MCQ Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 3

A solenoid (air core) ahs 400 turns, is 20 cm long and has a cross section of

$4 cm^2$ . Then the coefficient of self induction is approximately

0%
$4\times 10^{-5} H$
0%
$4\times 10^{1} H$
0%
$4\times 10^{-4} H$
0%
$4 H$

Explanation

$n=400 turns, l=20 cm, A=4 cm^2$

$L=?$

$L=\frac {1\cdot 25\times 10^{-6}Hm^{-1}\times 400\times 400\times 4\times 100}{100\times 100\times 20}$

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

Two coils of self inductances 2 mH and 8 mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is:

0%
10 mH
0%
6 mH
0%
4 mH
0%
16 mH

Explanation

Given,

$L_1=2mH$

$L_2=8mH$

The mutual inductance between coil is

$M=\sqrt{L_1L_2}$

$M=\sqrt{2\times 8}=\sqrt{16}mH$

$M=4mH$

The correct option is C.

A solenoid is connected to a source of constant e.m.f. for a long time. A soft iron piece is inserted into it. Then, which of the following is/are correct?

0%
Self-inductance of the solenoid gets increased
0%
Flux linked with the solenoid increases; hence, steady state current gets decreased
0%
Energy stored in the solenoid gets increased
0%
Magnetic moment of the solenoid gets increased

Explanation

The presence of iron core will increase the magnetic field produced by the solenoid due to high magnetic permeability of iron. The equation of solenoid with no core is

$B=\mu_0Ni/l$
and the equation of solenoid with core is

$B=\mu_0 \mu_rNi/l$
where

$\mu_r$ is the relative permeability of the material that the core is made of. In case of magnetic iron it is equal to 200.
Self-inductance increased with the increase of the magnetic permeability of core material. Magnetic moment and the energy will also increase as the magnetic field increases.

The current I in an induction coil varies with time t according to the graph shown in the figure. Which of the following graphs shows the induced emf

$(\varepsilon)$ in the coil with time

Magnetic flux

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

$10 \Omega$ varies with time t (in second) as

$\phi = 5{t^2} - 4t + 1$ . The induced electromotive force in the circuit at t = 0.2 second is

0%
0.4 V
0%
-0.4 V
0%
-2.0 V
0%
2.0 V

A circular disc of radius 0.2 metre is placed in a uniform magnetic field of induction

$\dfrac{1}{\pi }\left ( \dfrac{wb}{m^{2}} \right )$ in such a way that its axis makes an angle of 60 with

$\vec{B}$ The magnetic flux linked with the disc is:

0%
0.058 wb
0%
0.01 wb
0%
0.02 wb
0%
0.06 wb

Explanation

$A = \pi r^{2} = 0.04\pi$

$\Phi= B A cos \theta$

$=\dfrac{1}{\pi} \times 0.04 \pi \times cos 60 = 0.02 Wb$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

Explanation

An inductor, also called a coil, choke or reactor, is a passive two-terminal electrical component that stores energy in a magnetic field when electric current flows through it. An inductor typically consists of an insulated wire wound into a coil around a core.

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

where N is number of turns and L is distance.

When number of turns in a coil is doubled,

$L\propto (2N)^2=4N^2$

Assertion and Reason are correct and Reason is the correct explanation for Assertion

The frequency of A.C mains in India is

0%
$30 Hz$
0%
$50Hz$
0%
$60Hz$
0%
$120Hz$

Explanation

In India, the AC mains supply is referred to as single-phase alternating current and corresponds to a voltage of 230 V at a frequency of

$50 Hz$ , similar to most European countries. Whereas in the USA, AC mains supply uses

$60 Hz$

Read the following statements and answer whether the given statement is true or false.

The magnitude of induced current can be increased by increasing the number of turns in coil.

0%
True
0%
False

Explanation

We know that

$\phi\propto N$ where,

$\phi=$ magnetic flux and

$N=$ number of turns in coil

Hence,

$e.m.f \propto N \implies$ e.m.f. can be increased by increasing the number of turns in coil.

Hence, the induced current can be increased by increasing the number of turns in coil.

Hence, given statement is

$True$ .

A conducting disc of radius r spins about its axis with an angular velocity

$\omega$ . There is a uniform magnetic field of magnetude B perpendicular to the plane of the disc. C is the centre of the ring.

0%
No emf is induced in the disc.
0%
The potential difference between C and the rim is $\frac{1}{3} Br^2 \omega$ .
0%
C is at a higher potential than the rim.
0%
Current flows between C and the rim.

Explanation

emf induced =

$\int_0^r{ B dr r\omega} = B\omega \dfrac{r^2}{2}$
positive charges in the disc will experience a force n the direction given by

$q \vec v \times \vec B = q \vec{(r \omega)} \times \vec B$ = towards the center C.
Therefore C will be at a higher potential wrt to the rim.

Read the following statements and answer whether the given statement is true or false.
In a generator, magnitude of the induced current can be increased by decreasing the speed of rotation of armature.

0%
True
0%
False

Explanation

We know that in generator magnetic flux linked with coil is given by

$\phi=NBA\sin{\omega t}$ where

$\omega$ =angular speed of rotation of coil

Magnitude of e.m.f.=

$\implies E=\dfrac{d\phi}{dt}=NBA\omega \cos{\omega t}$

Since,

$E\propto \omega\implies I\propto \omega$

Hence, magnitude of induced current decreases by decreasing the speed of rotation of coil.

Hence, given statement is

$False$ .

Whenever current is changed in a coil, an induced e.m.f. is produced in the same coil, This property of the coil is due to

0%
mutual induction
0%
self induction
0%
eddy currents
0%
hysteresis

Explanation

The property of induction of e.m.f. in the same coil when there is a change in current in it is called

$self\hspace{2mm}induction.$

Hence, answer is option-(B).

Read the following statements and answer whether the given statement is true or false.

The induced e.m.f. depends only on the no of turns of the coil:

0%
True
0%
False

Explanation

Since, flux

$\phi=nBA\sin{\omega t}$ where

$n$ =number of turns and

$A$ =area of coil

Hence,

$e.m.f.$ does not depends solely on number of turns.

Hence, the given statement is

$False.$

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

Explanation

The induced emf ,

$E=-\dfrac{d\phi}{dt}$ where

$\phi$ =magnetic flux

Hence, emf will last as long as flux keeps changing.

Hence, answer is option-(D).

A conducting loop is placed in a uniform magnetic field with its plane perpendicular to the field. An emf is induced in the loop if

0%
it is translated
0%
it is rotated about its axis
0%
it is kept untouched for several minutes
0%
it is cut into two pieces

Explanation

e.m.f is induced in the coil when the flux linked with the coil changes.

And magnetic flux=

$\phi=\vec{B}.\vec{A}=BA\cos{\theta}$ where=

$\vec{A}$ =area vector of coil

and

$\theta$ =angle between

$\vec{A}$ and

$\vec{B}$

Hence, on changing

$\theta$ , emf can be induced.

Answer-(B).

To obtain maximum EMF from a number of cells, they must be connected in

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Series
0%
Parallel
0%
Both (a) and (b) above
0%
None of these

A solenoid of inductance

$L$ and length

$l$ and mass

$m$ , whose windings are made of material of density

$D$ and resistivity is

$\rho$ ,(the winding resistance is

$R$ ). Find out the inductance in terms of others.

0%
$\dfrac {\mu_0}{4\pi l}\dfrac {Rm}{\rho D}$
0%
$\dfrac {\mu_0}{4\pi R}\dfrac {lm}{\rho D}$
0%
$\dfrac {\mu_0}{4\pi l}\dfrac {R^2m}{\rho D}$
0%
$\dfrac {\mu_0}{2\pi R}\dfrac {lm}{\rho D}$

Explanation

For a solenoid,

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

$x$ is the length of the wire and

$a$ is the area of cross-section, then

Resistance

$R=\dfrac {\rho x}{a}$ and mass

$m=axD$

$Rm=\dfrac {\rho x}{a}axD, x=\sqrt {\dfrac {Rm}{\rho D}}$

Also,

$x=2\pi rN, N=\dfrac {x}{2\pi r} \left (\therefore L=\dfrac {\mu_0N^2A}{l}\right )$

$\therefore L=\mu_0\left (\dfrac {x}{2\pi r}\right )^2\dfrac {\pi r^2}{l}=\dfrac {\mu_0}{4\pi l}\dfrac {Rm}{\rho D}$

The self induction takes place when magnetic flux through a coil :

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Remains steady
0%
Decreases
0%
Increases
0%
Either (B) or (C)

In alternating current

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The direction of current is always positive
0%
The direction of current is always negative
0%
The direction of current changes constantly
0%
The direction of current is either positive or negative

A toroid is wound over a circular core with total no. of turns equal to

$N$ . The radius of each turn is

$r$ and radius of toroid is

$R (> > r)$ . The coefficient of self-inductance of the toroid is given by

0%
$L=\dfrac {\mu_0Nr^2}{2R}$
0%
$L=\dfrac {\mu_0Nr}{2R}$
0%
$L=\dfrac {\mu_0Nr^2}{R}$
0%
$L=\dfrac {\mu_0N^2r^2}{2R}$

Explanation

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

$B=\mu_0nI$
where

$n=\dfrac {N}{2\pi R}$

$\therefore \phi=N\pi r^2\left (\mu_0\dfrac {N}{2\pi R}I\right )$

$\phi=\dfrac {\mu_0N^2r^2I}{2R}$

$L=\dfrac {\phi}{I}=\dfrac {\mu_0N^2r^2}{2R}$

The emf produced in a wire by its motion across a magnetic field does not depend upon

0%
length of the wire
0%
composition of the wire
0%
speed of the wire
0%
orientation of the wire

Magnitude of e.m.f. produced in a coil when a magnet is inserted into it does not depend upon :

0%
Number of turns in the coil
0%
Speed of the magnet
0%
Magnetic strength of magnet
0%
Temperature of coil

A magnet is moved towards a coil, first quickly and then slowly. The induced e.m.f. produced is :

0%
Larger in first case
0%
Smaller in first case
0%
Equal in both cases
0%
Larger or smaller, depending upon resistance of the coil

Explanation

Emf induced in the coil

$\mathcal{E} = -\dfrac{d\phi}{dt}$
where

$\phi$ is the magnetic flux.
Thus emf induced is proportional to rate of change of magnetic flux.
When a magnet is moved towards the coil quickly, rate of change of flux is larger that that if the magnetic is moved slowly, thus larger emf is induced due to quick movement of the coil.

Magnetic flux during time interval

$\tau$ varies through a stationary loop of resistance

$R$ , as

$\phi_B = at (\tau - t)$ . Find the amount of heat generated during that time. Neglect the inductance of the loop.

0%
$\displaystyle \dfrac{a^2 \tau^3}{R}$
0%
$\displaystyle \dfrac{a^2 \tau^2}{2R}$
0%
$\displaystyle \dfrac{a^2 \tau^3}{3R}$
0%
$\displaystyle \dfrac{a^2 \tau^3}{4R}$

Explanation

$i = \displaystyle \frac{d\phi}{dt } \dfrac{1}{R} = \frac{a(\tau - 2t)}{R}$

Heat produced

$H = \displaystyle \int_0^{\tau} i^2 Rdt$

$H = \displaystyle \int_0^{\tau} \frac{a^2 (\tau - 2t)^2}{R} dt = \frac{a^2 \tau^3}{3R}$

The two rails of railway track, insulated from each other and from the ground, are connected to a mill voltmeter. What will be the reading of the milli voltmeter when a train travels on the track at a speed of

$180 km/hr$ and earth's magnetic field is

$0.2 \times 10^{-4}T$ and the rails are separated by 1m.

0%
1 m v
0%
2 m v
0%
3 m v
0%
4 m v

The length of a wire required to manufacture a solenoid of length

$l$ and self-induction

$L$ is (cross-sectional area is negligible)

0%
$\sqrt {\dfrac {2\pi Ll}{\mu_0}}$
0%
$\sqrt {\dfrac {\mu_0Ll}{4\pi}}$
0%
$\sqrt {\dfrac {4\pi Ll}{\mu_0}}$
0%
$\sqrt {\dfrac {\mu_0Ll}{2\pi}}$

Explanation

Hint: Use the formula of self-inductance of a solenoid.

Correction Option: C

Explanation for correct option:

$\textbf{Step1: Find number of turns in the solenoid and area of cross section of solenoid}$

Let is the length of the wire, then the length of the solenoid is . (Here N $\Rightarrow$ number of turns}

$\Rightarrow N = \dfrac{x}{{2\pi r}}$

Since, the wire is in the shape of a cylinder, the area of cross section is $A = \pi {r^2}$ (Here $r$ is radius of coil)

$\textbf{Step2: Find length of a wire required}$

The self-inductance of a solenoid is given by, $L = \dfrac{{{\mu _0}{N^2}A}}{l}$ , where ${\mu _0}$ is the magnetic permeability, $l$ is the length of the solenoid and $A$ is the area of each turn in the solenoid.

$\Rightarrow L = \dfrac{{{\mu _0}{{\left( {\dfrac{x}{{2\pi r}}} \right)}^2}\pi {r^2}}}{l}$

$\Rightarrow x = \sqrt {\dfrac{{4\pi Ll}}{{{\mu _0}}}}$

So, option C is correct.

In the process of electromagnetic induction, the magnitude of the induced emf does not depend on _______.

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

Faraday's Law states that,

$E = - N \dfrac{d \phi}{dt}$
Therefore, it depends only on the number of turns and the rate of change of magnetic flux. It may happen that rate of change depends on the area of cross section but this is always not the case. Consider the example,
A coil is placed in a uniform magnetic field but its area is shrinking at some given rate. This also results in electromagnetic induction.

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 the same wire is wound on a solenoid of length

$l/2$ and radius

$r/2$ , then the self-inductance will be

0%
$2L$
0%
$L$
0%
$4L$
0%
$8L$

Explanation

$L=\dfrac {\mu_0N^2\pi r^2}{l}$
Length of wire

$=N2\pi r= Constant (=C$ , suppose)

$\therefore L=\mu_0\left (\dfrac {C}{2\pi r}\right )^2\dfrac {\pi r^2}{l}$

$\therefore L\propto \dfrac {1}{l}$

$\therefore$ Self-inductance will become

$2L$ .

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 one cycle is

Two coils X and Y are linked such that emf

$E$ is induced in Y when the current in X is changing at the rate

$I'(=dI/dt)$ . If a current

$I_0$ is now made to flow through Y, the flux linked with X will be

0%
$EI_0I'$
0%
$\dfrac {I_0I'}{E}$
0%
$\left (\dfrac {E}{I_0}\right )I'$
0%
$\left (\dfrac {E}{I'}\right )I_0$

Explanation

Let

$M$ be the mutual inductance,
We can write

$E=MI'$
We know that

$E=-\dfrac{d\Phi_B}{dt}$

$\therefore \Phi_B=-\int Edt=-M\int I'dt=-MI$ where

$I$ is the current in the circuit.
Given

$I=I_0$ .
So

$\Phi_B=-MI_0=-(\dfrac{E}{I'})I_0$ .
Sign is indicative of the direction. So for magnitude only

$\Phi_B=(\dfrac{E}{I'})I_0$

The induction coil works on the principle of

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self induction
0%
mutual induction
0%
amperes rule
0%
fleming's right hand rule

The network shown in the figure is part of a complete circuit. If at a certain instant, the current

$I$ is

$5A$ and it is decreasing at a rate of

$10^{3} As^{-1}$ then

$V_{B}-V_{A}$ equals :

0%
$20 V$
0%
$15 V$
0%
$10 V$
0%
$5 V$

Explanation

We will write voltage drop equation as we go from A to B.

${ V }_{ A }-iR+V+L\dfrac { di }{ dt } ={ V }_{ B }$

${ V }_{ A }-5\times 1+15-5\times { 10 }^{ -3 }\times (-{ 10 }^{ 3 })={ V }_{ B }$

$V_B-V_A=15$

The time constant of an inductance coil is

$2\times 10^{-3}s$ . When a

$90\Omega$ resistance is joined in series, the same constant becomes

$0.5\times 10^{-3}s$ . The inductance and resistance of the coil are

0%
$30 mH; 30\Omega$
0%
$60 mH; 30\Omega$
0%
$30 mH; 60\Omega$
0%
$60 mH; 60\Omega$

Explanation

Initally the time constant can be representyed as:

$\dfrac{L}{R}=2\times 10^{-3}$ ..................(i)

When the resistance is increased, the new time constant is given as:

$\dfrac{L}{R+90}=0.5 \times 10^{-5}$ ..................(ii)

From (i) and (ii) , on solving , we get

$L=60 mH$ and

$R= 30 \Omega$

Two coils are at fixed locations. When coil 1 has no current and the current in coil 2 increases at the rate of

$15.0 As^{-1}$ , the emf in coil 1 is

$25 mV$ . When coil 2 has no current and coil 1 has a current of

$3.6A$ , the flux linkage in coil 2 is

0%
$16 mWb$
0%
$10 mWb$
0%
$4.00 mWb$
0%
$6.00 mWb$

Explanation

$|M|=\dfrac {e_1}{(di_2/dt)}=\dfrac {\phi_2}{i_1}$

$\therefore \phi_2=\dfrac {e_1i_1}{(di_2/dt)}=\dfrac {(25.0\times 10^{-3})(3.6)}{(15)}$

$=6\times 10^{-3}=6 mWb$

Calculate the inductance of a unit length of a double tape line as shown in figure. The tapes are separated by a distance

$h$ which is considerably less than their width

$b$ .

0%
$\dfrac {\mu_0h}{b}$
0%
$\dfrac {\mu_0h}{2b}$
0%
$\dfrac {2\mu_0h}{b}$
0%
$\dfrac {\sqrt 2\mu_0h}{b}$

Explanation

$\oint \vec B\cdot d\vec l=\mu_0I\Rightarrow B.2b=\mu_0I$

$\Rightarrow B=\dfrac {\mu_0I}{2b}\rightarrow$ due to one tape
Net field

$=2B$
Magnetic flux passing through this double tape

$\phi =2BA=2B(lh)\Rightarrow \phi=\dfrac {\mu_0I}{b}lh$

$L=\dfrac {\phi}{I}=\dfrac {\mu_0lh}{b}\Rightarrow \dfrac {L}{l}=\dfrac {\mu_0h}{b}$

The self inductance of the toroidal solenoid described in the passage is

0%
$\dfrac {\mu_0N^2h}{2\pi}$
0%
$\dfrac {\mu_0N^2h}{2\pi}log_e\dfrac {b}{a}$
0%
$\dfrac {\mu_0Nh}{2\pi}log_e\dfrac {b}{a}$
0%
$\dfrac {\mu_0N^2h}{2\pi}log_e\dfrac {a}{b}$

Explanation

Consider a small section of height

$h$ and width

$dr$ of the total toroidal cross-section.
Using Ampere's circuital law,

$\int B\cdot dl=N\mu_0i$
Or

$B\times 2\pi r = N\mu_0 i$

$\implies \ B=\dfrac {\mu_0iN}{2\pi r}$
Magnetic flux through this small section,

$d\phi=BdA$
where

$dA = hdr$
So,

$\phi_{per\: turn}$

$=\int d\phi=B\int dA$

$\phi_{per\: turn}$

$=\dfrac {\mu_0iNh}{2\pi}\int_a^b\dfrac {dr}{r}=\dfrac {\mu_0iNh}{2\pi}\ln \dfrac {b}{a}$

Total flux

$\phi_{total}=\text {Number of turns}\times \phi_{per\:turn}$

$\phi_{total}$

$=\dfrac {\mu_0N^2hI}{2\pi}\ln \dfrac {b}{a}$
Using

$\phi = Li$
We get

$L=\dfrac {\mu_0N^2h}{2\pi}\ln\dfrac {b}{a}$

The coil is wound on an iron core and looped back on itself so that core has two sets of closely wound coils carrying current in opposite directions. The self inductance is

0%
$0$
0%
$2L$
0%
$2L + M$
0%
$L + 2M$

Explanation

Let the inductance due to first loop alone be

$L$ .

Then inductance due to second loop alone is also

$L$ .

The effective self inductance can be calculate by:

$L_{eff}=L_1+L_2-2M$

$=L+L-2\sqrt{L.L}=0$

When a current of 5 A flows in the primary coil then the flux linked with the secondary coil is 200 weber. The value of coefficient of mutual induction will be

0%
1000 H
0%
40 H
0%
195 H
0%
205 H

Explanation

Coefficient of mutual induction will be the ratio of the flux linked with the secondary coil and the current primary coil.

$M = \dfrac{200}{5} = 40 H$

The electric flux through a certain area of dielectric is

$8.76 \times 10^3 t^4$ . The displacement current through the area is

$12.9 pA$ at

$t = 26.1 ms$ . Find the dielectric constant of the material.

0%
$2 \times 10^{-8}$
0%
$4 \times 10^{-8}$
0%
$8 \times 10^{-8}$
0%
$2 \times 10^{-7}$

Explanation

Displacement current is

$\displaystyle i_D = \varepsilon \dfrac{d\phi_E}{dt}$ or

$\displaystyle \varepsilon = \dfrac{i_D}{\dfrac{(d\phi_E}{dt})}$

$\displaystyle \varepsilon = \dfrac{12.9 \times 10^{-9}}{4 (8.76) \times 10^3 \times (26.1 \times 10^{-3})^3}$

$\simeq 2 \times 10^{-8}$

Find the approximate value of induced current assuming the resistance to the current is confined to the square.

0%
$\displaystyle \dfrac{BL \omega dt}{\rho}$
0%
$\displaystyle \dfrac{BL^2 \omega dt}{\rho}$
0%
$\displaystyle \dfrac{BL^2 \omega d}{\rho}$
0%
$\displaystyle \dfrac{BL \omega d^2}{\rho}$

Explanation

$\displaystyle emf,\ \varepsilon =\dfrac{d \phi}{dt} = BL^2 \omega$

$\displaystyle I = \dfrac{\varepsilon}{R}= \frac{BL^2 \omega}{\rho \dfrac{L}{dt}}= \dfrac{BL \omega dt}{\rho}$

A tank containing a liquid has turns of wire wrapped around it, causing it to act as an inductor. The liquid content of the tank can be measured by using its inductance to determine the height of the liquid level in the tank. The inductance of the tank changes from a value of

$L_0$ corresponding to a relative permeability of

$1$ when the tank is empty to value

$L_f$ corresponding to a relative permeability

$X_m$ (relative permeability of liquid) when the tank is full. The appropriate electronic circuit can determine the inductance correct upto

$5$ significant figures and thus the effective relative permeability of the combined air and liquid within the rectangular has height

$D$ . The height of the liquid level in the tank is

$d$ . Ignore the fringing effects. Assume tank is fitted with

$Hg X _{Hg} = 2.9 \times 10^5$ .

Express

$d$ as a function of

$L$ , inductance corresponding to a certain liquid height

$L_0, L_f$ and

$D$ .

0%
$\displaystyle d =\frac{(L - L_0)D}{L_f - L_0}$
0%
$\displaystyle d =\frac{LD}{L_f - L_0}$
0%
$\displaystyle d =\frac{(L - L_0)D}{L_f }$
0%
None of these

Explanation

Inductance of empty tank=

$L_0$

Relative permeability=1

Inductance of full tank=

$L_f$

Relative permeability=

$X_m$

Now,

$L_f-L_0 \quad \alpha D$

and

$L_f-L_0 \quad\alpha \dfrac{1}{d}$

therefore,

$(L_f-L_0) \quad \alpha \dfrac{D}{d}$

$(L_f-L_0)=k\dfrac{D}{d}$ [k is the probability constant]

Now when the tank is filled partly with liquid and partly empty,Inductance=

$L$

$K=L-L_0$

$L_f-L_0=(L-L_0)\dfrac{D}{d}$

$d=\dfrac{(L-L_0)D}{(L_f-L_0)}$

A conducting loop is placed in a uniform magnetic field with its plane perpendicular to the field. An emf is induced in the loop if :

0%
it is translated
0%
it is rotated about its axis
0%
it is rotated about a diameter
0%
it is deformed

Explanation

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

where

$\phi$ =magnetic flux,

$B$ =magnetic field and

$\vec{A}$ =area vector

The e.m.f. is induced only when the flux linked with the coil changes.

Since, field is constant , flux can be changed either by changing

$A$ or

$\theta$ .

Hence, answer are option-(C) and (D).

The self inductance of a motor of an electric fan is

$10 H$ . In order to impart maximum power at

$50 Hz$ , it should be connected to a capacitance of :

0%
$\displaystyle 4\mu F$
0%
$\displaystyle 8\mu F$
0%
$\displaystyle 1\mu F$
0%
$\displaystyle 2\mu F$

Explanation

Maximum power is transferred at resonance.

$\displaystyle \therefore f_0 = \dfrac{1}{2\pi \sqrt{LC}}$
or

$\displaystyle C = \dfrac{1}{4\pi^2 f^2_0 L} = \dfrac{1}{4\times 10\times(50)^2\times 10}$

$\displaystyle = 10^{-6}F = 1\mu F$

How many meters of a thin wire are required to design a solenoid of length

$1 m$ and

$L = 1mH$ ? Assume cross-sectional diameter is very small.

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$10 m$
0%
$40 m$
0%
$70 m$
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$100 m$
0%
$140 m$

Explanation

Length of the wire

$l = nl_0 2 \pi r$ and

$L = \mu_0 n^2 l_0 \pi r^2$
or

$\displaystyle n = \sqrt{\dfrac{L}{\mu_0 l_0 \pi r^2}}$
Thus,

$\displaystyle l = \sqrt{\dfrac{L}{\mu_0 l_0 \pi r^2}} l_0 2 \pi r$

$\displaystyle 2 \pi r = \sqrt{\dfrac{Ll_0 4 \pi}{\mu_0}}$

$= \displaystyle \sqrt{\dfrac{10^{-3} \times 1 \times 4 \pi}{4 \pi \times 10^{-7}}} = 100 m$

A small, conducting circular loop is placed inside a long solenoid carrying a current. The plane of the loop contains the axis of the solenoid. If the current in the solenoid is varied. The current induced in the loop is

0%
clockwise
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anti-clockwise
0%
zero
0%
clockwise or anti-clockwise depending on whether the resistance is increased or decreased

Explanation

As the plane of the loop contains the axis of the solenoid, the changing magnetic flux density will be along the plane of the loop and hence there will be no change in flux across loop

$( \vec{area} . \vec{B} = 0 )$ .
Therefore no emf is induced and then current in the loop is zero.

Two coils P and Q are lying parallels and very close to each other. Coil P is connected to an AC source whereas Q is connected to a sensitive galvanometer. On pressing key K

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small variations are observed in the galvanometer for applied 50 Hz voltage
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deflections in the galvanometer can be observed for applied voltage of 1 Hz to 2 Hz.
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no deflection in the galvanometer will be observed
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constant deflection will be observed in the galvanometer for 50 Hz supply voltage

When the number of turns per unit length in a solenoid is doubled then its coefficient of self induction will become

0%
half
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double
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four times
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unchanged

The number of turns in an air core solenoid of length 25 cm and radius 4 cm isIts self inductance will be

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$5 \times 10^{-4} H$
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$2.5 \times 10^{-4} H$
0%
$5.4 \times 10^{-3} H$
0%
$2.5 \times 10^{-3} H$

Explanation

Inducatnce =

$\mu N^2 \dfrac{ Area}{length}$

$= \mu 100^2 \dfrac{ \pi 0.4^2}{0.25}= 2.5 \times 10^{-4} H$

A cycle wheel with 64 spokes is rotating with N rotations per second at right angles to horizontal component of magnetic field. The induced emf generated between its axle and rim is E. If the number of spokes is reduced to 32 then the value of induced emf will be :

0%
E
0%
2E
0%
$\displaystyle \frac{E}{2}$
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$\displaystyle \frac{E}{4}$

Explanation

The emf induced is independent of the number of spoke and only depends on the lenght of spoke and the angular speed.
Therefore emf induced is same as before =

$E$

The maximum value of induced emf in a coil rotating in magnetic field does not depend on

0%
the resistance of coil
0%
the number of turns in the coil
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the area of the coil
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rotational frequency of the coil

Explanation

$emf$ induced

$= -$ rate of change of flux
flux is number of turns times the area times the magnetic flux density
rate is proportional to the rotational frequency

$emf$ does not depend on the resistance of the circuit

CBSE Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 3 - 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 3 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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