MCQ Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 1

Whenever the magnetic flux linked with a coil changes, then there is an induced emf in the circuit. This emf lasts :

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for a short time
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for a long time
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for ever
0%
so long as the change in the flux takes place

Explanation

$EMF\quad =-\dfrac { d\emptyset }{ dt } \quad$

So, till the flux is changing with respect to time EMF will be induced in the circuit.

When the flux linked with a coil changes :

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current is always induced
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an emf and a current are always induced
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an emf is induced but a current is never induced
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an emf is always induced and a current is induced when the coil is a closed one

Explanation

$EMF\quad =-\dfrac { d\phi }{ dt }$ , where

$\phi$ is flux
So, whenever flux changes, EMF is induced. But ,current will be induced only when the coil is a closed one.

The average self-induced emf in a

$25 m H$ solenoid when the current in it falls from

$0.2 A$ to

$0 A$ in

$0.01$ second, is

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$0.05 V$
0%
$0.5 V$
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$500 V$
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$50 V$

Explanation

$emf=L\dfrac { di }{ dt }$

$=25\times { 10 }^{ -3 }\times \dfrac { 0.2 }{ 0.01 }$

$= 0.5V$

The coefficient of self inductance and the coefficient of mutual inductance have

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same units but different dimensions
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different units but same dimensions
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different units and different dimensions
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same units and same dimensions

Explanation

$Emf=-L\dfrac { di }{ dt }$

$\phi =Mi$

$Emf\dfrac { d\phi }{ dt } =-M\dfrac { di }{ dt }$

So, units and dimension of coeffiecent of self inductance and coefficent of mutual inductance are same.

Assertion (A): Whenever the magnetic flux linked with a closed coil changes there will be an induced emf as well as an induced current.
Reason (R): According to Faraday, the induced emf is inversely proportional to the rate of change of magnetic flux linked with a coil.

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Both A and R are individually true and R is the correct explanation of A
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Both A and R are individually true but R is not the correct explanation of A
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A is true but R is false
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Both A and R are false

Explanation

$EMF\quad =-\dfrac {d\phi }{ dt } \quad$
So, EMF is directly proportional to the rate of change of magnetic flux linked with a coil. When closed coil is used current flows into it.

When current is coil changes from 5 A to 2 A in 0.1 s, an average of 50 V is produced. The self-inductance of the coil is.

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6 H
0%
0.67 H
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1.67 H
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3 H

Explanation

Inudctor opposes change in curren, as the current is decreased, the inductor will convert magnetic energy into electrical by creating a potential difference. The voltage drop across inductor is given by

$V=L\dfrac{di}{dt}$

$-50=L \dfrac{2-5}{.1}$ here negative sign is because voltage is increasing instead of drop across inductor

$-50=L \dfrac{2-5}{.1}$

$L=1.67 \;H$ .

In an AC generator, a coil with

$N$ turns, all of the same area

$A$ and total resistance

$R$ , rotates with frequency

$1 rad/s$ in a magnetic field

$B$ . The maximum value of emf generated in the coil is:

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$N.A.B$
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$N.A.B.R$
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$\dfrac{1}{10}N.A.B$
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$\dfrac{1}{10}N.A.B.R$

Explanation

Let at time

$t=0$ , the coil is vertical and at time

$t$ , plane of coil makes an angle

$\theta$ with the vertical, then

$\theta=\omega t$ , (

$\omega=$ uniform angular velocity),

in this position, magnetic flux linked with coil will be,

$\phi=NBA\cos \theta$ , (where,

$A=$ area of coil,)

$\phi=NBA \cos \omega t$ ,

now, differentiating this equation w.r.t. time, we get

$\dfrac{d\phi}{dt}=\dfrac{d}{dt}(NBA \cos\omega t)$ ,

$\dfrac{d\phi}{dt}=-NBA\omega\sin\omega t$ ,

$e$ $ is the emf induced in coil then by Faraday's law,

$e=-\dfrac{d\phi}{dt}=NBA\omega\sin\omega t$ ,

now if,

$\sin\omega t=1$ (maximum),

then

$e_{max}=NBA\omega$ ,

$\omega=1rad/s$ ,

then

$e_{max}=NBA$

$10 \,m$ long horizontal wire extends from North East to South West. It is falling with a speed of

$5.0 \,ms^{-1}$ , at right angles to the horizontal component of the earth's magnetic field, of

$0.3 \times 10^{-4} \,Wb/m^2$ . The value of the induced emf in wire is:

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$2.5 \times 10^{-3} V$
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$1.1 \times 10^{-3} V$
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$0.3 \times 10^{-3} V$
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$1.5 \times 10^{-3} V$

Explanation

Induced emf

$= Bv \ell sin45$

$= 0.3 \times 10^{-4} \times 5 \times 10 \times \dfrac{1}{\sqrt2}$

$= 1.1 \times 10^{-3} V$
1142597_1331253_ans_b833e99db5f447738b9f5573504c094b.PNG

At time

$t = 0$ magnetic field of

$1000$ Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to

$500$ Gauss, in nest

$5s$ , then induced EMF in the loop is:

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$36\mu V$
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$28\mu V$
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$56\mu V$
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$48\mu V$

Explanation

Area of loop

$AEBCFDA =$ area of rectangle

$ABCD \ -$ Area of triangle

$ABE\ -$ Area of triangle

$DCF$

$\therefore$ Required area

$=16cm \times 4 cm - \dfrac{1}{2}\times 4cm \times 2cm - \dfrac{1}{2}\times 4cm \times 2cm$

$= 64cm^2 - 4cm^2 - 4cm^2 = 56cm^2$

Magnetic flux passing through an area is given by

$\displaystyle \phi = \int B.dA = \int BdA \cos \theta$ .

Here,

$\theta = 0$ and

$B$ is constant throughout the area.

$\therefore \phi_{initial} = B_{initial}.A = 1000 \, Gauss \times 56cm^2$

$= 56000 \, gauss\, cm^2$

$= 5.6\times 10^{-4}Wb$

$\begin{bmatrix} 1 \, Gauss = 10^{-4}T \, and \, 1cm^2 = 10^{-4}m^2\\ \therefore 1\, Gauss\, cm^2 = 10^{-8}Tm^2 = 10^{-8}Wb \end{bmatrix}$

Also,

$\phi _{final} = B_{final}. A = 500\, Gauss \times 56cm^2$

$= 28000\, Gauss\, cm^2 = 2.8\times 10^{-4}Wb$

Since the flux changes linearly,

$\in = \left|\dfrac{d\phi}{dt}\right| = \dfrac{|\phi_{final} - \phi_{initial}|}{\Delta t} = \dfrac{(5.6-2.8)\times 10^{-4}}{5}V$

$= 56\times 10^{-6}V = 56\mu v$

1477772_1697215_ans_15a5588f82544ca7b5382a74b07d66cc.png

When the current changes from

$+2 A$ to

$- 2 A$ in

$0.05$ second, an e.m.f. of

$8 V$ is induced in a coil. The coefficient of self-induction of the coil is :

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$0.2 H$
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$0.4 H$
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$0.8 H$
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$0.1 H$

Explanation

lf $\mathrm{e}$ is the induced $\mathrm{e}$ .m.f. in the coil, then $\displaystyle \mathrm{e}=-\mathrm{L}\frac{\mathrm{d}\mathrm{i}}{\mathrm{d}\mathrm{t}}$

Therefore, $\displaystyle \mathrm{L}=-\frac{\mathrm{e}}{\mathrm{d}\mathrm{i}/\mathrm{d}\mathrm{t}}$

Substituting values, we get $\displaystyle \mathrm{L}=\frac{-8\times 0.05}{-4}=0.1\mathrm{H}$

A planar loops of wire rotates in a uniform magnetic field. Initially, at

$t = 0$ , the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of

$10 s$ about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at :

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$5.0 s$ and $7.5 s$
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$2.5 s$ and $7.5 s$
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$2.5 s$ and $5.0 s$
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$5.0 s$ and $10.0 s$

Explanation

Time period,

$T = 10 s$

Ref. image I

Axis of rotation

$\rightarrow$ y-axis.

$\vec{B} = B(-\hat{k})$

ref. image II

at some time t, area vector makes angle '

$\theta$ ' with

$\vec{B}$ .

flux,

$\phi = \vec{B}. \vec{A}$

$= BA \cos \theta$

emf,

$\varepsilon= \left|\dfrac{-d \phi}{dt} \right| = \left(BA \sin \theta \dfrac{d \theta}{dt}\right)$

$= BA \omega \sin \theta$

When

$\theta = \dfrac{\pi}{2} , \varepsilon = \varepsilon_{max}$

So,

$\theta = n \pi + \dfrac{\pi}{2}$ n = integes

When

$\theta = 0, \varepsilon = \varepsilon_{min} = 0$

So,

$\theta = n \pi$ , n = integes

for

$\varepsilon = \varepsilon_{max}, \theta = \omega t = n \pi + \dfrac{\pi}{2}$

$\Rightarrow t = \dfrac{n \pi}{\omega} + \dfrac{\pi}{2 \omega} = \dfrac{n \pi}{\left(\dfrac{2 \pi}{T} \right)} + \dfrac{\pi}{2 \times \left(\dfrac{2 \pi}{T} \right)}$

$t = \left(\dfrac{n \pi}{\dfrac{2 \pi}{T}} \right) + \dfrac{\pi}{2 \left(\dfrac{2 \pi}{T} \right)}$

$= \dfrac{n T}{2} + \dfrac{T}{4} = \dfrac{n \times 10}{2} + \dfrac{10}{4}$

$= 5n + 2.5$

$= 2.5 s, 5 + 2.5 , 5 \times 2 + 2.5 ...$

$= 2.5 s, 7.5 s, 12.5 s....$

for

$\varepsilon = \varepsilon_{min}, \theta = \omega t = n \pi$

$\Rightarrow t = \dfrac{n \pi}{\omega} = \dfrac{n \pi}{\left(\dfrac{2 \pi}{T} \right)} = \dfrac{n T}{2}$

$\Rightarrow t = \dfrac{n \times 10}{2} = 5n$

$= 0 s, 5s, 10s, ...$

So, possible option is option (C).

1477222_1697152_ans_1402de00da804a1b980e363bda7982af.png

Consider a thin metallic sheet perpendicular to the plane of the paper moving with speed

$'v'$ in a uniform magnetic field B going into the plane of the paper (See figure). If charge densities

$\sigma_1$ and

$\sigma_2$ are induced on the left and right surfaces, respectively, of the sheet then (ignore fringe effects) :

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$\sigma_1=\frac{-\epsilon_0vB}{2},\ \sigma_2=\frac{\epsilon_0vB}{2}$
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$\sigma_1=\epsilon_0vB,\ \sigma_2 = -\epsilon_0vB$
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$\sigma_1=\frac{\epsilon_0vB}{2},\ \sigma_2=\frac{-\epsilon_0vB}{2}$
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$\sigma_1=\sigma_2=\epsilon_0vB$

A source of constant voltage

$V$ is connected to a resistance

$R$ and two ideal inductors

$L_1$ and

$L_2$ through a switch

$S$ as shown. There is no mutual inductance between the two inductors. The switch

$S$ is initially open. At

$t = 0$ , the switch is closed and current begins to flow. Which of the following options is/are correct?

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At $t=0$ , the current through the resistance $R$ is $\dfrac{V}{R}$
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After a long time, the current through $L_2$ will be $\dfrac{V}{R} \dfrac{L_1}{L_1+L_2}$
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After a long time, the current through $L_1$ will be $\dfrac{V}{R} \dfrac{L_2}{L_1+L_2}$
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The ratio of the currents through $L_1$ and $L_2$ is fixed at all times $(t > 0)$

Explanation

$L = \dfrac{L_1L_2}{L_1+ L_2}$

$I = \dfrac{V}{R} (1- e^{\dfrac{-1}{2}})$ ,

$I = \dfrac{L_1L_2}{R(L_1+L_2)}$

$I _{L_2}= \dfrac{V}{R} \times \dfrac{L_1}{L_1+L_2}$

$I_{L_1}= \dfrac{V}{R} \times (\dfrac{L_2}{L_1+L_2})$

Applying KVL

$-IR - L \dfrac{dl}{dt} + V= 0$

$\Rightarrow _0\int^1 \dfrac{dl}{V-IR} = \dfrac{V}{R} (1- e^{\frac{-Rt}{L}})$

Hence B and C are correct.

A uniform magnetic field is restricted within a region of radius

$r$ . The magnetic field changes with time at a rate

$\dfrac{d\overrightarrow{B}}{dt}$ . Loop

$1$ of radius

$R > r$ enclose the region

$r$ and loop

$2$ of radius

$R$ is outside the region of magnetic field as shown in the figure below. Then the

$e.m.f$ . generated is :

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$-\dfrac{d\overrightarrow{B}}{dt} \pi r^2$ in loop $1$ and zero in loop $2$
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Zero in loop $1$ and zero in loop $2$
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$-\dfrac{d\overrightarrow{B}}{dt} \pi r^2$ in loop $1$ and $-\dfrac{d\overrightarrow{B}}{dt} \pi r^2$ in loop $2$
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$-\dfrac{d\overrightarrow{B}}{dt} \pi R^2$ in loop $1$ and zero in loop $2$

Explanation

Emf is induced in a coil due to change of flux through it.

In the loop 1, the flux passing through it is

$B\pi r^2$

Hence emf induced=

$-\dfrac{d\phi}{dt}=-\dfrac{d\vec{B}}{dt}\pi r^2$

Since the magnetic field does not change inside loop 2, no emf is induced in it.

A thin semicircular conducting the ring

$(PQR)$ of radius

$'r'$ is falling with its plane vertical in a horizontal magnetic field

$B$ , as shown in figure. The potential difference developed across the ring when its speed is

$v$ , is

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$Zero$
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$Bv\pi r^2/2$ and $P$ is at higher potential
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$\pi rBv$ and $R$ is at higher potential
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$2rBv$ and $R$ is at higher potential

Explanation

$emf= VBl_{eq}$

$= VB(2R)$

where R is at higher potential and P is at lower potential

A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced e.m.f. is

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Once per revolution
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Twice per revolution
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Four times per revolution
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Six times per revolution

A long solenoid has

$1000\$ turns. When a current of

$4\ A$ flows through it, the magnetic flux linked with each turn of the solenoid is

$4\times 10^{-3}$ Wb. The self-inductance of the solenoid is:

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$4$ H
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$3$ H
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$2$ H
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$1$ H

Explanation

Flux linked with each turn

$=4\times 10^{-3}\ Wb$

$\Rightarrow$ Total flux linked

$=1000\times 4\times 10^{-3}\ Wb$

$\Rightarrow \phi_{total}=4$

$\Rightarrow LI=4$

$\Rightarrow L=1 \ H$

A magnet is made to oscillate with a particular frequency, passing through a coil as shown in figure. The time variation of the magnitude of emf generated across the coil during on cycle is

Explanation

We know that electromagnetic e.m.f. induced,

$e=\displaystyle -\frac{d\phi}{dt}$

Initially,

$d\phi$ will be positive(during first

$\displaystyle\frac{T}{4}$ times period) then it becomes negative during the period from

$\displaystyle\frac{T}{4}$ to

$\displaystyle\frac{T}{2}$ .

During the period

$\displaystyle\frac{T}{2}$ to

$\displaystyle\frac{3T}{4}$ it is again positive and in the last

$\displaystyle\frac{T}{4}$ time it is negative.

Accordingly sign of emf produced will be changed. Figure

$(a)$ fits exactly in this change pattern. So this figure represents the answer.

In a coil of self inductance of

$5$ henry, the rate of change of current is

$2$ ampere per second, the e.m.f. induced in the coil is

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$5V$
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$-5V$
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$-10V$
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$10V$

Explanation

e.m.f.

$=\displaystyle -L\frac{di}{dt}$

$=-5\times 2$

$=-10V$

A coil of self-inductance L is connected in series with a bulb B and an AC source. Brightness of the bulb decreases when :

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frequency of the AC source is decreased
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number of turns in the coil is reduced
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a capacitance of reactance $X_C = X_L$ is included in the same circuit
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an iron rod is inserted in the coil

Explanation

Self-inductance of a coil is,

$L=\dfrac{\mu_0\mu_rN^2A}{l}$

Current flow through the circuit,

$I= \dfrac{E}{(w^2L^2+R^2)^{1/2}}$

Current in the circuit decrease with increasing self-inductance,

$L\alpha\mu_r$ .

When an iron rod is inserted in the coil, self-inductance increases, and the current decreases. The brightness of the bulb decreases.

A wire loop PQRSP formed by joining two semicircular wires of radii

$R_1$ and

$R_2$ carries a current I as shown in figure. The magnitude of magnetic induction at centre C is?

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$(\mu_0/4)l\left [ \frac{1}{R_2}-\frac{1}{R_1} \right ]$
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$\left ( \frac{\mu _0}{4} \right )l\left [ \frac{1}{R_1}-\frac{1}{R_2} \right ]$
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$\mu _0l\left [ \frac{1}{R_2}-\frac{1}{R_1} \right ]$
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$\mu _0l(1/R_1)$

Explanation

For a circular arc magnitude of magnetic field is

$\frac{\mu _0I}{2\pi R}(\theta )$

$\therefore$ for a semicircle with radius

$R_1$ magnetic field is

$\frac{\mu _0I}{4R_1}$ outward for semicircle with radius

$R_2$ magnetic field is

$\frac{\mu _0I}{4R_2}$ inward

$\therefore Total=\frac{\mu _0I}{4\pi }\left ( \frac{1}{R_1}-\frac{1}{R_2} \right )$

A varying current in a coil change from

$10A$ to

$0$ in

$0.5$ sec. If the average emf induced in the coil is

$220V$ , the self inductance of the coil is :

0%
$5H$
0%
$6H$
0%
$11H$
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$12H$

Explanation

$emf=L\dfrac { di }{ dt }$

$L=\dfrac { emf }{ (\dfrac { di }{ dt } ) }$

$L=\dfrac { 220 }{ (\dfrac { 10 }{ 0.5 }) }$

$=11H$

Two conducting circular loops F and G are kept in a plane on either side of a straight current-carrying wire as shown in the figure below.
If the current in the wire decreases in magnitude, the induced current in the loops will be

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clockwise in F and clockwise in G.
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anti-clockwise in F and clockwise in G.
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clockwise in F and anti-clockwise in G.
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anti-clockwise in F and anti-clockwise in G.

If ‘N’is the number of turns in a coil, the value of self inductance varies as

0%
$N^{0}$
0%
$N$
0%
$N^{2}$
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$N^{-2}$

Explanation

Formula for self-inductance of a coil:

$L = \frac{\mu_0 N^2 A}{l}$
Hence , it varies as

$N^2$

What is the SI unit of self-inductance ?

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Henry
0%
Tesla
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Weber
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Gauss

An induced e.m.f. is produced when a magnet is plunged into a coil. The strength of the induced e.m.f. is independent of

0%
the strength of the magnet
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number of turns of coil
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the resistivity of the wire of the coil
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speed with which the magnet is moved

Explanation

$\epsilon=-\dfrac {Nd\phi}{dt}\Rightarrow \epsilon \propto N$ and

$\epsilon \propto \dfrac {d\phi}{dt}$

If so the speed of magnet is fast then correspondingly rate of change of flux is fast &

$\epsilon$ increases. It does not depend on the resistance of coil.

An emf is induced in a coil when ______ linked with it changes.

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The magnetic flux
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Area
0%
No. of turns
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All of the above

The magnitude of induced emf increases with the increase in the number of turns in a closed coil.

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True
0%
False

In electromagnetic induction, the induced e.m.f. is independent of:

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Change of flux
0%
Resistance of circuit
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Number of turns of the coil
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None of the above

When the current through the electromagnet of a relay reaches a particular value

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It breaks the circuit
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It open the circuit by pulling in an iron contact
0%
It closes the circuit by pulling in an iron contact
0%
Both A or C

The phenomenon of electromagnetic induction was discovered by

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Lenz
0%
Maxwell
0%
Fleming
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Faraday

Whenever the magnetic flux linked with a coil changes, an induced e.m.f. is produced in the circuit. The e.m.f. lasts

0%
for a short time
0%
for a long time
0%
for ever
0%
so long as the change in flux takes place

The AC produced in India changes its direction in every :

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$\displaystyle \frac{1}{100}$ second
0%
100 second
0%
50 second
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None of these

Explanation

Answer is A.

In India, the frequency of AC voltage is

$50\ Hz$ .

It means

$50$ waves will be produced in

$1\ s$ .

In one wave, the direction is changed

$2$ times.

Thus, in 50 waves, the direction will be altered

$50\times 2=100$ times.

i.e. the direction is changed

$100$ times in

$1\ s$ .

Thus the direction is changed in every

$\dfrac{1}{100}\ s$ .

Alternating current is one which changes in its :

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direction
0%
magnitude
0%
magnitude and direction both
0%
none

Current is induced in a coil by electromagnetic induction when :

0%
Only the coil moves in a magnetic field.
0%
Only the magnet moves towards the coil.
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Coil and the magnet move with respect to each other.
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None of the above

Which of the following defines electromagnetic induction:

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When the magnetic field associated with a coil changes, an induced electric current flows through the coil.
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Electric current induces magnetic field near the wire carrying current.
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Two permanent magnets exert force on each other.
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Electrolyte disintegrates into ions in a battery.

In the process of electromagnetic induction, the magnitude of the induced emf depends on:

Select the correct options from the following

0%
The number of turns of the coil
0%
The magnetic flux linked with the coil
0%
The rate of change of magnetic flux linked with the coil
0%
Area of the coil

Explanation

In the process of electromagnetic induction, the magnitude of the induced emf depends on:

Magnetic strength of the core in the coil of wire. (stronger --> bigger emf)

Number of turns of wire in the coil (more turns --> bigger emf)

The cross-sectional area of the coil (bigger area --> bigger emf).

speed of magnet (fast you move the magnet) into/out of the coil. (Faster --> bigger emf)

But Faraday's law seems to not show all four of these:

induced emf

$E=N\dfrac{Δϕ}{Δt}.$

An emf is induced in an aeroplane during its ascent and descent in east-west direction due to

0%
The horizontal component of the earth's magnetic field
0%
The vertical component of the earth's magnetic field
0%
Both 1 and 2
0%
None of the above

Explanation

From the figure we have.An emf is induced in an aeroplane during its ascent and descent in east-west direction due to the horizontal component of the earth's magnetic field.

$H=B_ecosQ$

Magnet is moved towards a coil first slowly and then quickly, then the induced emf is:

0%
More in the first case
0%
Independent of the speed of the magnet
0%
Same in both the cases
0%
More in the second case

Explanation

According to Faraday's Law when the magnetic flux linking a circuit changes, an electromotive force is induced in the circuit proportional to the rate of change of the flux linkage

And the emf induced

$(E) =$

$-\dfrac{\mathrm{d\phi} }{\mathrm{d} t}$

Since the rate of changing of flux will be more in second case induced emf will also be more in second case.

Therefore, D is correct option.

An AC generator is a device which converts:

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Electrical energy to mechanical energy.
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Heat energy to electrical energy.
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Heat energy to light energy.
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Mechanical energy to electrical energy.

Check the incorrect statement: When a magnet is moved into a coil the strength of the current depends on:

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The number of turns in the coil
0%
The speed with which the magnet moves
0%
The resistance of the coil
0%
None of the above

Explanation

According to Faraday's Law when the magnetic flux linking a circuit changes, an electromotive force is induced in the circuit proportional to the rate of change of the flux linkage

And the emf induced

$(E) =$

$-\dfrac{N\mathrm{d\phi} }{\mathrm{d} t}$

$\Rightarrow current = \dfrac{E}{R}$

$=$

$-\dfrac{N\mathrm{d\phi} }{R\mathrm{d} t}$

$eq...1$

So when a magnet is moved into a coil the strength of the current depends on

A- The number of turns in the coil (N).So A is correct statement which can be seen from eq..1

B- The speed with which the magnet moves

$(\dfrac{\mathrm{d\phi} }{\mathrm{d} t})$ .So B is also correct statement which can be seen from eq..1

C- The resistance of the coil (R).So C is also correct statement which can be seen from eq..1

Therefore, D is correct option Since all are correct statements.

Production of electricity from magnetism is called

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Electric field
0%
Magnetic field lines
0%
Electromagnetic induction
0%
Magnetic induction

Which of following can induce the maximum induced voltage?

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1 amp dc.
0%
1 amp 1 Hz.
0%
1 amp 100 Hz.
0%
20 amp dc.

Explanation

If we apply a high-frequency supply of the same peak voltage to the coil, the current still is being delayed by 90

$^o$ but the time it requires to reach its maximum value has been reduced due to the increase in frequency. Because the frequency is inversely proportional to time (T). Hence, the rate of change of the flux within the coil has also increased due to the increase in frequency. Hence, the induced EMF is maximum in case of 1 amp 100 Hz supply source in the coil.

Electromagnetic induction is the :

0%
charging of a body with a positive charge
0%
production of current by relative motion between a magnet and a coil
0%
rotation of the coil of an electric motor
0%
generation of magnetic field due to a current carrying solenoid

Explanation

$\bullet$ Electromagnetic induction is the production of current by relative motion between a magnet and a coil.

$\bullet$ The phenomenon of producing electric current in a conductor by moving it perpendicular to a magnetic field or vice -versa, is called electro magnetic induction.

A coil of insulated copper wire is connected to a galvanometer. What would happen if a bar magnet is (i) Pushed into the coil?

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The galvanometer shows deflection
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The galvanometer do not show any deflection
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Current increases inside the coil
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none

When a straight wire is moved up and down rapidly between two poles of a horseshoe magnet then _______ is produced in the wire.

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magnetic field
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magnetic current
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electric current
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none

In the field of electromagnetism, the term 'EMI' stands for:

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Electromotive Impact
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Electromagnetic Induction
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Electromotive inertia
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none of these

Mark the incorrect statement.

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electric current produces magnetism.
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magnets can produce electric current.
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magnets can't produce electric current.
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a and b

Explanation

A magnet can generate electric current by the means of electromagnetic induction. When there is a relative motion between a coil and a magnet, the magnetic flux linked to the coil changes due to that relative motion. This change in magnetic flux will induce a current in the coil, which is known as

$induced\ current$ .

Which of the following uses the principle of electromagnetic induction?

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Electric generator
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Water Cooler
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Solar Cell
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Electrical Battery

Electric current produced by a moving straight wire in a magnetic field is?

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Magnetic Current
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Electrostatic Current
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Induced Current
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None

CBSE Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 1 - 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 1 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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