MCQ Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 12

An electron enters a magnetic field at right angles to it, as shown in the figure, the direction of force acting on the electron is

0%
Towards right
0%
Towards left
0%
Out of the page
0%
Into the page

If emf induced in a coil is

$2V$ by changing the current in it from

$8\ A$ to

$6\ A$ in

$2\times 10^{-3}s$ , then the coefficient of self induction is

0%
$2\times 10^{-3} H$
0%
$10^{-3}H$
0%
$0.5\times 10^{-3}H$
0%
$4\times 10^{-3}H$

Explanation

Induced emf

$e = 2V$

$i_{1} = 8A, i_{2} = 6A$

$\triangle t = 2\times 10^{-3}s$

Coefficient of self induction

$L = \dfrac {e}{\triangle i/ \triangle t} = \dfrac {-2}{(6 - 8)/ 2\times 10^{-3}}$

$= \dfrac {-2\times 2\times 10^{-3}}{-2}$

$= 2\times 10^{-3} H$

A coil of resistance 5 unit and inductance 4 H is connected to a 10 V battery. The energy stored in thecoil

0%
0.8 J
0%
8 J
0%
16 J
0%
4 J

An electric bulb has a rated power of

$50 W$ at

$100 V$ . If it is used on an AC source

$200 V$ ,

$50 Hz$ , a choke has to be used in series with it. This choke should have an inductance of

0%
$0.01 mH$
0%
$1 mH$
0%
$0.1 H$
0%
$1.1 H$

Explanation

Resistance of the bulb

$=\dfrac{V^2}{P}=200\Omega$

Let inductance of choke be

$L$ .

Then the impendence of inductor

$=2\pi\times50\times L$

Both the voltages will be perpendicular in the phasor diagram. Net voltage is vector addition of both.

The total is

$200V$ . For

$100V$ to be on resistor,

$100\sqrt{3}$ will be on inductor.

$\Rightarrow \dfrac{2\pi\times50\times L}{200}=\sqrt{3}$

$\Rightarrow L=1.1H$

Two concentric coils each of radius equal to

$2\pi$ cm are placed at right angles to each other.

$3$ ampere and

$4$ ampere are the currents flowing in each coil respectively. The magnetic induction in

$weber/m^2$ at the centre of the coils will be

$\left(\mu_0=4\pi\times 10^{-7}wb/ A.m\right)$

0%
$10^{-5}$
0%
$12\times 10^{-5}$
0%
$7\times 10^{-5}$
0%
$5\times 10^{-5}$

Explanation

$B_1=\displaystyle\frac{mu_0 i_1}{2(2\pi)}=\frac{\mu_0\times 3}{4\pi}$

$B_2=\displaystyle\frac{mu_0i_2}{2(2\pi)}=\frac{mu_0\times 4}{4\pi}$

$B=\sqrt{B^2_1+B^2_2}=\displaystyle \frac{mu_0}{4\pi}\cdot 5$

$\Rightarrow B=10^{-7}\times 5\times 10^2$

$\Rightarrow B=5\times 10^{-5}Wb/m^2$

Two coils have a mutual inductance

$0.005\ H$ . The current changes in the first coil according to the equation

$i = i_{m}\sin \omega t$ where

$i_{m} = 10\ A$ and

$\omega = 100\pi \ rad\ s^{-1}$ . The maximum value of the emf induced in the second coil is

0%
$2\pi$
0%
$5\pi$
0%
$\pi$
0%
$4\pi$

Explanation

Given :

$L = 0.005 H$

Variation of current in first coil

$i = i_m \sin wt$

$\therefore$

$\dfrac{di}{dt} = i_m w \cos wt$

Emf induced

$\mathcal{E} = -L\dfrac{di}{dt} = -i_m w L \cos wt$

Thus maximum emf induced

$\mathcal{E_{max}} = i_m wL$ when

$\cos wt =-1$

$\therefore$

$\mathcal{E_{max}} = 10\times 100\pi \times 0.005 = 5\pi$ volts

A coil of area 500 $cm^2$ having 1000 turns is placed such that the plane of the coil is perpendicular to a magnetic field of magnitude $4 \times 10^{-5}$ $weber/m^2$ . If it is rotated by 180 about an axis passing through one of its diameter in 0.1 sec, find the average induced emf.

0%
zero.
0%
30 mV
0%
40 mV
0%
50 mV

Explanation

0iven that :-

$N=1000, B=4\times 10^{-5}weber/m^2, A=500cm^2=0.05m^2$

Initial flux linked with the coil,

$\phi_1=1000\times 4\times 10^{-5}\times 0.05$

$\implies \phi_1=2\times 10^{-3}weber$

After rotation of

$180^{o}$ , B remains same but normal vector gets reversed, hence

$\phi_2=-\phi_1$

Average EMF=

$E=\dfrac{-\Delta \phi}{t}$

$\implies E=-\dfrac{\phi_2-\phi_1}{t}$

$\implies E=\dfrac{\phi_1-\phi_2}{t}$

$\implies E=\dfrac{2\phi_1}{t}$

$\implies E=\dfrac{4\times 10^{-3}}{0.1}V$

$\implies E=40mV$

Answer-(C)

A coil of area

$500 cm^2$ having 1000 turns is placed such that the plane of the coil is perpendicular to a magnetic field of magnitude

$4 10^{-5} weber/m^2$ . If it is rotated by

$180$ degree about an axis passing through one of its diameter in

$0.1$ sec, find the average induced emf.

0%
$zero.$
0%
$30 mV$
0%
$40 mV$
0%
$50 mV$

Explanation

Area of the coil

$A = 500 \ cm^2 = 0.05 \ m^2$
Number of coil

$N = 1000$
Magnetic field

$B = 4\times 10^{-5} \ Wb/m^2$
Magnetic flux passing through the coil initially

$(\theta = 0^o)$ ,

$\phi_i = NBA\cos\phi =1000\times 4\times 10^{-5}\times 0.05\times \cos 0^o =0.002 \ Wb$
Magnetic flux passing through the coil finally

$(\theta = 180^o)$ ,

$\phi_f = NBA\cos\phi =1000\times 4\times 10^{-5}\times 0.05\times \cos 180^o =-0.002 \ Wb$
Thus change of flux

$\Delta \phi =\phi_i-\phi_f = 0.002-(-0.002) = 0.004 \ Wb$
Change in time

$\Delta t = 0.1 \ s$
Thus average emf induced

$\mathcal{E} = \dfrac{\Delta \phi}{\Delta t} = \dfrac{0.004}{0.1} = 40 \ mV$

Radii of two conducting circular loops are

$b$ and

$a$ respectively, where

$b > > a$ . Centres of both loops coincide but planes of both loops are perpendicular to each other. The value of mutual inductance for these loops.

0%
$\dfrac {\mu_{0}\pi b^{2}}{2a}$
0%
zero
0%
$\dfrac {\mu_{0}\pi ab}{2(a + b)}$
0%
$\dfrac {\mu_{0}\pi a^{2}}{2b}$

Explanation

At the position of smaller loop, magnetic field due to larger loop is parallel to the plane of smaller loop. Due to larger loop, magnetic flux with smaller loop is zero. Hence, mutual induction is zero.

$\phi = M\ i_{2}$

$\Rightarrow 0 = Mi_{2}$

$\Rightarrow M = 0$ .

A generator with a circular coil of

$100$ turns of area

$2\times { 10 }^{ -2 }{ m }^{ 2 }$ is immersed in a

$0.01T$ magnetic field and rotated at a frequency of

$50Hz$ . The maximum emf which is produced during a cycle is

0%
$6.28V$
0%
$3.44V$
0%
$10V$
0%
$1.32V$

Explanation

Given :

$N = 100$ turns

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

$B = 0.01 \ T$

$\nu = 50 \ Hz$
So

$w =2\pi \nu = 2\times \dfrac{22}{7}\times 50$
Maximum emf induced

${ e }_{ 0 }=NBA\omega$

$e_0=100\times 0.01\times 2\times { 10 }^{ -2 }\times 2\times \cfrac { 22 }{ 7 } \times 50=6.28V$

A rod of length

$b$ moves with a constant velocity

$v$ in the magnetic field of a straight long conductor that carries a current

$i$ as shown in the figure. The emf induced in the rod is

0%
$\dfrac {\mu_{0}iv}{2\pi}\tan^{-1}\dfrac {a}{b}$
0%
$\dfrac {\mu_{0}iv}{2\pi} ln \left (1 + \dfrac {b}{a}\right )$
0%
$\dfrac {\mu_{0}iv\sqrt {ab}}{4\pi (a + b)}$
0%
$\dfrac {\mu_{0}iv(a + b)}{4\pi ab}$

A wire as a parabola

$y = 4x^2$ is located in a uniform magnetic field of inductance B perpendicular to the XY plane. At t=0 a connection starts translation wise from the parabola apex with constant acceleration

$\alpha$ . The induced emf in the loop, thus formed, as a function of y is:

0%
$e = \sqrt{2\alpha}\cdot{By}$
0%
$e = By\sqrt\frac{\alpha}{2}$
0%
$e = \frac{By\sqrt\alpha}{2\sqrt2}$
0%
$e = \frac{ By\sqrt\alpha}{4}$

Explanation

$\overrightarrow { B } =B\uparrow$

Velocity of rod after travelling

$dy$ distance

$u$ .

we know that,

$S=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }$

$y=0+\dfrac { 1 }{ 2 } \alpha { t }^{ 2 }$ (

$t=\sqrt{\dfrac{2y}{\alpha}}$ )

$\dfrac { dy }{ dt } =\alpha t$

$V=\alpha t=\sqrt { 2\alpha y }$

emf induced

$\varepsilon =VB(2x)$

$=\sqrt { 2\alpha y } \times 2xB$

$x=\dfrac { \sqrt { y } }{ 2 }$

$\varepsilon =\sqrt { 2\alpha } .By$

Magnetic field

$\vec { B } =-{ B }_{ 0 }x\hat { k }$ exists in a region of space. A particle of specific charge

$\pi$ enters this region of space. Its velocity and position at time t=0 are

$\vec { V } ={ V }_{ 0 }\hat { i }$ and (0, 0, 0). The maximum x-displacement of the particle is

0%
$$\sqrt { \dfrac { 2{ V }_{ 0 } }{ { B }_{ 0 }\pi }$$
0%
$\sqrt { \dfrac { 3 }{ { B }_{ 0 }\pi } } { V }_{ 0 }$
0%
$\sqrt { \dfrac { 1 }{ { 2B }_{ 0 }\pi } } { V }_{ 0 }$
0%
$\sqrt { \dfrac { 3 }{ { 2B }_{ 0 }\pi } } { V }_{ 0 }$

A superconducting rigid planar loop of area

$A$ and self-inductance

$L$ carrying a current is held motionless in a region of free spaces. Now a uniform magnetic field of induction

$B$ -pointing everywhere parallel to the magnetic moment

$m$ of the loop is switched on. Current in the loop after the magnetic field is switched on us given by

0%
$\dfrac {AB}{L}$
0%
$\dfrac {m}{A}$
0%
$\dfrac {m}{A}-\dfrac {AB}{L}$
0%
$\dfrac {m}{A}+\dfrac {AB}{L}$

When a coil rotated in magnetic field induced current in it :

0%
continuously changes
0%
remains same
0%
becomes zero
0%
becomes maximum

Explanation

We know that EMF induced=

$E=NBA\omega\cos\omega t$

$\implies I=\dfrac{E}{R}=\dfrac{NBA\omega}{R}\cos\omega t$

Hence, current continuously changes with time.

Answer-(A)

A wheel when spokes is rotated in a plane perpendicular to the magnetic filed of earth such that an emf e is induced between axle and rim of the wheel.In the same wheel, number of spokes is made 3 N and the wheel is rotated in the same manner in the same field then new emf is

0%
3e
0%
$\dfrac{3}{2}e$
0%
$\dfrac{e}{3}$
0%
2

An air-cored solenoid with length

$20 cm$ , area of cross-section is

$20 cm^2$ and the number of turns

$400$ carries a current

$2 A$ . The current is suddenly switched off within

$10^{-3}$ sec. The average back emf induced across the end of the open switch in the circuit is ( ignore variation in the magnetic field near the end of the solenoid)

0%
$2 V$
0%
$4 V$
0%
$3 V$
0%
$5 V$

Explanation

Given:

$l= 20 cm= 20 \times 10^{-2}m$

$A=20cm^2=20\times 10^{-4} m^2$

$N=400,$

$I_1= 2A,I_2=0, dt=10^{-3}s$

$\epsilon= \dfrac {d\phi}{dt}=\dfrac{d(BAN)}{dt}$

$=\dfrac{\mu_0 NdIAN}{ldt}$

$=\dfrac{\mu_0 N(I_1-I_2)AN}{ldt}$

$= \dfrac {d\phi}{dt}=\dfrac{d(BAN)} {dt}$

$\epsilon=\dfrac{4\pi\times\,10^{-7}\times(400)^2\times 2\times\times 10^{-4}}{20\times 10^{-2}\times 10^{-3}}$

$\epsilon=4$ V

A current

$i$ is flowing in a straight conductor of length

$L$ . The magnetic induction at the point on its perpendicular bisector at a distance

$L/4$ from its centre will be

0%
zero
0%
$\dfrac{\mu_{0}i}{\sqrt{2}\pi L}$
0%
$\dfrac{\mu_{0}i}{\sqrt{2}L}$
0%
$\dfrac{4\mu_{0}i}{\sqrt{5}\pi L}$

Figure shows a circular area of radius

$R$ where a uniform magnetic field

$\vec { B }$ is going into the plane of paper and increasing in magnitude at a constant rate. In rate case, which of the following graphs, drawn schematically, correctly shown the variation of the induced electric field

$E(r)$ ?

Two identical short bar magnets, each having magnetic moment M, are placed at a distance of '

$2$ d' with their axes perpendicular to each other in a horizontal plane. The magnetic induction at a point midway between them will be

0%
$\frac{\mu_0}{4\pi}\times \sqrt2 \frac{M}{d^3}$
0%
$\frac{\mu_0}{4\pi}\times \sqrt3 \frac{M}{d^3}$
0%
$\frac{2\mu_0}{4\pi}\times \frac{M}{d^3}$
0%
$\frac{\mu_0}{4\pi}\times \sqrt5 \frac{M}{d^3}$

A bulb of rated values 60 V and 10 W is connected in series with a source of 100 V and 50 Hz. The coefficient of self induction of a coil to be connected in series for its operation will be:

0%
1.53 H
0%
2.15 H
0%
3.27 H
0%
3.89 H

One of the two small circular coils, (none of them having any self-inductance) is suspended with a

$V-$ shaped copper wire, with plane horizontal. The other coil is placed just below the first one with plane horizontal. Both the coil are connected in series with a

$dc$ apply. The coils are found to attract each other with a force. Which one of the following statement is incorrect?

0%
Both the coils carry currents in the same direction.
0%
Coils will attract each other, even if the supply is an $ac$ source.
0%
Force is proportional to ${d}^{-1}, d=$ distance between the centers of the coil.
0%
Force is proportional to ${d}^{-2}$ .

Two coils,

$X$ and

$Y$ , are kept inclose vicinity of each other. When a varying current,

$l(t)$ , flows through coil

$X$ , the induced emf

$(V(t))$ in coil

$YY$ , varies in the manner shown here. The variation of

$l(t)$ , with time, can then represented by the graph labelled as graph.

A coil of inductance

$8.4\ mH$ and resistance

$6 \Omega$ is connected to a

$12\ V$ battery. The current in the coil is

$1.0\ A$ at the time, approximately

0%
$500\ s$
0%
$20\ s$
0%
$35\ ms$
0%
$1\ ms$

Explanation

In the given growth circuit as,

$i=\dfrac{V}{R}\left(1-e^{-\dfrac{Rt}{2}}\right)$

Here,

$V=12v$

$R=6\Omega$ and

$L=8.4\ mH$

$\Rightarrow i=2\left(1-e^{\dfrac{-10^3t}{1.4}}\right)$

For

$i=1\ A$ ,

$1=2\left(1-e^{\dfrac{-10^3t}{1.4}}\right)$

$1/2=1-e^{\dfrac{10^3t}{1.4}}$

$\Rightarrow e^{\dfrac{-10^3t}{1.4}}=\dfrac{1}{2}\rightarrow t=\dfrac{1.4(\ln 2)}{10^3}s$

$\Rightarrow t=(1.4)(0.69)\ ms=1\ ms$

option

$-D$ is correct.

1432661_1011202_ans_a13d3ef95bba4a7da2142729334c1e9e.png

An alternating current generator has an internal resistance

$R_{g}$ and an internal reactance

$X_{g}$ . It is used to supply power to a passive load consisting of a resistance

$R_{g}$ and a reactance

$X_{L}$ . For maximum power to be delivered from the generator to the load, the value of

$X_{L}$ is equal to

0%
Zero
0%
$X_{g}$
0%
$-X_{g}$
0%
$R_{g}$

Explanation

Generator

$R_g , X_g$

Passive load

$R_g , X_L$

The condition for the power to be maximum is the veactance should be zero.

$X_g + X_L = 0$

$X_L = -X_g$

The current through an inductor of

$1\ H$ is given by

$i=3t\ \sin { t }$
The voltage across the inductor of

$1\ H$ is:

0%
$3\ \sin { t } +3\ \cos { t }$
0%
$3\ \cos { t } +t\ \sin { t }$
0%
$3\ \sin { t } +3t\ \cos { t }$
0%
$3t\ \cos { t } +\ \sin { t }$

In figure shown, wire

$P_{1}Q_{1}$ and

$P_{2}Q_{2}$ , both are moving towards right with speed

$5\ cm/sec$ . Resistance of each wire is

$2\ \Omega$ . Then current through

$19\ \Omega$ resistor is:

0%
$0$
0%
$0.1\ mA$
0%
$0.2\ mA$
0%
$0.3\ mA$

Two conducting rings P and Q of radii and r and 2r rotate uniformly in opposite directions with centre of mass velocities 2 v and v respectively on a conducting surface S. There is a uniform magnetic field of magnitude B perpendicular to the plane of the rings. The potential difference between the highest points of the two rings is

0%
(a) zero
0%
(b) 4 Bvr
0%
(c) 8 Bvr
0%
(d)16 Bvr

Explanation

Effective length moving in magnetic field is

$2r$ and

$4r$ respectively.

$\therefore E_1 = B(2r)(2v) = 4 Bvr$

$E_r = B(4r)(v) = 4 Bvr$

To find the potential difference, we replace the induced emfs in the rings by cells.

$\therefore$ Potential Difference

$= E_1 + E_2 = 8 Bvr$

1000642_1030018_ans_c5da6b35656d49b0a2fb1e8f09138c1d.png

Current

$1$ in a long

$4y-$ axis is paced through a square metal frame of side

$2a$ oriented in the

$y-z$ p[lane a shown. The linear mass density of the frame is

$\lambda$ . A uniform magnetic field

$B$ is now switched on along

$x-$ axis. Then the instantaneous angular acceleration of the frame will be:

0%
$\dfrac {4IB}{\lambda a}$
0%
$\dfrac {12IB}{\lambda a}$
0%
$\dfrac {4IB}{3 \lambda a}$
0%
$zero$

Explanation

Linear mass density is

$\lambda$

Current I is along

$4y-axis$ through square metal frame of side

$2a$

Instantaneous acceleration :-

$\dfrac{IB}{\lambda a}=\dfrac{4IB}{\lambda q}$

1425510_1023987_ans_a432c802149f49d68af4c9c988aeac84.png

A thin circular ring of area

$A$ is held perpendicular to a uniform magnetic field of induction

$B$ . A small cut is made in the ring and a galvanometer is connected across the ends such that the total resistance of the circuit is

$R$ . When the ring is suddenly squeezed to zero area, the charge flowing through the galvanometer

0%
$\dfrac { BR }{ A }$
0%
$\dfrac { AB }{ R }$
0%
$ABR$
0%
$\dfrac { { B }^{ 2 }A }{ { R }^{ 2 } }$

The diagram below shows two coils A and B placed parallel to each other at a very small distance. Coil A is connected to an ac supply. G is a very sensitive galvanometer. When the key is closed -

0%
Constant deflection will be observed in the galvanometer for 50 Hz supply
0%
Visible small variations will be observed in the galvanometer for 50 Hz input
0%
Oscillations in the galvanometer may be observed when the input ac voltage has a frequency of 1 to 2 Hz
0%
No variation will be observed in the galvanometer even when the input ac voltage is 1 or 2 Hz

In the figure shown a square loop

$PQRS$ of side

$'a'$ and resistance

$'r'$ is placed in near an infinitely long wire carrying a constant current

$I$ . The sides

$PQ$ and

$RS$ are parallel to the wire. The wire and the loop are in the same plane. The loop is rotated by

$180^{\circ}$ about an axis parallel to the long wire and passing through the mid point of the side

$QR$ and

$PS$ . The total amount of charge which passes through any point of the loop during rotation is

0%
$\dfrac {\mu_{0}Ia}{2\pi} ln2$
0%
$\dfrac {\mu_{0}Ia}{\pi} ln2$
0%
$\dfrac {\mu_{0}Ia}{2\pi}$
0%
None of above

A conductor having two semicircular section are made to translate in the uniform magnetic filed

$B$ and

$2B$ as shown with a constant velocity

$v$ then the induced emf developed across the ends of conductor is

0%
$6vBR$
0%
$8vBR$
0%
$4vBR$
0%
$2vBR$

Th 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%
$\dfrac{\pi}{2\omega }$
0%
$\dfrac{\pi}{\omega }$
0%
$\dfrac{2\pi}{3\omega }$
0%
$\dfrac{3\pi}{2\omega }$

Explanation

Angular velocity=

$\omega$

$B.A=\phi, B.A=BA$

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

$=BA$

$=A\dfrac{dA}{dt}$

Let

$r$ be the radius.

$dA=\dfrac{r^{2}}{2}d\theta$

$dE=B.\dfrac{r^{2}}{2}\dfrac{d\theta}{dt}$

$\dfrac{Br^2\omega}{2}$

Time it will be therefore when there is charge in ares.

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

A uniform with resistance

$20\Omega/m$ in bent in the form of a circle as shown in the figure. If the equivalent resistance b/w P & Q is 13.8 Omega, then the length of the shown selection is

0%
2 m
0%
5 m
0%
1.8 m
0%
18 m

A semicircle loop

$PQ$ of radius

$'R'$ is moved with velocity

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

0%
$Bv\ (\pi r)$ , end $'P'$ at high potential
0%
$2\ BRv$ , end $P$ at high potential
0%
$2\ Brv$ , end $Q$ at high potential
0%
$B\dfrac {\pi R^{2}}{2}v$ , end $P$ at high potential

If number of turns of

$70cm^{2}$ coil is

$200$ and it is placed in a magnetic field of

$0.8\ Wb/m^{2}$ which is perpendicular to the plane of coil and it is rotated through an angle

$180^{\circ}$ in

$0.1\ sec$ , then induced emf in coil.

0%
$11.2\ V$
0%
$1.12\ V$
0%
$224\ V$
0%
$2.24\ V$

A 100 turns circular coil of area $100c{m^2}$ and a resistance 100 ohm is rotated about a diameter which is perpendicular to a uniform magnetic field of ${\text{0}}{\text{.1tesla}}$ . If the coil rotates at 200 rpm the amplitude of the alternating current induced in the coil is:

0%
20 mA
0%
5 mA
0%
6 mA
0%
6.67 mA

A conducting rod of length 2l is rotating with constant angular speed

$\omega$ about its perpendicular bisector.A uniform magnetic field

$\underset{B}{\rightarrow}$ exists parallel to the axis of rotation. The e.m.f., induced between two ends of the rod is

0%
$B\omega l^2$
0%
$\dfrac{1}{2}B\omega l^2$
0%
$\dfrac{1}{8}B\omega l^2$
0%
Zero

A rectangular loop is being pulled at a constant speed v, though a region of certian thickness d, in which a uniform magnetic filed B is set up. The graph between position x of the right hand edge of the loop and the induced emf E will be

Explanation

The question consists of two cases-

Case -1 when the loop entered into the magnetic field

Refer fig,1

let, the breadth of the loop is l then the induced e.m.f is written as-

e.m.f

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

$=-B\dfrac{d(lx)}{dt}$

e.m.f

$=-Bl\dfrac{dx}{dt}$

e.m.f

$\propto -\dfrac{dx}{dt}$

since

$x$ increases with time t. so

e.m.f

$\propto x$

case-2 - When the loop leaving the magnetic field

Refer fig, 2

e.m.f

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

$=-B\dfrac{d}{dt}(lx)$

e.m.f

$=-Bl \dfrac{dx }{dt}$

e.m.f

$\propto -\dfrac{dx}{dt}$

since x decreases with time t. so

e.m.f

$\propto -x$

from above two cases when can draw

Refer fig. 3

2033067_1096614_ans_0f8d06680b1f4472a6c89a414f974c6c.png

A conductor PQ with PQ=r, moves with a velocity v in a uniform magnetic filed of induction B. The emf induced in the rod is

0%
$(\underset{v}{\rightarrow} \times\underset{B}{\rightarrow} ).\underset{r}{\rightarrow}$
0%
$\underset{v}{\rightarrow} . (\underset{r}{\rightarrow} \times\underset{B}{\rightarrow} )$
0%
$\underset{B}{\rightarrow} . (\underset{r}{\rightarrow} \times\underset{v}{\rightarrow} )$
0%
$|\underset{r}{\rightarrow} \times (\underset{v}{\rightarrow} \times \underset{B}{\rightarrow} )|$

A conducting square loop of side l and resistance R moves in its plane with a uniform velocity v perpendicular to one of its sides. A magnetic induction B constant in time and space, pointing perpendicular and into the plane at the loop exists everywhere with half the loop outside the filed, as shown in figure. The induced e.m.f is

0%
Zero
0%
RvB
0%
vBl/R
0%
vBl

Two circuits, each consisting of a loop of wire are placed in the proximity of one another and one of them is carrying a current. The current is then suddenly stopped. Then

0%
There will be an induced e.m.f. in the other circuit
0%
There will be no induced e.m.f
0%
There will be an induced e.m.f. in both the circuits
0%
None of these

A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic induction B. At the position MNQ, the speed of the ring is v. The potential difference developed across the ring is

0%
zero
0%
$1/2 \,B \,v \pi R^2$ , and M is at a higher potential
0%
$\pi\,R\,Bv$ , and Q is at a higher potential
0%
$2\,R\,Bv$ , and Q at a higher potential

A square shaped metallic frame

$'PQRS'$ of dimensions,

$l\times l$ is placed inside a constant and uniform Magnetic Field

$\vec {B} = B_{0}\left (\dfrac {1}{\sqrt {2}} \hat {i} + \dfrac {1}{\sqrt {2}}\hat {j}\right )$ as shown in the figure such that it's sides are initially parallel to the

$X$ and

$Y$ axes.
In the square loop is now rotated with constant angular speed

$\omega_{0}$ about an axis parallel to the

$Y - axis$ and 'bisecting' the area of the loop in the above diagram, the maximum value of the EMF induced (instantaneous) in the loop due to induction will be

0%
$E = B_{0}\omega_{0}l^{2}$
0%
$E = \dfrac {B_{0}\omega_{0}l^{2}}{\sqrt {2}}$
0%
$E = \sqrt {2} B_{0}\omega_{0}l^{2}$
0%
$E = \dfrac {B_{0}\omega_{0}l^{2}}{2}$

$50\ Hz$

$AC$ current of crest value

$1\ A$ flows, through the primary of transformer. If the mutual inductance between the primary and secondary be

$0.5\ H$ , the crest voltage induced in the secondary is

0%
175 V
0%
150 V
0%
100 V
0%
157V

The coefficient of mutual inductance of two coils is

$6mH$ . If the constant current of

$2A$ is flowing in one coil , then the induced e.m.f in the second coil will be

0%
$3mV$
0%
$2mV$
0%
$3V$
0%
Zero

A conducting rings of radius 1 meter is placed in an uniform magnetic field B of 0.01 Tesla oscillating with frequency 100 Hz with its plane at right angles. What will be the induced electric field?

0%
$\pi$ volt /m
0%
2 volt/m
0%
10 volt/m
0%
62 volt/m

An electron originates at a point

$A$ lying on the axis of a straight solenoid and moves with velocity

$v$ at an angle

$\alpha$ to the axis. The magnetic induction of the field is equal to

$BA$ screen is oriented at right angles to the axis and is located at a distance

$1$ from the point

$a$ . Find the distance from the axis to the point on the screen into which the electron strikes.

0%
$d = 5r\sin \left (\dfrac {\theta}{2}\right )$ , Here $r = 2\dfrac {mv\sin \alpha}{eB}$ and $\theta = \dfrac {eBl}{mv\cos \alpha}$ .
0%
$d = 2r\sin \left (\dfrac {\theta}{2}\right )$ , Here $r = \dfrac {mv\sin \alpha}{eB}$ and $\theta = \dfrac {eBl}{mv\cos \alpha}$ .
0%
$d = 3r\sin \left (\dfrac {\theta}{2}\right )$ , Here $r = 3\dfrac {mv\sin \alpha}{eB}$ and $\theta = \dfrac {eBl}{mv\cos \alpha}$ .
0%
$d = 4r\sin \left (\dfrac {\theta}{2}\right )$ , Here $r = \dfrac {mv\sin \alpha}{eB}$ and $\theta = \dfrac {eBl}{mv\cos \alpha}$ .

$L$ -shaped conductor rod is moving in transverse magnetic field as shown in the figure. Potential difference between ends of the rod is maximum if the rod is moving with velocity :

0%
$3\hat{i} - 6\hat{j} \ m/s$
0%
$-4\hat{i} - 6\hat{j} \ m/s$
0%
$3\hat{i} - 2\hat{j} \ m/s$
0%
$\sqrt{13}\hat{i} \ m/s$

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

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.