# Practice questions for electromagnetic Induction

## Paragraph Based Questions

(A) A metal jumper of mass m can slide without friction along two parallel metal guides directed at an angle ? and separated by the distance a. The guides are connected at the bottom through an uncharged capacitor of capacitance C.

The entire system is placed in an upward magnetic field of induction B
At the initial moment ,the jumper is held at a distance b from the foot of the hump

Question 1

What is the acceleration of the jumper

Question 2
What is the time taken to reach bottom

Question 3
Find the instantaneous EMF as a function of time t

Question 4
What will be the velocity at the bottom
(a)$\sqrt {\frac {2bmg \sin \theta}{m + Ca^2B^2 \cos ^2 \theta}}$
(b)$\sqrt {\frac {2bmg \cos \theta}{m + Ca^2B^2 \cos ^2 \theta}}$
(c)$\sqrt {\frac {2amg}{m + Ca^2B^2 \cos ^2 \theta}}$
(d)$\sqrt {\frac {2bmg}{m + Ca^2B^2 \cos ^2 \theta}}$

During the motion of the Jumper , The magnetic Flux across the loop closed by the Jumper varies and as a result an emf is induced in the Loop
During the short interval of time where velocity of the jumper can be treated as constant ,the instantaneous value of the induced emf is

$E= -\frac {d\phi}{dt}=-vBa \cos \theta$ --(1)

Now current in the circuit at any point of time is given by
$I= \frac {dq}{dt}$

Where dq is the charge stored in the capacitor during the time dt

Now
q=CV
dq=CdV
or $dq= CBa \cos \theta dv$

Therefore Current at any time
$I= \frac {CBa \cos \theta dv}{dt}=CBa \cos \theta \frac {dv}{dt}$ ---(2)

Now Applying Newton's law on Jumper
$m \frac {dv}{dt}= mg \sin \theta - IaB \cos \theta$ ---(2)

Here $IaBcos \theta$ is the magnetic force acting on the jumper due to current
And Current will flow in such a way to oppose the motion

Substituting the value of I from equation (2) in equation (3)
$m \frac {dv}{dt}= mg \sin \theta - (CBa \cos \theta \frac {dv}{dt}) aB \cos \theta$
Or
$\frac {dv}{dt}=\frac {mg \sin \theta}{m + Ca^2B^2 \cos ^2 \theta}$

This is the acceleration of the Jumper

Now Time during which the jumper reach the bottom is given by
$b= \frac {1}{2} \frac {dv}{dt} t^2$
Or
$t_f=\sqrt { \frac {2b(m + Ca^2B^2 \cos ^2 \theta)}{mg \sin \theta}}$

Now velocity as a function of time is given by
$v= u + \frac {dv}{dt} t$

Or
$v=\frac {mg \sin \theta}{m + Ca^2B^2 \cos ^2 \theta} t$

Now induced EMF is given by
$E= -\frac {d\phi}{dt}=-vBa \cos \theta$
Or
$E=-\frac {(mg \sin \theta)(Ba \cos \theta)}{m + Ca^2B^2 \cos ^2 \theta} t$
Now velocity at the bottom is given
$v = \frac {dv}{dt} t_f = \sqrt {\frac {2bmg \sin \theta}{m + Ca^2B^2 \cos ^2 \theta}}$

(B) A square loop of wire of side (b) lies on table ,at a distance s from a very long straight wire which carries a current I

Question 5
Find the flux of magnetic field through the loop
(a) $\frac{\mu_0Ib}{4\pi}\frac{s+b}{s}$
(b) $\frac{\mu_0Ib}{2\pi}(\frac{s+b}{s})$
(c) $\frac{\mu_0Ib}{2\pi}ln{\frac{s}{s+b}}$
(d) $\frac{\mu_0Ib}{2\pi}ln{\frac{s+b}{s}}$

Let us assume a small strip of width dx at a distance x from the wire
Magnetic field at distance x is given by
$B=\frac{\mu_0I}{2\pi x}$

So, flux through this small strip
$d\varphi=\frac{\mu_0I}{2\pi x}bdx$
Total flux will be given by

$\varphi=\int_{s}^{s+b}{\frac{\mu_0Ib}{2\pi x}dx=\frac{\mu_0Ib}{2\pi}ln{\frac{s+b}{s}}}$

Question 6
If the loop is now pulled away from wire at a constant speed ,what emf will be generated in the loop
(a) $\frac{\mu_0Ib^2v}{2\pi}$
(b) $\frac{\mu_0Ib^2v}{2\pi bs}$
(c) $\frac{\mu_0Ib^2v}{2\pi b(s+b)}$
(d) $\frac{\mu_0Ib^2v}{2\pi b(s-b)}$

$E=-\frac{d\varphi}{dt}=-\frac{d}{dt}[\frac{\mu_0Ib}{2\pi}ln{\frac{s+b}{s}}]$
Or
$=-\frac{\mu_0Ib}{2\pi}[\frac{1}{s+b}\frac{ds}{dt}-\frac{1}{s}\frac{ds}{dt}]$
$=-\frac{\mu_0Ib^2v}{2\pi b(s+b)}$

Question 7
What if the loop is pulled to right at speed v ,instead of away. What is the induced EMF
(a) $\frac{\mu_0Ib^2v}{2\pi bs}$
(b) $\frac{\mu_0Ib^2v}{2\pi b(s-b)}$
(c) 0
(d) $\frac{\mu_0Ib^2v}{2\pi b(s+b)}$

As no flux changes in this case,So induced EMF would be zero

## Multiple Choice Questions

Question 8
The rectangular loop shown in below figure moves towards the origin at a velocity
$v=-250 \mathbf{j}$ m/s
There exists a Magnetic field in the region such that
$B=.80e^{-.50y}\mathbf{k}$

Find the current at the instant the coil sides are at y=.50m and .60m.
The value of R =2.5 ohm
(a) 3.04A
(b)2 A
(c)3 A
(d) 0

The only 1.0m side have induced voltages
Let the side at y=.50 m be 1
Then
Induced emf on side 1
$E=BvL=.80e^{-.5 \times.5}(250)(1)=155.8$V

Induced emf on side 2
$E=BvL=.80e^{-.5 \times .6}(250)(1)=148.2$V
Both the EMF induced will have same polarity. The instantaneous current
$i=\frac{155.8-148.2}{2.5}=3.04$A

Question 9
A square loop of side L ,Mass M and total resistance R falls vertically into uniform magnetic field B directed perpendicular to the plane of the coil.The height h through which the loop falls so that it attain terminal velocity on entering the region of magnetic field
(a) $\frac{MgR^2}{4B^3L^3}$
(b) $\frac{M^2gR^2}{2B^4L^4}$
(c) $\frac{M^2gR^2}{2B^2L^2}$
(d) $\frac{MgR}{2BL}$

Velocity after falling from height h
$v=\sqrt{2gh}$

Induced EMF
$E=BL\sqrt{2gh}$

Induced current will be given by
$I=\frac{BL\sqrt{2gh}}{R}$

The force due to current
$F=BIL=\frac{B^2L^2\sqrt{2gh}}{R}$

The loop will attain terminal velocity if this force equals mg i.e if
$\frac{B^2L^2\sqrt{2gh}}{R}=Mg$

Or
$h=\frac{M^2gR^2}{2B^4L^4}$

## Paragraph Based Questions

(C) A metal disk (radius =b and thickness =w) is placed in a solenoid with its axis coincident with the axis of the solenoid. The solenoid produces a magnetic field
$B=B_0sin{2}\pi ft$
Consider a small dotted ring of radius r as shown in fig

Question 10
calculate the EMF induced in the small ring
(a) $2\pi^2r^2fB_0sin{2}\pi ft$
(b) $2\pi r^2fB_0cos{2}\pi ft$
(c) $2\pi^2r^2B_0cos{2}\pi ft$
(d) $2\pi^2r^2fB_0cos{2}\pi ft$

Now
$E=-\frac{d\varphi}{dt}$

$\varphi=B.A=(B_0sin{2}\pi ft)(\pi r^2)$

So
$E=-2\pi^2r^2fB_0cos{2}\pi ft$

Question 11
If $\rho$ is the resistivity of the material and width of the ring is dr, what is the value of eddy current in the dotted ring
(a) $\frac{\pi f B_0wrdr}{\rho}$
(b) $\frac{\pi f B_0w(sin{2}\pi ft)rdr}{\rho}$
(c) $\frac{fB_0w(cos{2}\pi ft)rdr}{\rho\pi}$
(d) $\frac{\pi f B_0w(cos{2}\pi ft)rdr}{\rho}$

Resistance of the small dotted ring

$R=\frac{\rho L}{A}=\frac{\rho2\pi r}{wdr}$

Current is given by

$I=\frac{E}{R}=\frac{\pi f B_0w(cos{2}\pi ft)rdr}{\rho}$

Question 12
Find the total power loss due to eddy current in the disks
(a) $\frac{\pi^3wb^4f^2}{2\rho}{cos}^2{2}\pi ft$
(b) $\frac{\pi^3wb^4f^2}{\rho}{cos}^2{2}\pi ft$
(c) $\frac{\pi^3wf^2}{2\rho}{cos}^2{2}\pi ft$
(d) $\frac{\pi^3wb^4f^2}{2\rho}$

For small ring
$dP=I^2R$
Or
$dP=\frac{2\pi^3B_0^2f^2w{cos}^2{2}\pi ft}{\rho}r^3dr$
Total power
$P=\int_{0}^{b}{dP=\frac{\pi^3wb^4f^2B_0^2}{2\rho}}{cos}^2{2}\pi ft$

(D) A long solenoid of Radius r has n turns and it carries an alternating current
$I=I_0sin{\omega}t$
A concentric circular wire of Radius R( R> r) surrounds the solenoid

Question 13
Find the induced EMF generated in the wire
(a) $\mu_0\pi\omega nI_0r^2cos{\omega}t$
(b) $\mu_0\pi\omega nI_0R^2cos{\omega}t$
(c) $\mu_0\pi\omega nI_0R^2sin{\omega}t$
(d) $\mu_0\pi\omega nI_0(R-r)^2cos{\omega}t$

Outside the solenoid ,magnetic field is zero and it is along the axis inside the solenoid
And it is given by
$B=\mu_0nI$

So flux through the wire
$\varphi=B.A=(\mu_0nI)(\pi r^2)=\mu_0\pi nI_0r^2sin{\omega}t$
Induced EMF
$E=-\frac{d\varphi}{dt}=-\mu_0\pi\omega nI_0r^2cos{\omega}t$

Question 14
Find the induced Electric field at any point inside the wire
(a) $\frac{\mu_0\omega n I_0R^2cos{\omega}t}{2r}$
(b) $\frac{\mu_0\omega n I_0r^2cos{\omega}t}{R}$
(c) $\frac{\mu_0\omega n I_0r^2sin{\omega}t}{2R}$
(d) $\frac{\mu_0\omega n I_0r^2cos{\omega}t}{2R}$

Induced Electric field
$E=\frac{InducedEMF}{2\pi R}=\frac{\mu_0\omega n I_0r^2cos{\omega}t}{2R}$

(E)A rod AB of length L is placed along x -axis and it is pivoted at origin O such that A(-a,0) and B(b,0) . The magnetic field in the region is
$B=-B_0\mathbf{k}$
The rod is rotated in the x -y plane around the z axis with angular velocity $\omega$ in anti clockwise direction
Question 15
Find the polarity of the ends A and B
(a) Both are positive
(b) Both are negative
(c) A is positive and B is negative
(d) A is negative and B is positive
Question 16
Find the induced EMF between Ab
(a) $V_A-V_B=B_0\omega(b^2-a^2)$
(b) $V_A-V_B=\frac{1}{2}B_0\omega(a^2-b^2)$
(c) $V_A-V_B=\frac{1}{2}B_0\omega(b^2-a^2)$
(d) None of these

Motional emf =vBdl and since the rod is rotated in anticlockwise direction,As per lenz rule
For the part OB, polarity will O(+) and B(-)
For the part AO, polarity will O(+) and A(-)

Now Assume a small element dx at a distance x from origin
Induced EMF in the region
$e=\omega xB_0dx$
Total EMF for the region
$E_{OB}=\int_{0}^{b}{\omega x B_0dx=\frac{1}{2}B_0\omega b^2}$

Similary,
$E_{AO}=\frac{1}{2}B_0\omega a^2$
Taking polarity under consideration
$V_O-V_B=\frac{1}{2}B_0\omega b^2$
$V_O-V_A=\frac{1}{2}B_0\omega a^2$
So
$V_A-V_B=\frac{1}{2}B_0\omega(b^2-a^2)$

(F)A square loop side a and resistance R lies a at distance a from infinte straight wire that carries current I. The current varies as
$I(t)=(1-\alpha t)$ for 0$\le t\le1/\alpha$
$I(t)=0$ for $t>1/\alpha$

Question 17
Find the induced EMF in the loop
(a) $\frac{\mu_0a}{2\pi}\alpha$
(b) $\frac{\mu_0aln{2}}{\pi}\alpha$
(c) $\frac{\mu_0aln{a}}{2\pi}\alpha$
(d) $\frac{\mu_0aln{2}}{2\pi}\alpha$
Question 18
Find the induced current in the loop
(a) $\frac{\mu_0a\alpha}{2\pi R}$
(b) $\frac{\mu_0a\alpha l n{a}}{2\pi R}$
(c) $\frac{\mu_0a\alpha l n{2}}{\pi R}$
(d) $\frac{\mu_0a\alpha l n{2}}{2\pi R}$
Question 19
what total charge passes a given point in the loop during the time current flows
(a) $\frac{\mu_0aln{2}}{\pi R}$
(b) $\frac{\mu_0aln{4}}{2\pi R}$
(c) $\frac{\mu_0aln{2}}{2\pi R}$
(d) $\frac{\mu_0aln{2}}{R}$

Let us assume a small strip of width dx at a distance x from the wire
Magnetic field at distance x is given by
$B=\frac{\mu_0I}{2\pi x}$

So flux through this small strip
$d\varphi=\frac{\mu_0I}{2\pi x}adx$
Total flux will be given by

$\varphi=\int_{a}^{2a}{\frac{\mu_0Ia}{2\pi x}dx=\frac{\mu_0Ia}{2\pi}ln{2}}$

Now induced EMF
$E=-\frac{d\varphi}{dt}=-\frac{\mu_0aln{2}}{2\pi}\frac{dI}{dt}=\frac{\mu_0aln{2}}{2\pi}\alpha$

Now induced current

$I=\frac{E}{R}=\frac{\mu_0a\alpha l n{2}}{2\pi R}$

$\frac{dQ}{dt}=\frac{\mu_0a\alpha l n{2}}{2\pi R}$
Or
$Q=\int_{0}^{1/\alpha}{\frac{\mu_0a\alpha l n{2}}{2\pi R}dt=\frac{\mu_0aln{2}}{2\pi R}}$

(G) Conductors are joined together to form a circuit as shown below in figure.The resistance of a unit length of conductor is r.

The arrangement is placed in a magnetic field which is perpendicular to the plane of the circuit. And acting downward
The Magnetic field is varying with time as per the below expression
$B=B_0t$
Question 20
Find the current in the AD branch
(a) $\frac{3B_0b}{11r}$
(b) $\frac{7B_0b}{11r}$
(c) $\frac{B_0b}{22r}$
(d) $\frac{7B_0b}{22r}$
Question 21
Find the current in the BC branch
(a) $\frac{3B_0b}{11r}$
(b) $\frac{7B_0b}{11r}$
(c) $\frac{B_0b}{22r}$
(d) $\frac{7B_0b}{22r}$
Question 22
Find the current in the EF branch
(a) $\frac{3B_0b}{11r}$
(b) $\frac{7B_0b}{11r}$
(c) $\frac{B_0b}{22r}$
(d) $\frac{7B_0b}{22r}$

The emf of induction in the loop ABCD
$E_1=B_0b^2$

Similary the emf of induction in the loop BEFC
$E_2=\frac {B_0b^2}{2}$

These EMF will acts with accordance to the Lenz rule and will oppose change in flux
The circuit can be arranged with galvanic cell as the EMF of the induction as below

One the basis of ohm's law
$I_3br=E_1-I_13br$ --(1) $I_3br=I_22br-E_2$ --(2)

Also applying kirchoff law at the junction

$I_1=I_2+I_3$ --(3)

Solving equation (1), (2) and (3)
We get
$I_1=\frac{6E_1+2E_2}{22br}$
$I_2=\frac{2E_1+8E_2}{22br}$

$I_3=\frac{2E_1-3E_2}{11br}$

Substituting the values of $E_1$ and $E_2$ from above
$I_1=\frac{7B_0b}{22r}$
$I_2=\frac{3B_0b}{11r}$
$I_3=\frac{B_0b}{22r}$

## Multiple Choice questions

Question 23
Two concentric coplanar circular loops made of wire with resistance per unit length $10^{-4}$ O/m have diameters .2 m and 2 m respectively.
A time varying potential difference $V=(4+2.5 t)$ V is applied to larger loop.
Find the induced current in the small loop
(a) 3 A
(b) 2 A
(c) 1 A
(d) 1.25 A

The resistance of the Larger loop
$=2\pi R \times 10^{-4}=2\pi \times 1 \times 10^{-4}=2\pi10^{-4}$
Current in the larger loop will be
$I=\frac{V}{R}=\frac{(4+2.5t)}{2\pi10^{-4}}$

The magnetic Field due to this current in the larger loop at the common center will be
$B=\frac{\mu_0I}{2R}=\frac{\mu_0(4+2.5t)}{2*1*2\pi10^{-4}}=\frac{\mu_0(4+2.5t)}{4\pi10^{-4}}$

Since the smaller loop is very small compared to bigger loop,the magnetic field at the center can be assumed to be the magnetic field in the smaller loop
So flux through smaller loop
$\varphi=BA=\frac{\mu_0(4+2.5t)}{4\pi10^{-4}}(\pi10^{-2)}=\frac{\mu_0(4+2.5t)}{4}10^2$

Induced EMF in the smaller loop
$E=\frac{d\varphi}{dt}=62.5\mu_0$

Now resistance of the smaller loop
$=2\pi R*10^{-4}=2\pi*.1*10^{-4}=2\pi10^{-5}$
So induced current $I=\frac{E}{R}=\frac{62.5\mu_0}{2\pi10^{-5}}=1.25A$

Question 24
A coil PQR of Radius 'a' carries a current I and it is placed on the XY plane with origin as center .A small conducting ring ABC of radius b ( a> >>b) is also XY plane with center at $z=z_0$. The ring ABC is allowed to fall free over the ring PQR with velocity v

Find the induced EMF in the ring ABC as function of z
(a) $\frac{3\mu_0\pi I a^2b^2zv}{2(a^2+z^2)^{5/2}}$
(b) $\frac{3\mu_0\pi I a^2b^2zv}{2(a^2+z^2)^{3/2}}$
(c) $\frac{3\mu_0\pi I a^2zv}{2(a^2+z^2)^{5/2}}$
(d) $\frac{3\mu_0\pi I b^2zv}{2(a^2+z^2)^{5/2}}$

Magnetic Field due to coil on the axis is given by
$B=\frac{\mu_0IR^2}{2(R^2+x^2)^{3/2}}$

So Magnetic field at a distance z from coil PQR will be
$B=\frac{\mu_0Ia^2}{2(a^2+z^2)^{3/2}}$

Since the size of the coil ABC is very small,it may be assumed that magnetic field through it is uniform and equal to that on the axis

So flux through ABC
$\varphi=B.\pi b^2=\frac{\mu_0\pi I a^2b^2}{2(a^2+z^2)^{3/2}}$

Induced EMF
$E=\frac{d\varphi}{dt}=\frac{\mu_0I\pi a^2b^2}{2}\frac{d}{dt}\frac{1}{(a^2+z^2)^{3/2}}$
$=\frac{3\mu_0\pi I a^2b^2}{4(a^2+z^2)^{5/2}}2z\frac{dz}{dt}=\frac{3\mu_0\pi I a^2b^2zv}{2(a^2+z^2)^{5/2}}$

Question 25
A wire frame of Area S and resistance R is suspended freely from a thread of length L. There is a uniform magnetic field $B_0$ existing in space and plane of the wire frame is perpendicular to the magnetic field. The wire frame is made to oscillate under the force of gravity by displacing it through distance A from it initial position along the direction of the magnetic field.It is given A <<< Find the induced EMF in the wire frame as a function of time.
(a) $\frac{BSA^2\omega}{L^2}sin{2}\omega t$
(b) $\frac{BSA^2\omega}{2L^2}cos{\omega}t$
(c) $\frac{BSA^2\omega}{2L^2}sin{2}\omega t$
(d) None of these

The frame will oscillate Simple harmonic motion.
The equation of simple harmonic motion is given by
$x=Asin{(\omega t+\delta)}$

Assuming that frame is at its equilibrium position at time t=0
We have
$x=Asin(\omega t)$ ---(1)
Where
$\omega=\sqrt{\frac{g}{L}}$

Now the flux linked through wire frame at any instant is given by
$\varphi=BS cos{\alpha}$
Where $\alpha$ is the angle made by the wire at time t
$E=\frac{d\varphi}{dt}=BSsin{\alpha}\frac{d\alpha}{dt}$

Now as a is small we have $sin \alpha = \alpha$

(b)