Blog Log in Get Started Free

Electromagnetic Induction MCQs for NEET — Physics Questions with Answers

Practice free Electromagnetic Induction (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

Chapter Difficulty Show only questions I answered incorrectly Clear Register free for difficulty & keyword filters

● ● ● ● ●

The inductance of a coil is 60μH. A current in this coil increases from 1.0 A to 1.5 A in 0.1 second. The magnitude of the induced e.m.f. is

60 × 10^–6 V

300 × 10^–4 V

30 × 10^–4 V

3 × 10^–4 V

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$e = L \frac{d i}{d t} = 60 \times 10^{- 6} . \frac{(1 .5 - 1 .0)}{0 .1} = 3 \times 10^{- 4} v o l t$

● ● ● ● ●

The self inductance of a coil is L. Keeping the length and area same, the number of turns in the coil is increased to four times. The self inductance of the coil will now be

$\frac{1}{4} L$

4 L

16 L

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$L = \frac{μ_{0} N^{2} A}{l} L \propto N^{2} L_{f} = {(\frac{4 N}{N})}^{2} L = 16 L$

● ● ● ● ●

A coil has an inductance of 2.5 H and a resistance of 0.5 r. If the coil is suddenly connected across a 6.0 volt battery, then the time required for the current to rise 0.63 of its final value is

3.5 sec

4.0 sec

4.5 sec

5.0 sec

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$t = τ = \frac{L}{R} = \frac{2 .5}{0 .5} = 5 \sec$

● ● ● ● ●

If a current of 10 A flows in one second through a coil, and the induced e.m.f. is 10 V, then the self-inductance of the coil is

$\frac{2}{5} H$

$\frac{4}{5} H$

$\frac{5}{4} H$

1 H

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$| e | = L \frac{d i}{d t} \Rightarrow 10 = L \times \frac{10}{1} \Rightarrow L = 1 H$

● ● ● ● ●

An inductance L and a resistance R are first connected to a battery. After some time the battery is disconnected but L and R remain connected in a closed circuit. Then the current reduces to 37% of its initial value in

RL sec

$\frac{R}{L} s e c$

$\frac{L}{R} s e c$

$\frac{1}{L R} s e c$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

When battery disconnected current through the circuit start decreasing exponentially according to $i = i_{0} e^{- R t / L}$

⇒ $0 .37 i_{0} = i_{0} e^{- R t / L}$

⇒ $0 .37 = \frac{1}{e} = e^{- R t / L} \Rightarrow e = e^{\frac{L}{R t}} \Rightarrow \frac{L}{R t} = 1 \Rightarrow \frac{L}{R} = t$

● ● ● ● ●

In an LR-circuit, the time constant is that time in which current grows from zero to the value (where I₀ is the steady-state current)

0.63 I₀

0.50 I₀

0.37 I₀

I₀

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Current at any instant of time t after closing an L-R circuit is given by $I = I_{0} [1 - e^{\frac{- R}{L} t}]$

Time constant $t = \frac{L}{R}$

$∴ I = I_{0} [1 - e^{\frac{- R}{L} \times \frac{L}{R}}] = I_{0} (1 - e^{- 1}) = I_{0} (1 - \frac{1}{e})$

$= I_{0} (1 - \frac{1}{2 .718}) = 0 .63 I_{0} = 63 %$ of I₀

● ● ● ● ●

A copper rod of length l is rotated about one end perpendicular to the magnetic field B with constant angular velocity ω. The induced e.m.f. between the two ends is

$\frac{1}{2} B ω l^{2}$

$\frac{3}{4} B ω l^{2}$

$B ω l^{2}$

$2 B ω l^{2}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

● ● ● ● ●

Two conducting circular loops of radii R₁ and R₂ are placed in the same plane with their centres coinciding. If R₁ >> R₂, the mutual inductance M between them will be directly proportional to

R₁/R₂

R₂/R₁

$R_{1}^{2} / R_{2}$

$R_{2}^{2} / R_{1}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Mutual inductance between two coil in the same plane with their centers coinciding is given by

$M = \frac{μ_{0}}{4 π} (\frac{2 π^{2} R_{2}^{2} N_{1} N_{2}}{R_{1}})$ henry.

● ● ● ● ●

A circular loop of radius R carrying current I lies in the x-y plane with its centre at the origin. The total magnetic flux through the x-y plane is

Directly proportional to I

Directly proportional to R

Directly proportional to R²

Zero

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

The circular loop behaves as a magnetic dipole whose one surface will be N-pole and another will be S-pole. Therefore magnetic lines of force emerge from N will meet at S. Hence total magnetic flux through x-y plane is zero.

● ● ● ● ●

A small square loop of wire of side l is placed inside a large square loop of wire of side L (L > l). The loop are coplanar and their centre coincide. The mutual inductance of the system is proportional to

l / L

l² / L

L/l

L²/l

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Electromagnetic Induction question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.