NCERT solutions for Class 12 Physics Chapter 6 - Electromagnetic Induction

NCERT solutions for Class 12 Physics Chapter 6 - Electromagnetic Induction

Vishal kumarUpdated on 25 Aug 2025, 11:14 AM IST

Ever wondered how electric generators produce electricity or how a changing magnetic field induces current? That is the magic of Electromagnetic Induction. The Class 12 Physics Chapter 6 Solutions help you to master this topic with clear explanations for every concept and question. These Electromagnetic Induction Class 12 NCERT solutions make the tough theory and numericals simple and very easy.

This Story also Contains

  1. NCERT Solution for Class 12 Physics Chapter 6 Solutions: Download Solution PDF :
  2. Class 12 Physics Electromagnetic Induction Chapter: Exercise Questions
  3. NCERT Solutions for Class 12 Physics Chapter 6: Additional questions
  4. Class 12 Physics NCERT Chapter 6: Higher Order Thinking Skills (HOTS) Questions
  5. NCERT Class 12 Physics Chapter 6 Electromagnetic Induction: Important Topics
  6. Electromagnetic Induction Class 12: Important Formula
  7. Approach to Solve Questions of Electromagnetic Induction Class 12
  8. What Extra Should Students Study Beyond NCERT for JEE/NEET?
  9. Importance of NCERT Solutions for Class 12 Physics Chapter 6 in Exams
  10. NCERT Solutions for Class 12 Physics: Chapter-Wise
NCERT solutions for Class 12 Physics Chapter 6 - Electromagnetic Induction
Electromagnetic Induction

The NCERT solution for Class 12 Physics Electromagnetic Induction Chapter is essential for competitive tests like JEE and NEET, in addition to Class 12 exams, and helps in understanding its fundamentals, which makes it easier to understand how many of the gadgets we use on a daily basis—from big power systems to home appliances—operate. Essential information on how changing magnetic fields can produce electric currents, which power the technology that runs our world, is provided in the NCERT solutions Class 12 Physics of electromagnetic induction.

NCERT Solution for Class 12 Physics Chapter 6 Solutions: Download Solution PDF :

NCERT Solutions for Class 12 Physics Chapter 6: Electromagnetic Induction help in understanding well the phenomenon of how a changing magnetic field generates electrical currents. This article contains step-wise solutions to NCERT questions that help a student not only master a concept but also develop the skill of solving problems. You may also download a free PDF to revise at any time and prepare well in advance of board exams, JEE and NEET.

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Class 12 Physics Electromagnetic Induction Chapter: Exercise Questions

The Exercise Questions of Class 12 Physics Chapter 6 – Electromagnetic Induction are designed to test conceptual understanding and application of Faraday’s law, Lenz’s law, and induced EMF. Solving these NCERT exercises helps students build strong analytical skills and prepares them well for board exams as well as competitive exams like JEE and NEET.

Q 6.1(a) Predict the direction of induced current in the situations described by the following

Relative Motion Between Coil and Magnet

Answer:

To oppose the magnetic field current should flow in anti-clockwise, so the direction of the induced current is qrpq

Q 6.1 (b) Predict the direction of induced current in the situations described by the following Figs.

Answer:

Current in the wire in a way such that it opposes the change in flux through the loop. Here hence current will induce in the direction of p--->r--->q in the first coil and y--->z--->x in the second coil.

Q 6.1 (c) Predict the direction of induced current in the situations described by the following Figs.(c)

Answer:

When we close the key, the current will flow through the first loop, and suddenly, magnetic flux will pass through it such that the magnetic field will go from right to left of the first loop. Now, to oppose this change, current in the second loop will flow such that magnetic fields go from left to right, which is the direction yzxy

Q 6.1 (d) Predict the direction of induced current in the situations described by the following Fig. (d)

Answer:

When we increase the resistance of the rheostat, the current will decrease, which means flux will decrease, so current will be induced to increase the flux through it. Flux will increase if current flows in xyzx.

On the other hand, if we decrease the resistance, that will increase the current, which means flux will increase, so the current will induce to reduce the flux. Flux will be reduced if the current goes in the direction zyxz

Q 6.1 (e) Predict the direction of induced current in the situations described by the following Fig(e)

Answer:

As we release the tapping key, current will induce to increase in the flux. Flux will increase when current flows in the direction xryx.

Q 6.1 (f) Predict the direction of induced current in the situations described by the following Fig (f)

Answer:

The current will not induce as the magnetic field lines are parallel to the plane. In other words, since flux through the loop is constant (zero in fact), there won't be any induction of the current.

Q6.2 (a) Use Lenz’s law to determine the direction of induced current in the situations described by Fig. : a

A wire of irregular shape turning into a circular shape;

Answer:

By turning the wire from an irregular shape to a circle, we are increasing the area of the loop, so flux will increase and current will be induced in such a way that reduces the flux through it. By the right-hand thumb rule direction of current is adcba.

Q6.2 (b) Use Lenz’s law to determine the direction of induced current in the situations described by Fig. b :

A circular loop being deformed into a narrow straight wire.

Answer:

Here, by changing shape, we are decreasing the area or decreasing the flux, so the current will be induced in a manner such that it increases the flux. Since the magnetic field is coming out of the plane, the direction of the current will be adcba.

Q6.3 A long solenoid with 15 turns per cm has a small loop of area 2.0cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0A to 4.0A in 0.1s , what is the induced emf in the loop while the current is changing?

Answer:

Given in a solenoid,

The number of turns per unit length :

n=15turn/cm=1500turn/m

loop area :

A=2cm2=2104m

Current in the solenoid :

Initial current = Iinitial=2

Finalcurrent = Ifinal=4

Change in current :

ΔI =42=2

Change in time:

Δt=0.1s

Now, the induced emf :

e=dϕdt=d(BA)dt=d(μ0nIA)dt=μ0nAdIdt=μ0nAΔIΔt


e=4π1071500210420.1=7.54106V

Hence induced emf in the loop is 7.54106V .

Q6.4 A rectangular wire loop of sides 8cm and 2cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1cms1 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?

Answer:

Given:

Length of rectangular loop :

l=8cm=0.08m

Width of the rectangular loop:

b=2cm=0.02m

Area of the rectangular loop:

A=lb=(0.08)(0.02)m2=16104m2

Strength of the magnetic field

B=0.3T

The velocity of the loop :

v=1cm/s=0.01m/s

Now,

a) Induced emf in the long side wire of the rectangle:

e=Blv=0.30.080.01=2.4104V

This EMF will be induced till the loop gets out of the magnetic field, so the time for which the EMF will be induced:

t=distncevelocity=bv=21020.01=2s

Hence, a 2.4104V emf will be induced for 2 seconds.

b) Induced emf when we move along the width of the rectangle:

e=Bbv=0.30.020.01=6105V

time for which the emf will induce :

t=distncevelocity=lv=81020.01=8s

Hence, a 6105V emf will be induced for 8 seconds.

Q6.5 A 1.0m long metallic rod is rotated with an angular frequency of 400rads1 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.

Answer:

Given

length of metallic rod :

l=1m

Angular frequency of rotation :

ω=400s1

Magnetic field (which is uniform)

B=0.5T

Velocity: here velocity at each point of the rod is different. One end of the rod has zero velocity, and the other end has velocity ωl.

Hence, we take the average velocity of the rod, so,

Average velocity =0+ωl2=ωl2

Now,

Induce emf

e=Blv=Blwl2=Bl2ω2

e=0.5124002=100V

Hence emf developed is 100V.

Answer:

Given

Length of the wire l=10m

Speed of the wire v=5m/s

The magnetic field of the earth B=0.3104Wbm2

Now,

The instantaneous value of induced emf :

e=Blv=0.3104105=1.5103V

Hence instantaneous emf induced is 1.5103V.

Q6.7 Current in a circuit falls from 5.0A to 0.0A in 0.1s. If an average emf of 200V is induced, give an estimate of the self-inductance of the circuit.

Answer:

Given

Initial current Iinitial=5A

Final current Ifinal=0A

Change in time Δt=0.1s

Average emf e=200V

Now,

As we know, in an inductor

e=LdIdt=LΔIΔt=LIfinalIinitialΔt

L=eΔtIfinalIinitial=2000.150=4H

Hence self-inductance of the circuit is 4H.

Q6.8 A pair of adjacent coils has a mutual inductance of 1.5H. If the current in one coil changes from 0 to 20A in 0.5s, what is the change of flux linkage with the other coil?

Answer:

Given

Mutual inductance between two coils:

M=1.5H

Currents in a coil:

Iinitial=0

Ifinal=20

Change in current:

diI200=20

The time taken for the change

dt=0.5s

The relation between emf and mutual inductance:

e=MdIdt

e=dϕdt=MdIdt

dϕ=MdI dϕ=MdI=1.520=30Wb

Hence, the change in flux in the coil is 30Wb.

NCERT Solutions for Class 12 Physics Chapter 6: Additional questions

The Additional Questions of Class 12 Physics Chapter 6 Electromagnetic Induction offer extra practice outside of the textbook, centring on problems and application-oriented learning. These questions enhance the conceptual clarity of the topics such as mutual induction, eddy currents, and the generation of an alternating current, which are of great value for the board exams and entrance exams like JEE and NEET.

Q1: A jet plane is travelling towards the west at a speed of 1800km/h. What is the voltage difference developed between the ends of the wing having a span of 25m, if the Earth’s magnetic field at the location has a magnitude of 5×104T and the dip angle is 300.

Answer:

Given

Speed of the plane:

v=1800kmh1=180010006060=500m/s

Earth's magnetic field at that location:

B=5104T

The angle of dip is the angle made with the horizontal by Earth's magnetic field:

δ=300

Length of the wings

l=25m

Now, since only the vertical component of the magnetic field will cut the wings of the plane perpendicularly, only those will help in inducing an emf.

The vertical component of the Earth's magnetic field :

Bvertical=Bsinδ=5104sin30=51040.5=2.5104

So now, Induce emf :

e=Bverticallv=2.510425500=3.125V

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

Q2: 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 0.3T at the rate of 0.02Ts1. If the cut is joined and the loop has a resistance of 1.6Ω, how much power is dissipated by the loop as heat? What is the source of this power?

Answer:

Given,

Area of the rectangular loop which is held still:

A=lb=(0.08)(0.02)m2=16104m2

The resistance of the loop:

R=1.6Ω

The initial value of the magnetic field :

Binitial=0.3T

Rate of decrease of this magnetic field:

dBdt=0.02T/s

Induced emf in the loop :

e=dϕdt=d(BA)dt=AdBdt=161040.02=0.32104V

Induced Current :

Iinduced=eR=0.321041.6=2105A

The power dissipated in the loop:

P=Iinduced2R=(2105)21.6=6.41010W

The external force, which is responsible for changing the magnetic field, is the actual source of this power.

Q3: A square loop of side 12cm with its sides parallel to X and Y axes is moved with a velocity of 8cms1 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 103Tcm1 along the negative x-direction (that is it increases by 103Tcm1 as one moves in the negative x-direction), and it is decreasing in time at the rate of 103Ts1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50mΩ.

Answer:

Given,

Side of the square loop

l=12cm=0.12m

Area of the loop:

A=0.120.12m2=144104m2

The resistance of the loop:

R=4.5mΩ=4.5103Ω

The velocity of the loop in the positive x-direction

v=8cm/s=0.08m/s

The gradient of the magnetic field in the negative x-direction

dBdx=103T/cm=101T/m

Rate of decrease of magnetic field intensity

dBdt=103T/s

Now, here emf is being induced by means of both changing the magnetic field with time and changing with space. So let us find out the emf induced by both the change of space and time, individually.

Induced emf due to a field changing with time:

ewithtime=dϕdt=AdBdt=144104103=1.44105Tm2/s

Induced emf due to a field changing with space:

ewithspace=dϕdt=d(BA)dt=AdBdxdxdt=AdBdxv

ewithspace=1441041010.08=11.52105Tm2/s

Now, Total induced emf :

etotal=ewithtime+ewithspace=1.44105+11.52105=12.96105V

Total induced current :

I=eR=12.961054.5103=2.88102A

Since the flux is decreasing, the induced current will try to increase the flux through the loop along the positive z-direction.

Q4: 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 2cm2 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 900 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 7.5mC . The combined resistance of the coil and the galvanometer is 0.50Ω . Estimate the field strength of magnet.

Answer:

Given,

Area of search coil :

A=2cm2=2104m2

The resistance of the coil and the galvanometer

R=0.5Ω

The number of turns in the coil:

N=25

Charge flowing in the coil

Q=7.5mC=7.5103C

Now.

Induced emf in the search coil

e=Ndϕdt=Nϕfinalϕinitialdt=NBA0dt=NBAdt

e=iR=dQdtR=NBAdt

B=RdQNA=0.57.5103252104=0.75T

Hence magnetic field strength for the magnet is 0.75T.

Q6 An air-cored solenoid with length 30cm , area of cross-section 25cm2 and number of turns 500 , carries a current of 2.5A . The current is suddenly switched off in a brief time of 103s . 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.

Answer:

Given

Length of the solenoid l=30cm=0.3m

Area of the cross-section of the solenoid A=25cm2=25104m2

Number of turns in the solenoid N=500

Current flowing in the solenoid I=2.5A

The time interval for which current flows Δt=103s

Now.

Initial flux:

ϕinitial=NBA=N(μ0NIl)A=μ0N2IAl

ϕinitial=4π10750022.5251040.3=6.55103Wb

Final flux: since no current is flowing,

ϕfinal=0

Now

Induced emf:

e=dϕdt=ΔϕΔt=ϕfinalϕinitialΔt=6.551030103=6.55V

Hence, 6.55V of average back emf is induced.

Q7(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig.

Answer:

Here, let's take a small element dy in the loop at a distance y from the wire

Area of this element dy :

dA=ady

The magnetic field at dy (which is y distance away from the wire)

B=μ0I2πy

The magnetic field associated with this element dy

dϕ=BdA

dϕ=μ0I2πyady=μ0Ia2πdyy

ϕ=xa+xμ0Ia2πdyy=μ0Ia2π[lnx]aa+x=μ0Ia2πln[a+xx]

Now, as we know

ϕ=MI where M is the mutual inductance

so

ϕ=MI=μ0Ia2πln[a+xx]

M=μ0a2πln[a+xx]

Hence mutual inductance between the wire and the loop is:

μ0a2πln[a+xx]

Q7 (b) Now assume that the straight wire carries a current of 50A and the loop is moved to the right with a constant velocity, V=10m/s . Calculate the induced emf in the loop at the instant when X=0.2m . Take a=0.1m and assume that the loop has a large resistance.

Answer:

Given,

Current in the straight wire

I = 50A

Speed of the Loop, which is moving in the right direction

V=10m/s

Length of the square loop

a=0.1m

Distance from the wire to the left side of the square

x=0.2m

Now,

Induced emf in the loop :

E=Bxav=μ0I2πxav=(4π107502π0.2)0.110=5105V

Hence emf induced is 5105V.

Q8: 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. A uniform magnetic field extends over a circular region within the rim. It is given by,

B=B0k(ra;a<R)=0 (otherwise)


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

1594626697274

Answer:

Given,

The radius of the wheel =R

The mass of the wheel = M

Line charge per unit length when the total charge is Q

λ=Q2πr

Magnetic field :

B=B0K(ra;a<R)

Magnetic force is balanced by centrifugal force when v is the speed of the wheel that is

BQv=mv2r

B2πrλ=Mvr

v=B2πλr2M

Angular velocity of the wheel

w=vr=B2πλr2MR

when (ra;a<R)

w=B2πλa2MR

It is the angular velocity of the wheel when the field is suddenly shut off.

Class 12 Physics NCERT Chapter 6: Higher Order Thinking Skills (HOTS) Questions

The HOTS (Higher Order Thinking Skills) questions on Class 12 Physics Chapter 6 - Electromagnetic Induction are aimed at progressing the critical and analytical thinking skills. These tricky questions are beyond simple formulas so that students are ready to implement their knowledge of Faraday's law, Lenz's law, and energy conservation in real life and competitive exam interpretation exercises and JEE and NEET.

Q.1 A conducting circular loop is placed in a uniform magnetic field 0.4 T with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at 1 mm s1. The induced emf in the loop when the radius is 5 cm is:
a) 2π×105 V
b) 4π×105 V
c) π×103 V
d) 5π×104 V

Answer:

e=dϕdt=ddt(BA)

e=Bddt(πr2)=πB×2r(drdt)

e=π×0.4×2×5×102×103e=4π×105 V

Hence, the correct option is (2).

Q. 2 At time t=0 magnetic field of 1500 gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnet's field reduces linearly to 1000 gauss in the next 10 seconds. Then induced FMF in the loop is ____ μv.

Answer:

Area of loop = area of the rectangle - area of 2 triangle =32×82×12×8×4=25632=224 cm2

Magnetic flux passing through an area is given by

=BdA=BdAcosθ

ϕinitial =1500×224 cm2

=336000 gauss cm =3.36×105×108=3.36×103

ϕfinal =1000×224 cm2=224000 gauss cm2=2.24×103

since the film changes linearly,

ε=dϕdt=|ϕfinal ϕinitial |Δt=1.12×10310=1.12×104=112μv

Hence, the answer is 112.

Q.3 The configuration of a long straight wire and a wire loop with a = 15 cm and b = 20 cm is depicted in Fig. The wire's current fluctuates in accordance with the relationship i = 6t2 – 12 t, where t is measured in seconds and I is in amperes. What is the emf (in 10-7 volts) in the square loop at t = 6 s?

Answer:

The field (due to the current in the long straight wire) through the part of the rectangle above the wire is out of the page (by the right-hand rule), and below the wire, it is into the page. Thus, a strip below the wire (where the strip borders the long wire and reaches a distance of b - a away from it) has exactly the equal but opposite flux that cancels the contribution from the section above the wire, since the height of the part above the wire is b - a.


ϕB=BdA=baa{μ0i2πr}(bdr)=μ0ib2πln{aba}

As per Faraday's law:

ε=dϕBdt=ddt[μ0ib2πln(aba)]
ε=μ0b2πln(aba)didt
ε=μ0b2πln(aba)ddt(6t212t)
ε=μ0b(12t12)2πln(aba)

As, a=0.15 m,b=0.2 m,t=6 s
So,
ε=4π×107×0.2(12×612)2πln{0.150.200.15}=26.36×107 V

Hence, the answer is (26.36)

Q. 4 The self-inductance of a coil having 400 turns is 10 mH. The magnetic flux through the cross-section of the coil corresponding to a current of 2 mA is:

Answer:

Here, N=400, L=10mH=10×103H,

I=2 mA=2×103 A

Total magnetic flux linked with the coil,

ϕ=NLI=400×(10×103)×2×103=8×103 Wb


Magnetic flux through the cross-section of the coil = Magnetic flux linked with each turn

=ϕN=8×103200=4×105 Wb

Q.5 Two solenoids of an equal number of turns have their lengths and radii in the same ratio 1:2. The ratio of their self-inductances will be:
Answer:

Self-inductance of solenoids, L=μ0 N2 Al=μ0N2πr2l
Where l is the length of the solenoids, N is the total number of turns of the solenoid and A is the area of the cross-section of the solenoid.

L1 L2=(N1 N2)2(r1r2)2(l2l1) Here, N1=N2,l1l2=12,r1r2=12 L1 L2=(12)2(21)=12

NCERT Class 12 Physics Chapter 6 Electromagnetic Induction: Important Topics

The idea of electromagnetic induction is presented in a simple manner in the Class 10 NCERT book. However, a more thorough and in-depth description of electromagnetic induction may be found in Class 12 Physics, Chapter 6. Furthermore, a series of problems and their answers based on the subjects covered in this chapter are provided in the Chapter 6 Physics Class 12 NCERT Solutions. The Class 12 Physics Electromagnetic Induction NCERT Solutions cover questions from the following topics:

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Electromagnetic Induction Class 12: Important Formula

The important formulas of Class 12 Physics Chapter 6 – Electromagnetic Induction act as quick tools to solve numerical problems effectively. They include key relations from Faraday’s law, Lenz’s law, induced emf, mutual and self-induction, making them essential for board exams as well as JEE and NEET preparation.

1. Faraday's Law of Induction:

E=dΦBdt
Where:
E= induced EMF
ΦB= magnetic flux
t= time

2. Lenz's Law:

States that the direction of the induced current (or EMF) is such that it opposes the change in magnetic flux that produces it.
E=NdΦBdt (For a coil with N turns)

3. Magnetic Flux:

ΦB=BAcosθ
Where:
B= magnetic field strength
A= area through which the magnetic field passes
θ= angle between the magnetic field and the normal to the surface

4. Self-Induction:

EL=LdIdt
Where:
L= self-inductance
I= current

5. Mutual Induction:

EM=MdIdt
Where:
M= mutual inductance
I= current in the primary coil

Approach to Solve Questions of Electromagnetic Induction Class 12

  • Start with Faraday's Laws: Learn how changing magnetic fields can produce current. Focus on both laws; they are important in this chapter.
  • Know Lenz's Law: "Oppose the change" is Lenz's Law in simple words. Learn how it tells the direction of the induced current.
  • Magnetic Flux: It is like how much magnetic field passes through a surface. Learn the formula: ϕ=BAcos(θ)
  • EMF Formula: Learn how EMF is induced: emf=dϕdt
    The minus sign is from Lenz's Law.
  • Self-inductance (L): It is when a coil opposes the change in its own current. Formula:emf=Ldidt
  • Mutual Inductance (M): One coil affects another nearby coil by changing current. Formula:emf=Mdidt
  • Practice motion-based induction: Solve problems where a conductor moves in a magnetic field. This type of question usually comes in a competitive exam. Use:emf=Blv
  • Solve NCERT & PYQs: Focus on concept-based NCERT questions. Many numerical and theoretical questions come directly from them.

What Extra Should Students Study Beyond NCERT for JEE/NEET?

In addition to NCERT, students looking to appear in JEE/NEET in Electromagnetic Induction should practice in-depth applications such as AC generator, elaborate solutions to mutual and self-inductance, energy in inductors and conceptual problems including Lenz law. Solving higher-level numericals and previous JEE/NEET papers, one can develop good accuracy and speed.


Importance of NCERT Solutions for Class 12 Physics Chapter 6 in Exams

  • For both the CBSE Class 12 board exams and competitive tests like KVPY, NEET, and JEE Main, the NCERT answers for Class 12 Physics, Chapter 6 on Electromagnetic Induction, are essential.

  • Determining the direction of induced electromagnetic fields is a typical issue in tests like NEET and JEE Main. The Electromagnetic Induction part of the NCERT answers covers a lot of these difficulties.

  • The chapters on electromagnetic induction usually account for 10% of the questions on the CBSE board exam.

NCERT Solutions for Class 12 Physics: Chapter-Wise

The NCERT Solutions for Class 12 Physics (Chapter-wise) provide a well-structured approach to mastering every topic as per the latest CBSE curriculum. These step-by-step solutions help students strengthen concepts, practice problem-solving, and prepare effectively for both board exams and competitive exams like JEE and NEET.


Also, check NCERT Books and NCERT Syllabus here:

NCERT solutions subject-wise

Also, check

NCERT Exemplar Class 12 Solutions

Frequently Asked Questions (FAQs)

Q: What is Electromagnetic Induction in Class 12 Physics?
A:

The process by which a fluctuating magnetic field creates an electromotive force (EMF) in a conductor is known as electromagnetic induction.

Q: What is Lenz’s Law? Explain with an Example.
A:

According to Lenz's Law, the induced EMF is in opposition to the magnetic flux change.
For instance, bringing a magnet close to a coil causes current to be induced, which repels the magnet.

Q: What are Eddy Currents? What are their Applications?
A:

Eddy currents are circulating currents induced in conductors by changing magnetic fields.
Applications: Induction heating, magnetic braking, and electric meters.

Q: Is the class 12 chapter Electromagnetic Induction helpful for higher studies?
A:

Yes, the chapter is helpful for higher studies in the field of electronics and electrical engineering-related subjects and for scientific research.

Q: How important is the chapter Electromagnetic Induction for CBSE board exam?
A:

On an average 5 mark questions are asked for CBSE board exams. Follow the NCERT syllabus and NCERT book for a good score in the board exam. Along with the NCERT solutions exercises can also practice NCERT exemplar problems and CBSE board previous year papers.

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Yes, if you’re not satisfied with your marks even after the improvement exam, many education boards allow you to reappear as a private candidate next year to improve your scores. This means you can register independently, study at your own pace, and take the exams without attending regular classes. It’s a good option to improve your results and open up more opportunities for higher studies or careers. Just make sure to check the specific rules and deadlines of your education board so you don’t miss the registration window. Keep your focus, and you will do better next time.

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Yes, in the case that you appeared for the 2025 improvement exam and your roll number is different from what was on the previous year’s marksheet, the board will usually release a new migration certificate. This is because the migration certificate will reflect the most recent exam details, roll number and passing year. You can apply to get it from your board using the process prescribed by them either online or through your school/college.

Yes, if you miss the 1st CBSE exam due to valid reasons, then you can appear for the 2nd CBSE compartment exam.

From the academic year 2026, the board will conduct the CBSE 10th exam twice a year, while the CBSE 12th exam will be held once, as per usual. For class 10th, the second phase exam will act as the supplementary exam. Check out information on w hen the CBSE first exam 2026 will be conducted and changes in 2026 CBSE Board exam by clicking on the link .

If you want to change your stream to humanities after getting a compartment in one subject in the CBSE 12th Board Exam , you actually have limited options to qualify for your board exams. You can prepare effectively and appear in the compartment examination for mathematics again. If you do not wish to continue with the current stream, you can take readmission in the Humanities stream and start from Class 11th again, and continue studying for two more years to qualify for the 12th examination.

The GUJCET Merit List is prepared based on the Class 12th marks and GUJCET marks received by the students. CBSE students who are not from the Gujarat board can definitely compete with GSEB students, as their eligibility is decided based on the combined marks scored by them in GUJCET and the 12th board. The weightage of the GUJCET score is 40% and the weightage of the class 12 scores is 60%.