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}}$
Solutions 1-4
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) $\frac{\mu_0Ib}{2\pi}ln{\frac{s+b}{s}}$
Solution
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) $\frac{\mu_0Ib^2v}{2\pi b(s-b)}$
Solution
$E=-\frac{d\varphi}{dt}=-\frac{d}{dt}[\frac{\mu_0Ib}{2\pi}ln{\frac{s+b}{s}}]$
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) $\frac{\mu_0Ib^2v}{2\pi b(s+b)}$
Solution
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
Solution
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) $\frac{MgR}{2BL}$
Solution
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) $2\pi^2r^2fB_0cos{2}\pi ft$
Solution
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) $\frac{\pi f B_0w(cos{2}\pi ft)rdr}{\rho}$
Solution
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) $\frac{\pi^3wb^4f^2}{2\rho}$
Solution
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) $\mu_0\pi\omega nI_0(R-r)^2cos{\omega}t$
Solution
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) $\frac{\mu_0\omega n I_0r^2cos{\omega}t}{2R}$
Solution
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
Solution 15-16
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) $\frac{\mu_0aln{2}}{R}$
Solution 17-19
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) $\frac{7B_0b}{22r}$
Solution 20-22
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
Solution
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) $\frac{3\mu_0\pi I b^2zv}{2(a^2+z^2)^{5/2}}$
Solution
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
Solution
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 $sin \alpha = \alpha$
Also $\alpha=\frac {x}{L}
So
$E=BS\left(\frac{x}{L}\right)\frac{d}{dt}\left(\frac{x}{L}\right)=\frac{BSx}{L^2}\frac{dx}{dt}$
From equation (1)
$E=\frac{BS}{L^2}Asin{\omega}t\frac{d}{dt}Asin{\omega}t=\frac{BSA^2\omega}{2L^2}sin{2}\omega t$
Question 26
A ring of a rectangular cross-section is made of a material whose resistivity is $\rho$. The ring is placed in a magnetic field . The magnetic field is directed along the axis of the ring and increase directly with time
$B=B_0t$.
Find the induced current in the ring
(a) $\frac{B_0h}{4\rho}b^2$
(b) $\frac{B_0h}{2\rho}(b^2-a^2)$
(c) $\frac{B_0hab}{4\rho}$
(d) $\frac{B_0h}{4\rho}(b^2-a^2)$
Solution
Let us take a small ring of width dr at a distance r from the center.
The resistance of the small ring is
$R=\rho\frac{2\pi r}{hdr}R$
The EMF induced in this small ring due to changing magnetic field
$E=\frac{d\varphi}{dt}=B_0\pi x^2$
So current in this ring
$I=\frac{E}{R}=\frac{B_0hxdx}{2\rho}$
Total current through entire ring
$I=\int_{a}^{b}\frac{B_0hxdx}{2\rho}=\frac{B_0h}{4\rho}(b^2-a^2)$
Question 27
A rectangular loop is being pulled at a constant speed v, through a region of certain thickness d, in which a uniform magnetic field B is set up. The graph between position x of the right hand edge of the loop and the induced emf E will be
Solution
As x increase so $\frac {dB}{dt}$ increase i.e., induced emf (e) is negative. When loop completely entered in the magnetic field, emf=0
when it exit out x increase but $\frac {dB}{dt}$ decrease i.e., e is positive.
So correct option is (c)
Question 28
A conducting square loop of side l and resistance R moves in its plane with a uniform velocity v perpendicular to one of its sides. A magnetic induction B constant in time and space, pointing perpendicular and into the plane at the loop exists everywhere with half the loop outside the field, as shown in figure. The induced e.m.f. is
(a) vlB
(b) vlB/2
(c) 2vlB
(d) zero
Solution
Correct answer is (a)
Question 29
The coxial cable has an inner wire of Radius $R_1$ and an outer metal sheath with inner radius radius $R_2$ .The current I flows down the inner wire and back through the sheath.
What is the self inductance of the unit length of the cable
(a) $\frac{\mu_0}{2\pi}\frac{R_2}{R_1}$
(b) $\frac{\mu_0}{2\pi}\frac{R_1}{R_2}$
(c) $\frac{\mu_0}{2\pi}ln{\frac{R_2}{R_1}}$
(d) None of these
Solution
Between the inner wire and sheath only the inner wire contributes to B.
And we know Magnetic field due to inner wire is given by
$B=\frac{\mu_0I}{2\pi r}$
We consider a rectangular area between the surface of the inner wire and outer sheath and parallel to cable axis. We break this area into long parallel strips of width dr and length ds. Flux through sdr is Bsdr and total flux between $r=R_1$ and $r=R_2$ per unit length is given by
$\varphi=\frac{1}{s}\int_{R_1}^{R_2}{\frac{\mu_0Is}{2\pi r}=\frac{\mu_0I}{2\pi}ln{\frac{R_2}{R_1}}}$
Now self inductance is given by
$L=\frac{\varphi}{I}=\frac{\mu_0}{2\pi}ln{\frac{R_2}{R_1}}$
Question 30
Two neighboring coils X and Y have have 300 and 600 turns respectively .
It is find that the current 1.5 A in X causes $1.2 \times 10^{-4}$ wb to pass in X and $.9 \times 10^{-4}$ wb through Y
Which of the following is correct
(a) Mutual inductance of the coils X and Y is 36 mH
(b) Self inductance of the coil X is 24 mH
(c) Self inductance of the coil Y is 24 mH
(d) none of these
Solution
Self inductance is given by
$I_iL_i=N_i\varphi_i$
v
So self inductance for coil X
$L=\frac{300*1.2*10^{-4}}{1.5}=24mH$
We cannot find self inductance of coil B as data is insufficient
Now
Mutual inductance is given by
$I_xM=N_y\varphi_y$
So M will be
$M=\frac{600*.9*10^{-4}}{1.5}=36mH$
Question 31
A long solenoid X of diameter .1 m has $2 \times 10^4$ turns per meter .At the center of the solenoid a 100 turn coil Y of radius .01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid is decreased at a constant rate from +2A to -2A in .05 sec.The resistance of the coil is $10\pi^2\Omega$
Which of the following is true
(a) The EMF induced in the coil is $6.3 \times 10^{-2}$V
(b) The mutual inductance of the system is H
(c) The current flowing in the coil is $5 \times 10^{-4}$ A
(d) The flux in the small coil at current 2A is Weber
Solution
The mutual inductance of the solenoid and coil system is given by
$M=\mu_0n_1N_2A_2$
Where $n_1$ -> Number of turns per unit length in outer solenoid
$N_2$ -> Number of turns in the small inner solenoid
$A_2$ -> Area of the small solenoid
So here in this case
Substituting all the values
$M=8\pi^2 \times 10^{-5}$H
Now change in current
$\Delta I=-2-2=-4$A
$\Delta t=.05$
Now induced Emf is given by
$E=-M\frac{dI}{dt}$
Substituting the values
$E=6.3 \times 10^{-2}$V
Therefore current in the coil will be given by
$I=\frac {E}{R}$
Substituing the values from above
$I=6.4 \times 10^{-4}$ A
Now
Flux in the small can be calculated using Mutual inductance
$\varphi=MI$
Or
$\varphi=2 \times 8\pi^2 \times 10^{-5}=16\pi^2 \times 10^{-5}$ Weber
Question 32
The mutual inductance between the rectangular loop and long straight wire
(a) $\frac{\mu_0a}{2\pi}ln{(}1+\frac{b}{c})$
(b) $\frac{\mu_0a}{2\pi}ln{(}1+\frac{c}{b})$
(c) $\frac{\mu_0a}{2\pi}ln{\frac{b}{c}}$
(d) $\frac{\mu_0a}{2\pi}ln{\frac{c}{b}}$
Solution
We suppose that straight wire is carrying current I downward.
Now we know that Magnetic Field due to long straight wire at a distance x is given by
$B=\frac{\mu_0I}{2\pi x}$
Let us consider a small strip of width dx at a distance x from the long wire on the rectangular loop.
The flux through small strip will be given by
$d\varphi=\frac{\mu_0I}{2\pi x}adx$
Total flux through rectangular loop will be given by
$\varphi=\int_{c}^{c+b}{\frac{\mu_0I}{2\pi x}adx=\frac{\mu_0Ia}{2\pi}ln{(}1+\frac{b}{c})}$
Now mutual inductance is given by
$M=\frac{\varphi}{I}=\frac{\mu_0a}{2\pi}ln{(}1+\frac{b}{c})$
Question 33
A wire in the form of a circular loop of Diameter D lies with its plane normal to magnetic field B. The wire is pulled to take a equilateral shape in same plane in 1 sec.Find the Emf induced in the loop
(a) $E=\frac{B\pi D^2}{4}\left(1-\frac{\pi}{4}\right)$
(b) $E=\frac{B\pi D^2}{4}\left(1-\frac{\pi\sqrt3}{4}\right)$
(c) $E=\frac{B\pi D^2}{4}\left(1-\frac{\pi\sqrt3}{2}\right)$
(d) $E=\frac{B\pi D^2}{4}\left(1-\frac{\pi\sqrt3}{8}\right)$
Solution
Since circular loop is pulled to make triangle,
Perimeter of Circle=Perimeter of Triangle
$\pi D=3a\bigma=\frac{\pi D}{3}$
Now Area of circular Loop
$A_1=\pi\left(\frac{D}{2}\right)^2=\frac{\pi D^2}{4}$
Area of equilateral triangle
$A_2=\frac{\sqrt3}{4}a^2=\frac{\sqrt3}{4}\left(\frac{\pi D}{3}\right)^2=\frac{\pi^2D^2\sqrt3}{36}$
Now induced Emf
$E=B(A_1-A_2)$
$E=\frac{B\pi D^2}{4}\left(1-\frac{\pi\sqrt3}{4}\right)$
Question 34
A circular coil of Diameter D of metal wire is kept stationary in a non uniform magnetic field given by
$\mathbf{B}=B_x\mathbf{i}+B_y\mathbf{j}+B_z\mathbf{k}$
Let E and I be the Induced Emf and current in the coil
Statement 1: $E \ne 0$ ,$I \ne 0$
Statement 2: E=0 ,I=0
Statement 3: $E \ne 0$ , I=0
Statement 4: E=0 , $I \ne 0$
Which of the following is true
(a) Statement 2 is true
(b) Statement 3 is true
(c) statement 1 is true
(d) statement 4 is true
Solution
Since flux does not change with time
No induced Emf generated.
So a is true
Link Type comphrehension
An alternating current flows down a straight wire which runs along the axis of a toroidal coil with rectangular crosssection of inner radii a ,outer radii b ,height h and N turns) The coil is connected to the R ohm resistor
Question 35
What is the EMF induced in the torroid coil
(a) $\frac{\mu_0Nh\omega}{2\pi}I_0sin{\omega}t$
(b) $\frac{\mu_0Nh\omega}{2\pi}ln{\left(\frac{b}{a}\right)}I_0cos{\omega}t$
(c) $\frac{\mu_0Nh\omega}{2\pi}ln{\left(\frac{a}{b}\right)}I_0sin{\omega}t$
(d) $\frac{\mu_0Nh\omega}{2\pi}ln{\left(\frac{b}{a}\right)}I_0sin{\omega}t$
Solution
The magnetic Field at the distance s from long straight wire is given by
$B=\frac{\mu_0I}{2\pi s}$
Let us consider a small rectangular strip of width ds at a distance s from center on the rectangular crosssection of the toroid
Then
$d\varphi=\frac{\mu_0I}{2\pi s}hds$
Total flux
$\varphi=\int_{a}^{b}{\frac{\mu_0I}{2\pi s}hds=\frac{\mu_0Ih}{2\pi}ln{\frac{b}{a}}}$
Now we have N turns ,So total flux
$\varphi=\frac{\mu_0IhN}{2\pi}ln{\frac{b}{a}}$
Now $I=I_0cos{\omega}t$
So
$\varphi=\frac{\mu_0I_0(cos{\omega}t)hN}{2\pi}ln{\frac{b}{a}}$
Now EMF induced is
$E=-\frac{d\varphi}{dt}=\frac{\mu_0Nh\omega}{2\pi}ln{\left(\frac{b}{a}\right)}I_0sin{\omega}t$ ---(1)
Question 36
Find the Self Inductance of the toriod
(a) $L=\frac{\mu_0N^2h}{2\pi}ln{\frac{b}{a}}$
(b) $L=\frac{\mu_0N^2h}{2\pi}ln{\frac{q}{b}}$
(c) $L=\frac{\mu_0N^2h}{2\pi}$
(d) $L=\frac{\mu_0N^2}{2\pi}ln{\frac{b}{a}}$
Solution
The magnetic Field inside the toriod is given by
$B=\frac{\mu_0NI}{2\pi s}$
The flux through a single turn
$\varphi=\int_{a}^{b}{Bhds=\frac{\mu_0NIh}{2\pi}ln{\frac{b}{a}}}$
Total flux is N times,So self inductance would be
$L=\frac{\mu_0N^2h}{2\pi}ln{\frac{b}{a}}$
Question 37
Find the back EMF induced in the torriod due to changing current in the wire
(a) $\frac{\mu_0^2N^3h^2\omega I_0}{4\pi^2R}\left(ln{\frac{b}{a}}\right)^2cos{\omega}t$
(b) $\frac{\mu_0^2N^3h^2I_0}{4\pi^2R}\left(ln{\frac{b}{a}}\right)^2sin{\omega}t$
(c) $\frac{\mu_0^2N^3h^2\omega I_0}{4\pi^2R}\left(ln{\frac{b}{a}}\right)^2sin{\omega}t$
(d) $\frac{\mu_0^2N^3h^2\omega I_0}{4\pi^2}\left(ln{\frac{b}{a}}\right)^2sin{\omega}t$
Solution
From equation (1) ,induced EMF
$E=-\frac{d\varphi}{dt}=\frac{\mu_0Nh\omega}{2\pi}ln{\left(\frac{b}{a}\right)}I_0sin{\omega}t$
So induced current in the toriod
$I=\frac{E}{R}=\frac{\mu_0Nh\omega}{2\pi R}ln{\left(\frac{b}{a}\right)}I_0sin{\omega}t$
Now this current I changing,So Back EMF generated will be
$E_b=-L\frac{dI}{dt}=-\frac{\mu_0N^2h}{2\pi}ln{\frac{b}{a}}\frac{\mu_0Nh\omega}{2\pi R}I_0sin{\omega}tln{\frac{b}{a}}$
$E_b=\frac{\mu_0^2N^3h^2\omega I_0}{4\pi^2R}\left(ln{\frac{b}{a}}\right)^2sin{\omega}t$
Multiple Choice Questions
Question 38
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
Solution
(a), (b), (c)
Question 39
A coil of wire having finite inductance and resistance has a conducting ring placed coaxially within it. The coil is connected to a battery at time t = 0, so that a time-dependent current $I_1(t)$ starts flowing through the coil. If $I_2(t)$ is the current induced in the ring. and B(t) is the magnetic field at the axis of the coil due to $I_1(t)$, then as a function of time (t > 0), the product $I_2 (t) B(t)$
(a) Increases with time
(b) Decreases with time
(c) Does not vary with time
(d) Passes through a maximum
Solution
Correct option is (d)
Using $k_1$,$k_2$ etc, as different constants.
$I_1(t)=k_1[1-e^{-t/\tau}]$,$B(t)=k_2I_1(t)$
Now $I_2(t)$
$I_2(t)=k_3 \frac {dB(t)}{dt}=k_4 e^{-t/\tau}$
$I_2(t)B(t)=k_5[1-e^{-t/\tau}][e^{-t/\tau}]$
This quantity is zero for t=0 and t=$\infty$ and positive for other value of t. It must, therefore, pass through a maximum.
Question 40
Two circular coils P & Q are fixed coaxially & carry currents $I_1$ and $I_2$ respectively
(a) if $I_2$= 0 & P moves towards Q, a current in the same direction as $I_1$ is induced in Q
(b) if $I_1$= 0 & Q moves towards P, a current in the opposite direction to that of $I_2$ is induced in P.
(c) when $I_1 \ne 0$ and $I_2 \ne 0$ are in the same direction then the two coils tend to move apart .
(d) when $I_1 \ne 0$ and $I_2 \ne 0$ are in opposite directions then the coils tends to move apart.
Solution
(b)
Class 12 Maths
Class 12 Physics