Class 12 Physics Chapter 6 NCERT Solutions: Free PDF, Important Formulas and Questions

Physics Ncert Solutions Class 12th 2023

nitesh singh
Updated on Oct 28, 2025 17:20 IST

By nitesh singh, Senior Executive

Chapter 6 of Class 12 Physics covers Electromagnetic induction and how this concept helps understand the conversion of electric energy to magnetic energy and vice versa. Our experienced SMEs at Shiksha have created accurate and reliable class 12 study material, including NCERT Solutions, for the students preparing for the CBSE Board exams.

Practice the Chapter 6 Question and answers and resolve all your doubts with the Electromagnetic Induction Class 12 NCERT Solutions provided by Shiksha. Before solving the conceptual and numerical problems, you must have an in-depth conceptual understanding of the concepts. Learn all important concepts such as mutual induction, self-induction, AC generator, and more with Class 12 Physics NCERT notes for better preparation.

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  • Study offline with Chapter 6 Class 12 Physics NCERT Solutions PDF
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The key concepts are magnetic flux and changing magnetic flux, which are responsible for induced EMF. Solve the NCERT Textbook exercise and develop the foundation of competitive exam preparation. Access the complete solutions below:

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  • Class 12 Electromagnetic Induction Chapter NCERT Solution PDF: Download PDF for Free
  • Class 12 Physics Chapter 6 NCERT Solutions
  • Complete Chapter 6 Physics Class 12 Study Material
  • Chapter 6 Electromagnetic Induction Important Formulas
  • Class 12 Electromagnetic Induction: Key Topics, and Weightage
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  • Important Questions for CBSE, NEET and JEE on Electromagnetic Induction
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Class 12 Physics Chapter 6 NCERT Solutions

Q.6.1 Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f).

Ans.6.1

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

Using Lenz’s rule, the direction of the induced current in the given situation can be predicted as follows:

The direction of the induced current is along ‘qrpq’.

The direction of the induced current is along ‘prqp’.

The direction of the induced current is along ‘yzxy’.

The direction of the induced current is along ‘zyxz’.

The direction of the induced current is along ‘xryx’.

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

Q.6.2 Use Lenz’s law to determine the direction of induced current in the situations 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.6.2 As the loop changes from irregular to circular shape, its area increases. Hence the magnetic flux linked with it also increases. According to Lenz’s law, the induced current should produce magnetic flux in the opposite direction of the original flux. For this induced current should flow in the anti-clockwise direction, i.e. adcb.

As this circular loop is being deformed into a narrow straight wire, its area decreases. The magnetic flux linked also decreases. By Lenz’s law, the induced current should produce a flux in the direction of the original flux. For this the induced current should flow in the anti-clock wise direction, i.e. a’d’c’b’.

Q.6.3 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.6.3 Number of turns on the solenoid = 15 turns per cm = 1500 turns per m

Hence, number of turns per unit length, n = 1500

The loop area in the solenoid, A = 2.0  c m 2  = 2  × 10 - 4   m 2

Current carrying by the solenoid, changes from 2 .0 A to 4.0 A

So, change of current, di = 4 – 2 = 2 A

Change in time, dt = 0.1 s

According to Faraday’s law, the induced emf, e =  d φ d t  …………..(i)

Where  φ  = induced flux through the small loop = BA …………..(ii)

Magnetic field is given by B =  μ 0 n i  ……………………………..(iii)

Where  μ 0  = Permeability of free space = 4  π × 10 - 7  H/m

From equation (i),

e =  d d t ( B A )

= A  μ 0 n × d i d t

= 2  × 10 - 4 ×  4  π × 10 - 7 × 1500 × 2 0.1

= 7.54  × 10 - 6  V

Hence the induced emf while current is changing is 7.54  × 10 - 6  V

Q.6.4 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.6.4 Length of the rectangular wire loop, l = 8 cm = 0.08 m

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

Hence, the area of the rectangular loop, A = l  × b  = 0.08  × 0.02  = 1.6  × 10 - 3   m 2

Magnetic field strength, B= 0.3 T

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

Emf developed in the longer side:

Emf developed in the loop is given by e = Blv = 0.3  × 0.08 × 0.01  = 2.4  × 10 - 4  V

Time taken to travel the width =  D i s t a n c e t r a v e l l e d v e l o c i t y  =  b v  =  0.02 0.01  s= 2 s

Hence the induced voltage is 2.4  × 10 - 4  V, which lasts for 2 s.

Emf developed in the shorter side:

Emf developed in the loop is given by e = Bbv = 0.3  × 0.02 × 0.01  = 0.6  × 10 - 4  V

Time taken to travel the width =  D i s t a n c e t r a v e l l e d v e l o c i t y  =  l v  =  0.08 0.01  s= 8 s

Hence the induced voltage is 0.6  × 10 - 4  V, which lasts for 8 s.

 

Q&A Icon
Commonly asked questions
Q:  

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

Read more
A: 

6.1

The direction of the induced current in a closed loop is given by Lenz's law. The given pairs of figures show the direction of the induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively.

Using Lenz's rule, the direction of the induced current in the given situation can be predicted as follows:

The direction of the induced current is along 'qrpq'.

The direction of the induced current is along 'prqp'.

The direction of the induced current is along 'yzxy'.

The direction of the induced current is along 'zyxz'.

The direction of the induced current is along 'xryx'.

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

Q:  

6.2 Use Lenz’s law to determine the direction of induced current in the situations 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.

Read more
A: 

6.2 (a) As the loop changes from irregular to circular shape, its area increases. Hence the magnetic flux linked with it also increases. According to Lenz's law, the induced current should produce magnetic flux in the opposite direction of the original flux. For this induced current should flow in the anti-clockwise direction, i.e. adcb.

(b) As this circular loop is being deformed into a narrow straight wire, its area decreases. The magnetic flux linked also decreases. By Lenz's law, the induced current should produce a flux in the direction of the original flux. For this the induced current should flow in the anti-clock wise direction, i.e. a'd'c'b'.

Q:  

6.3 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?

Read more
A: 

6.3 Number of turns on the solenoid = 15 turns per cm = 1500 turns per m

Hence, number of turns per unit length, n = 1500

The loop area in the solenoid, A = 2.0 cm2 = 2 ×10-4 m2

Current carrying by the solenoid, changes from 2 .0 A to 4.0 A

So, change of current, di = 4 – 2 = 2 A

Change in time, dt = 0.1 s

According to Faraday’s law, the induced emf, e = dφdt …………. (i)

Where φ = induced flux through the small loop = BA …………. (ii)

Magnetic field is given by B = μ0ni ……………………………. (iii)

Where μ0 = Permeability of free space = 4 π×10-7 H/m

From equation (i),

e = ddt (BA)

= A μ0n×didt

= 2 ×10-4× 4 π×10-7×1500×20.1

= 7.54 ×10-6 V

Hence the induced emf while current is changing is 7.54 ×10-6 V

Q:  

6.4 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?

Read more
A: 

6.4 Length of the rectangular wire loop, l = 8 cm = 0.08 m

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

Hence, the area of the rectangular loop, A = l ×b = 0.08 ×0.02 = 1.6 ×10-3 m2

Magnetic field strength, B= 0.3 T

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

Emf developed in the longer side:

Emf developed in the loop is given by e = Blv = 0.3 ×0.08×0.01 = 2.4 ×10-4 V

Time taken to travel the width = Distancetravelledvelocity = bv = 0.020.01 s= 2 s

Hence the induced voltage is 2.4 ×10-4 V, which lasts for 2 s.

Emf developed in the shorter side:

Emf developed in the loop is given by e = Bbv = 0.3 ×0.02×0.01 = 0.6 ×10-4 V

Time taken to travel the width = Distancetravelledvelocity = lv = 0.080.01 s= 8 s

Hence the induced voltage is 0.6 ×10-4 V, which lasts for 8 s.

Q:  

6.5 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.

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A: 

6.5 Length of the rod, l = 1.0 m

Angular frequency,  ω = 400 rad/s

Magnetic field strength, B = 0.5 T

One end of the rod has zero linear velocity, while the other end has a linear velocity of l ω

The average linear velocity, v = 0+lω2 = lω2

emf developed between the centre and the ring is given by

e = Blv = Bl ( lω2) = Bl2ω2 = 0.5×12×4002 = 100 V

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

Q:  

6.6 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?

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A: 

6.6 Given,

Radius of the circular coil, r = 8.0 cm = 0.08 m

Area of the coil, A = πr2 = π×0.082 = 20.106 ×10-3 m2

Number of turns in the coil, n = 20

Angular speed, ω = 50 rad/s

Magnetic field strength, B = 3.0 ×10-2 T

Resistance of the loop, R = 10 Ω

Maximum induced emf is given as e = n ωAB = 20 ×50× 20.106 ×10-3× 3.0 ×10-2 = 0.603 V

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

Maximum current is given by, Imax = eR = 0.60310 = 0.0603 A

Average power loss due to Joule heating is given by, Ploss = eImax2 = 0.018 W

The current induced in the coil produces a torque opposing the rotation of the coil. The rotor is an external agent; it must supply a torque to counter this torque in order to keep the coil rotating uniformly. Hence, dissipated power comes from the external rotor.

Q:  

6.7 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?

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A: 

6.7 Length of the wire, l = 10 m

Falling speed of the wire, v = 5 m/s

Magnetic field strength, B = 0.30 *10-4 Wb/ m2

The instantaneous emf induced in the wire, e = Blv = 0.30 *10-4*10* 5 = 1.5 *10-3 V

Using Fleming's right hand rule, the direction of the induced emf is from West to East.

The eastern end of the wire is at a higher potential.

Q:  

6.8 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.

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A: 

6.8 Given:

Initial current,  I1 = 5.0 A

Final current,  I2 = 0 A

Change in current, di = I1-I2 = 5 A

Time taken for the change, dt = 0.1 s

Average emf, e = 200 V

For self-inductance (L) of the coil, L is given by

L = edidt = 20050.1 = 4 H

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

Q:  

6.9 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?

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A: 

6.9 Mutual inductance of a pair of coils, μ = 1.5 H

Initial current, I1 = 0 A

Final current, I2 = 20 A

Change in current, dI = I2 - I1 = 20 A

Time taken for change, dt = 0.5 s

From the relation of induced emf e = ddt , where d is the change in the flux linkage with the coil ……………(1)

The relation between emf and mutual inductance is e = μdIdt ………….. (2)

Combining equations (1) and (2), we get

ddt=μdIdt or d=μdI = 1.5 ×20 = 30 Wb

Hence, the change in the flux linkage is 30 Wb

Q:  

6.11 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?

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A: 

6.11 The area of the rectangular coil, A = 8 ×2= 16 cm2 = 16 ×10-4 m2

Initial value of the magnetic field, B1 = 0.3 T

Rate of decrease of the magnetic field, dBdt = 0.02 T/s

From the relation of induced emf e = ddt , where d is the change in the flux linkage with the coil = A ×B

Hence, e = d(AB)dt = A dBdt = 16 ×10-4×0.02 = 3.2 ×10-5 V

Resistance in the loop, R = 1.6 Ω

Hence I = eR = 3.2×10-51.6 = 2 ×10-5 A

Power dissipated in the form of heat is given by

P = I2 R = ( 2×10-5)2×1.6 = 6.4 ×10-10 W

The source of heat loss is an external agent, which is responsible for changing the magnetic field with time.

Q:  

6.12 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.

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A: 

6.12 Side of the square loop, s = 12 cm = 0.12 m

Area of the square loop, A = 0.12 ×0.12=0.0144m2

Velocity of the loop, v = 8 cm /s = 0.08 m/s

Gradient of the magnetic field along negative x – direction.

dBdx = 10-3 T/cm = 10-1 T/m

Rate of decrease of magnetic field

dBdt = 10-3 T/s

Resistance of the loop, R = 4.50 mΩ = 4.5 ×10-3 Ω

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

ddt = A ×dBdx ×v = 0.0144×10-1×0.08 = 1.152 ×10-4 T m2s-1

Rate of change of flux due to explicit time variation in field B is given as:

d'dt = A ×dBdt = 0.0144×10-3 = 1.44 ×10-5 T m2s-1

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

e = 1.152 ×10-4 + 1.44 ×10-5 = 1.296 ×10-4 V

The induced current, I = eR = 1.296×10-44.5×10-3 = 28.8 ×10-3 A

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

Q:  

6.13 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.

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A: 

6.13 Area of the coil, A = 2 cm2 = 2 ×10-4 m2

Number of turns, n = 25

Total charge flowing in the coil, Q = 7.5 mC = 7.5 ×10-3 C

Total resistance of the coil and galvanometer, R = 0.50 Ω

We know, Induced current in the coil, I = Inducedemf(e)R ……… (1)

Induced emf is given by the equation, e = - n ddt ………………… (2),

Where d  is the change of flux

Combining equations (1) and (2), we get

I = -nddtR or Idt = -nR d  ……… (3)

Initial flux through coil, i = BA, where B = Magnetic field strength

Final flux through coil, f = 0

Integrating equation (3), we get

Idt = -nRifd

Idt=Q=Totalchargeflowinginthecoil

Therefore

Q = -nRf-i=-nR0-i=niR=nBAR

B = QRnA = 7.5×10-3×0.525×2×10-4 = 0.75 T

Hence, the field strength is 0.75 T

Q:  

6.14 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 (=12cm 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?

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A: 

6.14 Length of the rod, l = 15 cm = 0.15 m

Magnetic field strength, B = 0.5 T

Resistance of the closed loop, R = 9 mΩ = 9 ×10-3 Ω

Induced emf = 9 mV

Polarity of the induced emf is such that end P shows positive while end Q shows negative ends.

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

Induced emf is given as, e = Bvl = 0.5 ×0.12×0.15 = 9 mV

Yes, when key K is closed, excess charge is maintained by the continuous flow of current. When the key K is open, there is excess charge built up at both ends of the rods.

Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rod. There is no net force on the electrons in rod PQ when key K is open and the rod is moving uniformly. This is because magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rods.

Retarding force exerted on the rod, F = IBl , where

I = Current flowing through the rod = eR = 9×10-39×10-3 = 1 A

F = 1 ×0.5×0.15 = 0.075 N

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

Power is given by P = Fv = 0.075 ×0.12=9×10-3 W = 9 mW

When key K is open, no power is expended.

Power dissipated by heat = I2R=1× 9 ×10-3 W = 9 mW

The source of this power is an external agent.

In this case, no emf is induced in the coil because the motion of the rod does not cut across the field lines.

Q:  

6.15 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.

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A: 

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

Area of cross-section, A = 25 cm2 = 25 ×10-4 m2

Number of turns on the solenoid, N = 500

Current in the solenoid, I = 2.5 A

Current flowing time, t = 10-3 s

We know average back emf, e = ddt …………………. (1)

Where d= Change in flux = NAB ……………… .(2)

B = Magnetic field strength = μ0NIl ………………... (3)

Where μ0 = Permeability of free space = 4 π×10-7 T m A-1

From equation (1) and (2), we get

e = NABdt = NAdt×μ0NIl = μ0N2AIlt = 4π×10-7×5002×25×10-4×2.50.3×10-3 = 6.544 V

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

Q:  

6.16 (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.

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A: 

6.16 Let us take a small element dy in the loop, at a distance y from the long straight wire.

 

Magnetic flux associated with element dy, d  = BdA, where

dA = Area of the element dy = a dy

Magnetic flux at distance y, B = μ0I2πy , where

I = current in the wire

μ0 = Permeability of free space = 4 π×10-7 T m A-1

Therefore,

 = μ0I2πy a dy = μ0Ia2π dyy

=μ0Ia2πdyy

Now from the figure, the range of y is x to x+a. Hence,

=μ0Ia2πxx+adyy = μ0Ia2πloge?yxa+x = μ0Ia2πloge?(a+xx)

For mutual inductance M, the flux is given as

=MI . Hence

MI=μ0Ia2πloge?(a+xx)

M = μ0a2πloge?(ax+1)

Emf induced in the loop, e = Bav

= ( μ0I2πx ) × av

For I = 50 A, x = 0.2 m, a = 0.1 m, v = 10 m/s

e = 4π×10-7×50×0.1×102π×0.2 = 5 ×10-5 V

Q:  

6.17 A line charge l 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?

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A: 

6.17 Line charge per unit length = λ = TotalchargeLength = Q2πr

Where r = distance of the point within the wheel

Mass of the wheel = M

Radius of the wheel = R

Magnetic field, B? = -B0k?

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

BQ v = Mv2r , where v = linear velocity of the wheel. Then,

×2λπr = Mvr

v = 2Bλπr2M

Angular velocity, ω=vR = 2Bλπr2MR

For r aR,weget ω=-2B0λπa2MRk?

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Complete Chapter 6 Physics Class 12 Study Material

Check the important related study tools, such as the formulas sheet, quick revision notes, and more, in the table.

Related Class 12 Physics Chapter 6 Study Material
Chapter 6 Class 12 Physics NCERT Exemplar Solutions
Electromagnetic Induction Class 12 Formulas
Class 12 Chapter 6 Physics Quick Revision Notes
Class 12 Electromagnetic Induction NCERT Notes
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Chapter 6 Electromagnetic Induction Important Formulas

Refer to the table below for important formulas and concepts of the Class 12 Chapter 6 Electromagnetic Induction:

Concept Formula
Magnetic Flux (Φ) Φ = B · A · cos(θ)
Faraday’s Second Law ε = N Δ Φ Δ t
Motional EMF ϵ = B l v sin ( θ )
Induced Current I = ϵ R
Self-Induction EMF
ϵ = L d I d t epsilon equals negative cap L the fraction with numerator d cap I and denominator d t end-fraction
Mutual Induction EMF ϵ = −M
d I d t
Self-Inductance of Long Solenoid
L = μ 0 N 2 A l cap L equals the fraction with numerator mu sub 0 cap N squared cap A and denominator l end-fraction
Self-Inductance of Long Solenoid (with materials permeability) L = μ 0 μ r N 2 A l
Self-Inductance of Toroid L = μ 0 N 2 h 2 π ln ( R 2 R 1 )
Mutual-Inductance of Long Solenoid (Coaxial) M = μ 0 μ r N 1 N 2 A l cap M equals the fraction with numerator mu sub 0 mu sub r cap N sub 1 cap N sub 2 cap A and denominator l end-fraction  
Energy Stored in an Inductor U = (1/2) · L · I²
AC Generator (Dynamo) E = NABω sin(ωt)
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Class 12 Electromagnetic Induction: Key Topics, and Weightage

The NCERT textbook has sections covering specific topics of electromagnetic induction. Below is a table with a section-wise topic list, along with the topic notes. These topics are fixed in a chronological order by NCERT to explain the how a normal change in magnetic flux is responsible for the most commonly used phenomenon in real life, that is, alternating current.

NCERT Topics Topics Covered
6.1 Introduction To Electromagnetic Induction
6.2 The Experiments Of Faraday And Henry
6.3 Magnetic Flux
6.4 Faraday’s Law Of Induction
6.5 Lenz’s Law And Conservation Of Energy
6.6 Motional Electromotive Force
6.7 Inductance: Self and Mutual Induction
6.8 AC Generator

Other Important Topics for CBSE, JEE, and NEET

Here are the other topics related to the Class 12 Physics Chapter Electromagnetic Induction.

  • Motional EMF due to the rotating rod and ring
  • Self-inductance of a long solenoid
  • Self-inductance of a toroid
  • Mutual inductance due to long coaxial solenoids
  • AC Voltage
  • Induced EMF and Current
  • Eddy Currents
  • Alternating current concept.

Try these practice questions

Q1:

A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be:

(Assume the coil to be short circ

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Q2:

A coil of inductance 1H and resistance 100 Ω is connected to a battery of 6.V. Determine approximately:

(A)The time elapsed before the current acquires half of its steady – state value.

(B)The energy stored in the magnetic field associated w

View Full Question

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Class 12 Physics Chapter 6 weightage in various Exams

Studying all the concepts and practicing numericals can be key to getting good marks. However, If you already have an idea, how important is this chapter for various exams, it motivates you learn in details and develop clear understanding. Check the weightage below.

Exam Name 

Weightage

Class 12 CBSE Boards 4-6 marks
JEE Main 3.33% ( (1 question)
NEET UG 3% (1 question)
CUET UG 8-10% (2-3 questions)
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Important Questions for CBSE, NEET and JEE on Electromagnetic Induction

Check the most important questions related to various exams below.

A coil has 100 turns, each of area 0.05 m² and total resistance 1.5 Ω. It is inserted at an instant in a magnetic field of 90 mT, with its axis parallel to the field. The charge induced in the coil at that instant is: [CBSE 2025] A) 3 mC B) 0.3 mC C) 0.45 mC D) 1.5 mC

Given:

Number of turns = 100 

Area = 0.05 m²

Magnetic field = 90 millitesla

Resistance = 1.5 ohms.

The total change in magnetic flux is:

ΔΦ = B × A × N = 0.09 × 0.05 × 100 = 0.45 Weber

 Induced charge:

Q = ΔΦ / R = 0.45 / 1.5 = 0.3 Coulombs = 0.3 mC

So, the induced charge is 0.3 milliCoulombs.

Correct answer: B) 0.3 mC

Assertion (A): The mutual inductance between two coils is maximum when the coils are wound on each other. Reason (R): The flux linkage between two coils is maximum when they are wound on each other. [CBSE 2025] Choose the correct option: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false and Reason (R) is also false.

When two coils are wound on each other, the magnetic field passes through the other coil completely. This is why flux linked with the coil will be maximum.

Correct Option (A): Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

A conducting rod of length l is kept parallel to a uniform magnetic field B. It is moved along the magnetic field with a velocity v. What is the value of the emf induced in the conductor? [CBSE 2020]

According to Faraday’s law:

Induced emf

A coil of 200 turns and area 0.20 m^2 is rotated at half a revolution per second in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation. The maximum voltage generated in the coil is 2π/β​ volts. Find the value of β. [JEE 2024 January]

Motional emf in a rotating coil formula:
Eₘₐₓ = N × B × A × ω

Given:
N = 200
B = 0.01 T
A = 0.20 m²
Frequency = 0.5 rev/sec
So, ω = 2π × frequency = 2π × 0.5 = π rad/s

So,
Eₘₐₓ = 200 × 0.01 × 0.20 × π = 0.4π volts

As per the question:
Eₘₐₓ = (2π) / β

So,
(2π) / β = 0.4π
β = 2 / 0.4 = 5

Correct Answer: β = 5

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NCERT Physics Chapter 6 Electromagnetic Induction – FAQs

Find below the frequently asked questions on Class 12 NCERT Physics Chapter 6 Electromagnetic Induction:

Q&A Icon
Commonly asked questions
Q:  

Explain the term Electromagnetic Induction.

A: 

The process of creating the electric current in a conductor by moving the conductor through the magnetic field or by changing the magnetic field around the conductor is known as the Electromagnetic Induction.

Q:  

Describe the principle of the electromagnet.

A: 

The basic principle of the electromagnet is the electromagnetism. When the ferromagnetic iron core is coiled by a wire and a current is passed through the core, it produces a strong and magnetic field, and turns the setup into a magnet. By changing the current, one can change the strength of the magnetism, and when the current is switched off, the magnetism disappears.

Q:  

According to Electromagnetic Induction Chapter 6 Class 12 Physics, What is Eddy current?

A: 

As per the Faraday's Law of Induction, when a conductor is exposed to a changing magnetic field, a circulating electric current is induced, it is called the Eddy current. The Eddy currents have many applications including - non-destructive testing and induction furnaces.

Q:  

What is Ohm's law in Electromagnetic Induction?

A: 

Ohm's Law states that when the temperature remains constant, the current (I) flowing through a conductor is directly proportional to voltage (v), and inversely proportional to resistance (R).

Q:  

Explain Magnetic Flux according to the Class 12 Physics Electromagnetic Induction.

A: 

The sum of a magnetic field that passes through a given surface area is called the magnetic flux. The mathematic formula is? = BA cos (? ). 

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