Physics Ncert Exemplar 2019 Solutions for Class 12 Science Physics Chapter 6 Electromagnetic Induction are provided here with simple step-by-step explanations. These solutions for Electromagnetic Induction are extremely popular among class 12 Science students for Physics Electromagnetic Induction Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Physics Ncert Exemplar 2019 Book of class 12 Science Physics Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Physics Ncert Exemplar 2019 Solutions. All Physics Ncert Exemplar 2019 Solutions for class 12 Science Physics are prepared by experts and are 100% accurate.

#### Page No 33:

#### Question 1:

*L*meters lies in the

*x-y*plane in a region, where the magnetic field is given by $\mathit{B}={B}_{0}\left(2\hat{\mathbf{i}}+3\hat{\mathbf{j}}+4\hat{\mathbf{k}}\right)\mathrm{T},$ where

*B*

_{o}is constant. The magnitude of flux passing through the square is

*B*

_{o }L^{2}Wb.

*B*

_{o }L^{2}Wb.

*B*

_{o }L^{2}Wb.

#### Answer:

Magnetic flux, $\varphi =\overrightarrow{B}.\overrightarrow{A}$

Given, *A* = ${L}^{2}\hat{k}$

$\overrightarrow{B}={B}_{o}(2\hat{i}+3\hat{j}+4\hat{k})\mathrm{T}\phantom{\rule{0ex}{0ex}}\varphi ={B}_{o}(2\hat{i}+3\hat{j}+4\hat{k}).{L}^{2}\hat{k}=4{B}_{o}{L}^{2}\mathrm{Wb}$

Hence, the correct answer is option (c).

#### Page No 33:

#### Question 2:

*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*

*).*A magnetic field $\mathit{B}={B}_{\mathrm{o}}\left(\hat{\mathit{i}}+\hat{\mathit{k}}\right)\mathrm{T}$ is present in the region. The flux passing through the loop ABCDEFA (in that order) is

_{o }L

^{2}Wb.

_{o }L

^{2}Wb.

(d) 4 B

_{o }L

^{2 }Wb.

#### 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, $\varphi =\overrightarrow{B}.\overrightarrow{A}$

$\mathrm{Area},\overrightarrow{A}={\overrightarrow{A}}_{1}+{\overrightarrow{A}}_{2}={L}^{2}(\hat{i}+\hat{k}){\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{and}\overrightarrow{B}={B}_{o}(\hat{i}+\hat{k})\mathrm{T}\phantom{\rule{0ex}{0ex}}\mathrm{\varphi}={B}_{\mathrm{o}}(\hat{i}+\hat{k}).{L}^{2}(\hat{i}+\hat{k})\phantom{\rule{0ex}{0ex}}\varphi =2{B}_{o}{L}^{2}\mathrm{Wb}$

Hence, the correct answer is option (b).

#### Page No 34:

#### Question 3:

*T*= 2π/ω.

#### 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).

#### Page No 34:

#### Question 4:

#### 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).

#### Page No 34:

#### Question 5:

*t*= 0) is counterclockwise and the coil A is as shown at this instant,

*t*= 0, is

\

#### 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).

#### Page No 34:

#### Question 6:

*L*of a solenoid of length

*l*and area of cross-section

*A*, with a fixed number of turns

*N*increases as

*l*and

*A*increase.

*l*decreases and

*A*increases.

*l*increases and

*A*decreases.

*l*and

*A*decrease.

#### Answer:

The self inductance, *L* = *μn*^{2}*Al.**L* = *μn*^{2}*Al*

Here, *n* = number of turns per unit length

$n=\frac{N}{l}\phantom{\rule{0ex}{0ex}}L=\frac{\mu {N}^{2}A}{l}$

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).

#### Page No 34:

#### Question 7:

#### 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).

#### Page No 35:

#### Question 8:

#### 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).

#### Page No 35:

#### Question 9:

*M*

_{12}of coil 1 with respect to coil 2

*M*

_{21}of coil 2 with respect to coil 1.

#### Answer:

The Mutual inductance *M*_{12} of coil increases when they are brought nearer and is same as *M*_{21} of Coil 2 with respect to coil 1.*M*_{12} = μ_{0}*n*_{1}*n*_{2}π*r*^{2}*l**M*_{21} = μ_{0}*n*_{1}*n*_{2}π*r*^{2}*l*.

So, *M*_{12} = *M*_{21}

Hence, the correct options are (a) and (d).

#### Page No 35:

#### Question 10:

#### 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).

#### Page No 35:

#### Question 11:

#### 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.

#### Page No 36:

#### Question 12:

#### 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.

#### Page No 36:

#### Question 13:

#### 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.

#### Page No 36:

#### Question 14:

Consider a metal ring kept on top of a fixed solenoid (say on a carboard) in the given figure. 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:

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.

#### Page No 36:

#### Question 15:

Consider a metal ring kept (supported by a cardboard) on top of a fixed solenoid carrying a current *I* in the given figure. 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:

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.

#### Page No 36:

#### Question 16:

#### 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.

#### Page No 36:

#### Question 17:

A magnetic field in a certain region is given by $\mathbf{B}={B}_{o}\mathrm{cos}\left(\omega t\right)\hat{\mathbf{k}}$ 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 in the given figure. Find the magnitude and the direction of the current at (*a*, 0, 0) at *t* = π/2ω, *t* = π/ω and *t* = 3π/2ω.

#### Answer:

Magnetic flux, Ï• = $\overrightarrow{B}.\overrightarrow{A}$ = *BA* cosθ

Here, Ï• = *B*_{o}(π*a*^{2}) cos*ωt*

$\mathrm{Induced}\mathrm{emf},E=\frac{d\varphi}{dt}={B}_{o}\left(\pi {a}^{2}\right)\omega \mathrm{sin}\omega t\phantom{\rule{0ex}{0ex}}\mathrm{Induced}\mathrm{current},\mathrm{I}=\frac{{E}_{0}}{R}=\frac{{B}_{o}\left(\pi {a}^{2}\right)\omega \mathrm{sin}\omega t}{R}$

$\mathrm{For},t=\frac{\mathrm{\pi}}{2\mathrm{\omega}}\phantom{\rule{0ex}{0ex}}I=\frac{{B}_{o}\left(\mathrm{\pi}{a}^{2}\right)\mathrm{\omega}}{R}\mathrm{along}+y\mathrm{axis}\phantom{\rule{0ex}{0ex}}\mathrm{For}t=\frac{\mathrm{\pi}}{\mathrm{\omega}}\phantom{\rule{0ex}{0ex}}I=0\phantom{\rule{0ex}{0ex}}\mathrm{For}t=\frac{3\mathrm{\pi}}{2\mathrm{\omega}}\phantom{\rule{0ex}{0ex}}I=\frac{B\left(\mathrm{\pi}{a}^{2}\right)\mathrm{\omega}}{R}\mathrm{along}-y\mathrm{axis}$

#### Page No 36:

#### Question 18:

Consider a closed loop C in a magnetic field in the given figure. The flux passing through the loop is defined by choosing a surface whose edge coincides with the loop and using the formula $\mathrm{\varphi}={\mathbf{B}}_{1}.{\mathbf{dA}}_{1}+{\mathit{B}}_{2}.{\mathbf{dA}}_{2}+....$ Now if we chose two different surfaces S_{1} and S_{2} having C as their edge, would we get the same answer

#### 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 (S_{1}and S_{2}) is the same. Hence, the magnetic flux passing through both the surface is same.

#### Page No 36:

#### Question 19:

*, the magnetic field is coming out of the paper.*

**B***θ*is a fixed angle made by PQ travelling smoothly over two conducting parallel wires seperated by a distance

*d*.

#### Answer:

Emf induced,

$E=Bvd$

Induced current,

$I=\frac{E}{R}=\frac{Bvd}{R}$

#### Page No 37:

#### Question 20:

A (current vs time) graph of the current passing through a solenoid is shown in the given 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 and 40 s. OA, AB and BC are straight line segments.

#### 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*

$\frac{dI}{dt}=\frac{1}{5}\mathrm{A}/\mathrm{s}\phantom{\rule{0ex}{0ex}}e=-L\frac{dI}{dt}\phantom{\rule{0ex}{0ex}}\Rightarrow L=-5e$

For other cases,*t* = 7 s,

$\frac{dI}{dt}=\frac{-2-1}{10-5}=\frac{-3}{5}\phantom{\rule{0ex}{0ex}}E\text{'}=-L\frac{dI}{dt}=5e\times \left(\frac{-3}{5}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow E\text{'}=-3e$

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.

#### Page No 37:

#### Question 21:

^{-2}Wb passes through B (no current through B). If no current passes through A and a current of 1 A passes through B, what is the flux through A?

#### 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,

${M}_{12}={M}_{21}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\varphi}_{1}}{1}=5\mathrm{mH}\phantom{\rule{0ex}{0ex}}\Rightarrow {\varphi}_{1}=5\mathrm{mWb}=5\times {10}^{-3}\mathrm{Wb}$

#### Page No 37:

#### Question 22:

*d*in the given 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

*, what is the current in the circuit. What is the force needed to keep the wire moving at constant velocity?*

**v**#### Answer:

As shown in diagram:

Length, AB = *d*

Velocity of the wire AB is *v*.

Magnetic field,

$\overrightarrow{B}=\left({B}_{0}\mathrm{sin}\omega t\right)\hat{k}$

Resistance of the wire AB is *R*.

Now, area OBAC,

A = *x* × d

flux, $\varphi =\left({B}_{0}\mathrm{sin}\omega t\right)\left(xd\right)$

E.m.f. induced,

${E}_{OBAC}=\frac{-d\varphi}{dt}=-{B}_{o}d\left[x\frac{d}{dt}\left(\mathrm{sin}\omega t\right)+\mathrm{sin}\omega t\frac{dx}{dt}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \left|{E}_{OBAC}\right|={B}_{o}d\left[x\omega \mathrm{cos}\omega t+\mathrm{sin}\omega t\times v\right]\left[\because \frac{d}{dt}\left(\mathrm{sin}\omega t\right)=\omega \mathrm{cos}\omega t\mathrm{and}\frac{dx}{dt}=v\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Now current in the wire,

$i=\frac{{B}_{o}d}{R}\left(\omega x\mathrm{cos}\omega t+v\mathrm{sin}\omega t\right)$ clockwise

So, the required force on the wire AB to keep the wire moving with uniform speed *v* is:

$\begin{array}{rcl}{F}_{AB}& =& i\times d\times B\\ & =& \frac{{B}_{o}d\left(\omega x\mathrm{cos}\omega t+v\mathrm{sin}\omega t\right)}{R}\times d\times \left({B}_{o}\mathrm{sin}\omega t\right)\\ & =& \frac{{{B}_{o}}^{2}{d}^{2}}{R}\left(\omega x\mathrm{cos}\omega t+v\mathrm{sin}\omega t\right)\mathrm{sin}\omega t\end{array}$

#### Page No 37:

#### Question 23:

A conducting wire XY of mass *m* and neglibile resistance slides smoothly on two parallel conducting wires as shown in the given figure. The closed circuit has a resistance *R* due to AC. AB and CD are perfect conductors. There is a magnetic field $\mathbf{B}=B\left(t\right)\hat{\mathbf{k}}.$

**B**is independent of time, obtain

*v*(

*t*) , assuming

*v*(0) =

*u*

_{0}.

*R*.

#### 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,

$\varphi \left(t\right)=B\left(t\right)\xb7\left[x\left(t\right)\times l\right]\phantom{\rule{0ex}{0ex}}\left[\because \mathrm{Area},A=x\left(t\right)\times l\right]$

Now, emf induced,

$\begin{array}{rcl}E& =& \frac{-d\varphi}{dt}=-\frac{d}{dt}\left[B\left(t\right)\xb7l\xb7x\left(t\right)\right]\\ & =& -lx\left(t\right)\frac{d}{dt}B\left(t\right)-lB\left(t\right)\frac{d}{dt}x\left(t\right)\end{array}$

So, current through the loop.

$\begin{array}{rcl}I& =& \frac{E}{R}=\frac{1}{R}\left[-lx\left(t\right)\frac{d}{dt}B\left(t\right)-lB\left(t\right)\frac{d}{dt}x\left(t\right)\right]\\ & =& \frac{1}{R}\left[-lx\left(t\right)\frac{dBt}{dt}-lB\left(t\right)v\left(t\right)\right]\end{array}$

So, force on the moving wire *XY*,

$F=ilB=\frac{lB\left(t\right)}{R}\left[-lx\left(t\right)\frac{dB\left(t\right)}{dt}-lB\left(t\right)\frac{d}{dt}x\left(t\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow ma=-\frac{{l}^{2}B\left(t\right)x\left(t\right)}{R}\frac{dB\left(t\right)}{dt}-\frac{{l}^{2}{B}^{2}\left(t\right)}{R}\frac{dx\left(t\right)}{dt}.....\left(2\right)$

$\Rightarrow a=\frac{{l}^{2}B\left(t\right)}{mR}\left[-x\left(t\right)\frac{dB\left(t\right)}{dt}-B\left(t\right)\frac{dx\left(t\right)}{dt}\right]$

(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, $m\times \frac{{d}^{2}x}{d{t}^{2}}=-\left(\frac{x{l}^{2}B}{R}\right)\frac{dB}{dt}-\left(\frac{{l}^{2}{B}^{2}}{R}\right)\frac{dx}{dt}\left[\mathrm{from}\mathrm{equation}\left(2\right)\right]$

$\left(\because a=\frac{{d}^{2}x}{d{t}^{2}}=\frac{dv}{dt}\mathrm{and}v=\frac{dx}{dt}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow m\frac{{d}^{2}x}{d{t}^{2}}=-\frac{{l}^{2}{B}^{2}}{R}v\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{dv}{dt}=-\frac{{l}^{2}{B}^{2}}{mR}v\phantom{\rule{0ex}{0ex}}\Rightarrow {\int}_{{u}_{o}}^{v}\frac{dv}{v}={\int}_{0}^{t}-\frac{{l}^{2}{B}^{2}}{mR}dt\left(\because \mathrm{at}t=0,v={u}_{o}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow v\left(t\right)={u}_{\circ}e.....\left(3\right)$

(iii) Now loss in heat energy in time* t*,

$={\int}_{0}^{t}{I}^{2}Rdt\left[\because P={I}^{2}R\right]$

${\int}_{0}^{t}\frac{{l}^{2}{B}^{2}{v}^{2}\left(t\right)}{{R}^{2}}\times Rdt\left(\because \frac{dB}{dt}=0\right)$

$={\int}_{0}^{t}\frac{{l}^{2}{B}^{2}}{{R}^{2}}{{u}_{o}}^{2}{e}^{\left(\frac{-2{l}^{2}{B}^{2}t}{mR}\right)}dt\phantom{\rule{0ex}{0ex}}=\frac{{l}^{2}{B}^{2}{{u}_{o}}^{2}}{R\left({\displaystyle \frac{-2{l}^{2}{B}^{2}}{mR}}\right)}\left[{e}^{\frac{-2{l}^{2}{B}^{2}t}{mR}}-1\right]\phantom{\rule{0ex}{0ex}}=-\frac{1}{2}m{{u}_{\circ}}^{2}\left[{e}^{\frac{-2{l}^{2}{B}^{2}t}{mR}}-1\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}m{{u}_{\circ}}^{2}-\frac{1}{2}m{{u}_{\circ}}^{2}{e}^{\left(\frac{-2{l}^{2}Bt}{mR}\right)}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}m{{u}_{\circ}}^{2}-\frac{1}{2}m{v}^{2}\left(t\right)\left[\mathrm{from}\mathrm{equation}\left(3\right)\right]$

= Initial kinetic energy – Final kinetic energy

⇒ Heat lost in time *t* = Decrease in kinetic energy.

#### Page No 38:

#### Question 24:

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

#### Answer:

Consider the rotation of the wire OP from $\theta =0\xb0$ to 45° as shown in diagram:,

θ = ωt

$\mathrm{tan}\theta =\frac{MD}{l}\phantom{\rule{0ex}{0ex}}\Rightarrow MD=l\mathrm{tan}\theta \phantom{\rule{0ex}{0ex}}=l\mathrm{tan}\left(\omega t\right)$

Now, area of ΔODM,

$=\frac{1}{2}\times l\times MD=\frac{1}{2}\times l\times l\mathrm{tan}\left(\omega t\right)=\frac{1}{2}{l}^{2}\mathrm{tan}\left(\omega t\right)$

So, flux through this area,

$\varphi =B\times \frac{1}{2}{l}^{2}\mathrm{tan}\left(\omega t\right)$

Now, emf induced will be,

${E}_{1}=\frac{-d\varphi}{dt}=-\frac{d}{dt}\left[\frac{B{l}^{2}\mathrm{tan}\left(\omega t\right)}{2}\right]\phantom{\rule{0ex}{0ex}}\left|{E}_{1}\right|=\frac{1}{2}B{l}^{2}\omega se{c}^{2}\left(\omega t\right)$

So, current through the wire,

${I}_{1}=\frac{{E}_{1}}{R}=\frac{{E}_{1}}{\lambda \times x}\left[\because \lambda =\mathrm{resistance}\mathrm{per}\mathrm{unit}\mathrm{length}\right]\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{1}{2}}B{l}^{2}\omega se{c}^{2}\left(\omega t\right)}{\lambda \times \left({\displaystyle \frac{l}{\mathrm{cos}\theta}}\right)}\left(\because \mathrm{cos}\theta =\frac{l}{x}\mathrm{and}\theta =\omega t\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times \frac{Bl\omega sec\left(\omega t\right)}{\lambda}.....\left(1\right)$

Now, for 45° < θ < 135° :

Area ODBQ

$={l}^{2}-\frac{1}{2}\times l\times \frac{l}{\mathrm{tan}\theta}\phantom{\rule{0ex}{0ex}}={l}^{2}-\frac{1}{2}\frac{{l}^{2}}{\mathrm{tan}\left(\omega t\right)}$

So, emf induced at time *t*,

${E}_{2}=-\frac{d\varphi}{dt}=-\frac{d}{dt}B\left[{l}^{2}-\frac{1}{2}\frac{{l}^{2}}{\mathrm{tan}\left(\omega t\right)}\right]\phantom{\rule{0ex}{0ex}}\left|{E}_{2}\right|=\frac{1}{2}wB{l}^{2}\frac{se{c}^{2}\left(\omega t\right)}{{\mathrm{tan}}^{2}\left(\omega t\right)}$

So, Current,

$\begin{array}{rcl}{I}_{2}& =& \frac{{E}_{2}}{R}\\ & =& \frac{{E}_{2}}{\lambda {\displaystyle \frac{l}{\mathrm{sin}\theta}}}=\frac{1}{2}\frac{B{l}^{2}\omega se{c}^{2}\left(\omega t\right)\times \mathrm{sin}\left(\omega t\right)}{\lambda l{\mathrm{tan}}^{2}\left(\omega t\right)}\\ & =& \frac{1}{2}\frac{Bl\omega \mathrm{cos}e{c}^{2}\left(\omega t\right)}{\lambda}\\ & =& \frac{1}{2}\frac{Blw}{\lambda \mathrm{sin}\left(\omega t\right)}.....\left(2\right)\end{array}$

Now, if 135° < θ < 180°:

Similarly from the steps taken to derive equation (1)

we get the required current,

${I}_{3}=\frac{1}{2}\frac{Bl\omega sec\left(\omega t\right)}{\lambda}$

#### Page No 38:

#### Question 25:

Consider an infinitely long wire carrying a current *I* (*t *), with $\frac{dI}{dt}=\lambda =$constant. Find the current produced in the rectangular loop of wire ABCD if its resistance is *R* in the given figure.

#### 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* )

$=\frac{{\mu}_{o}Il}{2\pi}{\int}_{{x}_{\circ}}^{x}\frac{1}{r}dr\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{o}Il}{2\pi}\times {\left[\mathrm{ln}r\right]}_{{x}_{o}}^{x}=\frac{{\mu}_{o}Il}{2\pi}\left[\mathrm{ln}x-\mathrm{ln}{x}_{\circ}\right]\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{o}Il}{2\pi}\times \mathrm{ln}\left(\frac{x}{{x}_{\circ}}\right)\left(\mathrm{as}\mathrm{ln}x\u2013\mathrm{ln}{x}_{\circ}=\mathrm{ln}\frac{x}{{x}_{\circ}}\right)$

So, emf induced:

$E=\frac{-d\varphi}{dt}=-\frac{d}{dt}\left[\frac{{\mu}_{o}Il}{2\pi}\mathrm{ln}\frac{x}{{x}_{o}}\right]\phantom{\rule{0ex}{0ex}}=-\frac{{\mu}_{o}l}{2\pi}\mathrm{ln}\frac{x}{{x}_{o}}\times \frac{dI}{dt}\phantom{\rule{0ex}{0ex}}\Rightarrow \left|E\right|=\frac{{\mu}_{o}l\lambda}{2\pi}\mathrm{ln}\frac{x}{{x}_{o}}\left(\because \frac{dI}{dt}=\lambda \right)$

So, current through the loop,

$i=\frac{E}{R}=\frac{{\mu}_{o}l\lambda}{2\pi R}\mathrm{ln}\frac{x}{{x}_{\circ}}$

#### Page No 38:

#### Question 26:

A rectangular loop of wire ABCD is kept close to an infinitely long wire carrying a current *I* (*t*) = *I _{o}* (1 –

*t*/

*T*) for 0 ≤

*t*≤

*T*and

*I*(0) = 0 for

*t*>

*T*in the given figure. Find the total charge passing through a given point in the loop, in time

*T*. The resistance of the loop is

*R*.

#### Answer:

Flux passing through the loop at a time *t* can be given as:

$\varphi \left(t\right)=\frac{{\mu}_{o}{L}_{1}}{2\pi}\mathrm{ln}\left(\frac{{L}_{2}+x}{x}\right)I\left(t\right)\phantom{\rule{0ex}{0ex}}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that},\phantom{\rule{0ex}{0ex}}I\left(t\right)={I}_{o}\left(1-\frac{t}{T}\right),t\le 0\le T\phantom{\rule{0ex}{0ex}}I\left({t}_{1}\right)=I\left(0\right)={I}_{o}\left(1-\frac{0}{T}\right)={I}_{o}\phantom{\rule{0ex}{0ex}}I\left({t}_{1}\right)=I\left(T\right)={I}_{o}\left(1-\frac{T}{T}\right)=0$

Induced, Current through the loop:

$i\left(t\right)=\frac{1}{R}\frac{d\varphi}{dt}\left(\theta =\mathrm{charge}\right)\phantom{\rule{0ex}{0ex}}\mathrm{So},\frac{dQ}{dt}=\frac{1}{R}\frac{{\mu}_{o}{L}_{1}}{2\pi}\mathrm{ln}\left(\frac{{L}_{2}+x}{x}\right)\frac{dI\left(t\right)}{dt}$

Integrating we, get:

$\Rightarrow \left|Q\left({t}_{1}\right)-Q\left({t}_{2}\right)\right|=\left|\frac{{\mu}_{o}{L}_{1}}{2\pi R}\mathrm{ln}\left(\frac{{L}_{2}+x}{x}\right)\left[I\left({t}_{1}\right)-I\left({t}_{2}\right)\right]\right|$

So, the total charge passing through a given point in the loop in time T$\left(0\le t\le T\right)$

$\u2206Q=\frac{{\mu}_{o}{L}_{1}}{2\pi R}\times \mathrm{ln}\left(\frac{{L}_{2}+x}{x}\right)\times \left({I}_{o}\right)\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{o}{L}_{1}{I}_{o}}{2\pi R}\mathrm{ln}\left(\frac{{L}_{2}+x}{x}\right)$

#### Page No 38:

#### Question 27:

**B**is confined to a region

*r*≤

*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 âˆ†

*t*. Find the angular velocity ω of the ring after the field vanishes.

#### 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,

$\left|E\right|=\left|-\frac{d\varphi}{dt}\right|=\left|\pi {a}^{2}\times \frac{\u2206\varphi}{\u2206t}\right|\phantom{\rule{0ex}{0ex}}=\left|\frac{\pi {a}^{2}\left(B-0\right)}{\u2206t}\right|\phantom{\rule{0ex}{0ex}}=\frac{\pi {a}^{2}B}{\u2206t}$

Now, if electric field across the charged ring be E, then, (Here, perimeter of the ring = $2\pi b$)

$2\pi b\times E=emf=\frac{\pi {a}^{2}B}{\u2206t}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{\pi {a}^{2}B}{2\pi b\u2206t}=\frac{B{a}^{2}}{2b\u2206t}$

Torque on the ring,

$\tau =b\times \left(\mathrm{force}\mathrm{on}\mathrm{the}\mathrm{ring}\right)=b\times QE=bQ\times \frac{B{a}^{2}}{2b\u2206t}\phantom{\rule{0ex}{0ex}}\Rightarrow \tau =\frac{BQ{a}^{2}}{2\u2206t}$

Now, change in angular momentum of the ring in time $\u2206t$ is given by,

$\u2206L=\tau \times \u2206t=\frac{BQ{a}^{2}}{2\u2206t}\times \u2206t=\frac{BQ{a}^{2}}{2}$

But, angular momentum,

L = Moment of inertia × Angular velocity

= I$\omega $

${L}_{\mathrm{initial}}=I\times {\omega}_{\mathrm{initial}}=m{b}^{2}\times 0=0\phantom{\rule{0ex}{0ex}}\left(\because {I}_{\mathrm{ring}}=m{b}^{2}\mathrm{and}\mathrm{initially}\omega =0\right)\phantom{\rule{0ex}{0ex}}{L}_{\mathrm{final}}=I\times {\omega}_{\mathrm{final}}=m{b}^{2}\times \omega \phantom{\rule{0ex}{0ex}}\u2206L={L}_{\mathrm{final}}-{L}_{\mathrm{initial}}=\frac{BQ{a}^{2}}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow m{b}^{2}\omega =\frac{BQ{a}^{2}}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \omega =\frac{BQ{a}^{2}}{2m{b}^{2}}$

#### Page No 39:

#### Question 28:

A rod of mass *m* and resistance *R* slides smoothly over two parallel perfectly conducting wires kept sloping at an angle θ with respect to the horizontal in the given 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:

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

$mg\mathrm{sin}\theta -{F}_{m}=ma\left(\because a=\frac{dv}{dt}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow mg\mathrm{sin}\theta -{F}_{m}=m\frac{dv}{dt}.....\left(1\right)$

Now, induced current,

$i=\left|\frac{E}{R}\right|=\frac{1}{R}\frac{d\varphi}{dt}=\frac{1}{R}\times Bvd\mathrm{cos}\theta $

So, magnetic force on the rod, along the incline

${F}_{m}=idB\mathrm{cos}\theta =\left(\frac{1}{R}Bvd\mathrm{cos}\theta \right)\times d\times B\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}=\frac{1}{R}{B}^{2}{d}^{2}v{\mathrm{cos}}^{2}\theta .....\left(2\right)$

Solving equation (1) and (2) we get:

$mg\mathrm{sin}\theta -\frac{1}{R}{B}^{2}{d}^{2}v{\mathrm{cos}}^{2}\theta =m\frac{dv}{dt}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{dv}{dt}+\frac{1}{mR}{B}^{2}{d}^{2}{\mathrm{cos}}^{2}\theta v=g\mathrm{sin}\theta $

Solving the above linear differential equation, using the end limit we get:

$v=\frac{g\mathrm{sin}\theta}{\left({\displaystyle \frac{{B}^{2}{d}^{2}{\mathrm{cos}}^{2}\theta}{mR}}\right)}\left[1-{e}^{\frac{-\left({B}^{2}{d}^{2}{\mathrm{cos}}^{2}\theta \right)t}{mR}}\right]$

This is the required expression for the velocity of the rod.

#### Page No 39:

#### Question 29:

Find the current in the sliding rod AB (resistance = *R*) for the arrangement shown in the given 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:

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

$=\frac{vBd}{R}-\frac{Q}{RC}\phantom{\rule{0ex}{0ex}}\left(\because \mathrm{Induced}\mathrm{current}=\frac{\mathrm{Emf}\mathrm{induced}}{\mathrm{Resistance}}=\frac{vBd}{R}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow I=\frac{dQ}{dt}=\frac{vBd}{R}-\frac{Q}{RC}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{dQ}{dt}+\frac{Q}{RC}=\frac{vBd}{R}$

Solving the above differential equation and taking the end limit at *t* = 0, *Q *= 0 we get:

$Q=vBCd\left[1-{e}^{\left(\frac{-t}{RC}\right)}\right]$

$\mathrm{Now},I=\frac{dQ}{dt}=\frac{d}{dt}\left[vBCd\left(1-{e}^{-\frac{t}{RC}}\right)\right]\phantom{\rule{0ex}{0ex}}=vBCd\left(0+\frac{1}{RC}\times {e}^{\frac{-t}{RC}}\right)\phantom{\rule{0ex}{0ex}}=\frac{vBCd}{RC}{e}^{\frac{-t}{RC}}\phantom{\rule{0ex}{0ex}}=\frac{vBd}{R}{e}^{\frac{-t}{RC}}$

This, is the required expression for the current in the sliding rod at time *t*.

#### Page No 39:

#### Question 30:

Find the current in the sliding rod AB (resistance = *R*) for the arrangement shown in the given figure. * B *is constant and is out of the paper. Parallel wires have no resistance.

**is constant. Switch S is closed at time**

*v**t*= 0.

#### 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:

$-L\frac{dI}{dt}+vBd=IR\phantom{\rule{0ex}{0ex}}\Rightarrow L\frac{dI}{dt}+Ir=vBd$

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*.

#### Page No 39:

#### Question 31:

*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

*B*=

_{z }*B*

_{o}(1 + λ

*z*). 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*, λ and acceleration due to gravity

*g*.

#### 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

$\Rightarrow mgv={\left(\frac{\pi {l}^{2}\lambda {B}_{o}v}{R}\right)}^{2}\times R\phantom{\rule{0ex}{0ex}}\Rightarrow v=\frac{mgR}{{\left(\pi {l}^{2}\lambda {B}_{o}\right)}^{2}}$

This is the required expression of the velocity of the ring falling downwards.

#### Page No 39:

#### Question 32:

*n*’ turns per meter, with diameter ‘

*a*’. At the centre of this coil we place a smaller coil of ‘

*N*’ turns and diameter ‘

*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 graph showing nature of variation in emf, if current varies as a function of

*mt*

^{2 }+ C.

#### 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,

$A=\pi {b}^{2}\left(\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}ba\right)$

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

$I=m{t}^{2}+C\phantom{\rule{0ex}{0ex}}\mathrm{So},\left|E\right|=\left(\pi Nn{\mu}_{o}{b}^{2}\right)\frac{d}{dt}\left(m{t}^{2}+C\right)\phantom{\rule{0ex}{0ex}}=\pi Nn{\mu}_{o}{b}^{2}\left(2mt\right)$

Hence, the emf induced in the coil is $2\pi Nn{\mu}_{o}{b}^{2}mt.$

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