MCQ Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 14

A flat coil of 500 turns, each of area

$50 cm^2$ , rotates in a uniform magnetic field of

$0.14 Wb/m^2$ about an axis normal to the field at an angular speed of

$150 rad/s$ . The coil has a resistance of

$5\Omega$ . The induced e.m.f. is applied to an external resistance of

$10\Omega$ . The peak current through the resistance is

0%
1.5 A
0%
2.5 A
0%
3.5 A
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4.5 A

Explanation

Hint: Write total flux through the loop and then use faraday law.

Solution:

Step-1: Find the induced e.m.f

For a coil of $\mathrm{N}$ number of turns and area $\mathrm{A}$ , rotating with the speed $\omega \mathrm{rad} / \mathrm{sec}$ in a magnetic field $\mathrm{B}$ the induced flux, $\phi=N B A \sin \omega t$

$\therefore$ Induced e.m.f $(\varepsilon)=-\frac{d \phi}{d t}=-N B A \omega \cos \omega t$

Step-2: Find the peak current

Induced e.m.f $(\varepsilon)=-N B A \omega \cos \omega t$

$\therefore$ Peak value of e. $m \cdot f=N B A \omega$

Therefore peak current through the resistance

$=\frac{\text { Peak e.m.f }}{\text { Total resistance }}=\frac{N B A \omega}{5+10}$ $\left(\begin{array}{l}\text { Given, } \mathrm{N}=500 \\ \mathrm{~B}=0.14 \mathrm{~Wb} / \mathrm{m}^{2} \\ A=50 \mathrm{~cm}^{2}=50 \times 10^{-4} \mathrm{~m}^{2} \\ \omega=150 \mathrm{rad} / \mathrm{sec}\end{array}\right)$

$=\frac{500 \times 0.14 \times\left(50 \times 10^{-4}\right) \times 150}{15}$

$=3.5 \mathrm{~A}$

Two coils X and Y are placed in a circuit such a current changes by 2 A in coil X and magnetic flux changes by 0.4 Wb in Y. The value of mutual inductance of the coils and its units are

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0.2 H
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5 H
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0.8 H
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0.4 H

The magnetic flux in a coil of 100 turns increases by

$12 \times { 10 }^{ 3 }$ Maxwell in 0.2 s due to the motion of a magnet The emf induced in the coil will be :-

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0.6 mV
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0.6 V
0%
6 V
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60 V

A circular arc of wire of radius of curvature r subtends an angle of

$\frac {\pi}{4}$ radian at its center. If i current is flowing in it then the magnetic induction at its centre will be:

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$\frac {\mu_o i}{8r}$
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$\frac {\mu_o i}{4r}$
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$\frac {\mu_o i}{16r}$
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zero

shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction

$B={ B }_{ e }{ e }^{ r }$ is established inside the coil.If the key (K) is closed, the electrical power developed right after the switch is equal to:-

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$\frac { { B }^{ 2 }{ \pi r }^{ 2 } }{ R }$
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$\frac { { B }_{ 0 }{ 10r }^{ 3 } }{ R }$
0%
$\frac { { B }_{ 0 }^{ 2 }{ \pi }^{ 2 }{ r }^{ 4 }R }{ 5 }$
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$\frac { { B }_{ 0 }^{ 2 }{ \pi }^{ 2 }{ r }^{ 4 } }{ R }$

A long straight wire is placed along the axis of a circular ring of radius R. The mutual inductance of this system is

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$\frac { { \mu }_{ 0 } R }{ 2 }$
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$\frac { { \mu }_{ 0 }\pi R }{ 2 }$
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$\frac { { \mu }_{ 0 } }{ 2 }$
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0

A semicircle loop

$PQ$ of radius

$R$ is moved with velocity

$'V'$ in the transverse magnetic field as shown in the figure. The value of induced emf between the ends of the loop is :

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$BV(\pi r)$ , end 'P' at high potential
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$2 BRV$ , end P at high potential
0%
$2 BRV$ , end Q at high potential
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Zero

The current is reduced from 3 amp to zero in 0.001 sec. in the primary coil. It induces an emf of 15000 volts in the secondary. The value of coefficient of mutual inductance in Henry will be -

0%
5
0%
0.5
0%
4.5
0%
50

A coil has self-inductance L. If number of turns per unit length is halved and radius is doubled, self inductance become

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2L
0%
4L
0%
L
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3L

An aeroplane is flying horizontally with a velocity of 360 km/h. The distance between the tips of the wings of aeroplane is 25 m. The vertical component of earth's magnetic field is

$4\times { 10 }^{ -4 }$

$Wb/{ m }^{ 2 }$ . The induced e.m.f. is

0%
1 V`
0%
100 V
0%
1 kV
0%
Zero

Two identical coils A and B are arranged coaxially as shown in the figure and the sign conversation adopted is that the direction of currents are taken as positive when they flow in the direction of arrows. Which of the following statements is correct.

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If A carries a steady positive current and A is moved towards B, than a positive current is induced in B.
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If A carries a steady positive current and B is moved towards A, then a negative current is induced in B
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If both coils carry positive current, then the coils repel each other.
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If a positive current flowing in A is switched off, then a negative current is induced momentarily in B

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 BB, constant in time and space, pointing perpendicular and into the plane of the loop exists everywhere, then the current induced in the loop is

0%
$\frac{BlV}{R}$ , clockwise
0%
$\frac{BlV}{R}$ anticlock wise
0%
$2\frac{BlV}{R}$ anticlock wise
0%
zero

Explanation

Given,

a conducting square loop of side = L

Resistance = R

Velocity = V

During its motion an emf induced

which is

$=\in =-BVL \sin \theta$

$=-BVL\sin \dfrac{\pi}{2}$

$\left[\theta=\text{ angle between } \overrightarrow{B}\text{ and } L.=\dfrac{\pi}{2}\right]$

Now according to Lenz's

Law this induced e.m.f leads to a current into the circuit in anticlockwise sense, to oppose the motion.

$\therefore$ Current into the circuit

$I=\dfrac{\in}{R}$

$\boxed{I=\dfrac{BVL}{R}},\boxed{anticlockwise}$

2133500_1477360_ans_772f09b68ab7474cae3e601c8e70fd7e.png

A coil of area

$500 cm^2$ and

$1000$ turns is perpendicular to a field of

$0.4$ gauss. The coil is turned through

$180^0$ in

$\frac{1}{10}$ sec. What is induced emf?

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$0.02$ V
0%
$0.04$ V
0%
$0.05$ V
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$0.06$ V

A loop of irregular shape made of flexible conducting wire carrying clockwise current is placed in uniform inward magnetic field , such that its plane is perpendicular to the field . then the loop

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experiences force
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develops induced current for a short time
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changes to circular loop
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all of these

A circular ring of diameter

$20 \,cm$ has a resistance of

$0.01 \Omega$ . The ring is turned from a position perpendicular to a uniform magnetic field of

$2T$ to a position parallel to the field. The amount of charge flowing through the ring in this process is:

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$2C$
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$\pi C$
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$2 \pi C$
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$8 \pi C$

In an indicator, the current varies with time t as

$I = 5A +16 (A/s)t$ . If induced emf in the inductor is 50mV, the self inductance of the indicator is

0%
$3.75 \times{ 10 }^{ -3 }H$
0%
$3.75 \times { 10 }^{ -4 }H$
0%
$3.125 \times { 10 }^{ -3 }H$
0%
$3.125 \times { 10 }^{ -4 }H$

A semicircular loop of radius R is rotated with an angular velocity

$\omega$ perpendicular to the plane of the magnetic field B as shown in the figure. Emf induced in the loop is

0%
B $\omega R ^2$
0%
$\dfrac {1}{2} B \omega R^2$
0%
$\dfrac {3}{2} B \omega R^2$
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$\dfrac {1}{4} B \omega R^2$

A square coil

$ABCD$ lying in

$x-y$ plane with it's centre at origin. A long straight wire passing through origin carries a current

$i=2t$ in negative

$z$ -direction. The induced current in the coil is

0%
Clockwise
0%
Anticlockwise
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Alternating
0%
Zero

A conducting square frame of side

$'a'$ and a long straight wire carrying current

$l$ are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity

$V$ . The emf induced in the frame will be proportional to

0%
$1/x^2$
0%
$1/(2x - a)^2$
0%
$1/(2x + a)^2$
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$1/((2x - a) (2x + a))$

Explanation

Hint: The induced emf in the square frame will be $\varepsilon = {B_1}av - {B_2}av$

Correct Option: (D)

Explanation for Correct Option:

Step 1: Find the magnetic induction.

As, Magnetic field $B = \dfrac{{{\mu _0}I}}{{2\pi r}}$ , where ${\mu _0}$ is a constant, $I$ is the current through the conductor, and $r$ is the distance between the conducting points.
${B_1} = \dfrac{{{\mu _o}I}}{{2\pi (x - \dfrac{a}{2})}}$ , where $I$ is the current flowing thriugh long wire; $x$ is the distance between the long wire and the middle axis of the square frame; $a$ is the side length of the square frame.
${B_2} = \dfrac{{{\mu _o}I}}{{2\pi (x + \dfrac{a}{2})}}$

Step 2: Find the induced emf.

Induced emf, $\varepsilon = {B_1}av - {B_2}av$

Here, $v$ is the velocity of the frame.

$\Rightarrow \varepsilon = ({B_1} - {B_2})av$

$\Rightarrow \varepsilon = (\dfrac{1}{{x - \dfrac{a}{2}}} - \dfrac{1}{{x + \dfrac{a}{2}}})\dfrac{{{\mu _o}Iav}}{{2\pi }}$

$\Rightarrow \varepsilon = \dfrac{{{\mu _o}Iav}}{{2\pi }}\dfrac{a}{{(2x - a)(2x + a)}}$

Therefore, the emf induced, $\varepsilon \alpha \dfrac{1}{{(2x - a)(2x + a)}}$

In a region of uniform magnetic induction

$B = 10 ^ { 2 } T ,$ a circular coil of radius

$30\mathrm { cm }$ and resistance

$\pi ^ { 2 } ohm$ is rotated about an axis which is perpendicular to the direction of

$\mathrm { B }$ and which forms a diameter of the coil, If the coil rotates at

$200 rpm,$ the amplitude of the alternating current induced in the coil is:

0%
$4\pi ^ { 2 } \mathrm { m } A$
0%
$30\mathrm { m }A$
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$6\mathrm { m }A$
0%
$200\mathrm { m }A$

Two coils

$A$ and

$B$ having turns

$300$ and

$600$ respectively are placed near each other, on passing a current of

$3.0$ ampere in

$A$ the flux linked with

$A$ is

$1.2 \times 10 ^ { - 4 } weber$ and with

$B$ it is

$9.0 \times 10 ^ { - 5 } weber.$ The mutual inductance of the system is

0%
$2 \times 10 ^ { - 5 }$ henry
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$3 \times 10 ^ { - 5 }$ henry
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$4 \times 10 ^ { - 5 }$ henry
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$6 \times 10 ^ { - 5 }$ henry

A vertical rod of length I is moved with constant velocity V towards east. The vertical component of earth magnetic field is B and angle of dip is

$\theta$ .The induced e.m.f. in the rod is

0%
$BIV sin \theta$
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$BIV tan \theta$
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$BIV cot \theta$
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$BIV cos \theta$

Three resistance of magnitude

$R$ each are connected in the form of an equilateral triangle of side

$a$ . The combination is placed in a magnetic field

$B=B_{0}e^{-\lambda t}$
perpendicular to the plane. The induced current in the circuit is given by:

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$\left(\dfrac{a^{2}\lambda}{2\sqrt{3R}}B_0\right)e^{-\lambda t}$
0%
$\left(\dfrac{a^{2}\lambda}{4(\sqrt{3})R}B_0\right)e^{-\lambda t}$
0%
$\left(\dfrac{a^{2}B_0}{\lambda 4\sqrt{3R}}\right)e^{-\lambda t}$
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$\left(\dfrac{a^{2}B_0 R}{\lambda 4\sqrt{3R}}\right)e^{-\lambda t}$

A wire is sliding as shown in the figure. The angle between the acceleration and velocity of the wire is

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$30^{o}$
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$40^{o}$
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$120^{o}$
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$90^{o}$

A circular current carrying coil has a radius R.The distance from the centre of the coil on the axis where the magnetic induction will be 1/8th to its value at the centre of the coil , is

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$R / \sqrt {3}$
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$R / {3}$
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$2 \sqrt {3} R$
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$2R \sqrt {3}$

Three long parallel wires carrying steady currents

$20\ A, 10\ A,10\ A$ are cut by a perpendicular plane in the vertices

$A$ and

$B$ and

$C$ of a triangle in which angles

$B$ and

$C$ are equal. The current of

$20\ A$ through

$A$ is in opposite direction through

$B$ and

$C$ , then:

0%
on the line through $A$ perpendicular to $BC$ , the only point at which the magnetic induction vanishes lies on the circumcircle of the triangle $ABC$
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on the line through $A$ perpendicular to $BC$ , the only point at which magnetic induction does not vanish lies on the circumcircle of the triangle $ABC$
0%
if the triangle $BAC$ is equilateral, each side being of length $10\ cm$ , the magnitude of the mechanical force per unit length on the wire through $A$ us zero
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if the triangle $ABC$ is equilateral, each side being of length $10\ cm$ , the magnitude of the mechanical force per unit length on the wire through $A$ is $1732$ dyne.

A bar magnet is moved along the axis of a copper ring placed far away from the magnet. Looking from the side of the magnet, an anticlockwise current is found to be induced in the ring. Which of the following may be true?

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The south pole faces the ring and the magnet moves towards it
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The north pole faces the ring and the magnet moves towards it
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The south pole faces the ring and the magnet moves away from it
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The north pole faces the ring and the magnet moves away from it

A solenoid of inductance

$L$ and resistance

$r$ is connected in parallel to a resistance

$R$ and a battery of emf

$E$ . Initially if the switch is closed for a long time and at

$t=0$ , then the :

0%
current through solenoid at any time $t$ , after opening the switch is $\dfrac{E}{r}e^{-(R+r)t/L}$
0%
induced emf across solenoid at time $t=0$ is $\dfrac{E(R+r)}{r}$
0%
amount of heat generated in solenoid is $\dfrac{E^2L}{2r(r+R)}$
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potential difference across solenoid at $t=0$ is $E$

An Indian ship with a vertical conducting mass navigates the Indian ocean in the latitude of magnetic equator. To induce the greatest emf in the mast, the ship should proceed:

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northward
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southward
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eastward
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none of these

A short solenoid (length

$l$ and radius

$r$ , with

$n$ turns per unit length) lies well inside and on the axis of a very long, coaxial solenoid (length

$L$ , radius

$R$ and

$N$ turns per unit length, with

$R > r$ ). Current

$I$ flows in the short solenoid.

Choose the correct statement.

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There is uniform magnetic field $\mu_0n I$ in the long solenoid.
0%
Mutual inductance of the solenoid is $\pi \mu_0 r^2 n Nl$ .
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Flux through outer solenoid due to current $I$ in the inner solenoid is proportional to the ratio $R/r$
0%
Mutual inductance of the solenoids is $\pi \mu_0 r RnNl L/(rR)^{1/2}$

Explanation

The magnetic flux linked with long solenoid of radius

$R$ due to short solenoid of radius

$r$ having current

$I$ is,

$\phi=MI$ . . . . .(1)

The magnetic field due to the short solenoid,

$B=\mu_0 nl I$

Magnetic flux linked is,

$\phi= BA$

$\phi= \mu_0nlI(\pi r^2)$ . . . . .(2)

From equation (1) and (2), we get,

$M=\pi \mu_0r^2nl$

For

$N$ number of turns,

$M=\pi \mu_0r^2nNl$ .

The correct option is B.

A rectangular coil ABCD is rotated anticlockwise with a uniform angular velocity about the axis shown in fig. The axis of rotation of the coil, as well as the magnetic field B, is horizontal. The induced emf in the coil would be minimum when the plane of the coil

0%
is horizontal
0%
makes an angle of $45 ^{\circ}$ with the direction of magnetic field
0%
is at right angle to the magnetic field
0%
makes an angle of $30 ^{\circ}$ with the magnetic field

Explanation

Let

$\theta$ is the angle between the area vector of the coil and the magnetic field. Then flux through coil

$\phi=BA cos \theta$

Induced emf=

$d=-\dfrac {d\phi}{dt}=BA sin \theta \dfrac {d\theta}{dt}$

$\Rightarrow e= BA \omega sin \theta$

e is minimum, when

$\theta=0^{\circ}$ , or when plane of coil is at right angle to magnetic field.

When the plane of the coil is perpendicular to the magnetic field, then ϕϕ is maximum and induce emf is minimum.

The e.m.f. induced in the closed loop is

0%
$-\dfrac{\mu_0I_0a}{2 \pi}ln(\dfrac{2a+l}{l})$
0%
$\dfrac{\mu_0I_0a}{2 \pi}ln(\dfrac{2a-l}{l})$
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$\dfrac{2\mu_0I_0a}{2 \pi}ln(\dfrac{a+l}{l})$
0%
$\dfrac{\mu_0I_0a}{ \pi}ln(\dfrac{a+l}{l})$

Explanation

$B=\dfrac{\mu_0I}{2\pi x}.....(I)$

$\Rightarrow d\phi=B\times A=\dfrac{\mu_0I}{2\pi x}\times adx$

$\int{^\phi_0}d\phi=\int{^{2a+l}_l}\dfrac{\mu_0Iadx}{2\pi x}$

$\Rightarrow \phi=\dfrac{\mu_0Ia}{2\pi}ln\left(\dfrac{2a+l}{l}\right)$

$\Rightarrow M=\dfrac{\phi}{I}=\dfrac{\mu_0a}{2\pi}ln(\dfrac{2a+l}{l})$

$E=-M\dfrac{dI}{dt}$

$E=-MI_0$

$\Rightarrow E=-\dfrac{\mu_0I_0a}{2\pi}ln(\dfrac{2a+l}{l})$

In fig a square loop PQRS of side a and resistance r is placed neat an infinitely long wire carrying a constant current l. the sides PQ and Rs parallel to the wire. The wore and the loop are in the same plane. The loop is rotated

$180^{\circ}$ about an axis parallel to the long wire and passing through the mid-points of the sides QR and PS.The total amount of charge which passes through any point of the loop during rotation is

0%
$\dfrac {\mu_0Ia}{2 \pi r}ln2$
0%
$\dfrac {\mu_0Ia}{\pi r}ln2$
0%
$\dfrac {\mu_0Ia^2}{2 \pi r}$
0%
Cannot be found because time of rotation is not give.

Explanation

As we know,

Magnetic flux density at a distance x from the infinitely long current-carrying wire is,

$B=\dfrac{\mu I }{2\pi x}$

Initial flux:

$\phi_i=- \int_{a}^{2a} \dfrac {\mu_0I_a}{2\pi} \dfrac {dx}{x}$

$=-\dfrac {\mu_0Ia}{2 \pi }ln2$

Final flux :

$\phi_f=\dfrac {\mu_0Ia}{2 \pi }ln2$

Charge flown:

$q=\dfrac {\Delta \phi}{r}=\dfrac {\phi_f-\phi_i}{r}=\dfrac {\mu_0Ia}{\pi r}ln2$

A metal bar is moving with a velocity of

$v=5 cm s^{-1}$ over a U-shaped conductor.AT

$t=0$ , the external magnetic field is 0.1 T directed out of the page and is increasing at a rate of

$0.2 T s^{-1}$ . Take

$l= 5 cm$ and

$t=0, x=5 cm$

The current flowing in the circuit is

0%
2.5 A
0%
5 A
0%
1 A
0%
2 A

Explanation

Let us take anticlockwise direction as positive, then the area vector in the upward direction will be positive.
now as we know the flux is

$\phi= BA=Blx$

$e= -{d\phi}{dt}=-l [B \dfrac {dx}{dt}+x \dfrac {dB}{dt}]$

$=-5 \times 10^{-2}[ 0.1 \times (-5 \times 10^{-2} + 5 \times 10^{-2} \times 0.2]$

$-250 \times 10^{-6} V= -250 \mu V$

emf is coming out to be negative, so it should be clockwise.

Current

$I= \dfrac {e}{R}= \dfrac {250 \times 10^{-6}}{10^{-4}}=2.5A$

The four wire loops shown in fig. Have vertical edge lengths of either L, 2L, 3L .They will move with the same speed into a region of uniform magnetic field

$\overrightarrow { B }$ directed out of the page. Rank them according to the maximum magnitude of the induced emf greatest to the least.

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1 and 2 tie, then 3 and 4 tie
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3 and 4 tie, then 1 and 2 tie
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4,2,3,1
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4 then, 2 and 3 tie and then 1

An aircraft is flying at a level height in presence the magnetic field of the Earth. If an electric bulb is connected between the two extreme ends of the wings,

0%
a voltage will induce across the wings and the bulb will glow.
0%
no voltage will induce across the wings but the bulb will glow.
0%
a voltage will induce across the wings but the bulb will not glow.
0%
no voltage will induce across the wings and the bulb will not glow.

A circuit ABCD is held perpendicular to the uniform magnetic field of

$B=5\times 10^{-2}T$ extending over the region PQRS and directed into the plane of the paper. The circuit is moving out of the field at a uniform speed of

$0.2 ms^{-1}$ for 1.5 s. During this time, the current in the 5

$\Omega$ resistor is

0%
0.6 mA from B to C
0%
0.9 mA from B to C
0%
0.9 mA from C to B
0%
0.6 mA from C to B

Explanation

$l=vt$

$=0.2\times1.5=0.3$

now induced current is

$I=\dfrac {Blv}{R}$

$I=\dfrac {5 \times 10^{-2}\times 0.3 \times 0.2}{5}A=0.6mA$

Area and flux are decreasing. So, current flows in increase the flux, Clearly, the current should be clockwise. So, it flows from B to C through

$5 \Omega$ .

The radius of the circular conducting loop shown in fig. is R. Magnetic field is decreasing at a constant rate

$\alpha$ Resistance per unit length of the loop is

$\rho$ then the current in wire AB is (AB is one of the diameter)

0%
$\dfrac {R \alpha}{2\rho}$ from A to B
0%
$\dfrac {R \alpha}{2\rho}$ from B to A
0%
$\dfrac {R \alpha}{\rho}$ from A to B
0%
0

A wire loop enclosing a semicircle of radius R is located on the boundry of a uniform magnetic field B. At the moment t=0, the loop set into rotation with constant angular acceleration

$\alpha$ about an axisThe clockwise emf direction is taken to be positive.

The variation of emf as a function of time is

0%
$\dfrac {1}{2} BR^2 \alpha t$
0%
$\dfrac {3}{2} BR^2 \alpha t\, A$
0%
$\sqrt {3} B R^2 \alpha t$
0%
$\dfrac {B R^2 \alpha t}{\sqrt 2}$

Explanation

emf is induced in the loop because area inside the magnetic field is continually changing.

From

$\theta=0$ to

$\pi, 2\pi\ to\ 3\pi,\ 4\pi\ to\ 5\pi$ the loop begins to enter the magnetic field passing through the loop is increasing. Hence, current in the loop is anticlockwise, and for

$\theta=0$ to

$\pi,\ 2\pi\ to\ 3\pi,\ 4\pi\ to\ 5\pi$ etc. The magnetic field passing through the loop is decreasing. Hence current in the loop is clockwise.

Let at any time, angle rotated is

$\theta$ then

$\theta=\dfrac {1}{2} \alpha t^2$

Area inside the magnetic field.

$A=\dfrac {1}{2}R^2 \theta= \dfrac {1}{2}R^2(\dfrac {1}{2} \alpha t^2)=\dfrac {1}{4} R^2\alpha t^2$

Flux in the loop:

$\phi=BA=\dfrac {B}{4}=\dfrac {B}{4} R^2\alpha t^2$

emf:

$e=-\dfrac {d\phi}{dt}=-\dfrac {B}{2} R^2 \alpha t$

1623580_1748084_ans_74b2c1f102a543839885ccd3c4dd4a4b.PNG

The heat produced in the loop in time t is

0%
$\dfrac{[\dfrac{\mu_0I_0}{2 \pi}ln(\dfrac{a+l}{l})]^2 at}{4 \lambda}$
0%
$\dfrac{[\dfrac{\mu_0I_0}{2 \pi}ln(\dfrac{2a+l}{l})]^2 at}{8 \lambda}$
0%
$\dfrac{[\dfrac{2\mu_0I_0}{2 \pi}ln(\dfrac{a+l}{l})]^2 at}{4 \lambda}$
0%
$\dfrac{[\dfrac{\mu_0I_0}{2 \pi}ln(\dfrac{3a+l}{l})]^2 at}{6 \lambda}$

Explanation

Let a small strip be

$dx$

The distance from wire to strip let be

$x$

The magnetic field

$B= \dfrac{(\mu_o)I}{2 (\pi)x}$ ..........(i)

Flux linked

$d\phi= B\times A$

$= \dfrac{(\mu_o)I}{2 (\pi)x}\times adx$ ..........(ii)

Now to find out the total flux we will integrate equation (ii) from o to

$\phi$ in L.H.S. and

$l$ to

$2a+l$ in R.H.S.

After integration we get,

$\phi= \dfrac{(\mu_o)I a}{ 2(\pi)} ln \dfrac{2a+l}{l}$

Mutual inductance

$M$ =

$\dfrac{\phi}{I}$

$M= \dfrac{(\mu_o) a}{ 2(\pi)} ln \dfrac{2a+l}{l}$ .......

$.(1)$

Now emf induced in the closed loop is,

$E= -M \dfrac{dI}{dt}$

Since it is an inner function,

So,

$E= -M\times I_o$

Now putting the value of

$M$ from euation

$(1)$ we get emf as,

$E= -\dfrac{(\mu_o)I a}{2( \pi)} ln \dfrac{2a+l}{l}$ ........

$(2)$

Now, heat produced in the loop is,

$H= \dfrac{E^2}{R}\times t$ .........

$(iii)$

Resistance

$R= 8a\times \lambda$

So, now putting the value of emf from equation

$(2)$ and, putting the value of resistance in equation

$(iii)$ we get heat inside the loop as,

$H= \dfrac{[\dfrac{\mu_0I_0}{2 \pi}ln(\dfrac{2a+l}{l})]^2 at}{8 \lambda}$

1629664_1749468_ans_6185763dabd04acfb6aa2e223d82845c.png

There is a conducting ring of radius R. Another ring having current i and radius

$r(r << R)$ is kept on the axis of bigger ring such that its centre lies in the axis of bigger ring at a distance x from the centre of bigger ring and its plane is perpendicular to that axis. The mutual inductance of the bigger ring due to the smaller ring is

0%
$\dfrac{\mu_0\pi R^2r^2}{(R^2+x^2)^{3/2}}$
0%
$\dfrac{\mu_0\pi R^2r^2}{4(R^2+x^2)^{3/2}}$
0%
$\dfrac{\mu_0\pi R^2r^2}{16(R^2+x^2)^{3/2}}$
0%
$\dfrac{\mu_0\pi R^2r^2}{2(R^2+x^2)^{3/2}}$

Explanation

Let current i flows in the bigger ring,then the magnetic field on its axis

$B=\dfrac{\mu_0iR^2}{2(R^2+x^2)^{3/2}}$

Flux linked with the smaller ring

$\phi=B \pi r^2$

$\phi =\dfrac{\mu_0iR^2}{2(R^2+x^2)^{3/2}}=M i$

$\therefore M=\dfrac{\mu_0 \pi R^2r^2}{2(R^2+x^2)^{3/2}}$

A small coil C with

$N = 200$ turns is mounted on one end of a balance beam and introduced between the poles of an electromagnet as shown in figure. The cross sectional area of coil is

$A= 1.0 cm$ length of arm OA of the balance beam is

$I=30 cm.$
When there is no current in the coil the balance is in equilibrium. On passing a current

$I = 22 mA$ through the coil the equilibrium is restored by putting the additional counter weight of mass

$\theta m = 60 mg$ on the balance pan. Find the magnetic induction at the spot where coil is located.

0%
$0.4 T$
0%
$0.3 T$
0%
$0.2 T$
0%
$0.1 T$

Explanation

On passing current through the coil, it acts as a magnetic dipole. Torque acting on magnetic dipole is counter balanced by the moment of additional weight about position O. Torque acting on a magnetic dipole.

$\tau = MB sin \theta = (Ni A )B sin 90^0 = NiAB$

Again

$\tau =Force \times lever arm = \Delta mg \times l$

$\Rightarrow NiAb = \Delta mgl$

$\Rightarrow B = \dfrac { \Delta mgl}{NiA} = \dfrac { 60 \times 10^{-3} \times 9.8 \times 30 \times 10^{-2} }{ 200 \times 22 \times 10^{-3} \times 1 \times 10^{-4} } = 0.4 T$

The magnetic field in the cylindrical region shown in figure increases at a constant rate of

$20\, mT/sec.$ Each side of the square loop

$ABCD$ has a length of

$1 \,cm$ and resistance of

$4\Omega$ . Find the current in the wire AB if the switch S is closed

0%
$1.25 \times 10^{-7} \,A,$ (anti-clockwise)
0%
$1.25 \times 10^{-7} \,A$ (clockwise)
0%
$2.5 \times 10^{-7} \,A$ (anti clockwise)
0%
$2.5 \times 10^{-7} \,A$ (clockwise)

Explanation

Given, Magnetic field in the cylindrical region increases at a constant rate of

$20mT/sec$ . Each side of square loop has length

$1$ cm and resistor

$4\Omega$ .

To find, current in the wire AB when switch is closed.

We know,

$e=n\Delta \Phi$

$\Rightarrow e=\dfrac{d\Phi}{dt}$ and

$\Phi =BA$

$\Rightarrow e=\dfrac{dB}{dt}(1\times 10^{-4})$

and

$e=IR$

$\Rightarrow \dfrac{dB}{dt}=IR$

As each side has resistance R

$\therefore R_{eq}=4+4+4+4$

$\Rightarrow R_{eq}=16\Omega$

Now,

$\dfrac{dB}{dt}\times (1\times 10^{-4})=I\times 16$

$\Rightarrow 20\times 10^{-3}\times 10^{-4}=I\times 16$

$\Rightarrow I =\dfrac{20\times 10^{-7}}{16}$

$\Rightarrow I=1.25\times 10^{-7}A$

By using Lenz's Law, Direction of current will be Anticlockwise.

Therefore, option A correct.

2033070_1818149_ans_5917070c524a4975860f3fec06e030f4.png

After half rotation _______ comes in contact with _________, and ______ comes in contact with ________ .

0%
ring R1, ring R2 & brush B1, brush B2
0%
ring R1, brush B1 & ring R2, brush B2
0%
ring R1, brush B2 & ring R2, brush B1
0%
ring R2, brush B2 & ring R1 , brush B2

Frequency of AC mains in India is:

0%
$20\ Hz$
0%
$50\ Hz$
0%
$100\ Hz$
0%
$240\ Hz$

Explanation

The alternating current reverses its direction periodically. In India, the AC changes direction after every

$1/100\ second$ , that is, the frequency of AC is

$50\ Hz$ .

The direction of the current in the DC generator is always the same and unidirectional because brush

$B1$ is always in contact with the wire.

0%
True
0%
False

In India the AC changes direction after every _________ second.

0%
$1/50$
0%
$1/150$
0%
$1/100$
0%
$1/1000$

Explanation

The alternating current reverses its direction periodically. Most power stations constructed these days produce AC. In India, the AC changes direction after every

$\dfrac {1}{100}\ second$ , that is, the frequency of AC is

$50\ Hz$ . An important advantage of AC over DC is that electric power can be transmitted over long distances without much loss of energy.

CBSE Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 14 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 14 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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