Electromagnetic Induction Questions — NEET Physics

Q1

A rectangular wire loop of sides 8 cm and 3 cm with a small cut, is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the plane of the loop. The emf developed across the cut, if the velocity of the loop is

2 \mathrm{~cm} \mathrm{~s}^{-1}

, in a direction normal to the shorter side of the loop, will be :

A

4.8 \times 10^{-4}

volt

B

1.2 \times 10^{-4}

volt

C

1.3 \times 10^{-4}

volt

D

1.8 \times 10^{-4}

volt

Correct Answer

Option D

Solution

Induced emf across the shorter side

\begin{aligned} & E_{\text{induced }}=B V l[\because \vec{B} \perp \vec{V} \perp \vec{I}] \\ & =0.3 \times 2 \times 10^{-2} \times 3 \times 10^{-2} \\ & =1.8 \times 10^{-4} \mathrm{~V} \\ & =1.8 \times 10^{-4} \text{ volt } \end{aligned}

Q2

A B

is a part of an electrical circuit (see figure). The potential difference

V_A-V_B

, at the instant when current

i=2 \mathrm{~A}

and is increasing at a rate of

1 \mathrm{amp} /

second is:

A 9 volt

B 10 volt

C 5 volt

D 6 volt

Correct Answer

Option B

Solution

\begin{aligned} &\text{ Given, } l=2 \mathrm{~A} \text{ and } \frac{d i}{d t}=+1 \mathrm{~A} / \mathrm{s}\\ &\begin{aligned} & V_A-L \frac{d i}{d t}-5-i \times 2=V_B \\ & \Rightarrow V_A-1 \times 1-5-2 \times 2=V_B \\ & \Rightarrow V_A-V_B=10 \text{ volt } \end{aligned} \end{aligned}

Q3

Let us consider two solenoids

A

and

B

, made from same magnetic material of relative permeability

\mu_r

and equal area of cross-section. Length of

A

is twice that of

B

and the number of turns per unit length in

A

is half that of

B

. The ratio of self inductances of the two solenoids,

L_A: L_B

is

A

1: 2

B

2: 1

C

8: 1

D

1: 8

Correct Answer

Option A

Solution

\begin{aligned} & L=\mu_0 \mu_r \times n \times A \times N \\ & L=\mu_0 \mu_r n \times A \times \frac{N}{I} \times I \\ & L=\mu_0 \mu_r \times n^2 \times A \times I \Rightarrow L \propto n^2 I \\ & \Rightarrow \frac{L_A}{L_B}=\frac{n_A^2}{n_B^2} \times \frac{I_A}{I_B} \\ & \Rightarrow \frac{L_A}{L_B}=\frac{1}{4} \times 2=\frac{1}{2} \end{aligned}

Q4

An emf is generated by an ac generator having 100 turn coil, of loop area

1 \mathrm{~m}^2

. The coil rotates at a speed of one revolution per second and placed in a uniform magnetic field of

0.05 \mathrm{~T}

perpendicular to the axis of rotation of the coil. The maximum value of emf is :-

A 3.14 V

B 31.4 V

C 62.8 V

D 6.28 V

Correct Answer

Option B

Solution

\omega=2\pi \frac{\mathrm{rad}}{\mathrm{sec}}

\mathrm{E_{max}=NBA\omega}

=100\times0.05\times1\times2\pi

=10\times\pi

=31.4\,\mathrm{V}

Q5

The net magnetic flux through any closed surface is :

A Positive

B Infinity

C Negative

D Zero

Correct Answer

Option D

Solution

Magnetic field exist in Closed Loops (Monopoles do not exist)

\phi \overrightarrow{\mathrm{B}} \cdot \mathrm{d}\overrightarrow{\mathrm{A}}=0

(Gauss law for magnetism)

Q6

The magnetic flux linked to a circular coil of radius R is

\phi = 2{t^3} + 4{t^2} + 2t + 5

Wb The magnitude of induced emf in the coil at t = 5 s is

A 192 V

B 108 V

C 197 V

D 150 V

Correct Answer

Option A

Solution

Given magnetic flux is

\phi = 2{t^3} + 4{t^2} + 2t + 5

Induced emf is given by

\varepsilon = {{ - d\phi } \over {dt}} = {{ - d} \over {dt}}(2{t^3} + 4{t^2} + 2t + 5)

|\varepsilon | = 6{t^2} + 8t + 2

At

t = 5

s,

|\varepsilon | = 6 \times 25 + 8 \times 5 + 2

= 150 + 40 + 2

= 192

V

Q7

A square loop of side 1 m and resistance 1

\Omega

is placed in a magnetic field of 0.5 T. If the plane of loop is perpendicular to the direction of magnetic field, the magnetic flux through the loop is

A 2 weber

B 0.5 weber

C 1 weber

D Zero weber

Correct Answer

Option B

Solution

Magnetic flux ( $\phi$ B ) =

\overrightarrow B

.

\overrightarrow A

\overrightarrow B

and

\overrightarrow A

are in same direction, therefore $\phi$ B = B . A = 0.5 $\times$ 1 2 = 0.5 Wb

Q8

Two conducting circular loops of radii R 1 and R 2 are placed in the same plane with their centres coinciding. If R 1 >> R 2 , the mutual inductance M between them will be directly proportional to :

A

{{R_2^2} \over {{R_1}}}

B

{{{R_1}} \over {{R_2}}}

C

{{{R_2}} \over {{R_1}}}

D

{{R_1^2} \over {{R_2}}}

Correct Answer

Option A

Solution

Two concentric coils are of radius R 1 and R 2 as shown Let current in outer loop be i Magnetic field at center = B =

{{{\mu _0}i} \over {2{R_1}}}

Magnetic flux through inner coil = B $\times$

\pi R_2^2

$\Rightarrow$ $\phi$ =

{{{\mu _0}i} \over {2{R_1}}} \times \pi R_2^2

As per definition, $\phi$ = Mi $\Rightarrow$ M =

\left( {{{{\mu _0}\pi } \over 2}} \right){{R_2^2} \over {{R_1}}}

$\therefore$ M $\propto$

{{R_2^2} \over {{R_1}}}

Q9

A 800 turn coil of effective area 0.05 m 2 is kept. perpendicular to a magnetic filed 5 × 10 –5 T. When the plane of the coil is rotated by 90 o around any of its coplanar axis in 0.1 s, the emf induced in the coil will be :

A 0.2 V

B 0.02 V

C 2 V

D 2

\times

10 -3 V

Correct Answer

Option B

Solution

Given N = 800, A = 0.05 m 2 , B = 5 × 10 –5 T

\Delta

t = 0.15 s We know, e =

- {{\left( {{\phi _f} - {\phi _i}} \right)} \over {\Delta t}}

=

- {{\left( {0 - NBA} \right)} \over {\Delta t}}

=

{{800 \times 5 \times {{10}^{ - 5}} \times 5 \times {{10}^{ - 2}}} \over {0.1}}

= 0.02 V

Q10

In which of the following devices, the eddy current effect is not used?

A magnetic braking in train

B electromagnet

C electric heater

D induction furnace

Correct Answer

Option C

Solution

Eddy current effect is not used in case of electric heater because it works on Joule's heating effect.