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Electromagnetic Induction NEET Questions — Free MCQ Practice

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A coil of wire having finite inductance and resistance has a conducting ring placed coaxially within it. The coil is connected to a battery at time t = 0 so that a time-dependent current I₁(t) starts flowing through the coil. If I₂(t) is the current induced in the ring and B(t) is the magnetic field at the axis of the coil due to I₁(t), then as a function of time (t > 0), the product I₂ (t) B(t)

A.

Increases with time

B.

Decreases with time

C.

Does not vary with time

D.

Passes through a maximum

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A long solenoid of diameter 0.1m has 2 $\times 10^{4}$ turns per meter. At the centre of the solenoid, a coil of 100 turns and radius 0.01m is placed with its axis coinciding with the solenoid's axis. The current in the solenoid reduces at a constant rate to 0 A from 4A in 0.05s. If the resistance of the coil is $10 π^{2} Ω$ , the total charge flowing through the coil during this time is

A.

32 $πμC$

B.

16 $μ C$

C.

32 $μ C$

D.

16 $πμC$

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Explanation

(c) Current induced in the coil given by

$i = \frac{1}{R} (\frac{d ϕ}{d t})$

$\Rightarrow \frac{∆ q}{∆ t} = \frac{1}{R} (\frac{∆ ϕ}{∆ t})$

Given, the resistance of the solenoid,

R=10 $π^{2} Ω$

The radius of the second coil r= $10^{- 2}$

$∆ t = 0.05 s, ∆ i = 4 - 0 = 4 A$

The charge flowing through the coil is given by

$∆ q = (\frac{∆ ϕ}{∆ t}) \frac{1}{R} (∆ t)$

= $μ_{0} N_{1} N_{2} π r^{2} (\frac{∆ i}{∆ t}) \frac{1}{R} ∆ t$

= $4 π \times 10^{- 7} \times 2 \times 10^{4} \times 100 \times π$

$\times {(10^{- 2})}^{2} \times (\frac{4}{0.05}) \times \frac{1}{10 π^{2}} \times 0.05$

= $32 \times 10^{- 6} C = 32 μ C$

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A long solenoid has 1000 turns. When a current of 4.0 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.

3H

B.

2H

C.

1H

D.

4H

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Explanation

(c) Given, Number of turns of solenoid, N=1000.

Current, I=4A

Magnetic flux, $ϕ_{B} = 4 \times 10^{- 3} Wb$

so, Self induction of solenoid is given by

$L = \frac{ϕ_{B} . N}{I}$ ...(1)

Substitue the given values in equation (1), we get

$L = \frac{4 \times 10^{- 3} \times 1000}{4} = 1 H$

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

A.

once per revolution

B.

twice per revolution

C.

four times per revolution

D.

six times per revolution

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Explanation

(b)

Frequency of change in direction of emf is double the frequency of rotation

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A coil of resistance 400 $Ω$ is placed in a magnetic field. If the magnetic flux $ϕ (W b)$ linked with the coil varies with time t (sec) as $ϕ = 50 t^{2} + 4 .$

The current in the coil at t=2s is

A.

0.5A

B.

0.1A

C.

2A

D.

1A

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Explanation

Induced emf of coil E = ${|- \frac{d ϕ}{d t}|}_{t}$

Given, $ϕ = 50 t^{2} + 4 a n d R = 400 Ω$

$E = {|- \frac{d ϕ}{d t}|}_{t = 2}$

$= {|100 t|}_{t = 2} = 200 V$

Current in the coil

$i = \frac{E}{R} = \frac{200}{400}$

= $\frac{1}{2} = 0.5 A$

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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 $m m s^{- 1}$ .The induced emf when the radius is 2cm, is

A.

$2 πμ V$

B.

$πμ V$

C.

$\frac{π}{2} μ V$

D.

$2 μ V$

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Explanation

Magnetic flux $ϕ = B \cdot A$

= $B \cdot {πr}^{2}$

Induced emf, $|e| = \frac{d ϕ}{d t} = B π 2 r \frac{d r}{d t}$

$= 0.025 \times π \times 2 \times 2 \times 10^{- 2} \times 1 \times 10^{- 3}$

= $πμV$

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A condenser of capacity C is charged to a potential difference of $V_{1} .$ The plates of the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to $V_{2}$ is

A.

${(\frac{C (V_{1} - V_{2})^{2}}{L})}^{\frac{1}{2}}$

B.

$\frac{C (V_{1}^{2} - V_{2}^{2})}{L}$

C.

$\frac{C (V_{1}^{2} + V_{2}^{2})}{L}$

D.

${(\frac{C (V_{1}^{2} - V_{2}^{2})}{L})}^{\frac{1}{2}}$

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Explanation

$Charge on the condenser at any instant q = q_{\circ} Cosωt where : ‐ ω = \frac{1}{\sqrt{LC}} Coswt = \frac{q}{q_{\circ}} = \frac{{CV}_{2}}{{CV}_{1}} Coswt = \frac{V_{2}}{V_{1}} q = q_{\circ} Cosωt I = \frac{dq}{dt} = - q_{\circ} Sinωt \times ω = - q_{\circ} \cdot ω \sqrt{q - {Cos}^{2} ωt} |I| = \frac{{CV}_{1}}{\sqrt{LC}} \cdot \sqrt{1 - \frac{V_{2}^{2}}{V_{1}^{2}}} = {[\frac{C (V_{1}^{2} - V_{2}^{2})}{L}]}^{1 / 2}$

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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, $\vec{v} = v \hat{i} .$ The magnetic field is directed along the negative z-axis direction. The induced emf, during the passage of these loops, out of the field region, will not remain constant for

A.

the rectangular, circular and elliptical loops

B.

the circular and the elliptical loops

C.

only the elliptical loop

D.

any of the four loops

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Explanation

Area coming out per second from the magnetic field is not constant for elliptical and circular loops, so induced emf, during the passage of these loops, out of the field region will not remain constant for the circular and the elliptical loops.

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a long solenoid has 500 turns. When a current of 2 A is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{- 3}$ Wh. The self-inductance of the solenoid is

A.

2.5 h

B.

2.0 H

C.

1.0 H

D.

4.0 H

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Explanation

Inductance of a coil is numerically equal to the emf induced in the coil when the current in the coil changes at the rate of 1 $A s^{- 1}$ . If I is the current flowing in the circuit, then flux linked with the circuit is observed to be proportional to i, i.e.,

$ϕ \propto I o r ϕ = L I . . . (i)$

where L is called the self-inductance or coefficient of self-inductance or simply inductance of the coil.

Net flux through solenoid,

$ϕ = 500 \times 4 \times 10^{- 3} = 2 W b o r 2 = L \times 2 [a f t e r p u t t i n g v a l u e s i n E q .] o r L = 1 H$

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A circular disc of radius 0.2 m is placed in a uniform magnetic field of induction $\frac{1}{π}$ $(\frac{W b}{m^{2}})$ in such a way that its axis makes an angle of $60^{°}$ with $\vec{B}$ . The magnetic flux linked with the disc is

A.

0.02 Wb

B.

0.06 Wb

C.

0.08 Wb

D.

0.01 Wb

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Electromagnetic Induction MCQs for NEET — Physics Questions with Answers

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