Electromagnetic Induction — Page 2 | NEET Physics

Q11

A long solenoid of diameter 0.1 m has 2

\times

10 4 turns per meter. At the centre of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is 10

\pi

2

\Omega

, the total charge flowing through the coil during this time is

A 16

\mu

C

B 32

\mu

C

C 16

\pi

\mu

C

D 32

\pi

\mu

C

Correct Answer

Option B

Solution

q =

- {{d\phi } \over R}

=

- {{NA\left( {{B_2} - {B_1}} \right)} \over R}

=

{{NA{\mu _0}n\left( {{i_1} - {i_2}} \right)} \over R}

=

{{100 \times \pi \times {{10}^{ - 4}} \times 4\pi \times {{10}^{ - 7}} \times 2 \times {{10}^4} \times 4} \over {10{\pi ^2}}}

= 32 $\times$ 10 -6 C = 32 $\mu$ C

Q12

A uniform magnetic field is restricted within a region of rafius r. The magnetic field changes with time at a rate

{{d\overrightarrow B } \over {dt}}

. Loop 1 of radius R > r encloses the region r and loop 2 of radius R is outside the region of magnetic field as shown in the figure. Then the e.m.f. generated is

A zero in loop 1 and zero in loop 2

B

- {{d\overrightarrow B } \over {dt}}\pi {r^2}

in loop 1 and

- {{d\overrightarrow B } \over {dt}}\pi {r^2}

in loop 2

C

- {{d\overrightarrow B } \over {dt}}\pi {R^2}

in loop 1 and zero in loop 2

D

- {{d\overrightarrow B } \over {dt}}\pi {r^2}

in loop 1 and zero in loop 2

Correct Answer

Option D

Solution

Emf generated in loop 1,

{\varepsilon _1} = - {{d\phi } \over {dt}} = - {d \over {dt}}\left( {BA} \right) = - A \times {{dB} \over {dt}}

$\Rightarrow$

{\varepsilon _1} = - \pi {r^2} \times {{dB} \over {dt}}

Emf generated in loop 2,

{\varepsilon _2} = - {{d\phi } \over {dt}} = - {d \over {dt}}\left( {BA} \right) = - {d \over {dt}}\left( {0 \times A} \right)

= 0

Q13

A long solenoid has 1000 turns. When a current of 4 A flows through it, the magnetic flux linked with each turn of the solenoid is 4

\times \,{10^{ - 3}}

Wb. The self-inductance of the solenoid is

A 2 H

B 1 H

C 4 H

D 3 H

Correct Answer

Option B

Solution

Total flux linked with the solenoid, $\phi$ = N $\phi$ 0 = 1000 $\times$ 4 $\times$ 10 -3 Wb = 4 Wb $\therefore$ Self-inductance of solenoid L =

{\phi \over I} = {4 \over 4}

= 1 H

Q14

An electron moves on a straight line path XY as shown. The abcd is a coil adjacent to the path of electron. What will be the direction of current, if any, induced in the coil ?

A The current will reverse its direction as the electron goes past the coil

B No current induced

C abcd

D adcb

Correct Answer

Option A

Solution

When e – is coming towards the loop magnetic flux of one type increased and when going away same magnetic flux decreased so induced current opposite to each other.

Q15

A thin semicircular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B, as shown in the figure. The potential difference developed across the ring when its speed is

v

, is

A zero

B

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

and P is at higher potential

C

\pi rBV

and R is at higher potential

D 2rBV and R is at higher potential

Correct Answer

Option D

Solution

Potential difference that is developed across ring when its speed is v :

\varepsilon

= B v (l i – l f ) where, l i – l f = displacement between end of semicircular ring = 2r Hence,

\varepsilon

= Bv(2r) where R is at high potential.

Q16

A current of 2.5 A flows through a coil of inductance 5 H. The magnetic flux linked with the coil is

A 0.5 Wb

B 12.5 Wb

C zero

D 2 Wb

Correct Answer

Option B

Solution

Here, I = 2.5 A, L = 5 H Magnetic flux linked with the coil is $\phi$ B = LI = (5 H)(2.5 A) = 12.5 Wb

Q17

A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced e.m.f. is

A four times per revoluation

B six times per revolution

C once per revolution

D twice per revolution

Correct Answer

Option D

Solution

This is the case of periodic EMI From graph, it is clear that direction is changing twice in 1 cycle.

Q18

In a coil of resistance 10

\Omega

, the induced current developed by changing magnetic flux through it, is shown in figure as a function of time. The magnitude of change in flux through the coil in weber is

A 8

B 2

C 6

D 4

Correct Answer

Option B

Solution

q = Area under i-t graph =

{1 \over 2} \times 4 \times 0.1

= 0.2 C As q =

{{\Delta \phi } \over R}

$\Rightarrow$

\Delta

$\phi$ = qR = 0.2 $\times$ 10 = 2 weber

Q19

A coil of resistance 400

\Omega

is placed in a magnetic field. If the magnetic flux

\phi

(Wb) linked with the coil varies with time t (sec) as

\phi = 50{t^2} + 4

. The current in the coil at t = 2 sec is

A 0.5 A

B 0.1 A

C 2 A

D 1 A

Correct Answer

Option A

Solution

According, to Faraday’s law of induction Induced e.m.f,

\varepsilon = - {{d\phi } \over {dt}}

= -(100t) Induced current i at t = 2 sec. =

\left| {{\varepsilon \over R}} \right| = + {{100 \times 2} \over {400}}

= +0.5 A

Q20

A conducting circular loop is placed in a uniform magnetic field, B = 0.025 T with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of 1 mm s

-

1 . The induced emf when the radius is 2 cm, is

A

2\pi \mu V

B

\pi \mu V

C

{\pi \over 2}\mu V

D 2

\mu

V

Correct Answer

Option B

Solution

Constant rate at which radius of the loop shrinks

{{dr} \over {dt}} = 1 \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}}

= 0.025 $\times$ $\pi$ $\times$ 2 $\times$ 2 $\times$ 10 -2 $\times$ 1 $\times$ 10 -3 = $\pi$ $\times$ 10 -6 V =

\pi \mu V