Physics Ncert Exemplar 2019 Solutions for Class 12 Science Physics Chapter 6 - Electromagnetic Induction

Answer:

Magnetic flux, $\varphi =\overrightarrow{B}.\overrightarrow{A}$
Given, A = $L^2 \hat{k}$
$$\begin{gathered} \overrightarrow{B}=B_o(2\hat{i}+3\hat{j}+4\hat{k}) \mathrm{T} \\ \varphi =B_o(2\hat{i}+3\hat{j}+4\hat{k}).L^2\hat{k}=4B_o{L}^2 \mathrm{Wb} \end{gathered}$$
Hence, the correct answer is option (c).

Answer:

Given points are A(0,0,0), B(L,0,0) C(L,L,0), D(0,L,0) E(0,L,L) and F(0,0,L) as drawn in the diagram below:

Magnetic flux, $\varphi =\overrightarrow{B}.\overrightarrow{A}$
$$\begin{gathered} \mathrm{Area}, \overrightarrow{A}={\overrightarrow{A}}_1+{\overrightarrow{A}}_2=L^2 (\hat{i}+\hat{k}) {\mathrm{m}}^2 \\ \mathrm{and} \overrightarrow{B}=B_o(\hat{i}+\hat{k}) \mathrm{T} \\ \mathrm{\varphi }=B_{\mathrm{o}}(\hat{i}+\hat{k}).L^2 (\hat{i}+\hat{k}) \\ \varphi =2B_o{L}^2 \mathrm{Wb} \end{gathered}$$
Hence, the correct answer is option (b).

Answer:

When cylindrical bar magnet is rotated about its axis no change in flux linked with the circuit takes place, consequently no emf induces and hence, no current flows through the ammeter A.

Hence, the correct answer is option (b).

Answer:

It can be inferred that the coil A must be carrying a constant current in counter clockwise direction. Because of that, when moves toward B, current induced in B is counterclockwise direction as per Lenz's law.
But the current in B would stop when A stops moving. If the current in coil A would be variable, there must be an induced emf in coil B even if coil A stops moving.
Hence, the correct answer is option (d).

Answer:

When the current is coil B (at t = 0) is anticlockwise and the coil A is considered above to it, the anticlockwise flow of the current in coil B is equivalent to north pole of the magnet and magnetic field lines are emanating upward to coil A. When coil A start rotating at t = 0, the current in coil A is constant along clockwise direction by Lenz's law.
Hence, the correct answer is option (a).

Answer:

The self inductance, L = μn²Al.
L = μn²Al
Here, n = number of turns per unit length
$$\begin{gathered} n=\frac{N}{l} \\ L=\frac{ \mu N^{2}A}{l} \end{gathered}$$
So, self inductance L  increases if length of the solenoid l decreases and area of cross-section A increases.
Hence, the correct answer is option (b).

Answer:

Any metal plate is getting heated when a DC or AC current is passed through the plate, due to the heating effect of current.When a metal plate in subjected to time varying magnetic field, the magnetic flux linked with the plate changes and eddy currents comes into existence which make the plate hot.
Hence, the correct options are (a), (b) and (d).

Answer:

Magnetic flux linked with the isolated coil changes when the coil being in a time varying magnetic field, the coil moving in a constant magnetic field or in a time varying magnetic field.
Hence, the correct options are (a), (b) and (c).

Answer:

The Mutual inductance M₁₂ of coil increases when they are brought nearer and is same as M₂₁ of Coil 2 with respect to coil 1.
M₁₂ = μ₀n₁n₂πr²l
M₂₁ = μ₀n₁n₂πr²l.
So, M₁₂ = M₂₁
Hence, the correct  options are (a) and (d).

Answer:

When a circular coil expand radially in a region of magnetic field, them emf induces. The magnitude of induced emf
$E=\frac{d\varphi }{dt}=\frac{d}{dt}\left(BA\right)$
Case 1:
When circular coil expands radially in a region of magnetic field such that the magnetic field is in the same plane as the circular coil or the direction of magnetic field is perpendicular to the direction of increasing area so that their dot product is always zero and hence change in magnetic flux is also zero.
Case 4:
The Magnetic has a perpendicular component whose magnitude is decreasing suitably in such a way that the dot product of magnetic field and surface area of plane of coil remain constant at any instant,
i.e., $E=\frac{d\varphi }{dt}=0$
Hence, the correct options are (b) and (c).

Answer:

Current induces due to change of magnetic flux linked to the circuit.
Magnetic flux, Ï• = BA cosθ
When the circuit becomes closed (from open position).
Neither the Magnetic field(B) nor Area (A) changes.
Hence, no electromotive force(emf) is produced and consequently no current will flow in the circuit.

Answer:

When the coil is stretched, the gaps between successive elements increases. The wires are pulled apart which lead to flux leakage through the gaps. From Lenz's law, the coil will oppose the flux leak/decrease in flux. This is done by increasing the current in the coil.

Answer:

When the core is inserted, the magnetic field increase and hence the magnetic flux also increases.
According to Lenz's law the solenoid will oppose the increase in flux. Hence, the current will decrease.

Answer:

Initially there is no flux was passing through the metal ring when the switch is off.
But, when the current is switched on magnetic flux starts passing through the ring. According to Lenz's law, this increase in flux will be opposed and it can happen if the ring move away from the solenoid.
This happen because the flux increase and cause anticlockwise current (seen from the top) in the ring i.e., in opposite direction to that of solenoid. This makes the ring and solenoid forming same magnetic pole in front of each other. Hence, they will repel each other and the ring will move upward due to repulsion.

Answer:

When the current is switched off magnetic flux linked with the coil decreases. Due to which emf induced in the ring and hence the current induced in the ring. According to Lenz's law the direction of induced current will be such that it opposes the change. When the current in the solenoid decreases a current flows in the same direction in the metal ring as in the solenoid. Thus there will be a downward force. This means the ring will remain on the cardboard. Hence, the upward reaction force of the cardboard on the ring will increase.

Answer:

Due to motion (downward) of cylindrical bar magnet, an eddy current produces in the metallic pipe. These currents oppose the motion of bar magnet through the coil, therefore the downward acceleration of magnet will be less than acceleration due to gravity (g). Hence, it will take relatively more time to come down.
But an unmagnetised iron bar will not produce eddy currents and will fall with an acceleration g. In this case the magnet will take less time to come down.

Answer:

Magnetic flux, Ï• = $\overrightarrow{B}.\overrightarrow{A}$ = BA cosθ
Here, Ï• = B_o(πa²) cosωt                       
$$\begin{gathered} \mathrm{Induced} \mathrm{emf}, E=\frac{d\varphi }{dt}=B_o\left(\pi a^2\right) \omega \sin \omega t \\ \mathrm{Induced} \mathrm{current}, \mathrm{I}=\frac{E_0}{R}=\frac{B_o\left(\pi a^2\right)\omega \sin \omega t}{R} \end{gathered}$$
$$\begin{gathered} \mathrm{For} , t=\frac{\mathrm{\pi }}{2\mathrm{\omega }} \\ I=\frac{B_o\left(\mathrm{\pi }{a}^2\right)\mathrm{\omega }}{R} \mathrm{along} +y \mathrm{axis} \\ \mathrm{For} t=\frac{\mathrm{\pi }}{\mathrm{\omega }} \\ I=0 \\ \mathrm{For} t=\frac{3\mathrm{\pi }}{2\mathrm{\omega }} \\ I=\frac{B\left(\mathrm{\pi }{a}^2\right)\mathrm{\omega }}{R} \mathrm{along} -y \mathrm{axis} \end{gathered}$$

Answer:

From the concept of continuity of magnetic field lines B cannot end or start in space, therefore the total number of lines passing through both the surface (S₁and S₂) is the same. Hence, the magnetic flux passing through both the surface is same.

Answer:

Emf induced,
$E=Bvd$
Induced current,
 $I=\frac{E}{R}=\frac{Bvd}{R}$

Answer:

The change in current is maximum between 5 s to 10 s.
Thus, the maximum back emf will be obtained between 5 s to 10 s.
Given, at t = 3 s
emf = e

$$\begin{gathered} \frac{dI}{dt}=\frac{1}{5} \mathrm{A}/\mathrm{s} \\ e=-L\frac{dI}{dt} \\ \Rightarrow L=-5e \end{gathered}$$
For other cases,
t = 7 s,
$$\begin{gathered} \frac{dI}{dt}=\frac{-2-1}{10-5}=\frac{-3}{5} \\ E\text{'}=-L\frac{dI}{dt}=5e\times \left(\frac{-3}{5}\right) \\ \Rightarrow E\text{'}=-3e \end{gathered}$$
Similarly, at t = 15 s the back emf is $\frac{e}{2}$
and at t = 40 s the back emf is 0 according to given graph.

Answer:

Suppose each coils of one turn $\left(N_1=N_2\right)$.
Mutual inductance of coil A with respect to coil B:
$M_{21}=\frac{N_2{\varphi }_2}{I_1}$
Putting the values,
 $M_{21}=\frac{{10}^{-2}}{2}=5 \mathrm{mH} ...\left(1\right)$
Mutual inductance of coil B with respect to coil A:
$M_{12}=\frac{N_1{\varphi }_1}{I_2}=\frac{{\varphi }_1}{I_2}=\frac{{\varphi }_1}{1} ...\left(2\right)$
But,
$$\begin{gathered} M_{12}=M_{21} \\ \Rightarrow \frac{{\varphi }_1}{1}=5 \mathrm{mH} \\ \Rightarrow {\varphi }_1=5 \mathrm{mWb}=5\times {10}^{-3} \mathrm{Wb} \end{gathered}$$

Answer:

As shown in diagram:
Length, AB = d
Velocity of the wire AB is v.
Magnetic field,
$\overrightarrow{B}=\left(B_0\sin \omega t\right)\hat{k}$
Resistance of the wire AB is R.


Now, area OBAC,
A = x × d
flux, $\varphi =\left(B_0\sin \omega t\right)\left(xd\right)$
E.m.f. induced,
$$\begin{gathered} E_{OBAC}=\frac{-d\varphi }{dt}=-B_{o}d\left[x\frac{d}{dt}\left(\sin \omega t\right)+\sin \omega t\frac{dx}{dt}\right] \\ \Rightarrow \left|E_{OBAC}\right|=B_{o}d\left[x\omega \cos \omega t+\sin \omega t\times v\right] \left[\because \frac{d}{dt}\left(\sin \omega t\right)=\omega \cos \omega t \mathrm{and} \frac{dx}{dt}=v \right] \end{gathered}$$
Now current in the wire,
$i=\frac{B_{o}d}{R}\left(\omega x\cos \omega t+v\sin \omega t\right)$ clockwise
So, the required force on the wire AB to keep the wire moving with uniform speed v is:
$\begin{array}{rcl}{F}_{AB}& =& i\times d\times B\\ & =& \frac{B_{o}d\left(\omega x\cos \omega t+v\sin \omega t\right)}{R}\times d\times \left(B_o\sin \omega t\right)\\ & =& \frac{{B_o}^2{d}^2}{R}\left(\omega x\cos \omega t+v\sin \omega t\right)\sin \omega t\end{array}$

Answer:

(i) Let, at time, t = 0 the wire XY is at position x = 0 and at time t,
its position is x(t) and velocity is $v\left(t\right)=\frac{d}{dt}x\left(t\right).$
It is given that,
 $\overrightarrow{B}=B\left(t\right)\hat{k} \mathrm{and} XY=l$
So, flux at time t be,
$$\begin{gathered} \varphi \left(t\right)=B\left(t\right)·\left[x\left(t\right)\times l\right] \\ \left[\because \mathrm{Area}, A=x\left(t\right)\times l\right] \end{gathered}$$



Now, emf induced,
$\begin{array}{rcl}E& =& \frac{-d\varphi }{dt}=-\frac{d}{dt}\left[B\left(t\right)·l·x\left(t\right)\right]\\ & =& -lx\left(t\right)\frac{d}{dt}B\left(t\right)-lB\left(t\right)\frac{d}{dt}x\left(t\right)\end{array}$
So, current through the loop.
$\begin{array}{rcl}I& =& \frac{E}{R}=\frac{1}{R}\left[-lx\left(t\right)\frac{d}{dt}B\left(t\right)-lB\left(t\right)\frac{d}{dt}x\left(t\right)\right]\\ & =& \frac{1}{R}\left[-lx\left(t\right)\frac{d Bt}{dt}-l B\left(t\right)v\left(t\right)\right]\end{array}$
So, force on the moving wire XY,
$$\begin{gathered} F=ilB=\frac{lB\left(t\right)}{R}\left[-lx\left(t\right)\frac{d B\left(t\right)}{dt}-l B\left(t\right)\frac{d}{dt}x\left(t\right)\right] \\ \Rightarrow ma=-\frac{l^{2}B\left(t\right)x\left(t\right)}{R}\frac{d B\left(t\right)}{dt}-\frac{l^2{B}^2\left(t\right)}{R}\frac{d x\left(t\right)}{dt} .....\left(2\right) \end{gathered}$$
$\Rightarrow a=\frac{l^{2}B\left(t\right)}{mR}\left[-x\left(t\right)\frac{dB\left(t\right)}{dt}-B\left(t\right)\frac{dx\left(t\right)}{dt}\right]$
(Here, a = acceleration, m = mass of the wire XY.)

(ii) Now of we assume, velocity of the wire XY at time t be v(t).
If B is independent of time 
it means, $\frac{dB}{dt}=0$
So, $m\times \frac{d^{2}x}{d{t}^2}=-\left(\frac{x{l}^{2}B}{R}\right)\frac{dB}{dt}-\left(\frac{l^2{B}^2}{R}\right)\frac{dx}{dt} \left[\mathrm{from} \mathrm{equation} \left(2\right)\right]$
$$\begin{gathered} \left(\because a=\frac{d^{2}x}{d{t}^2}=\frac{dv}{dt} \mathrm{and} v=\frac{dx}{dt}\right) \\ \Rightarrow m\frac{d^{2}x}{d{t}^2}=-\frac{l^2{B}^2}{R}v \\ \Rightarrow \frac{dv}{dt}=-\frac{l^2{B}^2}{mR}v \\ \Rightarrow {\int }_{u_o}^v\frac{dv}{v}={\int }_0^t-\frac{l^2{B}^2}{mR}dt \left(\because \mathrm{at} t=0, v=u_o\right) \\ \Rightarrow v\left(t\right)=u_{\circ }e .....\left(3\right) \end{gathered}$$

(iii) Now loss in heat energy in time t,
$={\int }_0^t{I}^{2}Rdt \left[\because P=I^{2}R\right]$
${\int }_0^t\frac{l^2{B}^2{v}^2\left(t\right)}{R^2}\times R dt \left(\because \frac{dB}{dt}=0\right)$
$$\begin{gathered} ={\int }_0^t\frac{l^2{B}^2}{R^2}{u_o}^2 e^{\left(\frac{-2l^2{B}^{2}t}{mR}\right)}dt \\ =\frac{l^2{B}^2{u_o}^2}{R\left({\displaystyle \frac{-2l^2{B}^2}{mR}}\right)}\left[e^{\frac{-2l^2{B}^{2}t}{mR}}-1\right] \\ =-\frac{1}{2}m{u_{\circ }}^2\left[e^{\frac{-2l^2{B}^{2}t}{mR}}-1\right] \\ =\frac{1}{2}m{u_{\circ }}^2-\frac{1}{2}m{u_{\circ }}^2{e}^{\left(\frac{-2l^{2}Bt}{mR}\right)} \\ =\frac{1}{2}m{u_{\circ }}^2-\frac{1}{2}m{v}^2\left(t\right) \left[\mathrm{from} \mathrm{equation} \left(3\right)\right] \end{gathered}$$
= Initial kinetic energy – Final kinetic energy
⇒ Heat lost in time t = Decrease in kinetic energy.

Answer:

Consider the rotation of the wire OP from $\theta =0°$ to 45° as shown in diagram:,
θ = ωt
$$\begin{gathered} \tan \theta =\frac{MD}{l} \\ \Rightarrow MD=l \tan \theta \\ =l \tan \left(\omega t\right) \end{gathered}$$

Now, area of ΔODM,
$=\frac{1}{2}\times l\times MD=\frac{1}{2}\times l\times l \tan \left(\omega t\right)=\frac{1}{2}{l}^2\tan \left(\omega t\right)$
So, flux through this area,
$\varphi =B\times \frac{1}{2}{l}^2\tan \left(\omega t\right)$
Now, emf induced will be,
$$\begin{gathered} E_1=\frac{-d\varphi }{dt}=-\frac{d}{dt}\left[\frac{B{l}^2\tan \left(\omega t\right)}{2}\right] \\ \left|E_1\right|=\frac{1}{2}B{l}^2\omega se{c}^2\left(\omega t\right) \end{gathered}$$
So, current through the wire,
$$\begin{gathered} I_1=\frac{E_1}{R}=\frac{E_1}{\lambda \times x} \left[\because \lambda =\mathrm{resistance} \mathrm{per} \mathrm{unit} \mathrm{length}\right] \\ =\frac{{\displaystyle \frac{1}{2}}B{l}^2\omega se{c}^2\left(\omega t\right)}{\lambda \times \left({\displaystyle \frac{l}{\cos \theta }}\right)} \left(\because \cos \theta =\frac{l}{x} \mathrm{and} \theta =\omega t\right) \\ =\frac{1}{2}\times \frac{Bl\omega sec\left(\omega t\right)}{\lambda } .....\left(1\right) \end{gathered}$$
 

Now, for 45° < θ < 135° :
Area ODBQ
$$\begin{gathered} =l^2-\frac{1}{2}\times l\times \frac{l}{\tan \theta } \\ =l^2-\frac{1}{2}\frac{l^2}{\tan \left(\omega t\right)} \end{gathered}$$                 

So, emf induced at time t,

$$\begin{gathered} E_2=-\frac{d\varphi }{dt}=-\frac{d}{dt}B\left[l^2-\frac{1}{2}\frac{l^2}{\tan \left(\omega t\right)}\right] \\ \left|E_2\right|=\frac{1}{2}wB{l}^2\frac{se{c}^2\left(\omega t\right)}{{\tan }^2\left(\omega t\right)} \end{gathered}$$
So, Current,
$\begin{array}{rcl}{I}_2& =& \frac{E_2}{R}\\ & =& \frac{E_2}{\lambda {\displaystyle \frac{l}{\sin \theta }}}=\frac{1}{2}\frac{B{l}^2\omega se{c}^2\left(\omega t\right)\times \sin \left(\omega t\right)}{\lambda l{\tan }^2\left(\omega t\right)}\\ & =& \frac{1}{2}\frac{Bl\omega \cos e{c}^2\left(\omega t\right)}{\lambda }\\ & =& \frac{1}{2}\frac{Blw}{\lambda \sin \left(\omega t\right)} .....\left(2\right)\end{array}$
 

 Now, if 135° < θ < 180°:

Similarly from the steps taken to derive equation (1)   
we get the required current,
$I_3=\frac{1}{2}\frac{Bl\omega sec\left(\omega t\right)}{\lambda }$
 

Answer:

Let small element of area l × dr at a distance r from the wire.
Magnetic field due to an infinitely long wire carrying current I at a distance r from the wire is,
$B=\frac{{\mu }_{o}I}{2\pi r}$(out of paper)
So, flux through the loop ABCD:
$\varphi ={\int }_{x_o}^x\frac{{\mu }_{o}I}{2\pi r}\times l\times dr$        (Area of small element l × dr )
$$\begin{gathered} =\frac{{\mu }_{o}Il}{2\pi }{\int }_{x_{\circ }}^x\frac{1}{r}dr \\ =\frac{{\mu }_{o}Il}{2\pi }\times {\left[\ln r\right]}_{x_o}^x=\frac{{\mu }_{o}Il}{2\pi }\left[\ln x-\ln x_{\circ }\right] \\ =\frac{{\mu }_{o}Il}{2\pi }\times \ln \left(\frac{x}{x_{\circ }}\right) \left(\mathrm{as} \ln x – \ln x_{\circ }=\ln \frac{x}{x_{\circ }}\right) \end{gathered}$$
So, emf induced:
$$\begin{gathered} E=\frac{-d\varphi }{dt}=-\frac{d}{dt}\left[\frac{{\mu }_{o}Il}{2\pi }\ln \frac{x}{x_o}\right] \\ =-\frac{{\mu }_{o}l}{2\pi }\ln \frac{x}{x_o}\times \frac{dI}{dt} \\ \Rightarrow \left|E\right|=\frac{{\mu }_{o}l\lambda }{2\pi }\ln \frac{x}{x_o} \left(\because \frac{dI}{dt}=\lambda \right) \end{gathered}$$
So, current through the loop,
$i=\frac{E}{R}=\frac{{\mu }_{o}l\lambda }{2\pi R}\ln \frac{x}{x_{\circ }}$
 

Answer:

Flux passing through the loop at a time t can be given as:
$$\begin{gathered} \varphi \left(t\right)=\frac{{\mu }_o{L}_1}{2\pi }\ln \left(\frac{L_2+x}{x}\right)I\left(t\right) \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that}, \\ I\left(t\right)=I_o\left(1-\frac{t}{T}\right), t\le 0\le T \\ I\left(t_1\right)=I\left(0\right)=I_o\left(1-\frac{0}{T}\right)=I_o \\ I\left(t_1\right)=I\left(T\right)=I_o\left(1-\frac{T}{T}\right)=0 \end{gathered}$$
Induced, Current through the loop:
$$\begin{gathered} i\left(t\right)=\frac{1}{R}\frac{d\varphi }{dt} \left(\theta =\mathrm{charge}\right) \\ \mathrm{So}, \frac{dQ}{dt}=\frac{1}{R}\frac{{\mu }_o{L}_1}{2\pi }\ln \left(\frac{L_2+x}{x}\right)\frac{dI\left(t\right)}{dt} \end{gathered}$$
Integrating we, get:
$\Rightarrow \left|Q\left(t_1\right)-Q\left(t_2\right)\right|=\left|\frac{{\mu }_o{L}_1}{2\pi R}\ln \left(\frac{L_2+x}{x}\right)\left[I\left(t_1\right)-I\left(t_2\right)\right]\right|$
So, the total charge passing through a given point in the loop in time T$\left(0\le t\le T\right)$
$$\begin{gathered} ∆Q=\frac{{\mu }_o{L}_1}{2\pi R}\times \ln \left(\frac{L_2+x}{x}\right)\times \left(I_o\right) \\ =\frac{{\mu }_o{L}_1{I}_o}{2\pi R}\ln \left(\frac{L_2+x}{x}\right) \end{gathered}$$
 

Answer:

Magnetic field confined within he $r\le a,$ 
So, flux through ring,
$\varphi =B.\left(\pi a^2\right)$
(Area of the circular region = $\pi a^2$)


Emf induced,
$$\begin{gathered} \left|E\right|=\left|-\frac{d\varphi }{dt}\right|=\left|\pi a^2\times \frac{∆\varphi }{∆t}\right| \\ =\left|\frac{\pi a^2\left(B-0\right)}{∆t}\right| \\ =\frac{\pi a^{2}B}{∆t} \end{gathered}$$
Now, if electric field across the charged ring be E, then, (Here, perimeter of the ring = $2\pi b$)
$$\begin{gathered} 2\pi b\times E=emf=\frac{\pi a^{2}B}{∆t} \\ \Rightarrow E=\frac{\pi a^{2}B}{2\pi b∆t}=\frac{B{a}^2}{2b∆t} \end{gathered}$$
Torque on the ring,
$$\begin{gathered} \tau =b\times \left(\mathrm{force} \mathrm{on} \mathrm{the} \mathrm{ring}\right)=b\times QE=bQ\times \frac{B{a}^2}{2b∆t} \\ \Rightarrow \tau =\frac{BQ{a}^2}{2∆t} \end{gathered}$$
Now, change in angular momentum of the ring in time $∆t$ is given by,
$∆L=\tau \times ∆t=\frac{BQ{a}^2}{2∆t}\times ∆t=\frac{BQ{a}^2}{2}$
But, angular momentum,
L = Moment of inertia × Angular velocity 
   = I$\omega$
$$\begin{gathered} L_{\mathrm{initial}}=I\times {\omega }_{\mathrm{initial}}=m{b}^2\times 0=0 \\ \left(\because I_{\mathrm{ring}}=m{b}^2 \mathrm{and} \mathrm{initially} \omega =0\right) \\ {L}_{\mathrm{final}}=I\times {\omega }_{\mathrm{final}}=m{b}^2\times \omega \\ ∆L=L_{\mathrm{final}}-L_{\mathrm{initial}}=\frac{BQ{a}^2}{2} \\ \Rightarrow m{b}^2\omega =\frac{BQ{a}^2}{2} \\ \Rightarrow \omega =\frac{BQ{a}^2}{2m{b}^2} \end{gathered}$$

Answer:

Let velocity of the rod at a particular instant t be v.
Angle between magnetic field $\left(\overrightarrow{B}\right)$ and the velocity $\left(\overrightarrow{v}\right)$ of the rod of length d is (180°– θ).
Let $F_m$ be the component of magnetic force acting on the rod along the incline.
Now, if the rod moves downward along the incline with acceleration a, then
$$\begin{gathered} mg \sin \theta -F_m=ma \left(\because a=\frac{dv}{dt}\right) \\ \Rightarrow mg\sin \theta -F_m=m\frac{dv}{dt} .....\left(1\right) \end{gathered}$$
Now, induced current, 
$i=\left|\frac{E}{R}\right|=\frac{1}{R}\frac{d\varphi }{dt}=\frac{1}{R}\times Bvd\cos \theta$
So, magnetic force on the rod, along the incline 
$$\begin{gathered} F_m=idB\cos \theta =\left(\frac{1}{R}Bvd\cos \theta \right)\times d\times B\cos \theta \\ =\frac{1}{R}{B}^2{d}^{2}v{\cos }^2\theta .....\left(2\right) \end{gathered}$$
Solving equation (1) and (2) we get:
$$\begin{gathered} mg\sin \theta -\frac{1}{R}{B}^2{d}^{2}v{\cos }^2\theta =m\frac{dv}{dt} \\ \Rightarrow \frac{dv}{dt}+\frac{1}{mR}{B}^2{d}^2{\cos }^2\theta v=g\sin \theta \end{gathered}$$
Solving the above linear differential equation, using the end limit we get:
$v=\frac{g\sin \theta }{\left({\displaystyle \frac{B^2{d}^2{\cos }^2\theta }{mR}}\right)}\left[1-e^{\frac{-\left(B^2{d}^2{\cos }^2\theta \right)t}{mR}}\right]$
This is the required expression for the velocity of the rod.

Answer:

Let charge on the capacitor at a time t be Q(t).
Now current due to the capacitor and the induced current will be opposite.
So, net current in the sliding rod AB be,
I = Induced current – Current due to capacitor 
$$\begin{gathered} =\frac{vBd}{R}-\frac{Q}{RC} \\ \left(\because \mathrm{Induced} \mathrm{current} = \frac{\mathrm{Emf} \mathrm{induced} }{\mathrm{Resistance}}=\frac{vBd}{R}\right) \\ \Rightarrow I=\frac{dQ}{dt}=\frac{vBd}{R}-\frac{Q}{RC} \\ \Rightarrow \frac{dQ}{dt}+\frac{Q}{RC}=\frac{vBd}{R} \end{gathered}$$
Solving the above differential equation and taking the end limit at t = 0, Q = 0 we get:
$Q=vBCd\left[1-e^{\left(\frac{-t}{RC}\right)}\right]$
$$\begin{gathered} \mathrm{Now}, I=\frac{dQ}{dt}=\frac{d}{dt}\left[vBCd\left(1-e^{-\frac{t}{RC}}\right)\right] \\ =vBCd\left(0+\frac{1}{RC}\times e^{\frac{-t}{RC}}\right) \\ =\frac{vBCd}{RC}{e}^{\frac{-t}{RC}} \\ =\frac{vBd}{R}{e}^{\frac{-t}{RC}} \end{gathered}$$
This, is the required expression for the current in the sliding rod at time t.

Answer:

Given, inductance of the inductor is L.
Resistance of the sliding rod is R.
Length of the rod is d.
As magnetic field (B) is perpendicular to the velocity (v) of the rod, 
so the emf induced across the rod AB be, 
E = vBd
So, taking the loop law, we get:
$$\begin{gathered} -L\frac{dI}{dt}+vBd=IR \\ \Rightarrow L\frac{dI}{dt}+Ir=vBd \end{gathered}$$
Solving the above differential equation and taking the limit at t = 0, 
current I = 0, we get:
$I=\frac{vBd}{R}\left(1-e^{\frac{-Rt}{L}}\right)$
This is the required expression for the current, in the rod AB at time t.

Answer:

Given, Z-component of magnetic field,
$B=B_{\circ }\left(1+\lambda Z\right)$
Resistance of the ring is R.
Velocity of the ring falling downwards is v.
Radius of the ring is l.
Let mass of the ring be m.
Flux passing through the area of the ring at time t be,
$\begin{array}{rcl}\varphi & =& B.\mathrm{Area}\\ & =& B_o\left(1+\lambda Z\right)\times \pi l^2 \left(\because \mathrm{Area}=\pi l^2\right)\end{array}$
Now, rate of change of flux
i.e., emf induced,
$\begin{array}{rcl}\left|E\right|& =& \left|-\frac{d\varphi }{dt}\right|=\pi l^2{B}_{o }\left(0+\lambda \frac{dz}{dt}\right)\\ & =& \pi l^2{B}_o\lambda \frac{dz}{dt}\\ & =& \pi l^2{B}_o\lambda v \left(\because v=\frac{dz}{dt}\right)\end{array}$

So, induced current in the ring:
$I=\frac{E}{R}=\frac{\pi l^2{B}_o\lambda v}{R}$
Now, rate of change of gravitational potential energy of the ring,
$mg\left(\frac{dz}{dt}\right)=mgv ....\left(1\right)$
Energy lost due to the heating effect of electric current per second 
$=I^{2}R={\left(\frac{\pi l^2{B}_o\lambda v}{R}\right)}^2\times R$
But, as the velocity of the ring becomes constant,
then the change in kinetic energy will be zero.
Thus the rate of change in gravitational potential energy = Energy lost per second due to heating effect
$$\begin{gathered} \Rightarrow mgv={\left(\frac{\pi l^2\lambda B_{o}v}{R}\right)}^2\times R \\ \Rightarrow v=\frac{mgR}{{\left(\pi l^2\lambda B_o\right)}^2} \end{gathered}$$
This is the required expression of the velocity of the ring falling downwards.
 

Answer:

Let, I current passes through the larger solenoid having 'n' number of turns per unit length.
Magnetic field at the centre of the solenoid be, 
$B={\mu }_{o}nI$
Now, area of the smaller solenoid having 'N' turns per unit length be,
$A=\pi b^{2 }\left(\mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} b < a\right)$
So, flux passing through the smaller coil,
$\varphi =NB.A=N\left({\mu }_{o}nI\right)\left(\pi b^2\right)$
So, emf induced in the solenoid
$\begin{array}{rcl}\left|E\right|& =& \left|-\frac{d\varphi }{dt}\right|=\frac{d}{dt}N\left({\mu }_{o}NI\right)\left(\pi b^2\right)\\ & =& \left(\pi Nn{\mu }_o{b}^2\right)\frac{dI}{dt}\end{array}$
It is given that 
$$\begin{gathered} I=m{t}^2+C \\ \mathrm{So}, \left|E\right|=\left(\pi Nn{\mu }_o{b}^2\right)\frac{d}{dt}\left(m{t}^2+C\right) \\ =\pi Nn{\mu }_o{b}^2\left(2mt\right) \end{gathered}$$
Hence, the emf induced in the coil is $2\pi Nn{\mu }_o{b}^{2}mt.$