MCQ Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 9

In a long hollow vertical metal pipe a magnet is dropped. During its fall, the acceleration of magnet

0%
will decrease linearly
0%
will decrease upto a value which is less than $g$ .
0%
will decrease to zero and will attain a terminal speed
0%
may increase or decrease

Explanation

The falling magnet will experience an upward force and

so the acceleration of magnet will decreases and then the magnet will a quire a terminal speed.

Hence

$C$ is the correct answer.

When current flowing in a coil changes from

$3A$ to

$2A$ is one millisecond,

$5\ volt$ emf is induced in it. The self-inductance of the coil will be

0%
$zero$
0%
$5kh$
0%
$5h$
0%
$5\ mh$

Explanation

As we know,

$EMF=L\times\dfrac{di}{dt}$

$5=L\times\dfrac{3-2}{0.001s}$

$L=\dfrac{5}{1000}$

$L=5mh$

The self induced e.m.f.in a 0.1 H coil when the current in it is changing at the rate of 200 ampere/second is

0%
$8\times 10^{-4}V$
0%
$8\times 10^{-5}V$
0%
$20 V$
0%
$125 V$

Explanation

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

$\Rightarrow e = 0.1 \times 200 = 20 \,V$

An electron beam is moving near to a conducting loop then the induced current in the loop:-

0%
clockwise
0%
anticlockwise
0%
both
0%
no current

A square conducting loop, whose plane is perpendicular to a uniform magnetic field, moves with velocity

$v$ normally to the magnetic field. If opposite sides of the loop, perpendicular to its velocity, remain in two mutually opposite uniform magnetic fields of strength

$B$ , the induced electromotive force in this coil will be ______ . Length of each side is

$l$ .

0%
$2Bvl$
0%
$Bvl$
0%
$0$
0%
$\dfrac{Bvl}{2}$

As shown in the figure, a magnet is moved with a fast speed towards a coil at rest. due to this induced e.m.f, induced current and induced charge in the coil is $e$ , $i$ and $q$ respectively .if the speed of the magnet is doubled , the incorrect statement is:

0%
$e$ increases
0%
$q$ remains same
0%
$i$ increases
0%
$q$ increases

A point charge of

$0.1\ C$ is placed on the circumference of a non-conducting ring of radius

$1m$ which is rotating with a constant angular acceleration of

$1\ rad/sec^{2}$ . If ring starts it's motion from rest at

$t = 0$ , the magnetic field at the centre of the ring at

$t =10\ sec$ , is

0%
$10^{-6}T$
0%
$10^{-7}T$
0%
$10^{-8}T$
0%
none

An atom is placed in a uniform magnetic induction

$B$ such that plane normal of the electron orbit makes an angle of

$30^\circ$ with the magnetic induction. The torque acting on the orbiting electron is:-

0%
$\dfrac{ehB}{8 \pi m}$
0%
$\dfrac{ehB}{4 \pi m}$
0%
$\dfrac{2ehB}{3 \pi m}$
0%
zero

A circular coil has

$500$ turns of wire and its radius is

$5cm$ . The self inductance of the coil is:-

0%
$25 \times 10^{-3} mH$
0%
$25 mH$
0%
$50 \times 10^{-3} H$
0%
$50 \times 10^{-3} mH$

Explanation

$\varphi = Li$

$\Rightarrow NBA=Li$ Since magnetic field at the center of circular coil

carring current is given by

$B = \dfrac{{{\mu _0}}}{{4\pi }}.\dfrac{{2\pi Ni}}{r}$

$\therefore$

$N.\dfrac{{{\mu _0}}}{{4\pi }}.\dfrac{{2\pi Ni}}{r} = Li$

$\Rightarrow L = \dfrac{{{\mu _0}{N^2}\pi r}}{2}$

Hence,

self conductance of coil is

$= \dfrac{{4\pi \times 10 \times 500 \times 500 \times \pi \times 0.05}}{2}$

$=25mH$

Hence,

self conductance of coil is

$25mH$

So option

$B$ is correct answer.

In an

$AC$ Generator, if the number of turns for the armature coil is doubled keeping all other specifications to be the same, which of the following applies to the voltage generated

0%
Peak Voltage will remain the same but the rms value of Voltage generated double
0%
Peak Voltage will double but the rms value of Voltage generated will remain the same
0%
Both the peak Voltage and the rms value of Voltage generated will double.
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Both the peak Voltage and the rms value of Voltage generated will reduce to half the original value.

The phenomena of a electromagnetic induction is

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The process of charging a body
0%
The process of generating magnetic field due to current passing through a coil
0%
Producing induced current in a coil due to relative motion between a magnet and a coil
0%
All the above

A long solenoid of diameter

$0.1\ m$ has

$2 \times {10^4}$ turns per metre.At the centre of the solenoid, a coil of

$100$ turns and radius

$0.01\ m$ is placed with its axis coinciding with the solenoid axis.The current in the solenoid reduces at a constant rate to

$0\ A$ from

$4\ A$ in

$0.05\ s$ . If the resistance of the coil is

$10 \ {\pi ^2}\Omega ,$ the total charge flowing through the coil during this time is.

0%
$32\ \pi \mu C$
0%
$16\ \mu C$
0%
$32\ \mu C$
0%
$16\ \pi \mu C$

Explanation

Given,

Number of turns,

$n=100$

Radius,

$r=0.01\,m$

Resistance,

$R=10\pi^2 \Omega$

As we know,

$\epsilon=-N\dfrac{d\phi}{dt}$

$=\dfrac{\epsilon}{R}=-\dfrac NR\dfrac{d\phi}{dt}$ ,

$\Delta I=-\dfrac NR\dfrac{d\phi}{dt}$

$\dfrac{\Delta}{\Delta t}=-\dfrac NR\dfrac{\Delta\phi}{\Delta t}\implies \Delta q=-[\dfrac NR(\dfrac{\Delta \phi}{\Delta t})]\Delta t$

$-$ ve sign shoes that induced emf opposes the change in flux.

$\Delta q=\dfrac{\mu_0 ni\pi r^2}{R}$

$\Delta q=\dfrac{4\pi\times 10^{-7}\times 100\times 4\times \pi\times (0.01)^2}{10\pi^2}=32\mu C$

The mutual inductance

$M_{12}$ of coil

$1$ with respect to coil

$2$ .

0%
Increases when they are bought nearer
0%
Depends on the current passing through the coils
0%
Increases when one of them is rotated about an axis
0%
Is not same as $M_{21}$ of coil $2$ with respect to coil $1$

Explanation

The mutual inductance

$M_{12}$ of coil

$1$ w.r.t. coil

$2$ increases when they are brought nearer and is the same as

$M_2$ of coil

$2$ with respect to coil

$1$ .

$M_{12}$ , i.e., mutual inductance of solenoid

$S_1$ with respect to solenoid

$S_2$ is given by

$M_{12} = \dfrac{\mu_0 N_1 N_2 \pi r_1^2}{l}$

Where signs are as usual.

Also,

$M_2$ , i.e., mutual inductance of solenoid

$S_2$ with respect to solenoid

$S_1$ is given by

$M_{21} = \dfrac{\mu_0 N_1 N_2 \pi r_1^2}{l}$

So, we have

$M_{12} = M_{21} = M$

1837751_1216130_ans_872e99b3cfb74263bf25a6ed17e6597f.png

The coefficients of self induction of two inductance coils arc 0.0 1H and 0.03H respectively. When they are connected in series so as to support each other. then the resultant self inductance becomes 0.06 Henry. The value of coefficient of mutual induction will be-

0%
0.02 H
0%
0.05 H
0%
0.01 H
0%
ZERO

Explanation

$L1=0.01$

$L2=0.03$

$L eff =0.06$

$Leff = L1+L2+2M$

$0.06=0.01+0.03+2M$

$2M = 0.06-0.03$

$M=0.03/2$

$=0.015$

When a coil is connected to a D.C source of emf 12V,a current of 4 amp flows in it.If same coil is connected to 12V,50Hz AC source, the current is 2.4 A. The self inductance of the coil is (in henry)

0%
$\frac { 1 }{ 2\pi }$
0%
$\frac { 1 }{ 10\pi }$
0%
$\frac { 1 }{ 25\pi }$
0%
$\frac { 1 }{ 5\pi }$

An average emf of

$20 V$ is induced in an inductor when the current in it is changed from

$2.5 A$ in one direction to the same value in the opposite direction in

$0.1$ s. The self-inuctance of the inductor is

0%
$zero$
0%
$0.2 H$
0%
$0.4 H$
0%
$1 H$

Explanation

Given that,

Average emf $V=20V$

Current $i=2.5\,A$

We know that,

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

$20=L\times \dfrac{5}{0.1}$

$L=\dfrac{20}{50}$

$L=0.4\,Henry$

Hence, the self-inductance is $0.4\ henry$

A magnetic field

$\vec {B}=B_{0}\hat {k}$ is present in elliptical region given by

$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$ . If conducting rod of length

$2b (< 2a)$ is kept parallel to

$y-$ axis, The rod starts moving with velocity

$v \hat {i}$ and at

$t=0$ centre of rod is at

$(-a, 0)$ . Find motional

$EMF$ in rod as function of time

$t.t < \dfrac{2a}{v}$ .

0%
$2Bbv\sqrt{1-\dfrac{(vt-a)^{2}}{a^{2}}}$
0%
$2bv^{2}\left[\dfrac{a^{2}t}{b^{3}}\right]$
0%
$2Bav\sqrt{1-\dfrac{(vt-a)^{2}}{a^{2}}}$
0%
$2Bbv$

An electron having kinetic energy T is moving in a circular orbit of radius R perpendicular to a uniform magnetic induction

$\vec { \mathrm { B } }$ If kinetic energy is doubled and magnetic induction tripled, the radius will

0%
$\frac { 3 R } { 2 }$
0%
$\frac{{\sqrt 2 }}{3}R$
0%
$\sqrt { \frac { 2 } { 9 } } R$
0%
$\sqrt { \frac { 4 } { 3 } } R$

Explanation

We know

$,$

$R = \frac{{\sqrt {2mk} }}{{qB}}$

$R' = \frac{{\sqrt {2m2k} }}{{q\left( {3B} \right)}}$

$= \frac{{\sqrt 2 }}{3}\frac{{\sqrt {2mk} }}{{qB}}$

$= \frac{{\sqrt 2 }}{3}R$

Hence,

option

$(B)$ is correct answer.

In electromagnetic induction, The induced e.m.f in a coil is
in depended of

0%
Change in flux
0%
Resistance of coil
0%
Time
0%
None

A thin copper wire of length 100 meters is wound as a solenoid of length

$\lambda$ and radius r. Its self-inductance is found to be L. Now if the same length of wire is wound as a solenoid of length

$\lambda$ but the radius r/2, then its self inductance will be:

0%
4 L
0%
2 L
0%
L
0%
L/2

Explanation

As we know that, self inductance =

${ \mu }_{ \circ }{ n }^{ 2 }lA$

where n= number of turns per unit length

l=length of solenoid

A=cross section

So, self inductance(L) =

${ \mu }_{ \circ }{ n }^{ 2 }l\pi { R }^{ 2 }$

$\Rightarrow L={ \mu }_{ \circ }{ \left( \frac { N }{ l } \right) }^{ 2 }l\pi { R }^{ 2 }$ (Where N= Total number of turns)

$\Rightarrow L=\frac { { \mu }_{ \circ }N^{ 2 }\pi { R }^{ 2 } }{ l }$

Now, it is given that when length is

$\lambda$ and radius is'r' then ,

${ { L } }={ \mu }_{ o }{ \left( \frac { 100 }{ 2\pi r } \right) }^{ 2 }\frac { \pi { r }^{ 2 } }{ \lambda }$

Now, in second case length is

$\lambda$ and radius is 'r/2' then,

${ { L } }'={ \mu }_{ o }{ \left( \frac { 100 }{ 2\pi \frac { r }{ 2 } } \right) }^{ 2 }\frac { \pi }{ \lambda } { \left( \frac { r }{ 2 } \right) }^{ 2 }$

$\Rightarrow{ { L } }'={ \mu }_{ o }{ \left( \frac { 100 }{ 2\pi r } \right) }^{ 2 }\frac { \pi { r }^{ 2 } }{ \lambda } =L$

$\Rightarrow L'=L$

Device/devices changing electrical energy into mechanical energy is/are _____
I. Electric generator II. Electric motor
III. Voltmeter IV. Ammeter

0%
I and II
0%
II and III
0%
II, III and IV
0%
Only II

Two coils of self inductance

$6 mH$ and

$8 mH$ are connected in series and are adjusted for highest coefficient of coupling. Equivalent self inductance

$L$ for the assenbly is approximately

0%
$50 mH$
0%
$36 mH$
0%
$28 mH$
0%
$18 mH$

A coil of inductance 300 mH and resistance

$2 \Omega$ is connected to a source of voltage 2V. The current reaches half of its steady state value in

0%
0.05 s
0%
0.1 s
0%
0.105 s
0%
0.3 s

Explanation

Hence, option

$B$ is the correct answer.
1380099_1362908_ans_fae33b8903854f51a4d72ee086b36b33.png

The inductance between

$A$ and

$D$ is:

0%
$3.66\ H$
0%
$9\ H$
0%
$0.66\ H$
0%
$1\ H$

The dimensions of self- induction are :

0%
$[ML{T^{ - 2}}]$
0%
$[M{L^2}{T^{ - 1}}{A^{ - 2}}]$
0%
$[M{L^2}{T^{ - 2}}{A^{ - 2}}]$
0%
$[M{L^2}{T^{ - 2}}{A^2}]$

Explanation

$\phi =LI$

and also

$\phi =B.A = B^{2}$

[B]

$\frac{MLT^{2}}{LI^{-1}A T}$

$[B]\frac{MLT^{-2}}{LA}$

$[B]MT^{-2}A^{-1}$

$[\phi ]=MT^{-2}L^{2}A^{-1}$

thee finally dimension of inductance

[L]=

$\frac{[\phi ]}{[I]}$

[L]=

$MT^{-2}C^{2}A^{-2}$

1243345_1331112_ans_94406114829c4f7c958119e9d4859de1.png

A conducting wire frame is placed in a magnetic field which is directed in to the paper. The magnetic field increasing at a constant rate. The directions of induced current in wires AB and CD are:

0%
B to A and D to C
0%
A to B and C to D
0%
A to B and D to C
0%
B to A and C to D

Explanation

$Solution:$ Inward magnetic field (')

increasing. Therefore, induced

current in both the loops should be

anticlockwise. But as the area of

loop on right side is more, induced

emf in this will be more compared

to the left side loop

$\left(e=-\frac{d \varphi}{d t}=-A . \frac{d B}{d t}\right)$

Therefore net current in the

complete loop will be in a direction

shown above.

$So,the$

$correct$

$option:A$

1997737_1352769_ans_4ac4d53e087b4cc9ba43caa1ba7b67bf.jpeg

Magnetic flux in the circuit containing a coil of resistance 2 ohm changes from 2.0 wb to 10 wb in 0.2 sec. the charge passed through the coil in this time is

0%
0.8 C
0%
1.0 C
0%
5.0C
0%
4.0C

Electromagnetic induction means ......... .

0%
charging of an electric conductor.
0%
production of magnetic field due to a current flowing through a coil.
0%
generation of a current in a coil due to relative motion between the coil and the magnet.
0%
motion of the coil around the axle in an electric motor.

A conducting ring of radius r is rolling without slipping with a constant angular velocity

$\omega$ ( given figure). if the magnetic field strength is B and is directed into the page then the emf induced across PQ is:

0%
$B \omega r^2$
0%
$\frac { B \omega r^2 }{2}$
0%
$4 B \omega r^2$
0%
$\frac { \pi^2r^3 B \omega }{8}$

Explanation

$\textbf{Solution}:$

emf induced is given by,

$\left | \vec{E} \right |=\dfrac{Bl^2}{2}\omega$

where

$l = \sqrt2 r$

$\therefore$

$\left | \vec{E} \right |= \dfrac{B\times (\sqrt 2 r)^2}{2}\omega$ =

$B\omega r^2$

$\textbf{Hence A is the correct option}$

A cylindrical magnet is kept along the axis of a circular coil. On rotating the magnet about its axis. the coil will have induced in it

0%
No current
0%
A current
0%
Only an e.m.f.
0%
Both an e.m.f. and a current

Explanation

No, Current will not be induced in the coil if the coil is rotated about it's axis

Formula for the flux is.

$\phi$ = NBA = constant

$\therefore$ i =

$\frac{d\phi}{dt}$ = 0

i = 0

A wire of fixed length is wound on a solenoid of length l and radius r. Its self inductance is found to be L. Now, if same wire is wound on a solenoid of length

$\dfrac{l}{2}$ and radius

$\dfrac{r}{2}$ , then self inductance will be

0%
2L
0%
$\dfrac{L}{2}$
0%
3L
0%
4L

Explanation

$L=\mu _{0}\times n^{2}\times V$ (n

$\rightarrow$ Number of turns per unit length , V

$\rightarrow$ Volume of cylinder)

$=\mu _{0}\times \frac{N^{2}}{l^{2}}\times \pi r^{2}\times l$

$L=\frac{\mu _{0}N^{2}\pi r^{2}}{l}$

Let length of wire be

$l_{1}$

$l_{1}=(2\pi r)\times N \Rightarrow N=(\frac{l_{1}}{2\pi r})$

$L=\frac{\mu _{0}\pi r^{2}}{l}\times \frac{l_{1}^{2}}{4\pi ^{2}r^{2}}=\frac{\mu _{0}l_{1}^{2}}{4\pi l}$

$\therefore L \propto \frac{1}{l}\Rightarrow \frac{L_{2}}{L_{1}}=\frac{l_{1}}{l_{2}}=\frac{l}{\frac{l}{2}}=2$

$L_{2}=2L_{1}$

1226283_1405901_ans_ad815d440f1e4d719df902ed6423a256.jpeg

Electro motive force induced by motion of conductor across magnetic field is called

0%
EMF
0%
motional EMF
0%
static EMF
0%
rotational EMF

The length of a thin wire require to manufacture a solenoid of length

$l=100$ cm and inductance

$L=1 mH$ , if the solenoid's cross-sectional diameter is considerably less than its length is:

0%
$1\ km$
0%
$0.10\ km$
0%
$0.010\ km$
0%
$10\ km$

Explanation

Let

$l_1$ is the length of wire,

$l_1=2\pi rN\Rightarrow Nr=\dfrac {l_1}{2\pi}\Rightarrow L=\dfrac{\mu_0\pi r^2N^2}{l}=\dfrac {\mu_0\pi}{l}\dfrac {l_1^2}{4\pi^2}$

$\Rightarrow l_1=\sqrt {\dfrac {Ll4\pi}{\mu_0}}=\sqrt {\dfrac {10^{-3}\times 1\times 4\pi}{4\pi\times 10^{-7}}}=10^2\ m=0.10 \ km$

A magnetic field of flux density

$10 T$ acts normal to a coil of

$50$ turns having

$100$

$cm^2$ area. The e.m.f. induced if the coil is removed from the magnetic field in

$0.1$ seconds is:

0%
$50 V$
0%
$60 V$
0%
$80 V$
0%
$40 V$

Explanation

In this case,

$B=10T\, ,N=50\,turns\, ,A=100\,cm^2=10^{-2}m^2$ and

$dt=0.1\,s$

We know that,

$\phi_1=NBA\,Cos\,\theta=NBA$ (as

$Cos\theta=Cos0=1$ )

Therefore,

$\phi_1=50\times 10\times 10^{-2}=5\,weber$

When the coil is removed from the magnetic field,

$B=0$

So,

$\phi_2=0$

Therefore,

$\phi_1-\phi_2=d\phi=0-5=-5\,weber$ .

We also know that

$e=-\dfrac{d\phi}{dt}=\dfrac{-5}{0.1}=50\,V$

Hence, the induced emf if the coil is removed from the magnetic field in

$0.1$ second is

$50\,V$ .

The inductance of a solenoid

$0.5\ m$ long of cross-sectional area

$20\ cm^2$ and with

$500$ turns is

0%
$12.5\ mH$
0%
$1.25\ mH$
0%
$15.0\ mH$
0%
$0.12\ mH$

If the magnetic flux linked up with a coil is changed by 40%, the energy storied in the coil changes by

0%
21%
0%
80%
0%
40%
0%
96%

Self inductance of coil measures

0%
electrical inertia
0%
electrical friction
0%
induced emf
0%
induced current

The magnetic flux linked with a coil varies as

$\phi =3t^2+4t+9$ . The magnitude of the emf induced at

$t=2$ seconds is?

0%
$8$ V
0%
$16$ V
0%
$32$ V
0%
$64$ V

Explanation

$\phi =\dfrac{d\theta}{dt}$

$=6t+4=6(2)+4$

$=16V$ .

Which type of current can be obtained from an electric generator?

0%
Only DC
0%
Only AC
0%
Both AC and DC at a time
0%
It is dependent on the type of an electric generator.

In a discharge tube at 0.02 mm, there is formation of

0%
crooke's dark space
0%
faraday's dark space
0%
both space partly
0%
none of these.

When the current of

$2$ amp increases to

$18$ amp in

$0.05$ second, the induced emf is

$20 V$ in a circular conducting coil. The self-inductance of the coil is

0%
$62.5 mH$
0%
$6.25 mH$
0%
$50.0 mH$
0%
None of these

Explanation

$| e | = L\dfrac{di}{dt}$

$\Rightarrow 20 = L \times \dfrac{(18 - 2)}{0.05}$

$\Rightarrow L = 62.5 \,mH$

Figure shows the cross-sectional view of the hollow cylindrical conductor with inner radius

$'R'$ and outer radius

$'2R'$ , cylinder carrying uniformly distributed current

$i$ along its axis. The magnetic induction at point

$'P'$ at a distance

$3R/2$ from the axis of the cylinder will be

0%
Zero
0%
$\dfrac{5 \mu_0i}{72 \pi R}$
0%
$\dfrac{7 \mu_0i}{18 \pi R}$
0%
$\dfrac{5 \mu_0i}{36 \pi R}$

Explanation

By using the formula for magnetic field due to cylindreical conductor:

$B=\dfrac{\mu_{0}i}{2\pi r}\left(\dfrac{r^{2}-a^{2}}{b^{2}-a^{2}}\right)$

here

$r=\dfrac{3R}{2}, a=R, b=2R$

$\Rightarrow B=\dfrac{\mu_{0}i}{2\pi \left(\dfrac{3R}{2}\right)}\times \left\{\dfrac{\left(\dfrac{3R}{2}\right)^{2}-R^{2}}{(2R^{2})-R^{2}}\right\}$

$=\dfrac{5.\mu_{0}i}{36\pi R}$

The current carrying wire and the rod AB are in same plane. The rod moves parallel to the wire with a velocity v. Which on of the following statement is true about induced emf in the rod?

0%
End A will be at lower potential with respect to B
0%
A and B will be at same potential
0%
There will be no induced emf in the rod.
0%
Potential at A be higher than that at B

A long straight wire is placed along the axis of a circular ring of radius

$R$ . The mutual inductance of this system is

0%
$\cfrac{{\mu}_{0}R}{2}$
0%
$\cfrac{{\mu}_{0}\pi R}{2}$
0%
$\cfrac{{\mu}_{0}}{2}$
0%
$0$

A constant current

$I_{0}$ is passing through a long straight wire (shown in figure).
A rectangular loop of total resistance

$R$ is moving parallel to the wire. Then:

0%
the heat generated in the loop is with constant rate.
0%
current in the loop is zero
0%
velocity of loop decreases according to Lenz's law
0%
none of the above

The magnetic flux

$\phi$ (in weber) in a closed circuit of resistance

$10\ ohm$ varies with time

$t$ (in second) according to equation

$\phi=6t^{2}-5t+1$ . The magnitude of induced current at

$t=0.25\ s$ is:

0%
$1.2\ A$
0%
$0.8\ A$
0%
$0.6\ A$
0%
$0.2\ A$

When current changes from

$13\ A$ to

$7\ A$ in

$0.5\ sec$ through a coil, the emf induced is

$3\times 10^{-4}\ V$ . The coefficient of self induction is:

0%
$25\times 10^{-3}\ H$
0%
$25\times 10^{-4}\ H$
0%
$25\times 10^{-5}\ H$
0%
$25\times 10^{-6}\ H$

A student wishes to demonstrate electromagnetic induction.
He has a magnet and connecting wires.
Which other apparatus does he need?

$\ \checkmark = needed$ ,

$\ X = not needed$

0%
Voltmeter - $\checkmark$ , battery - $\checkmark$
0%
Voltmeter - $\checkmark$ , battery - $X$
0%
Voltmeter - $X$ , battery - $\checkmark$
0%
Voltmeter - $X$ , battery - $X$

Explanation

$(B)$ EMI (electromagnetic induction ) is the production of voltage or electromotive force due to a change in the magnetic field.

Changing magnetic feild is created by moving magnet

emf is generated when the magnet is moved away or toward the coil so to detect it voltmeter will be required

A magnet is suspended from a spring so that it can move freely inside a stationary coil. The coil is connected to a sensitive centre-zero galvanometer.
The magnet repeatedly moves slowly up and down.
What does the galvanometer show?

0%
a constantly changing reading
0%
a steady reading to the left
0%
a steady reading to the right
0%
a steady reading of zero

Explanation

According to faraday law

The voltage is generated in a closed loop if the magnetic flux linked with it changes with respect to time.

$E \propto \dfrac{\partial \Phi _B}{\partial t}$

Since the flux linked with the coil changes with time so, there will be generation of voltage in the coil.

Therefore the reading of galvanometre constantly changes.

A fan blade of length

$2a$ rotates with frequency

$f$ cycle per second perpendicular to magnetic field

$B$ . Then potential difference between centre and end of blade is:

0%
$\pi Ba^{2}f$
0%
$4\pi Ba f$
0%
$4\pi a^{2} Bf$
0%
$2\pi a Bf$

CBSE Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 9 - 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 9 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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