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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

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.

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 = B_o (2i + 3j + 4k)T where Bo is constant. The magnitude of flux passing through the square is
(a) 2 B_o L^2 Wb.
(b) 3 B_o L^2 Wb.
(c) 4 B_o L^2 Wb.
(d) \sqrt{}2 9B_o L^2 Wb.

Answer:

For elementary area dA of a surface, flux linked d\phi=BdA\cos \theta\:or\: d\phi=\vec{B}.\vec{dA}
So-net flux through the surface
\phi = \oint \vec{B}\times \vec{dA}=BA \cos \theta
In this problem
A= L^2\hat{k}\:and\:B=B_0\left ( 2\hat{i} +3\hat{j}+4\hat{k}\right )T\\ \phi = \vec{B}.\vec{A}=B_0\left ( 2\hat{i} +3\hat{j}+4\hat{k}\right ).L^2\hat{k}= 4 B_o L^2 Wb.

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) \: and\: F(0,0,L). A magnetic field B = B_o (i + k) Tis present in the region. The flux passing through the loop ABCDEFA (in that order) is

(a) B_o\: L^2 Wb.
(b) 2 B_o L^2 Wb.
(c) \sqrt{}2 B_o L^2 Wb.
(d) 4 \: B_o L^2 Wb.

Answer:
In this problem first we have to analyse area vector, loop ABCDA lies in x-y plane whose area vector \vec{A_1}= L^2\hat{k} whereas loop ADEFA lies in y-z plane whose area vector \vec{A_2}= L^2\hat{i}
And magnetic flux is
\\ \phi _m=\vec{B}.\vec{A}\\ \vec{A}=\vec{A_1}+\vec{A_2}=\left ( L^2\hat{k} +L^2\hat{i}\right )\\ and\: \vec{B}=B_0\left ( \hat{i}+\hat{k} \right )\\ Now,\phi _m=\vec{B}.\vec{A}= B_0\left ( \hat{i}+\hat{k}\right ).\left ( L^2\hat{k} +L^2\hat{i}\right )\\ =2B_0L^2Wb


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=\frac{\left ( \mu_ r \: \mu_0\: n^2 \right )}{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 e_2=-N_2\frac{d\phi _2}{dt};e_2=-M\frac{di_1}{dt}

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

M_{12}= \frac{\mu_0 N_1 N_2 \pi {r_{1}}^{2}}{l}


Where signs are as usual,

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

M_{21}= \frac{\mu_0 N_1 N_2 \pi {r_{1}}^{2}}{l}

So we have M_{12}=M_{21}= 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=B_{0}\cos \left ( \omega t \right ) 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=\frac{\pi }{2\omega },t=\frac{\pi }{\omega }\; and\; t=\frac{3\pi }{2\omega }


Answer:

\phi _{m}=\vec{B}.\vec{A}=BA\; \cos \theta
And as we know both \vec{A} (area vector ) and \vec{B} (magnetic field vector) are directed along z-axis. So, angle between them is 0.
So, \cos \theta = 1 \left ( \because \theta =0 \right )
\Rightarrow \phi _{m}=BA
Area of coil of radius a=\pi a ^{2}
\varepsilon =B_{0}(\pi a^{2})\; \cos \omega t
By Faraday's law of electromagnetic induction, Magnitude of induced emf is given by
\varepsilon =B_{0}(\pi a^{2})\; \omega \sin \omega t
This causes flow of induced current, which is given by
I=\frac{B_{0}(\pi a^{2})\; \omega \sin \omega t}{R}
Now, the value of current at different instants,
(i) t=\frac{\pi }{2\omega }
I=\frac{B_{0}(\pi a^{2})\omega }{R} along\; \hat{j}
Because \sin \omega t=\sin \left ( \omega \frac{\pi }{2\omega } \right )=\sin \frac{\pi }{2}=1
(ii) t=\frac{\pi }{\omega },I=\frac{B(\pi a^{2})\omega }{R} =0
because, \sin \omega t=\sin \left ( \omega \frac{\pi }{2\omega } \right )=\sin \pi =0
(iii) t=\frac{3}{2}\frac{\pi }{\omega }
I = \frac{B(\pi a^{2})\omega }{R}\; along -\hat{j}
\sin \omega t=\sin \left ( \omega .\frac{3\pi }{2\omega } \right )=\sin \frac{3\pi }{2}=-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 \phi =B_{1}dA_{1},B_{2}dA_{2} …. Now, if we choose two different surfaces S_{1} and S_{2} having C as their edge, would we get the same answer for flux. Justify your answer.

Answer:

The magnetic flux is same as in S_{1} and S_{2} 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. \theta 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 F_{m}=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 F_{e}=F_{m}, i.e.,
eE=evB \; or\; E=vB\Rightarrow \text {Induced emf}e=El=Bvl\left [ E=\frac{V}{l} \right ]
If the rod is moved by making an angle \theta with the direction of the magnetic field or length. Induced emf,
e=Bvl\; \sin \theta

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 \vec{V}\; \text {and }\vec{B} and along \vec{V}\times \vec{B} ) =vB
Therefore, the motional emf along.
PQ=\left ( \text {length PQ} \right )\times \left ( \text {field along PQ} \right )
= \left ( \text {length PQ} \right )\times \left ( \text {vB} \sin \theta \right )
= \left ( \frac{d}{\sin \theta } \right )\times \left ( \text {vB} \sin \theta \right )=vBd

This induced emf make flow of current in closed circuit of resistnce R.
I=\frac{dvB}{R} and independent of \theta.

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=7\; s, 15\; s,40\; s. OA, AB and BC are straight line segments.

Answer:

\varepsilon =-\frac{d(N\phi _{B})}{dt}
\varepsilon =-L\frac{dl}{dt}
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 5\; s <t<10\; s.
Since, the back emf at t=3\; s is e.
Also, the rate of change of current at t=3,
and slope (s) of OA (from t=0\; s to t=5\; s) =\frac{1}{5}\frac{A}{s}
So , we have
If u=L\frac{1}{5}\left ( for\; t=3s,\frac{dI}{dt}=\frac{1}{5} \right ).
where, L is a constant (coefficient of self -induction).
and emf is \varepsilon =-L\frac{dI}{dt}
Similarly, we have for others values.
For 5s<t<10s, u_{1}=-L\frac{3}{5}=-\frac{3}{5}L=-3e
Thus, at \; t=7 s, u_{1}=-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 10^{-2} 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
M_{21}=\frac{N_{2}\phi _{2}}{I_{1}}
According to the problem, total flux N_{2}\phi _{2}= 10^{-2}Wb
Hence, the Mutual inductance =\frac{10^{-2}}{2}=5mH
As we know that mutual inductance is same for both the coils with respect to each other. So,
M_{21}=M_{12}
And let N_{1}\phi _{1} is the total flux through A if no current passes through A.
N_{1}\phi _{1}=M_{12}I_{2}=5mH\times 1 A=5\; m\; Wb

Question:6.22

A magnetic field B=B_{0}\sin\; (\omega t)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=-N\frac{d(\phi _{B})}{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
e_{1}=Blv
\Rightarrow e_{1}=(B_{0}\; \sin \omega t)vd (-\hat{j})
and emf due to change in field (along OBAC)
e_{2}=\frac{d(\phi _{B})}{dt}
\phi _{B}=(B_{0}\; \sin \omega t)(x(t)d) (Where,area A=xd)
e_{2}=-B_{0}\omega \cos \; \omega tx(t)d
Total emf in the circuit = emf due to change in field (along OBAC)+the motional emf across AB
e_{1}+e_{2}=-B_{0}d\left [ \omega x\; \cos (\omega t)+v\sin (\omega t) \right ]
The equivalent electrical diagram is shown in the diagram below.

Electric current in clockwise direction is given by
=\frac{B_{0}d}{R}(\omega x\; \cos \omega t+v \sin \omega t)
The force acting on the conductor is given by F=ilB\; \sin 90^{o}=ilB
Substituting the values,
\overrightarrow{F}_{m}=\frac{B_{0}d}{R}(\omega x \cos \omega t+v \sin \omega t)(d)(B_{0}\; \sin \omega t)(-\hat{i})
The external force needed on the wire is along the positive x-axis to keep moving it with constant velocity is given by,
\overrightarrow{F}_{m}=\frac{B_{0}^{2}d^{2}}{R}(\omega x \cos \omega t+v \sin \omega t)\sin \omega t(\hat{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
\phi _{m}=\vec{B}.\vec{A}=BA \cos \theta
and as we know both \vec{A} (area vector) and \vec{B} (magnetic field vector) are directed along z-axis. So angle between them is 0.
So, \cos \; 0^{o}=1\; \; \; \; \; \; (\because \theta =0^{o})
\Rightarrow \phi _{m}=BA
At any instant of time t,
Magnetic flux \phi _{m}=B(t)(1\times x(t))
Emf induced due to change in magnetic field
e_{1}=-\frac{d\phi }{dt}
\Rightarrow e_{1}=-\frac{dB(t)}{dt}lx(t)
Emf induced due to motion
e_{2}=Blv
e_{2}=B(t)lv(t)(-\hat{j})
Total emf in the circuit = emf due to change in field (along XYAC) + the motional emf across XY
E=-\frac{d\phi }{dt}=-\frac{dB(t)}{dt}lx(t)-B(t)lv(t)
Electric current in clockwise direction (as shown in equivalent diagram) is given by
I=\frac{E}{R}

The force acting on the conductor is given by F=ilB\; \sin 90^{o}=ilB
Substituting the vlues, we have
.Force =\frac{lB(t)}{R}\left [- \frac{B(t)}{dt}l x (t)-B(t)lv(t)\right ]\hat{i}
Applying Newton's second law of motion,
m\frac{d^{2}x}{dt^{2}}=\frac{l^{2}B(t)}{R}\frac{dB}{dt}x(t)-\frac{l^{2}B^{2}(t)}{R}\frac{dx}{dt}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(i)
which is the required equation.
If B is independent of time, i.e., B = Constant
or \frac{dB}{dt}=0
substituting the above value in Eq (i), we have
\frac{d^{2}x}{dt^{2}}+\frac{l^{2}B^{2}}{mR}\frac{dx}{dt}=0
or \frac{dv}{dt}+\frac{l^{2}B^{2}}{mR}v=0
Intergrating using variable separable form of differential equation, we have
v=A \; exp \left ( \frac{-l^{2}B^{2}t}{mR} \right )
Applying given conditions, at t=0, v=u_{0}
v(t)=u_{0}exp (-l^{2}B^{2}t/mR)
This is the required equation.
Since the power consumption is given by
p=I^{2}R
Here,
I^{2}R=\frac{B^{2}l^{2}v^{2}(t)}{R^{2}}\times R
=\frac{B^{2}I^{2}}{R}u_{0}^{2}exp (-2l^{2}B^{2}t/mR)
Now, energy consumed in time interval dt is given by energy consumed =Pdt=I^{2}Rdt. Therefore, toal energy consumed in time t.
=\frac{m}{2}u_{0}^{2}-\frac{m}{2}v^{2}(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 \omega (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 \lambda per unit length. Find the current in the rotating conductor, as it rotates by 180^{o}.

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<\theta <\frac{\pi }{4};\frac{\pi }{4}<\theta <\frac{3\pi }{4}\; and \frac{3\pi }{4}<\theta <\frac{\pi }{2}.
(i) Let us first assume the position of rotating conductor at the time interval
t=0\; to\; t=\frac{\pi }{4\omega }(or \frac{T}{8})

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<\frac{\pi }{4\omega }\; or\; 0<t<\frac{T}{8} be x. The flux through the area OQD is
\phi _{m}=BA=B\left ( \frac{1}{2}\times QD\times OD \right )=B\left ( \frac{1}{2}\times l \tan \theta \times l \right )
\Rightarrow \phi _{m}=\frac{1}{2}Bl^{2}\tan \theta , where \theta =\omega t
By applying Faraday's law of EMI,
Thus, the magnitude of the emf induced is \left | \varepsilon \right |=\left | \frac{d\phi }{dt} \right |=\frac{1}{2}Bl^{2}\omega \sec^{2 } \omega t
The current induced in the circuit will be I=\frac{\varepsilon }{R} where, R is the resistance of the rod in contact.
where, R\infty \lambda
R=\lambda x=\frac{\lambda l}{\cos \omega t}
\therefore I=\frac{1}{2}\frac{Bl^{2}\omega }{\lambda l}\sec ^{2}\omega t\; \cos\; \omega t =\frac{Bl\omega }{2\lambda \; \cos\; \omega t}
(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 \frac{\pi }{4\omega }<t<\frac{3\pi }{4\omega }\; or \; \frac{T}{8}<t<\frac{3T}{8} be x. The flux through the area OQBD is
\phi _{m}=\left ( l^{2}+\frac{1}{2}\frac{l^{2}}{\tan \theta } \right )B
Where, \theta =\omega t
Thus, the magnitude of emf induced in the loop is
\left | \varepsilon \right |=\left | \frac{d\phi }{dt} \right |=\frac{Bl^{2}\omega \; \sec ^{2}\omega t}{2\; \tan ^{2}\omega t}
The current induced in the circuit is I=\frac{\varepsilon }{R}=\frac{\varepsilon }{\lambda x}=\frac{\varepsilon \; \sin \omega t}{\lambda l}=\frac{1}{2}\frac{Bl\omega }{\lambda \sin \omega t}
(iii) Similarly, for time interval \frac{3\pi }{4\omega }<t<\frac{\pi }{\omega }or\frac{3T}{8}<t<\frac{T}{2}, the rod will be in touch with AC.

The flux through OQABD is given by
\phi _{m}=\left ( 2l^{2}-\frac{l^{2}}{2\; \tan \omega t} \right )B
And the magnitude of emf generated in the loop is given by
\varepsilon =\frac{d\phi }{dt}=\frac{B\omega l^{2}\; \sec ^{2}\omega t}{2\; \tan ^{2}\omega t}
I=\frac{\varepsilon }{R}=\frac{\varepsilon }{\lambda x}=\frac{1}{2}\frac{Bl\omega }{\lambda \sin \omega t}
These are the required expressions.

Question:6.25

Consider an infinitely long wire carrying a current I(t), with \frac{dl}{dt}=\lambda =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:
\vec{B}(r)=\frac{\mu _{0}I}{2\pi r} (Out of paper)
Area of the elementary strip is, dA=l.dr
So, total flux through the loop is
\phi _{m}=\vec{B}.\vec{A}=\frac{\mu _{0}I}{2\pi }l\int_{x_{0}}^{x}\frac{dr}{r}=\frac{\mu _{0}Il}{2\pi}ln\frac{x}{x_{0}}\; \; \; \; \; \; \; \; \; .....(i)
The emf induced can be obtained by differentiating the eq. (i) w.r.t. t and then applying Ohm's law
I=\frac{\varepsilon }{R} and \left | \varepsilon \right |=\frac{d\phi }{dt}
We have, induced current
I=\frac{1}{R}\frac{d\phi }{dt}=\frac{\mu _{0}l}{2\pi}\frac{\lambda }{R}ln\frac{x}{x_{0}}\left ( \because \frac{dI}{dt} =\lambda \right )

Question:6.26

A rectangular loop of wire ABCD is kept close to an infinitely long wire carrying a current I(t)=I_{0}(I-\frac{t}{T}) for 0\leq t\leq T and I(0) = 0\; for \; t >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=\frac{E}{R}=\frac{1}{R}\frac{d\phi }{dt}
According to the problem electric current is given as a function of time.
I(t)=\frac{dQ}{dt}\; or\; \frac{dQ}{dt}=\frac{1}{R}\frac{d\phi }{dt}
Integrating the variable separately in the form of the differential equation for finding the charge Q that passed in time t, we have
Q(t_{1})-Q(t_{2})=\frac{1}{R}\left [ \phi (t_{1})-\phi (t_{2}) \right ]
Q(t_{1})=L_{1}\frac{\mu _{0}}{2\pi}\int_{x}^{L_{2}+x}\frac{dx'}{x'}I(t_{1}) [Refer to the Eq. (i) of answer no.25]
=\frac{\mu _{0}L_{1}}{2\pi}I(t_{1})ln\frac{L_{2}+x}{x}
Therefore the magnitude of the charge is
Q=\frac{1}{R}\left [ \phi (T)-\phi (0) \right ]
=\frac{\mu _{0}L_{1}}{2\pi}I(t_{1})ln\frac{L_{2}+x}{x}[I(T)-I(0)]
Now I(T)=0\; and \; I(0)=1
\therefore Q=\frac{\mu _{0}L_{1}}{2\pi}I_{0}In \left ( \frac{L_{2}+x}{x} \right )

Question:6.27

A magnetic field B is confined to a region r\leq a 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 \Delta t. Find the angular velocity \omega 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 =
\text {Electric field E }\times \left ( 2\pi b\right )\left ( Because v=E\times d \right ) ....(i)
By Faraday's law of EMI
\left | \varepsilon \right |=\frac{d\phi }{dt}=A\frac{dB}{dt}
\left | \varepsilon \right |=\frac{B\pi a^{2}}{\Delta t}S ...(ii)
From Eqs. (i) and (ii), we have
2\pi b E=\varepsilon =\frac{B\pi a^{2}}{\Delta t}
As we know the electric force experienced by the changed ring, F_{e}=QE
This force try to rotate the coil, and the torque is given by
\text {Torque=b}\times\text {Force}
\tau =QEb=Q\left [ \frac{B\pi a^{2}}{2 \pi b \Delta t} \right ]b
\Rightarrow \tau =Q\frac{Ba^{2}}{2\Delta t}
If \Delta L is the change in angular momentum,
\Delta L=Torque \times \Delta t = Q\frac{Ba^{2}}{2}
Since initial angular momentum =0
and Torque \times \Delta t=\text {change in angular momentum}
Final angular momentum =
mb^{2}\omega =\frac{QBa^{2}}{2}
Where, mb^{2}=I (\text {moment of inertia of ring })
\omega =\frac{QBa^{2}}{2mb^{2}}

Question:6.28

A rod of mass m and resistance R slides smoothly over two parallel perfectly conducting wires kept sloping at an angle \theta 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 =B\sin \theta and other will be perpendicular i.e. =B\cos \theta. The conductor is moving perpendicular to =B\cos \theta. 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(B \cos \theta )d

This makes flow of induced current
i=\frac{v(B \cos \theta )d}{R}
where R is the resistance of rod. Now, current-carrying rod experience a magnetic force which is given by
F_{m}=iBd (horizontally in backward direction).
Now, the component of magnetic force parallels to the inclined plane along the upward direction.
F_{\parallel }=F_{m} \cos \theta =iBd \cos \theta =\left ( \frac{v(B \cos \theta )d}{R} \right )Bd \cos \theta
Where,
v=\frac{dx}{dt}
Also, the component of weight (mg) parallel to the inclined plane along downward direction = mg\; \sin \theta .
Now, by Newton's second law of motion
m\frac{d^{2}x}{dt^{2}}=mg \sin \theta -\frac{B \cos \theta d}{R}\left ( \frac{dx}{dt} \right )\times (BD)\; \cos \theta
\Rightarrow \frac{dv}{dt}=g\; \sin \theta -\frac{B^{2}d^{2}}{mR}(\cos \theta )^{2}v
\Rightarrow \frac{dv}{dt}+\frac{B^{2}d^{2}}{mR}(\cos \theta )^{2}v=g\; \sin \theta
But, this is the linear differential equation.
On solving, we get
v=\frac{g \sin \theta }{\frac{B^{2}d^{2}\cos^{2}\theta }{mR}}+ A\; exp\left ( -\frac{B^{2}d^{2}}{mR}(\cos ^{2}\theta )t \right )
A is a constant to be determined by initial conditions.
The required expression of velocity as a function of time is given by
\frac{mgR \sin \theta }{B^{2}d^{2}\cos^{2}\theta }\left ( 1-exp\left ( -\frac{B^{2}d^{2}}{mR}(\cos^{2}\theta )t \right ) \right )

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=\frac{dQ}{dt}=\frac{Bvd}{R}-\frac{Q}{RC}
\frac{Q}{RC}+\frac{dQ}{dt}=\frac{Bvd}{R}
Q+RC\frac{dQ}{dt}=vBCd\; \; \; (Let \; vBdC=A)
Q+RC\frac{dQ}{dt}=A
\frac{dQ}{A-Q}=\frac{1}{RC}dt
By integrating we have
\int_{0}^{Q}\frac{dQ}{A-Q}=\frac{1}{RC}\int_{0}^{t}dt
ln\frac{A-Q}{A}=-\frac{t}{RC}
\frac{A-Q}{A}=e^{-\frac{t}{RC}}
Q=A(1-e^{-\frac{t}{RC}})
Current in the rod
I=\frac{dQ}{dt}=\frac{d}{dt}\left [ A\left ( 1-e^{\frac{-t}{RC}} \right ) \right ]
=-A(e^{\frac{-t}{RC}})\left ( -\frac{1}{RC} \right )
I=\frac{vBd}{R}e^{\frac{-t}{RC}}

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;
\phi _{m}=B_{z}(\pi l^{2})=B_{0}(1+\lambda z)(\pi l)
Applying Faraday's law of EMI, we have emf induced given by
\frac{d\phi }{dt}= rate of change of flux. Also, by Ohm's law
B_{0}(\pi l^{2})\lambda \frac{dz}{dt}=IR
We have
I=\frac{\pi l^{2}B_{0}\lambda }{R}v
Energy lost/second =
I^{2}R=\frac{(\pi l^{2}\lambda )^{2}B_{0}^{2}v^{2}}{R}
Rate of change of
PE=mg\frac{dz}{dt}=mgv [ as kinetic energy is constant for v=constant]
According to the law of conservation of energy
Thus,
mgv=\frac{(\pi l^{2}\lambda B_{0})^{2}v^{2}}{R} or
v=\frac{mgR}{(\pi l^{2}\lambda B_{0})^{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 mt^{2}+C.

Answer:

Magnetic field caused by a solenoid is given by
B=\mu _{0}ni.
Magnetic flux in the smaller coil is
\phi _{m}=NBA
where, A=\pi b^{2}
Applying Faraday's law of EMI, we have
So, e=\frac{-d\phi }{dt}=\frac{-d}{dt}(NBA)
=-N\pi b^{2}\frac{d(B)}{dt}
Where, B=\mu _{0}Ni
\Rightarrow e=-N\pi b^{2}\; \mu _{0}n\frac{di}{dt}
Since, current varies as a function of time, so
i(t)=mt^{2}+C
\Rightarrow \; e=-Nn\pi \mu _{0}b^{2}\frac{d}{dt}(mt^{2}+C)

By solvinge=-\mu _{0}Nn\pi b^{2}2mt
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

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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.
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Frequently Asked Question (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.

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I hope this was helpful!

Good Luck

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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

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