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#### Question 1:

A square of side L meters lies in the x-y plane in a region, where the magnetic field is given by $\mathbit{B}={B}_{0}\left(2\stackrel{^}{\mathbf{i}}+3\stackrel{^}{\mathbf{j}}+4\stackrel{^}{\mathbf{k}}\right)\mathrm{T},$ where Bo is constant. The magnitude of flux passing through the square is
(a) 2 Bo L2  Wb.
(b) 3 Bo L2  Wb.
(c) 4 Bo L2  Wb.
(d)

Magnetic flux, $\varphi =\stackrel{\to }{B}.\stackrel{\to }{A}$
Given, A

Hence, the correct answer is option (c).

#### Question 2:

A loop, made of straight edges has six corners at 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  is present in the region. The flux passing through the loop ABCDEFA (in that order) is
(a) Bo L2 Wb.
(b) 2 Bo L2 Wb.
(c)
(d) 4
Bo L2 Wb.

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, $\varphi =\stackrel{\to }{B}.\stackrel{\to }{A}$

Hence, the correct answer is option (b).

#### Question 3:

A cylindrical bar magnet is rotated about its axis in the given figure. 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 = 2π/ω.
(d) a time varying non-sinosoidal current flows through the ammeter A.

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).

#### Question 4:

There are two coils A and B as shown in the given 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 counterclockwise. 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 counterclockwise direction in A.

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).

#### Question 5:

Same as problem 4 except the coil A is made to rotate about a vertical axis in the given figure. 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.

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).

#### Question 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.

The self inductance, L = μn2Al.
L = μn2Al
Here, n = number of turns per unit length

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).

#### Question 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.

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).

#### Question 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.

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).

#### Question 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.

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).

#### 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.

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).

#### Question 11:

Consider a magnet surrounded by a wire with an on/off switch S in the given 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. 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.

#### Question 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.

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.

#### Question 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.

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.

#### 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 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.

#### 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? 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.

#### Question 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 unmagnetised cylindrical iron bar dropped through the metallic pipe. Explain.

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.

#### Question 17:

A magnetic field in a certain region is given by  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 tπ/2ωtπ/ω and  t = 3π/2ω. Magnetic flux, Ï• = $\stackrel{\to }{B}.\stackrel{\to }{A}$ = BA cosθ
Here, Ï• = Bo(πa2) cosωt

#### 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 $\mathrm{\varphi }={\mathbf{B}}_{1}.{\mathbf{dA}}_{1}+{\mathbit{B}}_{2}.{\mathbf{dA}}_{2}+....$  Now if we chose two different surfaces S1 and S2 having C as their edge, would we get the same 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.

#### Question 19:

Find the current in the wire for the configuration shown in the given figure. Wire PQ has negligible resistance. B, the magnetic field is coming out of the paper. θ is a fixed angle made by PQ travelling smoothly over two conducting parallel wires seperated by a distance d. Emf induced,
$E=Bvd$
Induced current,
$I=\frac{E}{R}=\frac{Bvd}{R}$

#### 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. 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 For other cases,
t = 7 s,
$\frac{dI}{dt}=\frac{-2-1}{10-5}=\frac{-3}{5}\phantom{\rule{0ex}{0ex}}E\text{'}=-L\frac{dI}{dt}=5e×\left(\frac{-3}{5}\right)\phantom{\rule{0ex}{0ex}}⇒E\text{'}=-3e$
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.

#### Question 21:

There are two coils A and B seperated 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 1 A passes through B, what is the flux through A?

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,

Mutual inductance of coil B with respect to coil A:

But,

#### Question 22:

A magnetic field $\mathbf{B}={B}_{0}\mathrm{sin}\left(\mathrm{\omega }t\right)\stackrel{\mathbf{^}}{\mathbf{k}}$ covers a large region where a wire AB slides smoothly over two parallel conductors separated by a distance 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 v, what is the current in the circuit. What is the force needed to keep the wire moving at constant velocity? As shown in diagram:
Length, AB = d
Velocity of the wire AB is v.
Magnetic field,
$\stackrel{\to }{B}=\left({B}_{0}\mathrm{sin}\omega t\right)\stackrel{^}{k}$
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,

Now current in the wire,
clockwise
So, the required force on the wire AB to keep the wire moving with uniform speed v is:

#### 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 $\mathbf{B}=B\left(t\right)\stackrel{\mathbf{^}}{\mathbf{k}}.$

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

So, flux at time t be, 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.

So, force on the moving wire XY,

$⇒a=\frac{{l}^{2}B\left(t\right)}{mR}\left[-x\left(t\right)\frac{dB\left(t\right)}{dt}-B\left(t\right)\frac{dx\left(t\right)}{dt}\right]$
(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 is independent of time
it means, $\frac{dB}{dt}=0$
So,

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

= Initial kinetic energy – Final kinetic energy
⇒ Heat lost in time t = Decrease in kinetic energy.

#### 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°. Consider the rotation of the wire OP from $\theta =0°$ to 45° as shown in diagram:,
θ = ωt Now, area of ΔODM,

So, flux through this area,
$\varphi =B×\frac{1}{2}{l}^{2}\mathrm{tan}\left(\omega t\right)$
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, Now, for 45° < θ < 135° :
Area ODBQ
$={l}^{2}-\frac{1}{2}×l×\frac{l}{\mathrm{tan}\theta }\phantom{\rule{0ex}{0ex}}={l}^{2}-\frac{1}{2}\frac{{l}^{2}}{\mathrm{tan}\left(\omega t\right)}$

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,

Now, if 135° < θ < 180°: Similarly from the steps taken to derive equation (1)
we get the required current,
${I}_{3}=\frac{1}{2}\frac{Bl\omega sec\left(\omega t\right)}{\lambda }$

#### 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. 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,
(out of paper)
So, flux through the loop ABCD:
$\varphi ={\int }_{{x}_{ο}}^{x}\frac{{\mu }_{ο}I}{2\pi r}×l×dr$        (Area of small element l × dr )

So, emf induced:

So, current through the loop,
$i=\frac{E}{R}=\frac{{\mu }_{ο}l\lambda }{2\pi R}\mathrm{ln}\frac{x}{{x}_{\circ }}$

#### Question 26:

A rectangular loop of wire ABCD is kept close to an infinitely long wire carrying a current I (t) = Io (1 – t/Tfor 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. Flux passing through the loop at a time t can be given as:

Induced, Current through the loop:

Integrating we, get:
$⇒\left|Q\left({t}_{1}\right)-Q\left({t}_{2}\right)\right|=\left|\frac{{\mu }_{ο}{L}_{1}}{2\pi R}\mathrm{ln}\left(\frac{{L}_{2}+x}{x}\right)\left[I\left({t}_{1}\right)-I\left({t}_{2}\right)\right]\right|$
So, the total charge passing through a given point in the loop in time T$\left(0\le t\le T\right)$
$∆Q=\frac{{\mu }_{ο}{L}_{1}}{2\pi R}×\mathrm{ln}\left(\frac{{L}_{2}+x}{x}\right)×\left({I}_{ο}\right)\phantom{\rule{0ex}{0ex}}=\frac{{\mu }_{ο}{L}_{1}{I}_{ο}}{2\pi R}\mathrm{ln}\left(\frac{{L}_{2}+x}{x}\right)$

#### Question 27:

A magnetic field 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†tFind the angular velocity ω of the ring after the field vanishes.

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,
$\left|E\right|=\left|-\frac{d\varphi }{dt}\right|=\left|\pi {a}^{2}×\frac{∆\varphi }{∆t}\right|\phantom{\rule{0ex}{0ex}}=\left|\frac{\pi {a}^{2}\left(B-0\right)}{∆t}\right|\phantom{\rule{0ex}{0ex}}=\frac{\pi {a}^{2}B}{∆t}$
Now, if electric field across the charged ring be E, then, (Here, perimeter of the ring = $2\pi b$)
$2\pi b×E=emf=\frac{\pi {a}^{2}B}{∆t}\phantom{\rule{0ex}{0ex}}⇒E=\frac{\pi {a}^{2}B}{2\pi b∆t}=\frac{B{a}^{2}}{2b∆t}$
Torque on the ring,

Now, change in angular momentum of the ring in time $∆t$ is given by,
$∆L=\tau ×∆t=\frac{BQ{a}^{2}}{2∆t}×∆t=\frac{BQ{a}^{2}}{2}$
But, angular momentum,
L = Moment of inertia × Angular velocity
= I$\omega$

#### 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. Let velocity of the rod at a particular instant t be v.
Angle between magnetic field $\left(\stackrel{\to }{B}\right)$ and the velocity $\left(\stackrel{\to }{v}\right)$ 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

Now, induced current,
$i=\left|\frac{E}{R}\right|=\frac{1}{R}\frac{d\varphi }{dt}=\frac{1}{R}×Bvd\mathrm{cos}\theta$
So, magnetic force on the rod, along the incline

Solving equation (1) and (2) we get:
$mg\mathrm{sin}\theta -\frac{1}{R}{B}^{2}{d}^{2}v{\mathrm{cos}}^{2}\theta =m\frac{dv}{dt}\phantom{\rule{0ex}{0ex}}⇒\frac{dv}{dt}+\frac{1}{mR}{B}^{2}{d}^{2}{\mathrm{cos}}^{2}\theta v=g\mathrm{sin}\theta$
Solving the above linear differential equation, using the end limit we get:
$v=\frac{g\mathrm{sin}\theta }{\left(\frac{{B}^{2}{d}^{2}{\mathrm{cos}}^{2}\theta }{mR}\right)}\left[1-{e}^{\frac{-\left({B}^{2}{d}^{2}{\mathrm{cos}}^{2}\theta \right)t}{mR}}\right]$
This is the required expression for the velocity of the rod.

#### 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. 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

Solving the above differential equation and taking the end limit at t = 0, Q = 0 we get:
$Q=vBCd\left[1-{e}^{\left(\frac{-t}{RC}\right)}\right]$

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

#### 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. v is constant. Switch S is closed at time t = 0. 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:
$-L\frac{dI}{dt}+vBd=IR\phantom{\rule{0ex}{0ex}}⇒L\frac{dI}{dt}+Ir=vBd$
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.

#### Question 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 Bz = Bo (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.

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,

Now, rate of change of flux
i.e., emf induced,

So, induced current in the ring:
$I=\frac{E}{R}=\frac{\pi {l}^{2}{B}_{ο}\lambda v}{R}$
Now, rate of change of gravitational potential energy of the ring,

Energy lost due to the heating effect of electric current per second
$={I}^{2}R={\left(\frac{\pi {l}^{2}{B}_{ο}\lambda v}{R}\right)}^{2}×R$
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
$⇒mgv={\left(\frac{\pi {l}^{2}\lambda {B}_{ο}v}{R}\right)}^{2}×R\phantom{\rule{0ex}{0ex}}⇒v=\frac{mgR}{{\left(\pi {l}^{2}\lambda {B}_{ο}\right)}^{2}}$
This is the required expression of the velocity of the ring falling downwards.

#### Question 32:

A long solenoid ‘S’ has ‘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 mt2 + C.

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,
$B={\mu }_{ο}nI$
Now, area of the smaller solenoid having 'N' turns per unit length be,

So, flux passing through the smaller coil,
$\varphi =NB.A=N\left({\mu }_{ο}nI\right)\left(\pi {b}^{2}\right)$
So, emf induced in the solenoid
$\begin{array}{rcl}\left|E\right|& =& \left|-\frac{d\varphi }{dt}\right|=\frac{d}{dt}N\left({\mu }_{ο}NI\right)\left(\pi {b}^{2}\right)\\ & =& \left(\pi Nn{\mu }_{ο}{b}^{2}\right)\frac{dI}{dt}\end{array}$
It is given that

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

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