Electromagnetic Induction — Page 3 | NEET Physics

Q21

A rectangular, a square, a circular and an elliptical loop, all in the (x-y) plane, are moving out of a uniform magnetic field with a constant velocity.

\overrightarrow V = v\widehat i

. The magnetic field is directed along the negative z axis direction. The induced emf, during the passes of these loops, out of the field region, will not remain constant for

A The circular and the elliptical loops

B only the ellliptical loop

C any of the four loops

D the rectangular, circular and elliptical loops

Correct Answer

Option A

Solution

Once a rectangular loop or a square loop is being drawn out of the field, the rate of cutting the lines of field will be a constant for a square and rectangle, but not for circular or elliptical areas.

Q22

A conducting circular loop is placed in a uniform magnetic field 0.04 T with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at 2 mm/s. The induced emf in the loop when the radius is 2 cm is

A 4.8

\pi

\mu

V

B 0.8

\pi

\mu

V

C 1.6

\pi

\mu

V

D 3.2

\pi

\mu

V

Correct Answer

Option D

Solution

Constant rate at which radius of the loop shrinks

{{dr} \over {dt}} = 2 \times {10^{-3}}

m s -1 Magnetic flux linked with the loop is $\phi$ = BAcos $\phi$ = B( $\pi$ r 2 )cos0 o = B $\pi$ r 2 The magnitude of the induced emf is |e| =

{{d\phi } \over {dt}} = {d \over {dt}}\left( {B\pi {r^2}} \right) = B\pi 2r{{dr} \over {dt}}

= 0.04 $\times$ $\pi$ $\times$ 2 $\times$ 2 $\times$ 10 -2 $\times$ 2 $\times$ 10 -3 = 3.2 $\pi$ $\times$ 10 -6 V = 3.2

\pi \mu V

Q23

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

{1 \over \pi }\left( {{{Wb} \over {{m^2}}}} \right)

in such a way that its axis makes an angle of 60 o with

\overrightarrow B .

The magnetic flux linked with the disc is

A 0.08 Wb

B 0.01 Wb

C 0.02 Wb

D 0.06 Wb

Correct Answer

Option C

Solution

Given B =

{1 \over \pi }\left( {{{Wb} \over {{m^2}}}} \right)

$\theta$ = 60 o Area normal to the plane of the disc = $\pi$ r 2 cos60 o =

{{\pi {r^2}} \over 2}

Flux = B × normal area =

{{0.02 \times 0.02} \over 2}

= 0.02 Wb

Q24

A long solenoid has 500 turns. When a current of 2 ampere 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

A 1.0 henry

B 4.0 henry

C 2.5 henry

D 2.0 henry

Correct Answer

Option A

Solution

Total number of turns in the solenoid, N = 500 Current, I = 2A.

Magnetic flux linked with each turn = 4 × 10 –3 Wb Self inductance of coil : L =

{{N\phi } \over I} = {{500 \times 4 \times {{10}^3}} \over 2}

= 1 H

Q25

Two coils of self inductance 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

A 16 mH

B 10 mH

C 6 mH

D 4 mH

Correct Answer

Option D

Solution

Mutual inductance is given as, M = K

\sqrt {{L_1} \times {L_2}}

Now given, K = 1, L 1 = 2 mH, L 2 = 8 mH $\therefore$ M =

1 \times \sqrt {2 \times 8}

= 4 mH

Q26

As a result of change in the magnetic flux linked to the closed loop as shown in the figure, an e.m.f.

V

volt is induced in the loop. The work done (joule) in taking a charge Q coulomb once along the loop is

A QV

B 2QV

C QV/2

D zero

Correct Answer

Option A

Solution

Work done due to a charge W = QV

Q27

The magnetic flux through a circuit of resistance R changes by an amount

\Delta

\phi

in a time

\Delta

t. Then the total quantity of electric charge Q that passes any point in the circuit during the time

\Delta

t is represented by

A

Q = {1 \over R}.{{\Delta \phi } \over {\Delta t}}

B

Q = {{\Delta \phi } \over R}

C

Q = {{\Delta \phi } \over {\Delta t}}

D

Q = R.{{\Delta \phi } \over {\Delta t}}

Correct Answer

Option B

Solution

{{\Delta \phi } \over {\Delta t}}

= E = iR $\Rightarrow$

{\Delta \phi = \left( {i\Delta t} \right)R}

= QR $\Rightarrow$ Q =

{{\Delta \phi } \over R}

Q28

For a coil having L = 2 mH, current flow through it is

I = {t^2}{e^{ - t}}

then, the time at which emf become zero

A 2 sec

B 1 sec

C 4 sec

D 3 sec

Correct Answer

Option A

Solution

Given,

I = {t^2}{e^{ - t}}

So,

{{dI} \over {dt}} = 2t{e^{ - t}} - {t^2}{e^{ - t}}

\left| \varepsilon \right| = L{{dI} \over {dt}}

here emf is zero when

{{dI} \over {dt}}

= 0 $\therefore$

2t{e^{ - t}} - {t^2}{e^{ - t}}

= 0 $\Rightarrow$

2t{e^{ - t}} = {t^2}{e^{ - t}}

$\Rightarrow$

t{e^{ - t}}\left( {t - 2} \right)

= 0 As t $\ne$ $\infty$ , t $\ne$ 0 $\therefore$ t = 2 sec

Q29

Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon

A the rates at which currents are changing in the two coils

B relative position and orientation of the two coils

C the currents in the two coils

D the materials of the wires of the coils

Correct Answer

Option B

Solution

Mutual conductance depends on the relative position and orientation of the two coils.

Q30

An aeroplane, with its wings spread 10 m, is flying at a speed of 180 km/h in a horizontal direction. The total intensity of earth's field at that part is 2.5

\times

10

-

4 Wb/m2 and the angle of dip is 60

^\circ

. The emf induced between the tips of the plane wings will be __________.

A 88.37 mV

B 62.50 mV

C 54.125 mV

D 108.25 mV

Correct Answer

Option D

Solution

\varepsilon

ind = (Bv) LV and Bv = BTotal sin60o $\therefore$

\varepsilon

ind = (2.5 $\times$ 10 $-$ 4)(sin 60o) $\times$ 10 $\times$ 180 $\times$

{5 \over {18}}

= 108.25 mV