.
AC generators work at the precept of Faraday’s law of electromagnetic induction. When the armature rotates between the magnet’s poles upon an axis perpendicular to the magnetic field, the flux linkage of the armature adjusts continuously. Due to this, an emf is caused inside the armature. Students may refer to the NCERT Solutions Class 12 Physics Chapter 6 Electromagnetic Induction on Extramarks for detailed study notes.

List of NCERT Solutions Class 12 Physics Chapter 6 Electromagnetic Induction

Click on the below links to view NCERT Solutions for ch 6 Physics class 12:
NCERT Solutions Class 12 Physics Chapter 6: Exercise 6.1
NCERT Solutions Class 12 Physics Chapter 6: Exercise 6.2
NCERT Solutions Class 12 Physics Chapter 6: Exercise 6.3
NCERT Solutions Class 12 Physics Chapter 6: Exercise 6.4
NCERT Solutions Class 12 Physics Chapter 6: Exercise 6.5
NCERT Solutions Class 12 Physics Chapter 6: Exercise 6.6
NCERT Solutions Class 12 Physics Chapter 6: Exercise 6.7
NCERT Solutions Class 12 Physics Chapter 6: Exercise 6.8
NCERT Solutions Class 12 Physics Chapter 6: Exercise 6.9.1
NCERT Solutions Class 12 Physics Chapter 6: Exercise 6.9.2
NCERT Solutions Class 12 Physics Chapter 6: Exercise 6.10

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### Key Features of NCERT Solutions Class 12 Physics Chapter 6

The key features of the NCERT Solutions Class 12 Physics Chapter 6 offered by Extramarks are as follows

Class 12 Physics Chapter 6 is all about Electromagnetic induction. The notes give a detailed explanation of the relationship between magnetism and electricity. In addition, it also explains Faraday’s law and Lenz’s law which are considered to be vital topics. These notes are detailed to help students excel in their examinations. Students can expect some numerical questions based on this chapter. Hence, all problems come with an easy-to-understand solution. In addition, the Solutions also offer various practice exercises.

The NCERT Solutions Class 12 Physics Chapter 6 is curated by some of the most experienced subject matter experts to make the learning process easier. Students preparing for Class 12 or competitive exams like JEE or NEET can refer to solutions. The NCERT Solutions Class 12 Physics Chapter 6 also includes sample papers, revision notes, and important questions created by our highly experienced teachers. Extramarks provide solutions, notes and explanatory articles for all subjects and classes.

Q.1 Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f).

Ans.

Lenz’s law gives the direction of the induced current in a closed loop. The given pair of figures indicate the direction of induced current when the North pole of a bar magnet is moved towards and away from a closed loop.

According to Lenz’s law, the direction of induced current in the given situations can be given as follows:
(a) The induced current is directed along qrpq.
(b) The induced current is directed along prqp.
(c) The induced current is directed along yzxy.
(d) The induced current is directed along zyxz.
(e) The induced current is directed along xryx.
(f) No current is induced because the magnetic field lines are lying in the plane of the closed loop.

Q.2 Use Lenz’s law to determine the direction of induced current in the situation described by Fig. 6.19:

(a) A wire of irregular shape turning into a circular shape;

(b) A circular loop being deformed into a narrow straight wire.

Ans.

According to Lenz’s law, the direction of the induced current is such that it tends to produce a current that opposes the change in the magnetic flux that produced it.

(a)When a wire of irregular shape changes to a circular loop, the area of the loop tends to increase. Thus, magnetic flux associated with the loop increases. As per Lenz’s law, the direction of induced current must oppose the magnetic field. Therefore, the induced current is flowing in anticlockwise direction in the loop from adcb.

(b)When the shape of a circular loop is deformed into a narrow straight wire, the magnetic flux tends to decrease. Thus, the induced current flows along a’d’c’b’.

Q.3 A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?

Ans.

$Here, number of turns on the solenoid = 15 turns cm -1 = 1500 turns m -1 Number of turns per unit length, n = 1500 turns Area of the small loop of solenoid, A = 2 .0 cm 2 = 2 × 10 -4 m 2 As current in the solenoid changes from 2 A to 4 A, ∴Change in current , di = 4 A – 2 A = 2 A Time taken , dt = 0.1 s Rate of change of current = di dt = 2 0.1 = 20 As -1 According to Faraday’s law, induced emf in the solenoid is given as: e = df dt ….(i) Here, ϕ = Induced flux through the small loop ϕ = BA … ii B = Magnetic field = μ 0 ni μ 0 = Permeability of free space = 4π ×10 -7 Hm -1 ∴Equation i becomes: e = d dt BA = Aμ 0 n × di dt = 2×10 -4 m 2 × 4π ×10 -7 Hm −1 × 1500 × 2 A 0.1 s =7 .54 × 10 -6 V ∴The induced voltage in the loop = 7 .54 ×10 -6 V$

Q.4 A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s−1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

Ans.

$Here, length of rectangular wire,l = 8 cm = 0.08 m Width of rectangular wire,b = 2 cm = 0.02 m ∴Area of rectangular loop, A = lb = 0.08 m × 0.02 m = 16×10 -4 m 2 Magnetic field strength, B = 0.3 T Velocity of rectangular loop, v =1 cms -1 =0 .01 ms -1 a When velocity is normal to the longer side Emf developed in the loop is given by, e = Blv = 0.3 T × 0.08 m × 0 .01 ms -1 = 2 .4 × 10 -4 V Time taken to move along the width, t = Distance travelled Velocity = 2 × 10 -2 m 10 -2 ms -1 = 2 s Hence, the induced voltage is 2 .4 × 10 -4 V which lasts for 2 s. b When velocity is normal to the shorter side, Emf developed is given as: e = Bbv = 0.3 T × 0.02 m × 0 .01 ms -1 = 0 .6 × 10 -4 V Time taken to move along the length is given as: t = Distance travelled Velocity = 8 × 10 -2 m 10 -2 ms -1 = 8 s ∴The induced voltage is 0 .6 × 10 -4 V which lasts for 8 s.$

Q.5 A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s−1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

Ans.

$Here, length of the rod, l = 1 m Angular frequency of rotation,ω = 400 rads -1 Magnetic field strength, B = 0.5 T Linear velocity of one end of the rod has zero , and the linear velocity of the other end is lω. Average linear velocity,v = 0 + lω 2 = lω 2 Emf induced between the centre and the ring, e = Blv ∴e = Bl lω 2 = Bl 2 ω 2 = 0.5 T × 1 m 2 × 400 rads -1 2 = 100 V ∴EMF developed between the centre and the ring is 100 V.$

Q.6 A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s−1 in a uniform horizontal magnetic field of magnitude 3.0 × 10−2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω? Calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?

Ans.

$Here, radius of the circular coil, r = 0.08 m ∴Area of the circular coil, A = π r 2 = π × 0.08 2 m 2 Number of turns on the circular coil, N = 20 Angular speed of circular coil, ω = 50 rads -1 Magnetic field strength, B = 3 × 10 -2 T Resistance of the closed loop, R = 10 Ω Maximum induced emf is given by: e = NωAB = 20 × 50 rads -1 × π × 0.08 m 2 × 3 × 10 -2 T = 0.603 V Thus, maximum emf induced in the coil = 0.603 V. Over a full cycle, the average emf induced in the coil= 0 Maximum current is given by the relation: I = e R = 0.603 V 10 Ω = 0.0603 A Average power loss due to joule heating is given as: P = eI 2 = 0.603 V × 0.0603 A 2 = 0.018 W The induced current in the coil develops a torque opposing the rotation of the coil. The external agent (rotor) must supply a torque to counter this torque in order to keep the coil rotating uniformly. ∴The external rotor acts as the source of power dissipated as heat in the coil.$

Q.7 A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s−1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10−4 Wb m−2.

(a) What is the instantaneous value of the emf induced in the wire?

(b) What is the direction of the emf?

(c) Which end of the wire is at the higher electrical potential?

Ans.

$Here, length of wire, l = 10 m Speed of the wire, v = 5.0 ms −1 Magnetic field strength, B = 0.3 × 1 0 −4 Wbm −2 a Induced emf in the wire, e=Blv =0.3 ×1 0 −4 Wbm −2 ×10 m×5.0 ms −1 =1.5×1 0 −3 V (b) From Fleming’s right hand rule, it can be concluded that the direction of the induced emf is from West to East. (c) The western end of the wire must be at higher potential.$

Q.8 Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.

Ans.

$Here, initial current, I 1 = 5.0 A Final current, I 2 = 0.0 A Change in current, dI = I 1 – I 2 = 5 A Time taken for the change, dt = 0.1 s Rate of change of current = dI dt = 5 A 0.1 s = 50 As -1 Average induced emf, e = 200 V The self-inductance L is related to average induced emf as: e = L dI dt ∴L = e dI dt = 200 50 = 4 H ∴The self induction of the coil is 4 H.$

Q.9 A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?

Ans.

$Here, mutual inductance, μ = 1.5 H Initial current, I 1 = 0 A Final current, I 2 = 20 A Change in current,dI = I 2 – I 1 = 20 A – 0 A = 20 A Time taken, dt = 0.1 s Induced emf is given as: e = dϕ dt →(i) Here, dϕ = Change in the flux linkage with coil Relation between emf and mutual inductance is given as: e = μ dI dt →(ii) Comparing equation (i) and (ii), we obtain: dϕ dt = μ dI dt ∴dϕ = μdI ∴dϕ = 1.5 H × 20 A = 30 Wb ∴The change in the flux linkage is 30 Wb.$

Q.10 A jet plane is travelling towards west at a speed of 1800 kmh-1. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 ×10−4 T and the dip angle is 30°.

Ans.

$Here, speed of jet plane, v = 1800 kmh −1 = 500 ms −1 Wing span of the jet plane, l = 25 m Magnetic field strength of Earth, B = 5.0 × 1 0 −4 T Angle of dip, δ=30° Vertical component of Earth’s magnetic field is given as: B V = B sinδ = 5×1 0 −4 sin 30° = 2.5×1 0 −4 T Voltage difference between the ends of the wing is given as: e = B V × l × v = 2.5 × 1 0 −4 T× 25 A × 500 ms −1 = 3.125 V ∴Voltage difference developed between the ends of the wings = 3.125 V.$

Q.11 Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s−1. If the cut is joined and the loop has a resistance of 1.6 Ω, how much power is dissipated by the loop as heat? What is the source of this power?

Ans.

$Here, sides of the rectangular loop are 8 cm and 2 cm. ∴Area of the rectangular loop,A = length × width = 8 cm × 2 cm = 16 cm 2 = 16 × 10 -4 m 2 Initial magnetic field, B 1 = 0.3 T Rate of decrease of magnetic field, dB dt = 0.02 Ts -1 Induced emf developed in the rectangular loop is given as: e = dϕ dt Here, dϕ = Change in the flux through the loop area = AB ∴e = d AB dt = A dB dt = 16 × 10 -4 m 2 × 0 .02 Ts -1 = 0 .32 × 10 -4 V Resistance of loop, R = 1.6 Ω Induced current, i = e R = 0 .32 ×10 -4 V 1.6 Ω = 0 .2 × 10 -4 A Power dissipated as heat is given as: P = i 2 R = 2 × 10 -5 A 2 ×1.6 Ω = 6 .4 × 10 -10 W The source of this heat loss is an external agent, which causes the changing the magnetic field with time.$

Q.12 A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s−1 in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10−3 T cm−1 along the negative x-direction (that is it increases by 10− 3 T cm−1 as one moves in the negative x-direction), and it is decreasing in time at the rate of 10−3 T s−1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.

Ans.

$Here, side of square loop, is = 12 cm = 0.12 m ∴Area of square loop, A = 0.12 m × 0.12 m = 0 .0144 m 2 Velocity of square loop, v = 8 cms -1 = 0 .08 ms -1 Gradient of magnetic field along negative x-direction: dB dx = 10 -3 Tcm -1 =10 -1 Tm -1 Rate of decrease of the magnetic field: dB dt = 10 -3 Ts -1 Resistance of loop, R = 4.5 mΩ = 4 .5 ×10 -3 Ω Rate of change of magnetic flux because of motion of loop in a non-uniform magnetic field: dϕ dt = A × dB dx × v = 0 .0144 m 2 × 10 -1 Tm -1 × 0 .08 ms -1 =11 .52 × 10 -5 Tm 2 s -1 Rate of change of magnetic flux due to explicit time change in ϕ =A dB dt = 0 .0144 × 10 -3 Wbs -1 = 1 .44 ×10 -5 Wbs -1 Since, rate of change of magnetic flux = induced emf ∴Total induced emf, e = 1 .44 × 10 -5 Wbs -1 + 11 .52 × 10 -5 Tm 2 s -1 = 12 .96 ×10 -5 V ∴ Induced current, i = e R = 12 .96 × 10 -5 V 4 .5 ×10 -3 Ω i = 2 .88 × 10 -2 A ∴The direction of induced current is such that there is an increase in the magnetic flux through the loop along positive z-direction.$

Q.13 It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm2 with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 Ω. Estimate the field strength of magnet.

Ans.

$Here, area of the small flat search coil, A = 2 cm 2 = 2 × 10 -4 m 2 Number of turns on the coil, N = 25 Total charge flowing through the coil, Q = 7.5 mC = 7 .5 × 10 -3 C Total resistance of the coil and galvanometer, R = 0.50 Ω Induced current through the coil is given as: I = e R →(i) Induced emf, e = -N dϕ dt →(ii) Here, dϕ = Change in flux Comparing equations i and ii , we obtain: I = – N dϕ dt R ∴Idt = – N R dϕ →(iii) Initial flux through coil is given as: ϕ i = BA Here, B = Strength of magnetic field Final flux through coil, ϕ f = 0 Integrating equation (iii) on both sides, we obtain: ∫ Idt =- N R ∫ ϕ i ϕ f dϕ Total charge, Q = ∫ Idt ∴Q = – N R ϕ f – ϕ i = – N R 0 – ϕ i = N ϕ i R Q = NBA R ∴B = QR NA = 7 .5 ×10 -3 C × 0.5 Ω 25 × 2 ×10 -4 m 2 = 0.75 T ∴The magnetic field strength of the magnet is 0.75 T.$

Q.14 Figure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod= 9.0 mΩ. Assume the field to be uniform.

(a) Suppose K is open and the rod is moved with a speed of 12 cm s−1 in the direction shown. Give the polarity and magnitude of the induced emf.

(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?

(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.

(d) What is the retarding force on the rod when K is closed?

(e) How much power is required (by an external agent) to keep the rod moving at the same speed (=12 cm s−1) when K is closed? How much power is required when K is open?

(f) How much power is dissipated as heat in the closed circuit? What is the source of this power?

(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?

Ans.

$Here, length of rod, l = 15 cm= 0.15 m Strength of magnetic field , B = 0.50 T Resistance of the closed loop containing rod, R= 9 mΩ = 9 × 10 -3 Ω a Here, speed of rod, v = 12 cms -1 = 0 .12 ms -1 Magnitude of induced emf is given as:e = Bvl = 0.5 × 0.12 × 0.15= 9 mV The polarity of the induced emf is such that P acquires positive polarity while Q acquires negative polarity. b Yes; when key K is open, excess positive charge develops at P and an equal negative excess charge develops at Q. When key K is closed, continuous flow of induced current maintains the excess charge. c When key K is open and the rod is moving uniformly, there is no net force on the electron in the rod PQ. This is because, the effect of Lorentz magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rod. d Here, induced current flowing through the rod, I = e R = 9 × 10 -3 9 × 10 -3 = 1 A Retarding force on the rod, F = BIl F = 0.50 T ×1 A × 0.15 m = 7 .5 × 10 -2 N e Here, speed of rod, v = 12 cms -1 = 0 .12 ms -1 Retarding force on the rod, F = 7 .5 ×10 -2 N ∴Power = F×v P = 7 .5 × 10 -2 N× 0 .12 ms -1 = 9 × 10 -3 W. f Power dissipated as heat: P = I 2 R= 1 A 2 × 9 × 10 -3 Ω = 9 × 10 -3 W = 9 mW Source of power is external agent in this case. g As magnetic field is parallel to the rails, θ = 0 o ∴Induced emf = Bvlsinθ = Bvlsin0 o = 0 In this case, motion of the rod does not cross the field lines so that flux change = 0 and therefore, e = 0.$

Q.15 An air-cored solenoid with length 30 cm, area of cross-section 25 cm2 and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10−3 s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.

Ans.

$Here, length of solenoid, l = 30 cm = 0.3 m Cross-sectional area, A = 25 cm 2 = 25 × 10 -4 m 2 Number of turns, N = 500 Current flowing through the solenoid, I = 2.5 A Time for which current flows, t = 10 -3 s Average back emf is given as: e = dϕ dt →(i) Here, dϕ = Change in flux = NAB → 2 Here, B = Magnetic field strength = μ o NI l →(iii) Here, μ o = Permeability of free space = 4π × 10 -7 TmA -1 Applying equations ii and iii in equation i , we obtain: e = μ o N 2 IA lt = 4π × 10 -7 TmA -1 × 500 2 × 2 .5 A × 25 ×10 -4 m 2 0 .3 m ×10 -3 s = 6.5 V ∴Average back emf induced in the solenoid is 6.5 V.$

Q.16 (a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. 6.21.

(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s. Calculate the induced emf in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large resistance.

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{Take a small strip of width dy in the loop at a}\\ \text{distance y from the wire carrying current.}\\ \text{Here, length of small strip = a}\\ \text{Area of small strip, dA = a × dy}\\ \text{Let current flowing through the wire = I}\\ \text{Magnetic field due to current carrying wire at a }\\ \text{distance y from the wire is given as: B =}\frac{{\text{μ}}_{\text{o}}\text{I}}{\text{2}\mathrm{\pi }\text{y}}\\ \text{Magnetic flux associated with the small strip, d}\mathrm{\varphi }\text{= BdA}\\ \therefore \text{d}\mathrm{\varphi }\text{ = }\frac{{\text{μ}}_{\text{o}}\text{I}}{\text{2}\mathrm{\pi }\text{y}}\left(\text{a × dy}\right)\\ \text{= }\frac{{\text{μ}}_{\text{o}}\text{Ia}}{\text{2}\mathrm{\pi }}\frac{\text{dy}}{\text{y}}\\ {\text{Here, μ}}_{\text{o}}\text{= Permeability of free space}\\ \text{= 4}\mathrm{\pi }{\text{× 10}}^{\text{-7}}{\text{T m A}}^{\text{-1}}\\ \text{Magnetic flux associated with the square loop is given as:}\\ \mathrm{\varphi }\text{=}\frac{{\text{μ}}_{\text{o}}\text{Ia}}{\text{2}\mathrm{\pi }}\underset{\text{y = x}}{\overset{\text{y = a+x}}{\int }}\frac{\text{dy}}{\text{y}}\\ \mathrm{\varphi }\text{=}\frac{{\text{μ}}_{\text{o}}\text{Ia}}{\text{2}\mathrm{\pi }}{\left[{\text{log}}_{\text{e}}\text{y}\right]}_{\text{x}}^{\text{a+x}}\\ \mathrm{\varphi }\text{=}\frac{{\text{μ}}_{\text{o}}\text{Ia}}{\text{2}\mathrm{\pi }}{\text{log}}_{\text{e}}\left(\frac{\text{a+x}}{\text{x}}\right)\\ \text{For mutual inductance M, the magnetic flux is given as:}\\ \mathrm{\varphi }\text{= MI}\\ \therefore \text{MI =}\frac{{\text{μ}}_{\text{o}}\text{Ia}}{\text{2}\mathrm{\pi }}{\text{log}}_{\text{e}}\left(\frac{\text{a+x}}{\text{x}}\right)\\ \text{M =}\frac{{\text{μ}}_{\text{o}}\text{a}}{\text{2}\mathrm{\pi }}{\text{log}}_{\text{e}}\left(\frac{\text{a}}{\text{x}}\text{+1}\right)\\ \left(\text{b}\right)\text{ Given, I = 50 A}\\ \text{x = 0.2 m}\\ \text{a = 0.1 m}\\ {\text{v = 10 ms}}^{\text{-1}}\\ \text{Induced emf in the loop is given as:}\\ \text{e = Bav}\\ \therefore \text{e =}\left(\frac{{\text{μ}}_{\text{o}}\text{I}}{\text{2}\mathrm{\pi }\text{x}}\right)\text{av}\\ \text{e =}\frac{\text{4}\mathrm{\pi }{\text{× 10}}^{\text{-7}}{\text{T m A}}^{\text{-1}}{\text{× 50 A × 0.1 m × 10 ms}}^{\text{-1}}}{\text{2}\mathrm{\pi }\text{× 0.2 m}}\\ {\text{e = 5 ×10}}^{\text{-5}}\text{ V}\end{array}$

Q.17 A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. 6.22). A uniform magnetic field extends over a circular region within the rim. It is given by,

B = – B0 k (r ≤ a; a < R)
= 0 (otherwise)

What is the angular velocity of the wheel after the field is suddenly switched off?

Ans.

$\begin{array}{l}\text{Here, line charge per unit length = λ}\\ \text{Mass of wheel = M}\\ \text{Radius of wheel = R}\\ \text{Line charge per unit length is given as: λ =}\frac{\text{Net charge}}{\text{Length}}\text{=}\frac{\text{Q}}{\text{2}\mathrm{\pi }\text{r}}\\ \text{Here, r = Distance of the point within the wheel}\\ \text{Magnetic field strength is given as: }\stackrel{\to }{\text{B}}{\text{= -B}}_{\text{o}}\stackrel{^}{\text{k}}\\ \text{At distance r, the centripetal force balances the }\\ \text{magnetic force i.e.,}\\ \text{BQv =}\frac{{\text{Mv}}^{\text{2}}}{\text{r}}\\ \text{Here, v = linear velocity of wheel}\\ \therefore \text{B2}\mathrm{\pi }\text{rλ =}\frac{\text{Mv}}{\text{r}}\\ \text{v =}\frac{\text{B2}\mathrm{\pi }{\text{λr}}^{\text{2}}}{\text{M}}\\ \therefore \text{Angular velocity is given as:}\\ \text{ω =}\frac{\text{v}}{\text{R}}\text{=}\frac{\text{B2}\mathrm{\pi }{\text{λr}}^{\text{2}}}{\text{MR}}\\ \text{In case of r}\le \text{a and a < R, we have:}\\ \text{ω = –}\frac{\text{2}\mathrm{\pi }{\text{B}}_{\text{o}}{\text{a}}^{\text{2}}\text{λ}}{\text{MR}}\stackrel{^}{\text{k}}\end{array}$