# NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction

## NCERT Solutions for Class 12 Physics Chapter 6 – Free PDF Download

The NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction is crucial for the students of 12th standard. The NCERT Solutions for Class 12 Physics Chapter 6 PDF is provided here to help students understand the chapter in an easy and interesting way.Â In order to understand the topic thoroughly and to sort out Class 12 Physics Chapter 6 Electromagnetic Induction NCERT notes, students must practise these solutions regularly.

Class 12 PhysicsÂ NCERT Solutions for Chapter 6, Electromagnetic Induction, give comprehensive answers to the questions provided in textbooks, previous years’ question papers and sample papers.Â  The NCERT Solutions for Class 12 Physics for all the chapters are created by subject experts according to the latest CBSE Syllabus 2023-24 and its guidelines.

## NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction

### Class 12 Physics NCERT Solutions Electromagnetic Induction Important Questions

Q 1. Predict the direction of induced current in the situations described by the following figures.

Lenzâ€™s law shows the direction of the induced current in a closed loop. The given two figures show the direction of induced current when the North pole of a bar magnet is moved towards and away from a closed loop, respectively.

We can predict the direction of induced current in different situations by using Lenzâ€™s rule:

(i) The direction of the induced current is along qrpq.

(ii) The direction of the induced current is along prq along yzx.

(iii) The direction of the induced current is along yzxy.

(iv) The direction of the induced current is along zyxz.

(v) The direction of the induced current is along xryx.

(vi) No current is induced since the field lines are lying in the plane of the closed loop.

Q 2. We are rotating a 1 m long metallic rod with an angular frequency of 400 red

$$\begin{array}{l}s^{-1}\end{array}$$
with an axis normal to the rod passing through its one end. And on to the other end of the rod, it is connected with a circular metallic ring. There exist a uniform magnetic field of 0.5 T, which is parallel to the axis everywhere. Find out the emf induced between the centre and the ring.

Ans:

Length of the rod = 1m

Angular frequency,

$$\begin{array}{l}\omega = 400 \; rad/s\end{array}$$

Magnetic field strength, B = 0.5 T

At one of the ends of the rod, it has zero liner velocity, while on its other end, it has a linear velocity of

$$\begin{array}{l} I \omega\end{array}$$

Average linear velocity of the rod,

$$\begin{array}{l}v = \frac{I \omega + 0}{ 2 } = \frac{I \omega}{2}\end{array}$$

Emf developed between the centre and ring.

$$\begin{array}{l}\\e = Blv = Bl\left ( \frac{I \omega }{2} \right ) = \frac{B l^{2 } \omega}{2} \\ \\ = \frac{0.5 \times \left ( 1 \right )^{2} \times 400}{2} = 100 V\end{array}$$

Hence, the emf developed between the centre and the ring is 100 V.

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:

Number of turns on the solenoid â€“ 15 turns/cm = 1500 turns / m

Number of turns per unit length, n = 1500 turns

The solenoid has a small loop of area, A = 2.0 cm2 = 2 Ã— 10âˆ’4 m2

The current carried by the solenoid changes from 2 A to 4 A.

Therefore, Change in current in the solenoid, di = 4 â€“ 2 = 2 A

Change in time, dt = 0.1 s

According to Faradayâ€™s law, induced emf in the solenoid is given by:

$$\begin{array}{l}e = \frac{d\phi }{dt} \;\;\;\;\;\;\;\;\; . . . (1)\end{array}$$

Where,

$$\begin{array}{l}\phi\end{array}$$
= Induced flux through the small loop

= BA Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  . . . (2)

B = Magnetic field

=

$$\begin{array}{l}\mu _{0} ni\end{array}$$

$$\begin{array}{l}\mu _{0}\end{array}$$
= Permeability of free space

=

$$\begin{array}{l}4\pi \times 10 ^{-7} \; H/m\end{array}$$

Hence, equation (1) can be reduced to:

$$\begin{array}{l}e = \frac{d}{dt}\left ( BA \right )\end{array}$$

$$\begin{array}{l}e = A\, \mu _{0}\, n \times \left (\frac{di}{dt} \right ) \\ \\ = 2 \times 10^{-4} \times 4 \pi \times 10 ^{-7} \times 1500 \times \frac{2}{0.1}\\ \\ = 7.54 \times 10^{-6}\; V\end{array}$$

Hence, the induced voltage in the loop is

$$\begin{array}{l}7.54 \times 10^{-6}\; V\end{array}$$

Q 4. A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of the 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 sâ€“1 in a direction normal to the (a) longer side and (b) shorter side of the loop? For how long does the induced voltage last in each case?

Ans:

Length of the wired loop, l = 8 cm = 0.08 m

Width of the wired loop, b = 2 cm = 0.02 m

Since the loop is a rectangle, the area of the wired loop,

A = lb

= 0.08 Ã— 0.02

=

$$\begin{array}{l}16 \times 10 ^{-4} \, m^{2}\end{array}$$

Strength of magnetic field, B = 0.3 T

Velocity of the loop, v = 1 cm / s = 0.01 m / s

(i) Emf developed in the loop is given as:

e = Blv

= 0.3 Ã— 0.08 Ã— 0.01 =

$$\begin{array}{l}2.4\times 10^{-4} \; V\end{array}$$

Time taken to travel along the width ,

$$\begin{array}{l}t = \frac{Distance \; travelled}{Velocity} = \frac{b}{v} \\ \\ = \frac{0.02}{0.01} = 2s\end{array}$$

Hence, the induced voltage is

$$\begin{array}{l}2.4\times 10^{-4} \; V\end{array}$$
, which lasts for 2 s.

(ii) Emf developed, e = Bbv

= 0.3 Ã— 0.02 Ã— 0.01 =

$$\begin{array}{l}0.6 \times 10^{-4} \, V\end{array}$$

The time taken to travel along the length,

$$\begin{array}{l}t = \frac{Distance \; travelled}{Velocity} = \frac{l}{v} \\ \\ = \frac{0.08}{0.01} = 8s\end{array}$$

Hence, the induced voltage is

$$\begin{array}{l}0.6 \times 10^{-4} \, V\end{array}$$
, which lasts for 8 s.

Â

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 sâ€“1 in a uniform horizontal magnetic field of magnitude 3.0 Ã— 10â€“2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 â„¦, 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:

Maximum emf induced = 0.603 V

Average emf induced = 0 V

Maximum current in the coil = 0.0603 A

Power loss (average) = 0.018 W

(Power which is coming from external rotor)

Circular coil radius, r = 8 cm = 0.08 m

Area of the coil,

$$\begin{array}{l}A = \pi r^{2} = \pi \times \left ( 0.08 \right )^{2}m^{2}\end{array}$$

Number of turns on the coil, N = 20

Angular speed,

$$\begin{array}{l}\omega = 50 \; rad/s\end{array}$$

Strength of magnetic,

$$\begin{array}{l}B = 3 \times 10 ^{-2} \; T\end{array}$$

Total resistance produced by the loop,

$$\begin{array}{l}R = 10 \Omega\end{array}$$

Maximum emf induced is given as:

$$\begin{array}{l}e = N\omega \; AB\\ = 20 \times 50 \times \pi \times \left ( 0.08 \right )^{2} \times 3 \times 10^{-2} \\ = 0.603 \; V\end{array}$$

The maximum emf induced in the coil is 0.603 V.

Over a full cycle, the average emf induced in the coil is zero.

The maximum current is given as:

$$\begin{array}{l}I = \frac{e}{R} \\ = \frac{0.603}{10} = 0.0603\; A\end{array}$$

Average power because of the Joule heating:

$$\begin{array}{l}P = \frac{el}{2} \\ = \frac{0.603 \times 0.0603}{2} = 0.018\; W\end{array}$$

The torque produced by the current induced in the coil opposes the normal rotation of the coil. To keep the rotation of the coil continuous, we must find a source of torque which opposes the torque by the emf, so here the rotor works as an external agent. Hence, dissipated power comes from the external rotor.

Q 7. A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m sâ€“1, at right angles to the horizontal component of the earthâ€™s magnetic field, 0.30 Ã— 10â€“4 Wb mâ€“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:

Wireâ€™s length, l = 10 m

Speed of the wire with which it is falling, v = 5.0 m/s

Strength of magnetic field, B =

$$\begin{array}{l}0.3 \times 10 ^{-4} Wb \, m^{-2}\end{array}$$

(a) EMF induced in the wire, e = Blv

$$\begin{array}{l}\\ =0.3 \times 10 ^{-4} \times 5 \times 10 \\ = 1.5 \times 10 ^{-3} \, V\end{array}$$

(b) We can determine the direction of the induced current by using Flemingâ€™s right-hand thumb rule; here, the current is flowing in the direction from West to East.

(c) In this case, the eastern end of the wire will have a 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 is induced, give an estimate of the self-inductance of the circuit.

Ans:

Current at initial point,

$$\begin{array}{l}I_{1}\end{array}$$
:Â  5.0 A

Current at final pint,

$$\begin{array}{l}I_{2}\end{array}$$
: 0.0 A

Therefore, change in current is, dI =

$$\begin{array}{l}I_{1} – I_{2} \end{array}$$
= 5 A

Total time taken, t = 0.1 s

Average EMF, e = 200 V

We have the relation for selfâ€“inductance (L) and average emf of the coil:

$$\begin{array}{l}\\e = L \frac{di}{dt} \\ \\ L = \frac{e}{\left (\frac{di}{dt} \right )} \\ \\ = \frac{200}{\frac{5}{0.1}} = 4 H\end{array}$$

Hence, the self-induction of the coil is 4 H.

Â

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?

Â Ans:

Given,

Mutual inductance,

$$\begin{array}{l}\mu = 1.5 \; H\end{array}$$

Current at initial point,

$$\begin{array}{l}I_{1}\end{array}$$
:Â  0 A

Current at final point,

$$\begin{array}{l}I_{2}\end{array}$$
: 20 A

Therefore, change in current is, dI =

$$\begin{array}{l}I_{2} – I_{1} \end{array}$$
= 20 â€“ 0 = 20 A

The time taken for the change, t = 0.5s

Emf induced,

$$\begin{array}{l}e = \frac{d\phi }{dt}\end{array}$$
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  . . . (1)

$$\begin{array}{l}d\phi\end{array}$$
= change in the flux linkage with the coil.

The relation of emf and inductance is:

$$\begin{array}{l}e = \mu \frac{dl}{dt}\end{array}$$
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  . . . (2)

On equating both the equation, we get

$$\begin{array}{l}\frac{d\phi }{dt} = \mu \frac{dl}{dt}\end{array}$$
$$\begin{array}{l}\frac{d\phi }{dt} = 1.5 \times \left ( 20 \right ) \\ \\ = 30 \;Wb\end{array}$$

Therefore, the change in flux linkage is 30 Wb.

Q 10. A jet plane is travelling towards the west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earthâ€™s magnetic field at the location has a magnitude of 5 Ã— 10â€“4 T and the dip angle is 30Â°?

Ans:

Speed of the plane with which it is moving,Â  v = 1800 km/h = 500 m/s

Wing span of the jet, l = 25 m

Magnetic field strength by earth, B =

$$\begin{array}{l}5 \times 10 ^{-4} \; T\end{array}$$

Dip angle,

$$\begin{array}{l}\delta = 30^{\circ}\end{array}$$

The vertical component of Earthâ€™s magnetic field,

$$\begin{array}{l}\\B_{v} = B \sin \delta \\ = 5 \times 10^{-4} \sin 30^{\circ} \\ = 2.5 \times 10^{-4} T\end{array}$$

The difference in voltage between both ends can be calculated as:

$$\begin{array}{l}\\e = \left (B_{v} \right ) \times l \times v \\ \\ = 2.5 \times 10^{-4} \times 25 \times 500 \\ \\ = 3.125 V\end{array}$$

Hence, the voltage difference developed between the ends of the wings is 3.125 V.

Q 11. Let us assume that the loop in question number 4 is stationary or constant, but the current source, which is feeding the electromagnet which is producing the magnetic field, is slowly decreased. It had an initial value of 0.3 T, and the rate of reducing the field was 0.02 T / sec. If the cut is joined to form the loop having a resistance of

$$\begin{array}{l}1.6 \Omega\end{array}$$
, calculate how much power is lost in the form of heat. What is the source of this power?

Â

Ans:

The rectangular loop has sides of 8 cm and 2 cm.

Therefore, the area of the loop will be, A = L Ã— B

= 8 cm Ã— 2 cm

=

$$\begin{array}{l}16 cm^{2}\end{array}$$

=

$$\begin{array}{l}16 \times 10^{-4} cm^{2}\end{array}$$

Value of magnetic field at the initial phase, Bâ€™ = 0.3 T

Magnetic fields decreasing rate,

$$\begin{array}{l}\frac{dB}{dt} = 0.02 T/s\end{array}$$

Emf induced in the loop is:

$$\begin{array}{l}e = \frac{d\phi }{dt}\end{array}$$
$$\begin{array}{l}d\phi\end{array}$$
= change in flux in the loop area

= AB

$$\begin{array}{l}âˆ´ e = \frac{d\left ( AB \right )}{dt} = \frac{AdB}{dt} \\ \\ = 16 \times 10^{-4} \times 0.02 = 0.32 \times 10^{-4} V\end{array}$$

Resistance in the loop will be, R =

$$\begin{array}{l}1.6 \Omega\end{array}$$

The current developed in the loop will be:

$$\begin{array}{l}i = \frac{e}{R} \\ \\ = \frac{032 \times 10 ^{-4}}{1.6} = 2 \times 10^{-5} A\end{array}$$

Power loss in the loop in the form of heat is:

$$\begin{array}{l}P = i^{2}R \\ \\ = \left ( 2 \times 10^{-5} \right )^{2} \times 1.6 \\ \\ = 6.4 \times 10^{-10} W\end{array}$$

An external agent is a source of this heat loss, which is responsible for the change in the magnetic field with time.

Â Â

Â Q 12. We have a square loop having a side of 12 cm, and its sides parallel to the x and the y-axis is moved with a velocity of 8 cm/s in the positive x-direction in a region which have a magnetic field in the direction of the positive z-axis. Â The field is not uniform, whether in the case of its space or in the case of time. It has a gradient of 10âˆ’3 T cmâˆ’1 along the negative x-direction(i.e. its value increases by

$$\begin{array}{l}10 ^{-3} \; T \; cm^{-1}\end{array}$$
as we move from positive to negative direction ), and it is reducing in the case of time with the rate ofÂ
$$\begin{array}{l}10 ^{-3} \; T \; s^{-1}\end{array}$$
. Calculate the magnitude and direction of induced current in the loop (Given: Resistance =
$$\begin{array}{l}4.50 \; m \Omega\end{array}$$
).

Â

Ans:

Side of the Square loop, s = 12cm = 0.12m

Area of the loop, A = s Ã— s = 0.12 Ã— 0.12 = 0.0144

$$\begin{array}{l}m^{2}\end{array}$$

Velocity of the loop, v = 8

$$\begin{array}{l} cm \;s^{-1}\end{array}$$
= 0.08
$$\begin{array}{l} m \;s^{-1}\end{array}$$

The gradient of the magnetic field along the negative x-direction,

$$\begin{array}{l}\frac{dB}{dx} = 10^{-3} \; T\,cm^{-1} = 10^{-1} \; m^{-1}\end{array}$$

And the rate of decrease of the magnetic field,

$$\begin{array}{l}\frac{dB}{dt} = 10^{-3} \; T \, s^{-1}\end{array}$$

Resistance, R =

$$\begin{array}{l}4.50 \; m \Omega = 4.5 \times 10 ^{-3} \; \Omega\end{array}$$

The rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:

$$\begin{array}{l}\frac{dB}{dt} = A \times \frac{dB}{dx} \times v \\ \\ =144 \times 10 ^{-4} m ^{2} \times10^{-1} \times 0.08 \\ \\ = 11.52 \times 10 ^{-5} \; Tm^{2} s^{-1}\end{array}$$

The rate of change of the flux due to explicit time variation in field B is given as:

$$\begin{array}{l}\frac{d\phi }{dt} = A \times \frac{dB}{dt} \\ \\ = 144 \times 10^{-4} \times 10 ^{-3} \\ \\ = 1.44 \times 10^{-5} T\,m^{2}s^{-1}\end{array}$$

Since the rate of change of the flux is the induced emf, the total induced emf in the loop can be calculated as:

$$\begin{array}{l}e = 1.44 \times 10^{-5} + 11.52 \times 10^{-5}\\ \\ = 12.96 \times 10^{-5} V \\ \\ âˆ´ Induced \; current, \; i = \frac{e}{R}\\ \\ = \frac{12.96 \times 10 ^{-5}}{4.5 \times 10 ^{-3}} \\ \\ i = 2.88 \times 10^{-2}\; A\end{array}$$

Hence, the direction of the induced current is such that there is an increase in the flux through the loop along the positive z-direction.

Â Â

Q 13. We have a powerful loudspeaker magnet and have to measure the magnitude of the field between the poles of the speaker. And a small search coil is placed, normal to the field direction and then quickly removed out of the field region; the coil is of

$$\begin{array}{l}2 \, cm^{2}\end{array}$$
area and has 25 closely wound turns. Similarly, we can give the coil a quick
$$\begin{array}{l}90 ^{\circ}\end{array}$$
turn to bring its plane parallel to the field direction. We measure the total charge flown in the coil by using a ballistic galvanometer, and it comes to 7.5 mC. Total resistance after combining the coil and the galvanometer is
$$\begin{array}{l}0.50 \; \Omega\end{array}$$
. Estimate the field strength of the magnet.

Ans:

Given,

Coilâ€™s Area, A =

$$\begin{array}{l}2 \, cm^{2} = 2 \times 10^{-4} m^{2}\end{array}$$

Number of turns on the coil, N = 25

Total Charge in the coil, Q = 7.5 mC =

$$\begin{array}{l}7.5 \times 10 ^{-3}\; C\end{array}$$

Total resistance produced by the combo of coil and galvanometer,

$$\begin{array}{l}R = 0.50 \; \Omega\end{array}$$

Current generated in the coil,

$$\begin{array}{l}I = \frac{Induced \; emf\; \left ( e \right )}{R}\;\;\;\;\;\;\;\; . . . (1)\end{array}$$

The EMF induced is shown as:

$$\begin{array}{l}e = -N\frac{d\Phi}{dt} \;\;\;\;\;\;\;\;\; . . . (2)\end{array}$$

Where,

$$\begin{array}{l}d \phi\end{array}$$
= Change in flux

From equations (1) and (2), we have

$$\begin{array}{l}I = -\frac{N\frac{d\Phi }{dt}}{R} \\ \\ Idt = -\frac{N}{R}d \phi \;\;\;\;\;\;\;\; . . . (3)\end{array}$$

Flux through the coil at the initial phase,

$$\begin{array}{l}\phi _{i} = BA\end{array}$$

Where B = Strength of the magnetic field

Flux through the coil at the final phase,

$$\begin{array}{l}\phi _{f} = 0\end{array}$$

After integrating equation (3) on both sides, we get

$$\begin{array}{l}\int Idt = \frac{-N}{R}\int_{ \phi_{i} }^{ \phi_{f} } d \phi\end{array}$$

Total Charge,

$$\begin{array}{l}Q = \int Idt\end{array}$$
$$\begin{array}{l}\\âˆ´ Q = \frac{-N}{R}\left ( \phi_{f} – \phi_{i} \right )= \frac{-N}{R}\left ( – \phi_{i} \right ) = + \frac{N \phi_{i}}{R} \\ \\ Q = \frac{NBA}{R}\\ \\ âˆ´ B = \frac {QR}{NA} \\ \\ = \frac {7.5 \times 10^{-3} \times 0.5}{25 \times 2 \times 10^{-4}} = 0.75 \; T\end{array}$$

Hence, the field strength is 0.75 T.

Q 14. In the given figure, we have a metal rod PQ which is put on the smooth rails AB, and they are kept in between the two poles of permanent magnets. All three (rod, rails and the magnetic field ) are in a mutually perpendicular direction. There is a galvanometer â€˜Gâ€™ connected through the rails by using a switch â€˜Kâ€™.Given, Rodâ€™s length = 15 cm , Magnetic field strength, B = 0.50 T, Resistance produced by the closed-loop =

$$\begin{array}{l}9.0\; m\Omega\end{array}$$
. Letâ€™s consider the field is uniform.

(i) Determine the polarity and the magnitude of the induced emf if we will keep the K open, and the rod will be moved with the speed of 12 cm/s in the direction shown in the figure.

Â (ii) When the K was open, was there any excess charge built up? Assuming that K is closed, then what will happen after it?

Â (iii) When the rod was moving uniformly, and the K was open, then on the electron in the rod PQ, there was no net force even though they did not experience any magnetic field because of the motion of the rod. Explain.

Â (iv) After closing the K, calculate the retarding force.

Â (v) When the K will be closed, calculate the total external power which will be required to keep moving the rod at the same speed (12 cm/s), and also calculate the power required when K will be closed.

Â (vi) What would be the power loss ( in the form of heat) when the circuit is closed? What would be the source of this power?

Â (vii) Calculate the emf induced in the moving rod if the direction of the magnetic field is changed from perpendicular to parallel to the rails?

Ans:

Length of the rod, lÂ  = 15 cm = 0.15 m

Strength of the magnetic field, B = 0.50 T

Resistance produced by the closed-loop, R =

$$\begin{array}{l}9.0\; m\Omega = 9 \times 10^{-3} \Omega \end{array}$$

(i)

The polarity of the emf induced is in such a way that its P end is showing positive, while the other end, i.e. Q, is showing negative.

Since, speed, v = 12cm/s = 0.12 m/s

Emf induced is: e = Bvl

= 0.5 Ã— 0.12 Ã— 0.15

=

$$\begin{array}{l}9 \times 10 ^{-3}\; V\end{array}$$

= 9 mV

Here, the polarity of the emf is induced in a way that the P end shows +ve and the Q end shows -ve.

(ii) Yes, when the key K was opened, then at both ends, there was excess charge built up.

An excess charge was also built up when the key K was closed, and that charge was maintained by the continuous flow of current.

(iii) Because of the electric charge set-up, there was an excess charge of the opposite nature at both ends of the rod. Because of that, the Magnetic force was cancelled up.

When the key K was opened, then there was no net force on the electrons in the rod PQ, and the rod was moving uniformly. It is because of the cancelled magnetic field on the rod.

(iv) Regarding force exerted on the rod, F = IBl

Where,

I = current flowing through the rod

$$\begin{array}{l}= \frac {e}{R} = \frac{9 \times 10^{-3}}{9 \times 10^{-3}} = 1 \; A\\ \\ âˆ´F = 1 \times 0.5 \times 0.15\\ \\ = 75 \times 10^{-3} \; N\end{array}$$

(v)

No power will be expended when the key K is opened.

Speed of the rod, v = 12 cm/s = 0.12 m /s

Hence,

Power, P = Fv

$$\begin{array}{l}\\= 75 \times 10^{-3} \times 0.12\\ \\ = 9 \times 10^{-3} \; W \\ \\ = 9 \; mW\end{array}$$

When the key K is opened, no power is expended.

(vi) 9mW,

Power is provided by an external agent.

Power loss in the form of heat is given as follows:

=

$$\begin{array}{l}I^{2}R\end{array}$$

=

$$\begin{array}{l}1^{2} \times 9 \times 10^{-3}\end{array}$$

= 9 mW

(vii) Zero (0)

There would be no emf induced in the coil. As the emf induces if the motion of the rod cuts the field lines. But in this case, the movement of the rod does not cut across the field lines.

Q 15. We have an air-cored solenoid having a length of 30 cm, whose area is

$$\begin{array}{l}25 \; cm^{2}\end{array}$$
and the number of turns is 500. And the solenoid has carried a current of 2.5 A. Suddenly the current is turned off, and the time taken for it is
$$\begin{array}{l}10 ^{-3}\end{array}$$
s. What would be the average value of the induced back-emf by the ends of the open switch in the circuit? (Neglect the variation in the magnetic fields near the ends of the solenoid.)

Â

Ans:

Given,

Length of the solenoid, l = 30 cm = 0.3 m

Area of the solenoid, A =

$$\begin{array}{l}25 \; cm^{2} = 25 \times 10^{-4} \; m^{2}\end{array}$$

Number of turns on the solenoid, N = 500

Current in the solenoid, I = 2.5 A

Time duration for the current flow, t =

$$\begin{array}{l}10^{-3}\end{array}$$

Average back emf,

$$\begin{array}{l}e = \frac{d \phi}{dt} \;\;\;\;\;\; . . . (1)\\ \\Where, \\ \\ d \phi =\end{array}$$
change in flux

= NABÂ Â Â Â  Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  . . . . (2)

Where,

B = Strength of the magnetic field

$$\begin{array}{l}= \mu_{0} \frac{NI}{l} \;\;\;\;\;\;\;\; . . . (3)\end{array}$$

Where,

$$\begin{array}{l}\mu _{0}\end{array}$$
= Permeability of free space =Â
$$\begin{array}{l}4 \pi \times 10 ^{-7} \; T\; m \; A^{-1}\end{array}$$

Using equations (2) and (3) in equation (1), we get

$$\begin{array}{l}\\e = \frac{\mu _{0}N ^{2} I A}{lt} \\ \\ = \frac{4 \pi \times 10^{-7} \times \left ( 500 \right )^{2} \times 2.5 \times 25 \times 10^{-4}}{0.3 \times 10^{-3}} = 6.5 V\end{array}$$

Hence, the average back emf induced in the solenoid is 6.5 V.

Â

Â Q 16.Â (i) We are given a long straight wire and a square loop of a given size (refer to figure). Find out an expression for the mutual inductance between both.

(ii) Now, consider that we passed an electric current through the straight wire of 50 A, and the loop is then moved to the right with constant velocity, v = 10 m/s. Find the emf induced in the loop at an instant where x = 0.2 m. Take a = 0.01 m and assume that the loop has a large resistance.

Ans:

(i) Take a small element dy in the loop at a distance y from the long straight wire (as shown in the given figure).

Magnetic flux associated with element

$$\begin{array}{l}dy, \, d \phi = a \, dy\end{array}$$

B = Magnetic field at distance

$$\begin{array}{l}y = \frac{\mu _{0}I}{2 \pi y}\end{array}$$

I = Current in the wire

$$\begin{array}{l}\mu _{0}\end{array}$$
= Permeability of free space =
$$\begin{array}{l}4 \pi \times 10^{-7}\\ \\ âˆ´ d\phi = \frac{\mu _{0}Ia}{2 \pi}\frac{dy}{y}\\ \\ \phi = \frac{\mu _{0}Ia}{2 \pi} \int \frac{dy}{y} \\ \\ y \; tends\; from \; x \; to \; a + x\\ \\ âˆ´ \phi = \frac{\mu_{0}Ia}{2 \pi} \int_{x}^{a + x} \frac{dy}{y} \\ \\ = \frac{\mu_{0}Ia}{2 \pi} \left [ \log_{e} y \right ]_{x}^{a + x} \\ \\ = \frac{\mu_{0}Ia}{2 \pi} \log_{e} \frac{a + x}{x}\end{array}$$

For mutual inductance M, the flux is given as:

$$\begin{array}{l}\phi = MI \\ \\âˆ´ MI = \frac {\mu _{0} I a}{2 \pi} \log _{e}\left ( \frac{a}{x} + 1 \right ) \\ \\ \; \;M = \frac {\mu _{0} a}{2 \pi} \log _{e}\left ( \frac{a}{x} + 1 \right )\end{array}$$

(ii) EMF induced in the loop, e = Bâ€™av

$$\begin{array}{l}= \frac {\mu _{0} I}{2 \pi x} av\end{array}$$

Given, I = 50 A

x = 0.2 m

a = 0.1 m

v = 10 m/s

$$\begin{array}{l}e = \frac {4 \pi \times 10 ^{-7} \times 50 \times 0.1 \times 10}{2 \pi \times 0.2} \\ \\ e = 5 \times 10^{-5}\; V\end{array}$$

Q 17.A line charge Î» 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,

$$\begin{array}{l}B = -B _{0} \, k \left ( r \leq a ; a < R \right )\end{array}$$

= 0 (otherwise)

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

Ans:

Line charge per unit length =

$$\begin{array}{l}\lambda = \frac{Total \; charge}{Length} = \frac{Q}{2 \pi r}\end{array}$$

Where,

r = Distance of the point within the wheel

Mass of the wheel = M

The radius of the wheel = R

Magnetic field,

$$\begin{array}{l}\vec{B} = -B_{0} \dot{k}\end{array}$$

At distance r, the magnetic force is balanced by the centripetal force, i.e.

$$\begin{array}{l}BQv = \frac{Mv^{2}}{r}\end{array}$$

Where,

v = linear velocity of the wheel

$$\begin{array}{l}\\âˆ´ B2 \pi r \lambda = \frac {Mv}{r} \\ \\ v = \frac{B2 \pi \lambda r^{2}}{M}\\ \\ âˆ´ Angular \; Velocity, \omega = \frac{v}{R} \\ \\ = \frac{B2 \pi \lambda r^{2}}{M R} \\ \\ For\; r \leq a \; and \; a < R, we\; get \\ \\ \omega = – \frac{2 \pi B_{0} a^{2} \lambda}{M R} \hat{k}\end{array}$$

## Class 12 Physics NCERT Solutions for Chapter 6 Electromagnetic Induction

Chapter 6 Electromagnetic Induction of Class 12 Physics, is prepared according to the current CBSE Syllabus 2023-24. The NCERT Solutions for Class 12 Physics of this chapter discusses magnetism and electricity together. Faradayâ€™s law should be well understood. Faradayâ€™s law states that any change in the magnetic field of a current-carrying conductor results in voltage (emf) being induced in the conductor. Lenzâ€™s law states that there is always a counter-force opposing the induced emf. Both laws are important from an exam point of view.

Numerical problems from these concepts are often asked. While solving the numerical problems, students should keep in mind the direction of induced emf and magnetic field. One can get a good understanding of these problems by solving practice problems of NCERT Class 12 Physics Electromagnetic Induction and by referring to NCERT Solutions for Class 12 Chapter 6 of Physics to understand the correct method of solving problems.

### Topics covered in Class 12 Physics Electromagnetic Induction are:

 Section Number Topic 6.1 Introduction 6.2 The Experiments of Faraday and Henry 6.3 Magnetic Flux 6.4 Faradayâ€™s Law of Induction 6.5 Lenzâ€™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

### What makes BYJUâ€™S important?

BYJU’S has a unique approach to teaching students, which includes using lots of animations. These animations help students understand the concepts much faster as their imagination gets a vision. Also, regular and updated notes are provided by the subject experts, and these notes are available for free download. Class 12 Physics Chapter 6 Electromagnetic Induction NCERT Solutions is also available for free download, along with theÂ CBSE Sample Papers.

Disclaimer –Â

Dropped Topics –Â

6.7 Energy Consideration: A Quantitative Study
6.8 Eddy Currents
Exercises 6.6, 6.10â€“6.17

## Frequently Asked Questions on NCERT Solutions for Class 12 Physics Chapter 6

Q1

### Does BYJUâ€™S give the most reliable answers in Chapter 6 of NCERT Solutions for Class 12 Physics?

The most accurate and reliable NCERT Solutions for Class 12 Physics Chapter 6 are available at BYJUâ€™S. Students can easily download the solutions, which are present in PDF format and access them during their board exam preparation. The solutions are framed and compiled by a set of expert faculty who possess numerous years of experience in their respective fields. The most detailed solutions to the exercise-wise problems are curated with the aim of helping students ace the board exam confidently.
Q2

### Can I get the NCERT Solutions for Class 12 Physics Chapter 6 in PDF format?

Yes, students can download the NCERT Solutions for Class 12 Physics Chapter 6 in PDF format, at BYJUâ€™S. The solutions for the exercise-wise problems are designed by the highly experienced faculty based on the latest CBSE Syllabus 2023-24. Students can also cross-check their answers with the solutions PDF in order to get a clear idea about the other methods of solving complex problems effortlessly.
Q3

### What are the topics covered under each exercise of NCERT Solutions for Class 12 Physics Chapter 6?

The main topics covered under NCERT Solutions for Class 12 Physics Chapter 6 are:
6.1 Introduction
6.2 The Experiments of Faraday and Henry
6.3 Magnetic Flux