NCERT Exemplar Class 12 Physics Solutions Chapter 6 Electromagnetic Induction

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 respect 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: 6.10

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 makes flow of current in closed circuit of resistance 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 values, 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, $R\propto \lambda$
$R=\lambda x=\frac{\lambda l}{\cos \omega t}$
$\therefore I=\frac{1}{2}\frac{Bl^{2}\omega }{\lambda l}\sec ^{2}\omega t\; \cos\; \omega t =\frac{Bl\omega }{2\lambda \; \cos\; \omega t}$
(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:6.30

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
$-L\frac{dI}{dt}+vBd=IR\; or\; L\frac{dI}{dt}+IR=vBd$
This is the linear differential equation. On solving,
$I=\frac{vBd}{R}+Ae^{-\frac{Rt}{ L}}$
$At \; t=0, I=0$
$\Rightarrow A=-\frac{vBd}{R}\Rightarrow I=\frac{vBd}{R}\left ( 1-e^{\frac{-Rt}{L}} \right )$

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 solving$e=-\mu _{0}Nn\pi b^{2}2mt$
The negative sign signifies the opposite nature of induced emf.