Careers360 Logo
NCERT Exemplar Class 12 Physics Solutions Chapter 6 Electromagnetic Induction

NCERT Exemplar Class 12 Physics Solutions Chapter 6 Electromagnetic Induction

Edited By Safeer PP | Updated on Sep 14, 2022 12:52 PM IST | #CBSE Class 12th
Upcoming Event
CBSE Class 12th  Exam Date : 10 Mar' 2025 - 10 Mar' 2025

NCERT Exemplar Class 12 Physics solutions chapter 6 explores the relation and familiarity between electricity and magnetism by studying moving electric charges producing magnetic fields. NCERT Exemplar Class 12 Physics chapter 6 solutions answer questions like - How has been a relation between electricity and magnetism established? Is it possible for moving magnets to produce electric current? It throws light on the experiments done by various scientists to prove electromagnetic induction, not merely a topic of theoretical or academic interest but also of practical utility. Class 12 NCERT Exemplar Physics solutions chapter 6 given on this page would help comprehend theoretical and practical concepts of Electromagnetic induction and efficiently summarise the essential portions that could be used to perform well in 12 Boards and competitive exams. NCERT Exemplar Class 12 Physics solutions chapter 6 PDF download can also be used by students.

This Story also Contains
  1. NCERT Exemplar Class 12 Physics Solutions Chapter 6 Electromagnetic Induction-Main subtopics
  2. NCERT Exemplar Class 12 Physics Solutions Chapter 6- What Will The Students Learn?
  3. NCERT Exemplar Class 12 Physics Chapter Wise Links
  4. Some applications of electromagnetic Induction
LiveCBSE Board Exam 2025 Live: Class 12 mathematics paper 'moderate but lengthy'; memory-based questionsMar 8, 2025 | 9:47 PM IST

The Central Board of Secondary Education (CBSE) chairperson Rahul Singh clarified that the scheme of subjects and the number of academic subjects will remain the same for the board exams 2026. 

Read More: No change in CBSE Class 10 subjects for 2026, board will offer 45 languages: Chairman clarifies

Read More

Also check - NCERT Solutions for Class 12 Physics

Question:6.1

A square of side L meters lies in the x-y plane in a region, where the magnetic field is given by B=Bo(2i+3j+4k)T where Bo is constant. The magnitude of flux passing through the square is
(a)2BoL2Wb.
(b)3BoL2Wb.
(c)4BoL2Wb.
(d)29BoL2Wb.

Answer:

For elementary area dA of a surface, flux linked dϕ=BdAcosθordϕ=B.dA
So-net flux through the surface
ϕ=B×dA=BAcosθ
In this problem
A=L2k^andB=B0(2i^+3j^+4k^)Tϕ=B.A=B0(2i^+3j^+4k^).L2k^=4BoL2Wb.

Question:6.2

A loop, made of straight edges has six corners at A(0,0,0),B(L,O,0)C(L,L,0),D(0,L,0)E(0,L,L)andF(0,0,L). A magnetic field B =Bo(i+k)Tis present in the region. The flux passing through the loop ABCDEFA (in that order) is

(a)BoL2Wb.
(b)2BoL2Wb.
(c)2BoL2Wb.
(d)4BoL2Wb.

Answer:
In this problem first we have to analyse area vector, loop ABCDA lies in x-y plane whose area vector A1=L2k^ whereas loop ADEFA lies in y-z plane whose area vector A2=L2i^
And magnetic flux is
ϕm=B.AA=A1+A2=(L2k^+L2i^)andB=B0(i^+k^)Now,ϕm=B.A=B0(i^+k^).(L2k^+L2i^)=2B0L2Wb


Question:6.3

A cylindrical bar magnet is rotated about its axis. A wire is connected from the axis and is made to touch the cylindrical surface through a contact. Then

(a) a direct current flows in the ammeter A.
(b) no current flows through the ammeter A.
(c) an alternating sinusoidal current flows through the ammeter A with a time period T=2π/ω.
(d) a time-varying non-sinusoidal current flows through the ammeter A.

Answer:

b)
When the cylindrical bar magnet is rotated about its axis, there is no change in flux linked with the circuit. So no current flows through the ammeter A

Question:6.4

There are two coils A and B as shown in figure. A current starts flowing in B as shown when A is moved towards B and stops when A stops moving. The current in A is counter clockwise. B is kept stationary when A moves. We can infer that
(a) there is a constant current in the clockwise direction in A.
(b) there is a varying current in A.
(c) there is no current in A.
(d) there is a constant current in the counter clockwise direction in A.

Answer:

The answer is the option (d)
The problem infers that there is a constant current flowing in coil A, as when it is moved relative to coil B, the flux linked to B due to A is changed. This induces an emf in B, and hence current starts flowing. But when A is stopped there is no change in flux and hence no current would be induced in the coil B. Whereas, if there were a changing current in the coil A, then due electromagnetic induction, there would have been an emf generated in B.

Question:6.5

Same as problem 4 except the coil A is made to rotate about a vertical axis. No current flows in B if A is at rest. The current in coil A, when the current in B (at t = 0) is counterclockwise and the coil A is as shown at this instant, t = 0, is
(a) constant current clockwise.
(b) varying current clockwise.
(c) varying current counterclockwise.
(d) constant current counterclockwise.

Answer:

The answer is the option (a)
Here Lenz’s law is followed which states that, the direction of induced emf (current) is in such that it opposes the cause that produced it. It is derived from the law of conservation of energy.
The current of coil B is given to be counter-clockwise, so if this current is induced so as to oppose its cause than the current in the coil A should be in constant clockwise direction. And it is also supported by the fact that when A is stationary then there is no current induced in coil B.

Question:6.6

The self-inductance L of a solenoid of length l and area of cross-section A, with a fixed number of turns N increases as
(a) l and A increase.
(b) l decreases and A increases.
(c) l increases and A decreases.
(d) both l and A decrease.

Answer:

The answer is the option (b)
The formula for self-inductance of a solenoid is given by:

L=(μrμ0n2)I
So, from the equation, the inductance should increase as l decreases and A increases. As, L is inversely proportional to l and directly proportional to A.

Question:6.7

A metal plate is getting heated. It can be because
(a) a direct current is passing through the plate
(b) it is placed in a time varying magnetic field
(c) it is placed in a space varying magnetic field, but does not vary with time
(d) a current (either direct or alternating) is passing through the plate

Answer:

The correct answers are the options (a,b,c)
The heating of the plate can happen due to the following reasons; either it is placed in a magnetic field that possesses a relative motion or a direct current is passed through it.
Here, (a) is the option where DC current is passed through the plate. While b and c are the options where the magnetic field is having a relative motion with the plate.

Question:6.8

An e.m.f is produced in a coil, which is not connected to an external voltage source. This can be due to
(a) the coil being in a time varying magnetic field.
(b) the coil moving in a time varying magnetic field.
(c) the coil moving in a constant magnetic field.
(d) the coil is stationary in external spatially varying magnetic field, which does not change with time.

Answer:

The correct answers are the options (a,b,c)
EMF is induced in a conductor only if it is placed in a magnetic field which is moving relative to the conductor. The first three options i.e. a, b and c justify the above statement.

Question:6.9

The mutual inductance M12 of coil 1 with respect to coil 2
(a) increases when they are brought nearer
(b) depends on the current passing through the coils
(c) increases when one of them is rotated about an axis
(d) is the same as M21 of coil 2 with respect to coil 1

Answer:

The correct answers are the options (a, d)

According to Faraday's second law emf induces in secondary e2=N2dϕ2dt;e2=Mdi1dt

The mutual inductance M12 of coil 1 wrt coil 2 increases when they are brought nearer and is the same as M21 of coil with respect to coil 1.
M12 e.g. mutual inductance of solenoid S1 with respected to solenoid S2 is given by

M12=μ0N1N2πr12l


Where signs are as usual,

Also, M21, i.e. mutual inductance of solenoid S2 with respect to solenoid S1 is given by:

M21=μ0N1N2πr12l

So we have M12=M21=M

Question:10

A circular coil expands radially in a region of magnetic field and no electromotive force is produced in the coil. This can be because
(a) the magnetic field is constant.
(b) the magnetic field is in the same plane as the circular coil and it may or may not vary.
(c) the magnetic field has a perpendicular (to the plane of the coil) component whose magnitude is decreasing suitably.
(d) there is a constant magnetic field in the perpendicular (to the plane of the coil) direction.

Answer:

Answer: The correct answers are the options (b, c)
When the magnetic flux cutting a cross sectional area of a conductor changes w.r.t. to the conductor only then it induces an EMF. In this particular case there can be two of possibilities which are fulfilled by b and c.

Question:6.11

Consider a magnet surrounded by a wire, with an on/off switch S (figure). If the switch is thrown from the off position (open circuit) to the on position (closed circuit), will a current flow in the circuit? Explain.

Answer:

As shown in the figure, the magnet bar and the coil are stationary, therefore there would be no change in the magnetic flux linked to the projected area of the coil. Hence there would be no EMF generate and if there is no EMF, there is not going to be any current flowing in the circuit when it is closed.

Question:6.12

A wire in the form of a tightly wound solenoid is connected to a DC source and carries a current. If the coil is stretched so that there are gaps between successive elements of the spiral coil, will the current increase or decrease? Explain.

Answer:

This solution for this problem can be explained using Lenz’s law. The states that, in a conductor, the direction of induced EMF is such that it opposes it because that produced it. Here if the current-carrying tightly wound solenoid is stretched, the flux would try to leak through the gaps. Therefore, the system will try to oppose this decrease in magnetic flux and hence the current will increase.

Question:6.13

A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, will the current increase or decrease? Explain.

Answer:

The iron core, when inserted in the coil, will increase the magnetic field due to the magnetization of the iron bar. So according to Lenz’s law, this increase in the magnetic field should be opposed, therefore, the current will decrease.

Question:6.14

Consider a metal ring kept on top of a fixed solenoid such that the centre of the ring coincides with the axis of the solenoid. If the current is suddenly switched on, the metal ring jumps up. Explain.

Answer:

Here, initially, there was no flux linked to the ring. When a sudden current passes through the solenoid, a magnetic field is induced. This magnetic field is now linked to the ring, so a current would get induced in the ring which would oppose the magnetic field. Hence ring will jump, both the magnetic fields (solenoid and ring induced) will interact.

Question:6.15

Consider a metal ring kept (supported by a cardboard) on top of a fixed solenoid carrying a current I. The centre of the ring coincides with the axis of the solenoid. If the current in the solenoid is switched off, what will happen to the ring?

Answer:

Here, the ring will now experience a downward force. This is followed from the Lenz’s law, as the current stops flowing in the solenoid; the magnetic flux will also get decreases. A current in the ring will be induced so as to oppose that and hence it will experience a downward force.

Question:6.16

Consider a metallic pipe with an inner radius of 1 cm. If a cylindrical bar magnet of radius 0.8 cm is dropped through the pipe, it takes more time to come down than it takes for a similar un-magnetised cylindrical iron bar dropped through the metallic pipe. Explain.

Answer:

The magnetized cylindrical bar carries a magnetic field with it. So, when it moves inside the metallic bar, the magnet field lines change w.r.t. the metallic hollow cylinder. This change in magnetic flux linked to the metallic pipe is opposed by it. Therefore, a force is applied on the magnetic cylinder; hence it takes more time to travel through the pipe.

Question:6.17

A magnetic field in a certain region is given by B=B0cos(ωt) and a coil of radius a with resistance R is placed in the x-y plane with its centre at the origin in the magnetic field (figure). Find the magnitude and the direction of the current at (a, 0, 0) at
t=π2ω,t=πωandt=3π2ω


Answer:

ϕm=B.A=BAcosθ
And as we know both A (area vector ) and B (magnetic field vector) are directed along z-axis. So, angle between them is 0.
So, cosθ=1(θ=0)
ϕm=BA
Area of coil of radius a=πa2
ε=B0(πa2)cosωt
By Faraday's law of electromagnetic induction, Magnitude of induced emf is given by
ε=B0(πa2)ωsinωt
This causes flow of induced current, which is given by
I=B0(πa2)ωsinωtR
Now, the value of current at different instants,
(i) t=π2ω
I=B0(πa2)ωRalongj^
Because sinωt=sin(ωπ2ω)=sinπ2=1
(ii) t=πω,I=B(πa2)ωR=0
because, sinωt=sin(ωπ2ω)=sinπ=0
(iii) t=32πω
I=B(πa2)ωRalongj^
sinωt=sin(ω.3π2ω)=sin3π2=1

Question:6.18

Consider a closed loop C in a magnetic field (figure). The flux passing through the loop is defined by choosing a surface whose edge coincides with the loop and using the formula ϕ=B1dA1,B2dA2 …. Now, if we choose two different surfaces S1 and S2 having C as their edge, would we get the same answer for flux. Justify your answer.

Answer:

The magnetic flux is same as in S1 and S2 as magnetic field lines do not start anywhere in the space because of continuity. As the numbers of magnetic field lines passing through the circuit are same, therefore the flux would be same for both of them.

Question:6.19

Find the current in the wire for the configuration shown in figure. Wire PQ has negligible resistance. B, the magnetic field is coming out of the paper. θ is a fixed angle made by PQ travelling smoothly over two conducting parallel wires separated by a distance d.

Answer:


Conducting electrons experience a magnetic force Fm=evB. So they move from P to Q within the rod. The end P of the rod becomes positively charged while end Q becomes negatively charged, hence an electric field is set up within the rod which opposes the further downward movement of electrons, i.e., an equilibrium Fe=Fm, i.e.,
eE=evBorE=vBInduced emfe=El=Bvl[E=Vl]
If the rod is moved by making an angle θ with the direction of the magnetic field or length. Induced emf,
e=Bvlsinθ

Emf induced across PQ due to its motion or change in magnetic flux linked with the loop change due to the change of enclosed area. The induced electric field E along the dotted line CD (perpendicular to both Vand B and along V×B ) =vB
Therefore, the motional emf along.
PQ=(length PQ)×(field along PQ)
=(length PQ)×(vBsinθ)
=(dsinθ)×(vBsinθ)=vBd

This induced emf make flow of current in closed circuit of resistnce R.
I=dvBR and independent of θ.

Question:6.20

A (current versus time) graph of the current passing through a solenoid is shown in figure. For which time is the back electromotive force (u) a maximum. If the back emf at t=3 s is e, find the back emf at t=7s, 15s,40s. OA, AB and BC are straight line segments.

Answer:

ε=d(NϕB)dt
ε=Ldldt
Thus, negative sign indicates that induced emf (e) opposes any change (increase or decrease) of current in the coil.
When the rate of change of current is maximum, then back emf in solenoid is (u) a maximum. This occurs in AB part of the graph. So maximum back emf will be obtained between 5s<t<10s.
Since, the back emf at t=3s is e.
Also, the rate of change of current at t=3,
and slope (s) of OA (from t=0s to t=5s) =15As
So , we have
If u=L15(fort=3s,dIdt=15).
where, L is a constant (coefficient of self -induction).
and emf is ε=LdIdt
Similarly, we have for others values.
For 5s<t<10s,u1=L35=35L=3e
Thus, att=7s,u1=3e

Question:6.21

There are two coils A and B separated by some distance. If a current of 2 A flows through A, a magnetic flux of 102 Wb passes through B (no current through B). If no current passes through A and a current of 1A passes through B, what is the flux through A ?

Answer:

The mutual inductance of coil B with respect to coil A is
M21=N2ϕ2I1
According to the problem, total flux N2ϕ2=102Wb
Hence, the Mutual inductance =1022=5mH
As we know that mutual inductance is same for both the coils with respect to each other. So,
M21=M12
And let N1ϕ1 is the total flux through A if no current passes through A.
N1ϕ1=M12I2=5mH×1A=5mWb

Question:6.22

A magnetic field B=B0sin(ωt)k covers a large region where a wire AB slides smoothly over two parallel conductors separated by a distance d (figure). The wires are in the x-y plane. The wire AB (of length d) has resistance R and the parallel wires have negligible resistance. If AB is moving with velocity v, what is the current in the circuit? What is the force needed to keep the wire moving at constant velocity?

Answer:

Due to motion : e=Blv
Due to change in magnetic flux: e=Nd(ϕB)dt
First we have to analyse the situation as shown in the figure let the parallel wires are at y=0 and y=d and are placed along x-axis. Wire AB is along y-axis.
Let us redraw the diagram as shown below.

At t=0,wire AB starts from x=0 and moves with a velocity v. Let at time t, wire is at x(t)=vt
(Where, x(t) is the displacement as a function of time).
Now, the motional emf across AB is
e1=Blv
e1=(B0sinωt)vd(j^)
and emf due to change in field (along OBAC)
e2=d(ϕB)dt
ϕB=(B0sinωt)(x(t)d) (Where,area A=xd)
e2=B0ωcosωtx(t)d
Total emf in the circuit = emf due to change in field (along OBAC)+the motional emf across AB
e1+e2=B0d[ωxcos(ωt)+vsin(ωt)]
The equivalent electrical diagram is shown in the diagram below.

Electric current in clockwise direction is given by
=B0dR(ωxcosωt+vsinωt)
The force acting on the conductor is given by F=ilBsin90o=ilB
Substituting the values,
Fm=B0dR(ωxcosωt+vsinωt)(d)(B0sinωt)(i^)
The external force needed on the wire is along the positive x-axis to keep moving it with constant velocity is given by,
Fm=B02d2R(ωxcosωt+vsinωt)sinωt(i^)
This is the required expression for force.

Question:6.23

A conducting wire XY of mass m and negligible resistance slides smoothly on two parallel conducting wires as shown in figure. The closed circuit has a resistance R due to AC. AB and CD are perfect conductors. There is a magnetic field

(i) Write down an equation for the acceleration of the wire XY
(ii) If B is independent of time, obtain v(t), assuming v(0) = u0
(iii) For (ii), show that the decrease in kinetic energy of XY equals the heat lost in.

Answer:

Upon analyzing the situation:
The wire is parallel with y-axis at y=0 and y=L. It placed along the x-axis.
At t=0, the wire starts from x=0 and is moving with a velocity of v. At certain time wire is at x(t)=vt.(x(t) is the displacement function_). The diagram is redrawn as:

The magnetic flux linked with the loop is given by
ϕm=B.A=BAcosθ
and as we know both A (area vector) and B (magnetic field vector) are directed along z-axis. So angle between them is 0.
So, cos0o=1(θ=0o)
ϕm=BA
At any instant of time t,
Magnetic flux ϕm=B(t)(1×x(t))
Emf induced due to change in magnetic field
e1=dϕdt
e1=dB(t)dtlx(t)
Emf induced due to motion
e2=Blv
e2=B(t)lv(t)(j^)
Total emf in the circuit = emf due to change in field (along XYAC) + the motional emf across XY
E=dϕdt=dB(t)dtlx(t)B(t)lv(t)
Electric current in clockwise direction (as shown in equivalent diagram) is given by
I=ER

The force acting on the conductor is given by F=ilBsin90o=ilB
Substituting the vlues, we have
.Force=lB(t)R[B(t)dtlx(t)B(t)lv(t)]i^
Applying Newton's second law of motion,
md2xdt2=l2B(t)RdBdtx(t)l2B2(t)Rdxdt....(i)
which is the required equation.
If B is independent of time, i.e., B = Constant
or dBdt=0
substituting the above value in Eq (i), we have
d2xdt2+l2B2mRdxdt=0
or dvdt+l2B2mRv=0
Intergrating using variable separable form of differential equation, we have
v=Aexp(l2B2tmR)
Applying given conditions, at t=0,v=u0
v(t)=u0exp(l2B2t/mR)
This is the required equation.
Since the power consumption is given by
p=I2R
Here,
I2R=B2l2v2(t)R2×R
=B2I2Ru02exp(2l2B2t/mR)
Now, energy consumed in time interval dt is given by energy consumed =Pdt=I2Rdt. Therefore, toal energy consumed in time t.
=m2u02m2v2(t)
= decrease in kinetic energy
This proves that the decrease in kinetic energy of XY equals the heat lost in R.

Question:6.24

ODBAC is a fixed rectangular conductor of negligible resistance (CO is not connected) and OP is a conductor which rotates clockwise with an angular velocity ω (figure). The entire system is in a uniform magnetic field B whose direction is along the normal to the surface of the rectangular conductor A’BDC. The conductor OP is in electric contact with ABDC. The rotating conductor has a resistance of λ per unit length. Find the current in the rotating conductor, as it rotates by 180o.

Answer:

When the conductor OP is rotated, then the rate of change of area and hence the rate of change of flux can be considered uniform from
0<θ<π4;π4<θ<3π4and3π4<θ<π2.
(i) Let us first assume the position of rotating conductor at the time interval
t=0tot=π4ω(orT8)

The rod OP will make contact with the side BD. Let the length OQ of the contact after some time interval t such that 0<t<π4ωor0<t<T8 be x. The flux through the area OQD is
ϕm=BA=B(12×QD×OD)=B(12×ltanθ×l)
ϕm=12Bl2tanθ, where θ=ωt
By applying Faraday's law of EMI,
Thus, the magnitude of the emf induced is |ε|=|dϕdt|=12Bl2ωsec2ωt
The current induced in the circuit will be I=εR where, R is the resistance of the rod in contact.
where, Rλ
R=λx=λlcosωt
I=12Bl2ωλlsec2ωtcosωt=Blω2λcosωt
(ii) Now let the rod OP will make contact with the side AB. And the length of OQ of the contact after some time interval t such that π4ω<t<3π4ωorT8<t<3T8 be x. The flux through the area OQBD is
ϕm=(l2+12l2tanθ)B
Where, θ=ωt
Thus, the magnitude of emf induced in the loop is
|ε|=|dϕdt|=Bl2ωsec2ωt2tan2ωt
The current induced in the circuit is I=εR=ελx=εsinωtλl=12Blωλsinωt
(iii) Similarly, for time interval 3π4ω<t<πωor3T8<t<T2, the rod will be in touch with AC.

The flux through OQABD is given by
ϕm=(2l2l22tanωt)B
And the magnitude of emf generated in the loop is given by
ε=dϕdt=Bωl2sec2ωt2tan2ωt
I=εR=ελx=12Blωλsinωt
These are the required expressions.

Question:6.25

Consider an infinitely long wire carrying a current I(t), with dldt=λ=constant.. Find the current produced in the rectangular loop of wire ABCD if its resistance is R (figure).


Answer:

Let us consider a strip of length l and width dr at a distance r from an infinite long current carrying wire. The magnetic field in the strip is given by:
B(r)=μ0I2πr (Out of paper)
Area of the elementary strip is, dA=l.dr
So, total flux through the loop is
ϕm=B.A=μ0I2πlx0xdrr=μ0Il2πlnxx0.....(i)
The emf induced can be obtained by differentiating the eq. (i) w.r.t. t and then applying Ohm's law
I=εR and |ε|=dϕdt
We have, induced current
I=1Rdϕdt=μ0l2πλRlnxx0(dIdt=λ)

Question:6.26

A rectangular loop of wire ABCD is kept close to an infinitely long wire carrying a current I(t)=I0(ItT) for 0tT and I(0)=0fort>T(figure). Find the total charge passing through a given point in the loop, in time T. The resistance of the loop is R.

Answer:

I=ER=1Rdϕdt
According to the problem electric current is given as a function of time.
I(t)=dQdtordQdt=1Rdϕdt
Integrating the variable separately in the form of the differential equation for finding the charge Q that passed in time t, we have
Q(t1)Q(t2)=1R[ϕ(t1)ϕ(t2)]
Q(t1)=L1μ02πxL2+xdxxI(t1) [Refer to the Eq. (i) of answer no.25]
=μ0L12πI(t1)lnL2+xx
Therefore the magnitude of the charge is
Q=1R[ϕ(T)ϕ(0)]
=μ0L12πI(t1)lnL2+xx[I(T)I(0)]
Now I(T)=0andI(0)=1
Q=μ0L12πI0In(L2+xx)

Question:6.27

A magnetic field B is confined to a region ra and points out of the paper (the z-axis), r=0 being the centre of the circular region. A charged ring (charge = Q) of radius b,b>a and mass m lies in the x-y plane with its centre at the origin. The ring is free to rotate and is at rest. The magnetic field is brought to zero in time Δt. Find the angular velocity ω of the ring after the field vanishes.

Answer:

The magnetic field decreases which induce an emf hence electric field around the ring. And therefore, the ring experiences a torque which produces a change in angular momentum.
As the magnetic field is brought to zero, the magnetic flux linked reduces to zero which is linked to the ring. This induces an emf in-ring and in turn an electric field E around the ring.
The induced emf =
Electric field E ×(2πb)(Becausev=E×d)....(i)
By Faraday's law of EMI
|ε|=dϕdt=AdBdt
|ε|=Bπa2ΔtS...(ii)
From Eqs. (i) and (ii), we have
2πbE=ε=Bπa2Δt
As we know the electric force experienced by the changed ring, Fe=QE
This force try to rotate the coil, and the torque is given by
Torque=b×Force
τ=QEb=Q[Bπa22πbΔt]b
τ=QBa22Δt
If ΔL is the change in angular momentum,
ΔL=Torque×Δt=QBa22
Since initial angular momentum =0
and Torque×Δt=change in angular momentum
Final angular momentum =
mb2ω=QBa22
Where, mb2=I(moment of inertia of ring )
ω=QBa22mb2

Question:6.28

A rod of mass m and resistance R slides smoothly over two parallel perfectly conducting wires kept sloping at an angle θ with respect to the horizontal (figure). The circuit is closed through a perfect conductor at the top. There is a constant magnetic field B along the vertical direction. If the rod is initially at rest, find the velocity of the rod as a function of time.

Answer:

The component of magnetic field along the inclined plane will be =Bsinθ and other will be perpendicular i.e. =Bcosθ. The conductor is moving perpendicular to =Bcosθ. It is the vertical component of the magnetic field. The movement will cause motional emf across the two ends of the rod.
given by =v(Bcosθ)d

This makes flow of induced current
i=v(Bcosθ)dR
where R is the resistance of rod. Now, current-carrying rod experience a magnetic force which is given by
Fm=iBd (horizontally in backward direction).
Now, the component of magnetic force parallels to the inclined plane along the upward direction.
F=Fmcosθ=iBdcosθ=(v(Bcosθ)dR)Bdcosθ
Where,
v=dxdt
Also, the component of weight (mg) parallel to the inclined plane along downward direction = mgsinθ.
Now, by Newton's second law of motion
md2xdt2=mgsinθBcosθdR(dxdt)×(BD)cosθ
dvdt=gsinθB2d2mR(cosθ)2v
dvdt+B2d2mR(cosθ)2v=gsinθ
But, this is the linear differential equation.
On solving, we get
v=gsinθB2d2cos2θmR+Aexp(B2d2mR(cos2θ)t)
A is a constant to be determined by initial conditions.
The required expression of velocity as a function of time is given by
mgRsinθB2d2cos2θ(1exp(B2d2mR(cos2θ)t))

Question:6.29

Find the current in the sliding rod AB (resistance = R) for the arrangement shown in the figure. B is constant and is out of the paper. Parallel wires have no resistance, v is constant. Switch S is closed at time t = 0.

Answer:

Due to the motion of the conductor, motional emf is induced in the conductor. It will give rise to a current.
I=dQdt=BvdRQRC
QRC+dQdt=BvdR
Q+RCdQdt=vBCd(LetvBdC=A)
Q+RCdQdt=A
dQAQ=1RCdt
By integrating we have
0QdQAQ=1RC0tdt
lnAQA=tRC
AQA=etRC
Q=A(1etRC)
Current in the rod
I=dQdt=ddt[A(1etRC)]
=A(etRC)(1RC)
I=vBdRetRC

Question:6.30

Find the current in the sliding cod AB (resistance = R) for the arrangement shown in figure. B is constant and is out of the paper. Parallel wires have no resistance, v is constant. Switch S is closed at time t = 0.


Answer:

By applying KVL in the given circuit, we have
LdIdt+vBd=IRorLdIdt+IR=vBd
This is the linear differential equation. On solving,
I=vBdR+AeRtL
Att=0,I=0
A=vBdRI=vBdR(1eRtL)

Question:6.31

A metallic ring of mass m and radius l (ring being horizontal) is falling under gravity in a region having a magnetic- field. If z is the vertical direction, the z-component of magnetic field is . If R is the resistance of the ring and if the ring falls with a velocity v, find the energy lost in the resistance. If the ring has reached a constant velocity, use the conservation of energy to determine v in terms of m, B, and acceleration due to gravity g.

Answer:

Here a relation is established between the current induced, velocity of free falling ring and power lost. Let the mass of ring be m and radius l;
ϕm=Bz(πl2)=B0(1+λz)(πl)
Applying Faraday's law of EMI, we have emf induced given by
dϕdt= rate of change of flux. Also, by Ohm's law
B0(πl2)λdzdt=IR
We have
I=πl2B0λRv
Energy lost/second =
I2R=(πl2λ)2B02v2R
Rate of change of
PE=mgdzdt=mgv [ as kinetic energy is constant for v=constant]
According to the law of conservation of energy
Thus,
mgv=(πl2λB0)2v2R or
v=mgR(πl2λB0)2
This is the required expression of velocity.

Question:6.32

A long solenoid S has n turns per metre, with radius a. At the centre of this coil, we place a smaller coil of N turns and radius b (where b<a). If the current in the solenoid increases linearly with time, what is the induced emf appearing in the smaller coil. Plot a graph showing nature of variation in emf, if current varies as a function of mt2+C.

Answer:

Magnetic field caused by a solenoid is given by
B=μ0ni.
Magnetic flux in the smaller coil is
ϕm=NBA
where, A=πb2
Applying Faraday's law of EMI, we have
So, e=dϕdt=ddt(NBA)
=Nπb2d(B)dt
Where, B=μ0Ni
e=Nπb2μ0ndidt
Since, current varies as a function of time, so
i(t)=mt2+C
e=Nnπμ0b2ddt(mt2+C)

By solvinge=μ0Nnπb22mt
The negative sign signifies opposite nature of induced emf.

NCERT Exemplar Class 12 Physics Solutions Chapter 6 Electromagnetic Induction-Main subtopics

  • Introduction
  • The Experiments Of Faraday And Henry
  • Magnetic Flux
  • Faraday's Law Of Induction
  • Lenz's Law And Conservation Of Energy
  • Motional Electromotive Force
  • Energy Consideration: A Quantitative Study
  • Eddy Currents
  • Inductance
  • Mutual inductance
  • Self-inductance
  • Ac Generator
Background wave

NCERT Exemplar Class 12 Physics Solutions Chapter 6- What Will The Students Learn?

Class 12 Physics NCERT Exemplar solutions chapter 6 is a collection of laws deduced with the help of a long series of experiments carried out by scientists like Faraday, Henry, and Lenz, a brief study about magnetic flux and its properties, topics concerning magnetic induction, electromotive force, eddy currents, the types and applications of inductance and the working and characteristics of an AC generator.

NCERT Exemplar Class 12 Physics Chapter Wise Links

Some applications of electromagnetic Induction

  • NCERT Exemplar Solutions For Class 12 Physics chapter 6 explores the concept of Induction. Induction is used in power generation and transmission; it is used in designing the graphical pen to draw on individual graphical tablets and smart devices·
  • Electromagnetic induction is used in rock instruments like electric guitars and electric and hybrid vehicles.
  • Another crucial area of research in which electromagnetic induction is implemented successfully is transcranial magnetic stimulation (TMS) to trace disorders, including depression and hallucinations. In TMS, a rapidly altering and very localised magnetic field is placed close to sites identified in the brain, and the results are printed out.
  • The NCERT Exemplar Class 12 Physics Solutions Chapter 6 also discusses some application level problems. The NCERT Class 12 Physics book also discusses applications like generators and transformers.
Pearson | PTE

Trusted by 3,500+ universities and colleges globally | Accepted for migration visa applications to AUS, CAN, New Zealand , and the UK

NCERT Exemplar Class 12 Solutions

Also, check the NCERT solutions of questions given in book

Also read NCERT Solution subject wise

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook

Must read NCERT Notes subject wise

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Questions (FAQs)

1. Is this chapter crucial for boards in Class 12 Physics?

Yes, this chapter is one of the important parts of the Physics Class 12 syllabus, which will help in scoring well in the board exams.

2. How to use the solutions for exam preparation?

You can use the NCERT exemplar Class 12 Physics solutions chapter 6 as a reference while solving questions. The solutions are also helpful for knowing the pattern that is the CBSE board exam suitable.

3. How many questions are solved in this chapter?

All 32 questions from the exercise of Class 12 Physics NCERT exemplar solutions chapter 6 of the electromagnetic induction chapter is solved in complete and thorough detail.

Articles

Explore Top Universities Across Globe

Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.
  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.
  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

View All

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Back to top