MCQ Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 8

The equivalent quantity of mass in electricity is :

0%
current
0%
self inductance
0%
potential
0%
charge

Explanation

B. self-inductance

The equivalent quantity of mass in electricity is self- inductance. Self-inductance of an inductor is referred to as the "inertia of electricity" .

Kinetic energy in Mechanics,

$\dfrac{1}{2}mv^2$

where m= mass of an object.

v= velocity of the object.

The magnetic energy stored in an inductor,

K'=

$\dfrac{1}{2}LI^2$

where L= self-inductance of the inductor and,

I= current through the inductor.

So, from the above equation, we can conclude that self-inductance is equivalent to mass in electricity.

A jet plane is travelling towards west at a speed of

$1800\ km/h$ . What is the voltage difference developed between the ends of the wing having a span of

$25\ m$ , if the Earth's magnetic field at the location has a magnitude of

$5 \times {10}^{-4}\ T$ and the dip angle is

$30^o$ .

0%
0.31 V
0%
3.1 V
0%
5 V
0%
10 V

Explanation

$v=1800\quad km/h=1800\times \dfrac { 5 }{ 18 } =500m{ s }^{ -1 }$

$l\rightarrow$ distance between the ends of the wings=

$25m$

$B=5\times { 10 }^{ -4 }T$

$\\ \theta =30°$

Motional emf

$\varepsilon ={ B }_{ v }vl\\$ [

$B_v$ -verticle component of earth's magnetic field]

$\varepsilon =5\times { 10 }^{ -4 }\times \sin { 30°\times 500\times 25=3.1V }$

Mutual inductance of two coils can be increased by

0%
decreasing the number of turns in the coils
0%
increasing the number of turns in the coils
0%
winding the coils on wooden cores
0%
none of these.

Explanation

As M =

$\dfrac{\mu_0N_1N_2A}{l}$
i.e M can be increased by increasing the numbers of turns in the coils.

A circular copper disc 10 cm in diameter rotates at 1800 revolution per minute about an axis through its centre and at right angles to disc. A uniform field of induction B of 1 Wb

$m^2$ is perpendicular to disc. What potential difference is developed between the axis of the disc and the rim ?

0%
0.023 V
0%
0.23 V
0%
23 V
0%
230 V

Explanation

Here,

$l = r = 5\, cm = 5 \times 10^{-2} m,$
B = 1 Wb

$m^{-2}$

$\omega = 2 \pi \left( \dfrac{1800}{60} \right) \, rad \, s^{-1} = 60 \pi \, rad \, s^{-1},$

$\epsilon \, = \, \dfrac{1}{2} Bl^2 \omega \, =\, \dfrac{1}{2} \times 1 \times (5 \times 10^{-2})^2 \times 60 \pi = 0.23 V$

A small square loop of wire of side

$l$ is placed inside a large square loop of wire of side

$L(L>>l)$ .
The loops are co-planar and their centres coincide. The mutual inductance of the system is proportional to:

0%
$l/L$
0%
${l}^{2}/L$
0%
$L/l$
0%
${L}^{2}/l$

Explanation

The correct answer is option (B).

Hint: Calculate the magnetic flux and use it to find out the mutual inductance.

Step 1: Calculate magnetic flux of smaller loop.

Let the length of the larger square and the smaller square be $L$ and $l$ respectively. Accordingly, the current in these two square loops are $I$ and $i$ .

According to Faraday’s first law of electromagnetic radiation, whenever magnetic flux passes through surface changes with respect to time, an induced EMF is generated.
Magnetic field at a distance $\dfrac{L}{2}$ from the centre of the current carrying wire of length $L$ is given by:

$B_1=\dfrac{\mu _0}{4\pi}\dfrac{i}{\dfrac{L}{2}}2sin 45$

$B_1=\dfrac{\mu _0}{4\pi}\dfrac{i\times 2\times 2sin \times \dfrac{1}{\sqrt{2}}}{L}$

Magnetic field in the centre of square due to four wires of length $L$ will be:

$B_1=\dfrac{\mu _0}{4\pi}\dfrac{4\times i\times 2\times 2sin \times \dfrac{1}{\sqrt{2}}}{L}$

$B_1=\dfrac{\mu_0}{\pi}\dfrac{2\sqrt2i}{L}$

Since, the smaller loop is at the centre of the larger loop, so $L>>l$

In a small square, the magnetic field is uniform and is equal to $B_1$

Flux in the smaller loop will be:

$\phi =B_1l^2$

$\phi =\dfrac{\mu_0}{\pi}\dfrac{2\sqrt2il^2}{L}$

Step 2: Calculate the mutual inductance.

Mutual inductance is given by –

$M=\dfrac{\phi}{i}$

$M =\dfrac{\mu_0}{\pi}\dfrac{2\sqrt2l^2}{L}$

$M\propto \dfrac{l^2}{L}$

Hence, the mutual inductance of the system is proportional to $\dfrac{l^2}{L}$ .

The correct answer is option (B).

Who gave the principle of Electro-magnetic induction?

0%
Volta
0%
Oerstead
0%
Ampere
0%
Faraday

This

$Sl$ unit of inductance, the henry can be written as:

0%
weber/ampere
0%
volt-second/ampere
0%
joule/ $(ampere)^{2}$
0%
ohm-second

Explanation

$(A)$

The inductance of an inductor is given by:

$L=\dfrac{\phi}{I}$

So, its unit can be written as

$weber/Ampere$ .

$(B)$

Inductance can also be written as:

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

Here, the unit of inductance can be written as:

$L=\dfrac{volt}{\dfrac{ampere}{second}}$ i.e.

$\dfrac{volt-second}{ampere}$

$(C)$

The energy of inductor is given by:

$U=\dfrac{1}{2}LI^2$

The unit of inductance can be written as:

$L=\dfrac{Joule}{(ampere)^2}$

$(D)$

The energy can also be given as:

$U=I^2Rt=\dfrac12LI^2$

Here, unit can be given as:

$L=Ohm-second$

If 1 A of current is passed through CuSO

$_{4}$ solution for 10 seconds,then the number of copper ions deposited at the cathode will be about

0%
$1.6\times10 ^{19}$
0%
$3.1\times10 ^{19}$
0%
$4.8\times10 ^{19}$
0%
$6.2\times10 ^{19}$

Explanation

1A of current is passed through the solution for 10 seconds .

Then the total charge supplied

$1\times 10=10C\therefore 2$ electronic charge

$(3.2\times 10^{-19}C)$ liberates one

$Cu^{++}$

$\therefore$ Number of

$Cu^{++}$ ions liberates by 10 C charge

$=\dfrac{1}{3.2\times 10^{-19}}\times 10=3.1\times 10^{19}$

Two circular coils can be arranged in any of the three situations shown in the figure. Their mutual inductance will be

0%
maximum in situation $(a)$
0%
maximum in situation $(b)$
0%
maximum in situation $(c)$
0%
the same in all situations

Explanation

Reffer image.

Mutual induction

$\propto$ flux linkage between coils

1) flux linkage in coil (a) is maximum because both coils have same axis.

2) Flux linkage between coils is

$0$ because thus axis are perpendicular to each other.

3) Flux linkage between coils is

$0$ because their axis are perpendicular to each other .

So, mutual inductance between coils is maximum in case (a).

The self inductance of two solenoids A & B having equal length are same. If the number of turns in two solenoids A & B are

$100$ and

$200$ respectively. The ratio of radii of their cross-section will be __________.

0%
$2:1$
0%
$1:2$
0%
$1:4$
0%
$4:1$

Explanation

Self inductance of a solenoids is given by

$\dfrac{\mu_0 N^2A}{l}$ .

Given.

$\dfrac{\mu_0 N_1^2A_1}{l_1}$ =

$\dfrac{\mu_0 N_2^2A_2}{l_2}$

$\Rightarrow$

$\dfrac{100\times 100\mu_0 A_1}{l_1}$ =

$\dfrac{ 200\times200\mu_0 A_2}{l_2}$

$\Rightarrow$

$A_1 = 4A_2(\because l_1 =l_2)$

$\Rightarrow$

$\dfrac{r_1}{r_2}= 2$

Therefore, correct answer is

$2$ .

A magnetic flux of 5 microweber is linked with a coil when a current of 1 mA flows through it. Calculate self-inductance of the coil.

0%
5 mA
0%
10 mH
0%
15 mH
0%
20 mH

Explanation

The correct answer is A

Magnetic flux

$B=5mw=5\times 10^{-6}mw$

$current I= 1mA=10^{-3}A$

$self-inductanceL=\frac{Q}{I}$

$L=\frac{5\times 10^6}{10^{-3}}$

$L=5\times 10^{-3}$ Hen.

$L=5mh$

A source of 220 V is applied in an A C circuit . The value of resistance is 220

$\Omega$ . Frequency & inductance are 50Hz & 0.7 H then wattless current is

0%
0.5 amp
0%
0.7 amp
0%
1.0 amp
0%
None

Explanation

A source=

$220V$

The value of resistance=

$220 \Omega$

Frequency=

$50 Hz$

Inductance=

$0.7H$

Find the wattless current= ?

Wattless component of current is

$i=i_v\sin \theta$

$=\cfrac {Ev}{z}\sin \theta$

where,

$z=$ impedance of

$L-R$ circuit

$=\sqrt {R^2+L^2W^2}$ so,

$i=\cfrac {220}{\sqrt {R^2+L^2+W^2}}\sin \theta$ from impedance triangle,

$\sin \theta= \cfrac {LW}{\sqrt {R^2+L^2W^2}}$

$\Rightarrow i=\cfrac {220}{\sqrt {R^2+L^2W^2}}\cfrac {LW}{\sqrt {R^2+L^2W^2}}$

$=\cfrac {220}{R^2+L^2W^2}LW$

$=\cfrac {220 \times 0.7 \times 2 \Pi \times 50}{(220)^2+(0.7\times 2\Pi \times 50)^2}$

$=\cfrac {220 \times 220}{(220)^2+(220)^2}$

$=0.5 A$ .

A 60 volt - 10 watt bulb is operated at 100 volt - 60 Hz a.c. The inductance required is?

0%
2.56 H
0%
0.32 H
0%
0.64 H
0%
1.28 H

A simple pendulum with bob of mass

$m$ and conducting wire of length

$L$ swings under gravity through an angle

$2\theta$ . The earth's magnetic field component in the direction perpendicular to swing is

$B$ .The maximum potential difference induced across the pendulum is:

0%
$2BL\sin { \left( \cfrac { \theta }{ 2 } \right) } \sqrt { gL }$
0%
$BL\sin { \left( \cfrac { \theta }{ 2 } \right) } \sqrt { gL }$
0%
$BL\sin { \left( \cfrac { \theta }{ 2 } \right) } { \left( gL \right) }^{ 3/2 }$
0%
$BL\sin { \left( \cfrac { \theta }{ 2 } \right) } { \left( gL \right) }^{ 2 }$

Explanation

We have

$h=L(1-Cos\theta)$

Maximum velocity at equilibrium is given by,

$v^2=2gh=2gL(1-Cos\theta)$

$=2gL(2Sin^2(\dfrac{\theta}{2}))$

$v=2\sqrt{gL}sin(\dfrac{\theta}{2})$

Thus Maximum potential difference,

$V_{max}=BvL$

$=B\times 2\sqrt{gL}sin(\dfrac{\theta}{2}) L$

$=2BLsin (\dfrac{\theta}{2}) \sqrt{gL}$

A flat coil of 500 turns, each of area

$50cm^2$ , 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
0%
4.5 A

Two long parallel wires are at a distance

$2d$ apart. They carry steady equal current flowing out of the plane of the paper as shown. The variation of the magnetic field

$B$ along the line

$XX$ ' is given by

A solid conducting sphere of radius

$R$ is moved with a velocity

$V$ in a uniform magnetic field of strength

$B$ such that

$\bar { B }$ is perpendicular to

$\bar {V}$ . The maximum e.m.f. induced between two points of the sphere is :

0%
$RBV$
0%
$\sqrt { 2 }\ RBV$
0%
$\dfrac { RBV }{ 2 }$
0%
$2\ RBV$

An e.m.f. of 5 volt is produced by a self inductance, when the current changes at a steady rate from 3 A to 2 A in 1 millisecond. The value of self inductance is

0%
zero
0%
5 H
0%
5000 H
0%
5 mH

Explanation

Hint:- Use formula of self emf induced in inductor

Step 1: Note the given values

$Emf induced, e=5v$
Change in current dI= (2-3) A=-1A

$Time,dt=1ms$

$Inductace,L=?$

Step 2: Calculate self inductance

we know that em.f develop by the inductor due to change in current

$E=-L\dfrac{\mathrm{dI} }{\mathrm{d} t}=-L\dfrac{(2-3)}{10^{-3}}=L\times10^3$

$\Rightarrow L\times10^3=5$

$\Rightarrow L=5mH$

Hence the correct option is D

A uniform rod of resistance

$10\Omega$ is bent into a ring of radius

$20\ cm$ . The ring is placed in a uniform magnetic field of induction

$2.0T$ directed parallel to its plane. If ideal battery of emf

$5.0\ V$ is connected between two points of the ring to obtain maximum force on the ring, then the value of this maximum force is

0%
$0.4\ N$
0%
$0.8\ N$
0%
$1.2\ N$
0%
$1.6\ N$

The device that does not work on the principle of mutual induction is

0%
Induction coil
0%
Motor
0%
Tesla coil
0%
Transformer

Two long cylindrical coils with uniform winding of the same length and nearly the same radius having inductances

${ L }_{ 1 }$ and

${ L }_{ 2 }$ . The coils are coaxially inserted and connected to a current source as shown in the figure.The inductance L of the composite system is

0%
${ L }_{ 1 }+{ L }_{ 2 }$
0%
${ L }_{ 1 }{ L }_{ 2 }({ L }_{ 1 }+{ L }_{ 2 })$
0%
${ L }_{ 1 }+{ L }_{ 2 }+\sqrt { { L }_{ 1 }{ L }_{ 2 } }$
0%
${ L }_{ 1 }+{ L }_{ 2 }+2\sqrt { { L }_{ 1 }{ L }_{ 2 } }$

Two concentric co-planar circular loops of radii

$r_1$ and

$r_2$ carry currents of respectively

$i_1$ and

$i_2$ in opposite directions (one clockwise and the other anticlockwise.) The magnetic induction at the centre of loops is half that due to

$i_1$ alone at the centre. If

$r_2 = 2r_1$ . the value of

$i_2 / i_1$ is

0%
$2$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$
0%
$1$

Explanation

$\begin{array}{l} B=\frac { { { \mu _{ o } }{ i_{ 1 } } } }{ { 2{ r_{ 1 } } } } -\frac { { { \mu _{ o } }{ i_{ 2 } } } }{ { 2\left( { 2{ r_{ 1 } } } \right) } } \\ =\frac { { { \mu _{ o } }{ i_{ 1 } } } }{ { 2{ r_{ 1 } } } } -\frac { { { \mu _{ o } }{ i_{ 2 } } } }{ { 4{ r_{ 2 } } } } \\ { B^{ ' } }=\frac { { { \mu _{ o } }{ i_{ 1 } } } }{ { 2{ r_{ 1 } } } } \\ \Rightarrow B=\frac { { { B^{ ' } } } }{ 2 } \\ \Rightarrow \frac { { { \mu _{ o } }{ i_{ 1 } } } }{ { 2{ r_{ 1 } } } } -\frac { { { \mu _{ o } }{ i_{ 2 } } } }{ { 4{ r_{ 1 } } } } =\frac { { { \mu _{ o } }{ i_{ 1 } } } }{ { 4{ r_{ 1 } } } } \\ \Rightarrow \frac { { { \mu _{ o } }{ i_{ 1 } } } }{ { 4{ r_{ 1 } } } } =\frac { { { \mu _{ o } }{ i_{ 2 } } } }{ { 4{ r_{ 1 } } } } \\ \Rightarrow \frac { { { i_{ 1 } } } }{ { { i_{ 2 } } } } =1 \\ Hence, \\ option\, \, D\, \, is\, \, correct\, \, answer. \end{array}$

A long solenoid has

$500\ turns$ . When a current of

$2\ A$ is passed through it, the resulting magnetic flux linked with each turn of the solenoid is

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

0%
$4.0\ H$
0%
$2.5\ H$
0%
$2.0\ H$
0%
$1.0\ H$

Explanation

Given:-

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

$N=$ Number of turns

$=500$

$I=$ Current flowing

$=2A$

$L=$ Self-inductance

$=?$

Formula:-

$N\times \Phi=L\times I$

$\therefore 500\times 4\times 10^{-3}=L\times 2$

$\therefore L=\dfrac{500\times 4\times 10^{-3}}{2}$

$\therefore L=1H$

$\therefore$ Self-inductance of solenoid is

$1.0H$

The armature of a generator of resistance

$1\Omega$ is rotated at its rated speed and produces

$125V$ without load and

$115V$ with full load. The current in the armature coil is

0%
$240A$
0%
$10A$
0%
$1A$
0%
$2A$

Explanation

$R=1\Omega$

$E=125V$

$e=115V$ (Back emf)

we know,

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

$I=\dfrac{125-115}{1}$

$I=10 A$

So, the current in armature coil is 10 ampere

An alternating current is that current which changes regularly its

0%
Magnitude
0%
Direction
0%
Both a and b
0%
None of the above

A circular coil of mean radius of 7 cm and having 4000 tums is rotated at the rate of 1800 revolutions per minute in the earth's magnetic field (B = 0.5 Gauss ) , The peak value of emf induced is

0%
1 . 158 V
0%
0 . 58 V
0%
0 . 29 V
0%
5 . 8 V

Explanation

Given,

Angular frequency $\omega =2\pi f=\dfrac{2\pi 1800}{60}=60\pi$

Magnetic field, $B=0.5\ G=0.5\times {{10}^{-4}}\ T$

Area, $A=\pi \times {{0.07}^{2}}=0.0154\,{{m}^{2}}$

Flux $\phi =nBA\cos \theta =nB(\pi {{r}^{2}})\cos \omega t$

The induced emf is

$\varepsilon =\dfrac{-d\phi }{dt}=nBA\omega \sin \omega t$

For maximum E.M.F, $\varepsilon ={{\varepsilon }_{o}},\ \sin \omega t=1$

${{\varepsilon }_{o}}=n\omega BA$

${{\varepsilon }_{o}}=4000\times 60\pi \times 0.5\times {{10}^{-4}}\times 0.0154$

${{\varepsilon }_{o}}=0.58\,V$

A body enters in MRI machine in

$10\ sec$ . If the magnetic field is

$1.5\ T$ and circumference of MRI machine is

$0.9\ m$ then find out emf induced in the body.

0%
$0.96\ V$
0%
$9.6\ V$
0%
$9.6\ mV$
0%
$96\ mV$

Explanation

circumference

$2\pi r = 0.9\,m$

$\Rightarrow$ radius

$= \dfrac{0.9}{2\pi } = \dfrac{0.9}{0.28} \,meter$

area

$A = \pi r^{2} = 3.14\times \dfrac{0.81}{6.28\times 6.28} m^{2}$

initial flux = 0

flux after 10 second is

$\phi = BA$

$\phi =1.5\times 0.064$ weber

$\Delta \phi =$ charge in flux =

$\phi -0 = 0.096\, weber$

emf

$= \dfrac{\Delta \phi }{\Delta t} = \dfrac{0.096\, weber}{10sec} = .0096$ volt

$emf = 9.6\,mv$

1184732_1070256_ans_b9c5bb688db7477f8b235be62495940e.jpg

A rod of length

$50\ cm$ moves with a speed of

$10\ cm/s$ , in a uniform magnetic field of strength

$10\ G$ at an angle of

$30^{o}$ with the field. The emf induced across the ends of the rod is :

0%
$5000\ CGS\ unit$
0%
$2500\ CGS\ unit$
0%
$7500\ CGS\ unit$
0%
$1000\ CGS\ unit$

Explanation

Given,

Velocity, $V=10\,cm/s$

Length of rod, $L=50\,cm$

Magnetic Field, $B=10\,G$

Motional $emf=BlV\sin \theta =10\times 50\times 10\sin {{30}^{o}}=2500\,\ CGS\ unit$

A conducting disc of radius R is rotating with angular velocity $\omega$ .Mass of electron is and charged e. If electrons are the current carries in a conductor, the potential difference between the center and the edge of the disc is:

0%
$\dfrac{{m{\omega ^2}{R^2}}}{e}$
0%
$\dfrac{{m{\omega ^2}{R^2}}}{{4e}}$
0%
$\dfrac{{m{\omega ^2}{R^2}}}{{3e}}$
0%
$\dfrac{{m{\omega ^2}{R^2}}}{{2e}}$

Explanation

Due to rotations, the electron experfence a centripctual force

$F_{e}$ given by :-

$F_{e} = m\omega^{2}r$

Due to the fonce, a radical electric field is

generated which is :-

$E = \dfrac{F_{e}}{e} \,\,\,\, \left [ as F = qE \right ]$

$\Rightarrow E = \dfrac{m\omega^{2}r}{e}$

So, the potential difference generated between the centre

(r = 0) and edge of dise (r = R) is obtained as :-

V =

$\displaystyle \int E. dr = \int_{o}^{R} \dfrac{m\omega^{2}r}{e}dr = \dfrac{m\omega^{2}}{e} \int_{o}^{R} r dr = \dfrac{m\omega^{2}}{e} \left [ \dfrac{r^{2}}{2} \right ]_{o}^{R}$

$\Rightarrow V = \dfrac{m\omega^{2}R^{2}}{2l}$

1101125_1071784_ans_e9b9358b4ec34f83b23934a27a315a2c.JPG

Flux linked through following coils changes with respect to time then for which coil an e.m.f. is not induced:

0%
Copper coils
0%
Wood coils
0%
Iron coils
0%
None

A long magnet moves with constant velocity along the axis of a fixed metal ring. It starts from a large distance from the ring, passes through the ring and then moves away far from the ring. The current in the ring is plotted against time. Which of the following best represents the resulting curve?

Current in a coil of self-inductance

$2.0 H$ is increasing as

$i=2\sin { { t }^{ 2 } }$ . The amount of energy spent during the period when the current changes from

$0$ to

$2A$ is:

0%
$1\ J$
0%
$2\ J$
0%
$3\ J$
0%
$4\ J$

Explanation

Given that,

of coil = $2.0 H$

Let the current flowing through the coil be 2A at time ${{t}^{2}}$

Now,

$2=2\sin {{t}^{2}}$

$t=\sqrt{\dfrac{\pi }{2}}$

Now, the amount of energy spent during the period when the current changes from $0\ to\ 2\ A$

Now, Self-induced e. m. f

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

Now, the work done is

$dW=L\left( \dfrac{di}{dt} \right)dq$

$dW=L\times \dfrac{di}{dt}\times idt$

$dW=Lidi$

Now, on integrate

$\int{dW=\int\limits_{0}^{t}{Li}}di$

$W=\int\limits_{0}^{t}{L2\sin {{t}^{2}}d\left( 2\sin {{t}^{2}} \right)}$

$W=\int\limits_{0}^{t}{8L\sin {{t}^{2}}\cos {{t}^{2}}tdt}$

$W=4L\int\limits_{0}^{t}{\sin 2{{t}^{2}}dt}$

Now, let

$\theta =2{{t}^{2}}$

$d\theta =4tdt$

$dt=\dfrac{d\theta }{4t}$

Now, put the value of $2{{t}^{2}}$ and $dt$

$W=4L\int\limits_{0}^{t}{\dfrac{\sin \theta d\theta }{4}}$

$W=L\left[ \left( -\cos \theta \right) \right]_{0}^{t}$

$W=-L\left[ \cos 2{{t}^{2}} \right]_{0}^{t}$

$W=-L\left[ \cos 2{{t}^{2}} \right]_{0}^{\sqrt{\dfrac{\pi }{2}}}$

$W=2L$

$W=2\times 2$

$W=4\,J$

is $4\ J$

A metal disc rotates freely, between the poles of a magnet in the direction indicated. Brushes P and Q make contact with the edge of the disc and the metal axle. What current, if any, flows through R ?

0%
a current from P to Q
0%
a current from Q to P
0%
no current, because the emf in the disc is opposed by the back emf
0%
no current, because the emf induced in one side of the disc is opposed by the emf induced in the other side.
0%
no current , because no radial emf is induced in the disc.

Explanation

Given,

The disc is rotating in magnetic field from North to South.

If the direction of the magnetic field is perpendicular to the plane of the disc:

Viewing disk from the top.

Force on each proton is toward the inside, and force on each electron is outside.

So, the negative charge will develop, in the outskirt of the disk, and the positive charge will develop at the center.

Hence, Current will flow from P to Q.

$Alternatively:$

If the magnetic field is perpendicular to the normal vector of the plane of the disc:

The emf induced will not be radial, as governed by flaming's right hand rule, thus no current will flow through the resistance R.

979584_1067647_ans_5745555d1d3f4b1c8c3547a820270af5.png

Magnetic flux linked through the coil changed with respect to tine according to following graph, then induced emf

$\dfrac{v}{s}$ time graph for coil is:

A uniform rod of mass

$6M$ and length

$6l$ is bent to make an equilateral hexagon. Its mutual inductance about an axis passing through the centre of mass and perpendicular to the plane of hexagon is

0%
$5ml^2$
0%
$6ml^2$
0%
$4ml^2$
0%
$12ml^2$

Explanation

M.I

$=6\times \left[\cfrac { ML^2 }{ 12 }+M\left(\cfrac{l}{\sqrt{3}}\right)^2\right]$

$6\times \left[ \cfrac { ML^{ 2 } }{ 12 } +M\left( \cfrac { l }{ \sqrt { 3 } } \right) ^{ 2 } \right]$

$=6\left[\cfrac{Ml^2+4ml^2}{12}\right]$

$=5ML^2$

976062_1073851_ans_ade628691cdc442f9689e60bd9dec94f.png

A coil and a magnet moves with their constant speeds

$5 \,{m}/{sec}$ and

$3\, {m}/{sec}$ respectively, towards each other, then induced emf in coil is

$16 mV$ . If both moves in the same direction, then induced emf in the coil:

0%
$15 mV$
0%
$4 mV$
0%
$64 mV$
0%
$zero$

Explanation

When both coil and magnet moves in same direction, then there will be no induced emf in the coil because there is no change in the magnetic flux. If both are moves in same direction, then induced emf in coil is zero. The induced emf is given by:

$emf=\dfrac{-d\phi }{dt}=\dfrac{-d(BA)}{dt}$

A conducting rod of length

$l$ is moving in a transverse magnetic field of strength

$B$ with velocity

$v$ . The resistance of the rod is

$R$ . The current in the rod is

0%
$\dfrac{Blv}{R}$
0%
$Blv$
0%
$Zero$
0%
$\dfrac{B^2v^2l^2}{R}$

Explanation

Les consider,

$l=$ length of a conducting rod

$B=$ transverse magnetic field

$v=$ velocity

$R=$ resistance

The motional emf induced in the conducting rod is given by

$E=Blv$

The current in the rod is

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

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

The correct option is A.

A straight line conductor of length

$0.4\ m$ is moved with a speed of

$7\ m/s$ perpendicular to a magnetic filed of intensity

$0.9\, Wb/m^2$ . The induced e.m.f. across the conductor is

0%
$5.04\ V$
0%
$1.26\ V$
0%
$2.52\ V$
0%
$25.2\ V$

Self inductance of two oils connected in series are

$0.01$ and

$0.03\ H$ if the windings in the coils are in opposite sense and

$M=0.01\ H$ , then resultant self-inductance will be

0%
$2H$
0%
$0.2\ H$
0%
$0.02\ H$
0%
$Zero$

Two inductance's connected in parallel are equivalent to a single inductance of

$1.5 \ H$ , and when connected in series are equivalent to a single inductance of

$8 \ H$ . The difference in their inductance is:-

0%
$3 \ H$
0%
$7.5 \ H$
0%
$2 \ H$
0%
$4 \ H$

Explanation

$4 H$

Inductance in series combination,

$L= L_{1}+L_{2}$

$8= L_{1}+L_{2}$

Inductance in parallel combination,

$\dfrac{1}{L}$

$= \dfrac{1}{L_{1}}$ +

$\dfrac{1}{L_{2}}$

$\dfrac{1}{1.5}$

$=\dfrac{L_{1}+L_{2}}{L_{1}.L_{2}}$

$\dfrac{1}{1.5}$

$=\dfrac{8}{L_{1}.L_{2}}$

$L_{1}.L_{2}= 12$

$L_{1}=$

$\dfrac{12}{L_{2}}$

Put value of

$L_{1}$ in the series combination

$\dfrac{12}{L_{2}}+L_{2}= 8$

$L_{2}^2-8L_{2}+12=0$

$L_{2}= 2, 6$

$L_{1}=$

$\dfrac{12}{L_{2}}$

$=6,2$

So, the difference between the two inductance is 4.

1415761_1079373_ans_0649c4cad48e4cf2b77eae8568119882.jpeg

The time constant of a circuit is 10 sec, When a resistance of

$100 \Omega$ is connected in series in a previous circuit then time constant becomes 2 second,then the self inductance of the circuit is;-

0%
$250 H$
0%
$50H$
0%
$150 H$
0%
$25 H$

Explanation

In LR circuit,

The time constant

$\tau=\dfrac{L}{R}$

$10=\dfrac{L}{R}$

$L=10R$ . . . . . . .(1)

When Resistance

$100\Omega$ is connect in series, than the time constant is

$\tau'=\dfrac{L}{R+100}=2s$

$L=2R+200$ . . . . . . .(2)

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

$2R+200=10R$

$8R=200$

$R=25\Omega$

From equation (1),

$L=10R=10\times 25$

$L=250H$

The current through choke coil increases from zero to

$6A$ in

$0.3\ sec$ and an induced e.m.f of

$30\ V$ is produced. The inductance of the coil of choke is

0%
$5\ H$
0%
$2.5\ H$
0%
$1.5\ H$
0%
$2\ H$

A conducting rod AB moves parallel to x-axis in a uniform magnetic field, pointing in the positive z-direction. The end A of the rod gets positively charged. Is this statement true?

0%
Yes
0%
No
0%
Not defined
0%
Any answer is right

The circular arc (in x-y plane) shown in figure rotates (about z-axis) with a constant angular velocity

$\omega$ . Time in a cycle for which there will be induced emf. in the loop is:

0%
$\cfrac { \pi }{ 2\omega }$
0%
$\cfrac { \pi }{ \omega }$
0%
$2\cfrac { \pi }{ 3\omega }$
0%
$3\cfrac { \pi }{ 2\omega }$

the number of turn of the primary and secondary coil of the transformer is

$5$ and

$10$ respectively ad the mutual inductance is

$25 H.$ if the number f turns of the primary and secondary is made

$10$ and

$5$ , then the mutual inductance of the coils will be

0%
$6.25 H$
0%
$12.5 H$
0%
$25 H$
0%
$50 H$

Explanation

$M=\mu_o\mu_r \cfrac {N_1N_2}{l}A$

$M \propto N_1N_2$

Since

$N_1N_2=10 \times 5= 5 \times 10=50$ in both cases.

Mutual inductance will remain same.

There is a uniform (in spatial districution) magnetic filed B in a circular region of radius R as shown in the figure whose magnitude varies uniformly a the rate

$\beta$ w.r.t. time. The emf induced across the ends of a circular concentric conducting arc of radius

$R_1$ having an agle

$\theta$ as shown

$( < OAO'=\theta)$ is

0%
$\dfrac{\theta}{2 \pi}R^2_1. \beta$
0%
$\dfrac{\theta}{2 }R^2. \beta$
0%
$\dfrac{\theta}{2 \pi}R^2. \beta$
0%
Zero

Explanation

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

where

$\dfrac{dB}{dt} = \beta$

$\phi = \displaystyle \int B. ds$

$= B \dfrac{1}{2} r^2 . \dfrac{\theta}{\pi}$

$= \dfrac{Br^2 \theta}{2 \pi}$

$Emf = \dfrac{dB}{dt} \dfrac{r^2 \theta}{2 \pi}$

$= \dfrac{\beta r^2 \theta}{2 \pi}$

$Emf = \dfrac{\theta}{2 \pi} r^2 . \beta$

1336788_1115953_ans_cc616fb3f9f748288bbb669d80780d0f.png

The resistance that must be connected in series with inductance of 0.2 H in order thta the phase difference between current and e.m.f. may be

$45^0$ when the frequency is 50 Hz is:-

0%
6.28 ohm
0%
62.8 ohm
0%
628 ohm
0%
31.4 ohm

The length of a solenoid is 0.3 m and the number of turns isthe area of cross-section of the solenoid is

$1.2\times10^{-3} m^2$ . another coil of 300 turns is wrapped over the solenoid. a current of 2A is passed through the solenoid and its direction is changed in 0.25 sec. then the induced emf in coil :

0%
$4.8\times10^{-2} V$
0%
$4.8\times10^{-3} V$
0%
$3.2\times10^{-4} V$
0%
$3.2\times10^{-2} V$

Explanation

$N_1=2000$

$N_2=300$

$A=1.2\times 10^{-3}m^2$

$l=0.3\ m$

$t=0.25\ s$

$I=0.2\ A$

we know,

Mutual inductance

$(M)=\dfrac{\mu_0N_1N_2A}{l}$

$\Rightarrow M=\dfrac{\mu_0N_1N_2A}{l}$

also,

induced emf

$(\varepsilon)=M.\dfrac{dI}{dt}$

and

$\dfrac{dI}{dt}=\dfrac{2-(-2)}{0.25}=\dfrac{4}{0.25}=16$

$\varepsilon =M\dfrac{dI}{dt}=\dfrac{\mu_0N_1N_2A}{l}.\dfrac{dI}{dt}$

$\varepsilon=\dfrac{4\pi \times 10^{-7}\times 2000\times 300\times 1.2\times 10^{-3}\times 16}{0.3}$

$\boxed{\varepsilon=0.048\ V=4.8\times 10^{-2}\ V}$ hence

$(A)$ option is correct

There is a uniform magnetic field B normal to the XY plane.A conductor ABC has length

$AB = I_p$ parallel to the X-axis, and length

$BC = I_2$ , parallel to the y-axis . ABC moves in the XY plane with velocity

$V_{ x }\hat { i } +V\_ y\hat { j }$ .The potential difference between A and C is proportional to

0%
$V_x l_1 + V_y l_2$
0%
$V_x l_2 + V_y l_1$
0%
$V_x l_2 - V_y l_1$
0%
$V_x L_1 - V_y L_2$

A rectangular loop of sides 'a' and 'b' is placed in the XY plane. A very long wire is also placed in xy plane such that side of length 'a' of the loop is parallel to the wire. The distance between the wire and the nearest edge of the loop is 'd'. The mutual inductance of this system is proportional to?

0%
a
0%
b
0%
$1/d$
0%
Current in wire

CBSE Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 8 - 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 8 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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