Electromagnetic Induction — Page 4 | NEET Physics

Q31

A wire of length 1m moving with velocity 8 m/s at right angles to a magnetic field of 2T. The magnitude of induced emf, between the ends of wire will be __________.

A 20 V

B 8 V

C 16 V

D 12 V

Correct Answer

Option C

Solution

Induced emf across the ends = Bv $l$ = 2 × 8 × 1 = 16 V

Q32

A conducting circular loop of radius

\dfrac{10}{\sqrt\pi}

cm is placed perpendicular to a uniform magnetic field of 0.5 T. The magnetic field is decreased to zero in 0.5 s at a steady rate. The induced emf in the circular loop at 0.25 s is :

A emf = 10 mV

B emf = 5 mV

C emf = 100 mV

D emf = 1 mV

Correct Answer

Option A

Solution

$\begin{aligned} & \mathrm{EMF}=\dfrac{\mathrm{d} \phi}{\mathrm{dt}}=\dfrac{\mathrm{BA}-0}{\mathrm{t}} \\\\ & \mathrm{A}=\pi \mathrm{r}^2=\pi\left(\dfrac{0.1^2}{\pi}\right)=0.01 \\\\ & \mathrm{~B}=0.5 \\\\ & \mathrm{EMF}=\dfrac{(0.5)(0.01)}{0.5}=0.01 \mathrm{~V}=10~ \mathrm{mV}\end{aligned}$

Q33

The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1s, the change in the energy of the inductance is -

A 740 J

B 637.5 J

C 540 J

D 437.5 J

Correct Answer

Option D

Solution

L{{di} \over {dt}} = 25

L \times {{15} \over 1} = 25

L = {5 \over 3}H

\Delta E = {1 \over 2} \times {5 \over 3} \times ({25^2} - {10^2})

= {5 \over 6} \times 525 = 437.5

J

Q34

A metal conductor of length

1

m

rotates vertically about one of its ends at angular velocity

5

radians per second. If the horizontal component of earth's magnetic field is

0.2 \times {10^{ - 4}}T,

then the

e.m.f.

developed between the two ends of the conductor is

A

5mV

B

50\mu V

C

5\mu V

D

50mV

Correct Answer

Option B

Solution

\ell = 1m,\,\,\omega = 5\,rad/s,\,\,B = 0.2 \times {10^{ - 4}}T

\varepsilon = {{B\omega {\ell ^2}} \over 2} = {{0.2 \times {{10}^{ - 4}} \times 5 \times 1} \over 2} = 50\mu V

Q35

Which of the following units denotes the dimension

{{M{L^2}} \over {{Q^2}}}

, where

Q

denotes the electric charge?

A

Wb/{m^2}

B Henry

(H)

C

H/{m^2}

D Weber

(Wb)

Correct Answer

Option B

Solution

Mutual inductance

= {\phi \over I} = {{BA} \over I}

[Henry]

= {{\left[ {M{T^{ - 1}}{Q^{ - 1}}{L^2}} \right]} \over {\left[ {Q{T^{ - 1}}} \right]}} = M{L^2}{Q^{ - 2}}

Q36

A boat is moving due east in a region where the earth's magnetic fields is

5.0 \times {10^{ - 5}}

N{A^{ - 1}}\,{m^{ - 1}}

due north and horizontal. The best carries a vertical aerial

2

m

long. If the speed of the boat is

1.50\,m{s^{ - 1}},

the magnitude of the induced

emf

in the wire of aerial is :

A

0.75

mV

B

0.50

mV

C

0.15

mV

D

1

mV

Correct Answer

Option C

Solution

Induced

emf

= v{B_H}l = 1.5 \times 5 \times {10^{ - 5}} \times 2

= 15 \times {10^{ - 5}} = 0.15\,mV

Q37

A circular loop of radius

0.3

cm

lies center of the small loop is on the axis of the bigger loop. The distance between their centers is

15

cm.

If a current of

2.0

A

flows through the smaller loop, than the flux linked with bigger loop is

A

9.1 \times {10^{ - 11}}\,

weber

B

6 \times {10^{ - 11}}\,

weber

C

3.3 \times {10^{ - 11}}\,

weber

D

6.6 \times {10^{ - 9}}\,

weber

Correct Answer

Option A

Solution

As we know, Magnetic flux,

\phi = B.A

{{{\mu _0}\left( 2 \right){{\left( {20 \times {{10}^{ - 2}}} \right)}^2}} \over {2\left[ {{{\left( {0.2} \right)}^2} + {{\left( {0.15} \right)}^2}} \right]}} \times \pi {\left( {0.3 \times {{10}^{ - 2}}} \right)^2}

On solving

= 9.216 \times {10^{ - 11}} = 9.2 \times {10^{ - 11}}\,\,

weber

Q38

Two coaxial solenoids of different radius carry current

I

in the same direction.

\overrightarrow {{F_1}}

be the magnetic force on the inner solenoid due to the outer one and

\overrightarrow {{F_2}}

be the magnetic force on the outer solenoid due to the inner one. Then :

A

\overrightarrow {{F_1}}

is radially in wards and

\overrightarrow {{F_2}} = 0

B

\overrightarrow {{F_1}}

is radially outwards and

\overrightarrow {{F_2}} = 0

C

\overrightarrow {{F_1}} = \overrightarrow {{F_2}} = 0

D

\overrightarrow {{F_1}}

is radially inwards and

\overrightarrow {{F_2}}

is radially outards

Correct Answer

Option C

Solution

\mathop {F{}_1}\limits^ \to = \mathop {F{}_2}\limits^ \to = 0

because of action and reaction pair

Q39

A small circular loop of wire of radius a is located at the centre of a much larger circular wire loop of radius b. The two loops are in the same plane. The outer loop of radius b carries an alternating current I = Io cos (

\omega

t). The emf induced in the smaller inner loop is nearly :

A

{{\pi {\mu _o}{I_o}} \over 2}.{{{a^2}} \over b}\,\omega \sin \left( {\omega t} \right)

B

{{\pi {\mu _o}{I_o}} \over 2}.{{{a^2}} \over b}\,\omega \cos \left( {\omega t} \right)

C

\pi {\mu _o}{I_o}\,{{{a^2}} \over b}\omega \sin \left( {\omega t} \right)

D

{{\pi {\mu _o}{I_o}\,{b^2}} \over a}\omega \cos \left( {\omega t} \right)

Correct Answer

Option A

Solution

Mutual inductance, M =

{{{\mu _0}\pi {N_1}{N_2}\,{a^2}} \over {2b}}

here

{{N_1}}

= N2 = 1

\therefore\,\,\,

M =

{{{\mu _0}\pi {a^2}} \over {2b}}

Current I = I0 cos ( $\omega$ t) According to Faraday's law, e = $-$ M

{{dI} \over {dt}}

= $-$

{{{\mu _0}\pi {a^2}} \over {2b}}

{d \over {dt}}

(I0 cos $\omega$ t) = +

{{{\mu _0}\pi {a^2}} \over {2b}}

I0 $\omega$ sin $\omega$ t =

{{\pi {\mu _0}{I_0}} \over 2}

.

{{{a^2}} \over b}

$\omega$ sin $\omega$ t

Q40

A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity

\omega ,

the maxium e.m.f. induced in the coil will be:

A 3 nBA

\omega

B

{3 \over 2}

nBA

\omega

C nBA

\omega

D

{1 \over 2}

nBA

\omega

Correct Answer

Option C

Solution

Flux in the coil, $\phi$ = nBA sin( $\omega$ t) When n = no. of turns

\,\,\,\,\,\,\,\,\,\,\,\,\,\,

A = Area of coil

\,\,\,\,\,\,\,\,\,\,\,\,\,\,

$\omega$ = angular speed Induced emf,

\left| e \right| = {{d\phi } \over {dt}}

= nBA $\omega$ cos $\omega$ t

\therefore\,\,\,

emax = nBA $\omega$