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Home/JEE MAINS/physics/electromagnetic induction
Curated PYQ Collection

Top 50 Most Repeated ELECTROMAGNETIC INDUCTION PYQs | JEE MAINS

A curated collection of the most important questions from ELECTROMAGNETIC INDUCTION, fully solved with step-by-step concepts to prepare for JEE MAINS.

Question #1

Practice Question

Concept Applied

Faraday's law: $\mathcal{E} = -N \frac{d\Phi}{dt} = -N A \frac{dB}{dt}$. $\frac{dB}{dt} = 0.02$, so $\mathcal{E} = 50 \times 0.1 \times 0.02 = 0.1\,\t...

Read Full Step-by-Step Solution →

Question #2

Practice Question

Concept Applied

Use $\mathcal{E} = -N \frac{d\phi}{dt}$; differentiate $\phi = 0.02\sin(100t)$ and evaluate at $t=0.01$ s....

Read Full Step-by-Step Solution →

Question #3

Practice Question

A.0.012 V
B.0.024 V
C.0.006 V
D.0.0012 V

Concept Applied

The changing magnetic field induces an emf given by $\varepsilon = -N A\,\frac{dB}{dt}$. Substituting $N=500$, $A=2.0\times10^{-4}\,\text{m}^2$ and $\...

Read Full Step-by-Step Solution →

Question #4

Practice Question

Concept Applied

Energy = 0.5 × L × I² = 0.5 × 2 × 25 = 25 J....

Read Full Step-by-Step Solution →

Question #5

Practice Question

A.Opposes the change in magnetic flux that produced it.
B.Aids the change in magnetic flux that produced it.
C.Is always clockwise.
D.Is always counter-clockwise.

Concept Applied

Lenz's law is a consequence of conservation of energy, stating induced current opposes the cause....

Read Full Step-by-Step Solution →

Question #6

Practice Question

Concept Applied

Use $\mathcal{E} = -N \frac{\Delta \phi}{\Delta t} = -50 \cdot \frac{0.06}{0.1} = 30$ V....

Read Full Step-by-Step Solution →

Question #7

Practice Question

A.A) 0.1 V
B.B) 0.5 V
C.C) 1 V
D.D) 2 V

Concept Applied

The induced emf is calculated using the formula ε = N(dΦ/dt), where N is the number of turns, Φ is the magnetic flux, and t is the time....

Read Full Step-by-Step Solution →

Question #8

Practice Question

Concept Applied

Using $\mathcal{E} = -L \frac{dI}{dt}$, $L = \frac{\mathcal{E} \cdot dt}{dI} = \frac{10 \times 0.1}{2} = 0.5$ H....

Read Full Step-by-Step Solution →

Question #9

Practice Question

Concept Applied

$\mathcal{E}_{\text{peak}} = N A \omega B_0 = 50 \times 0.1 \times 100 \times 0.02 = 10\,\text{V}$...

Read Full Step-by-Step Solution →

Question #10

Practice Question

A.A) 1.57 V
B.B) 3.14 V
C.C) 5 V
D.D) 10 V

Concept Applied

The maximum emf induced in the coil is given by ε = NABω, where N is the number of turns, A is the area of the coil, B is the magnetic field, and ω is...

Read Full Step-by-Step Solution →

Question #11

Practice Question

A.10 V
B.20 V
C.30 V
D.40 V

Concept Applied

The emf induced in the coil is given by the equation ε = nABω sin(ωt), where n is the number of turns, A is the area of the coil, B is the magnetic fi...

Read Full Step-by-Step Solution →

Question #12

Practice Question

A.A) 10 V
B.B) 20 V
C.C) 50 V
D.D) 100 V

Concept Applied

The maximum emf induced in the coil is given by the equation ε = NABω, where N is the number of turns, A is the area of each turn, B is the magnetic f...

Read Full Step-by-Step Solution →

Question #13

Practice Question

A.A. 0.1 V
B.B. 0.2 V
C.C. 0.5 V
D.D. 1 V

Concept Applied

The induced emf in the coil can be calculated using Faraday's law of electromagnetic induction....

Read Full Step-by-Step Solution →

Question #14

Practice Question

A.A) 0.5 V
B.B) 1 V
C.C) 0.1 V
D.D) 0.05 V

Concept Applied

The induced emf is calculated using the formula ε = N(dΦ/dt), where ε is the induced emf, N is the number of turns, and dΦ/dt is the rate of change of...

Read Full Step-by-Step Solution →

Question #15

Practice Question

A.A) 0.05 V
B.B) 0.5 V
C.C) 1.0 V
D.D) 5.0 V

Concept Applied

The average induced emf is calculated using the formula ε = N * ΔΦ / Δt, where N is the number of turns, ΔΦ is the change in magnetic flux, and Δt is ...

Read Full Step-by-Step Solution →

Question #16

Practice Question

A.A. Zero
B.B. Increasing
C.C. Decreasing
D.D. Constant

Concept Applied

The magnetic flux linked with the coil is zero because there is no relative motion between the coil and the bar magnet....

Read Full Step-by-Step Solution →

Question #17

Practice Question

A.2 mV
B.5 mV
C.10 mV
D.20 mV

Concept Applied

The coil is placed near a long straight conductor, which induces an electromotive force (emf) in the coil. The emf induced is given by Faraday's law o...

Read Full Step-by-Step Solution →

Question #18

Practice Question

A.A fundamental principle of Physics.
B.A complex derivation in JEEMains syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Electromagnetic Indu...

Read Full Step-by-Step Solution →

Question #19

Practice Question

Concept Applied

Faraday's law: $\mathcal{E} = -N \frac{d\Phi}{dt} = -N A \frac{dB}{dt}$. $\frac{dB}{dt} = 0.02$, so $\mathcal{E} = 50 \times 0.1 \times 0.02 = 0.1\,\t...

Read Full Step-by-Step Solution →

Question #20

Practice Question

Concept Applied

Use $\mathcal{E} = -N \frac{d\phi}{dt}$; differentiate $\phi = 0.02\sin(100t)$ and evaluate at $t=0.01$ s....

Read Full Step-by-Step Solution →

Question #21

Practice Question

A.0.012 V
B.0.024 V
C.0.006 V
D.0.0012 V

Concept Applied

The changing magnetic field induces an emf given by $\varepsilon = -N A\,\frac{dB}{dt}$. Substituting $N=500$, $A=2.0\times10^{-4}\,\text{m}^2$ and $\...

Read Full Step-by-Step Solution →

Question #22

Practice Question

Concept Applied

Energy = 0.5 × L × I² = 0.5 × 2 × 25 = 25 J....

Read Full Step-by-Step Solution →

Question #23

Practice Question

A.Opposes the change in magnetic flux that produced it.
B.Aids the change in magnetic flux that produced it.
C.Is always clockwise.
D.Is always counter-clockwise.

Concept Applied

Lenz's law is a consequence of conservation of energy, stating induced current opposes the cause....

Read Full Step-by-Step Solution →

Question #24

Practice Question

Concept Applied

Use $\mathcal{E} = -N \frac{\Delta \phi}{\Delta t} = -50 \cdot \frac{0.06}{0.1} = 30$ V....

Read Full Step-by-Step Solution →

Question #25

Practice Question

A.A) 0.1 V
B.B) 0.5 V
C.C) 1 V
D.D) 2 V

Concept Applied

The induced emf is calculated using the formula ε = N(dΦ/dt), where N is the number of turns, Φ is the magnetic flux, and t is the time....

Read Full Step-by-Step Solution →

Question #26

Practice Question

Concept Applied

Using $\mathcal{E} = -L \frac{dI}{dt}$, $L = \frac{\mathcal{E} \cdot dt}{dI} = \frac{10 \times 0.1}{2} = 0.5$ H....

Read Full Step-by-Step Solution →

Question #27

Practice Question

Concept Applied

$\mathcal{E}_{\text{peak}} = N A \omega B_0 = 50 \times 0.1 \times 100 \times 0.02 = 10\,\text{V}$...

Read Full Step-by-Step Solution →

Question #28

Practice Question

A.A) 1.57 V
B.B) 3.14 V
C.C) 5 V
D.D) 10 V

Concept Applied

The maximum emf induced in the coil is given by ε = NABω, where N is the number of turns, A is the area of the coil, B is the magnetic field, and ω is...

Read Full Step-by-Step Solution →

Question #29

Practice Question

A.10 V
B.20 V
C.30 V
D.40 V

Concept Applied

The emf induced in the coil is given by the equation ε = nABω sin(ωt), where n is the number of turns, A is the area of the coil, B is the magnetic fi...

Read Full Step-by-Step Solution →

Question #30

Practice Question

A.A) 10 V
B.B) 20 V
C.C) 50 V
D.D) 100 V

Concept Applied

The maximum emf induced in the coil is given by the equation ε = NABω, where N is the number of turns, A is the area of each turn, B is the magnetic f...

Read Full Step-by-Step Solution →

Question #31

Practice Question

A.A. 0.1 V
B.B. 0.2 V
C.C. 0.5 V
D.D. 1 V

Concept Applied

The induced emf in the coil can be calculated using Faraday's law of electromagnetic induction....

Read Full Step-by-Step Solution →

Question #32

Practice Question

A.A) 0.5 V
B.B) 1 V
C.C) 0.1 V
D.D) 0.05 V

Concept Applied

The induced emf is calculated using the formula ε = N(dΦ/dt), where ε is the induced emf, N is the number of turns, and dΦ/dt is the rate of change of...

Read Full Step-by-Step Solution →

Question #33

Practice Question

A.A) 0.05 V
B.B) 0.5 V
C.C) 1.0 V
D.D) 5.0 V

Concept Applied

The average induced emf is calculated using the formula ε = N * ΔΦ / Δt, where N is the number of turns, ΔΦ is the change in magnetic flux, and Δt is ...

Read Full Step-by-Step Solution →

Question #34

Practice Question

A.A. Zero
B.B. Increasing
C.C. Decreasing
D.D. Constant

Concept Applied

The magnetic flux linked with the coil is zero because there is no relative motion between the coil and the bar magnet....

Read Full Step-by-Step Solution →

Question #35

Practice Question

A.2 mV
B.5 mV
C.10 mV
D.20 mV

Concept Applied

The coil is placed near a long straight conductor, which induces an electromotive force (emf) in the coil. The emf induced is given by Faraday's law o...

Read Full Step-by-Step Solution →

Question #36

Practice Question

A.A fundamental principle of Physics.
B.A complex derivation in JEEMains syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Electromagnetic Indu...

Read Full Step-by-Step Solution →

Question #37

Practice Question

Concept Applied

Faraday's law: $\mathcal{E} = -N \frac{d\Phi}{dt} = -N A \frac{dB}{dt}$. $\frac{dB}{dt} = 0.02$, so $\mathcal{E} = 50 \times 0.1 \times 0.02 = 0.1\,\t...

Read Full Step-by-Step Solution →

Question #38

Practice Question

Concept Applied

Use $\mathcal{E} = -N \frac{d\phi}{dt}$; differentiate $\phi = 0.02\sin(100t)$ and evaluate at $t=0.01$ s....

Read Full Step-by-Step Solution →

Question #39

Practice Question

A.0.012 V
B.0.024 V
C.0.006 V
D.0.0012 V

Concept Applied

The changing magnetic field induces an emf given by $\varepsilon = -N A\,\frac{dB}{dt}$. Substituting $N=500$, $A=2.0\times10^{-4}\,\text{m}^2$ and $\...

Read Full Step-by-Step Solution →

Question #40

Practice Question

Concept Applied

Energy = 0.5 × L × I² = 0.5 × 2 × 25 = 25 J....

Read Full Step-by-Step Solution →

Question #41

Practice Question

A.Opposes the change in magnetic flux that produced it.
B.Aids the change in magnetic flux that produced it.
C.Is always clockwise.
D.Is always counter-clockwise.

Concept Applied

Lenz's law is a consequence of conservation of energy, stating induced current opposes the cause....

Read Full Step-by-Step Solution →

Question #42

Practice Question

Concept Applied

Use $\mathcal{E} = -N \frac{\Delta \phi}{\Delta t} = -50 \cdot \frac{0.06}{0.1} = 30$ V....

Read Full Step-by-Step Solution →

Question #43

Practice Question

A.A) 0.1 V
B.B) 0.5 V
C.C) 1 V
D.D) 2 V

Concept Applied

The induced emf is calculated using the formula ε = N(dΦ/dt), where N is the number of turns, Φ is the magnetic flux, and t is the time....

Read Full Step-by-Step Solution →

Question #44

Practice Question

Concept Applied

Using $\mathcal{E} = -L \frac{dI}{dt}$, $L = \frac{\mathcal{E} \cdot dt}{dI} = \frac{10 \times 0.1}{2} = 0.5$ H....

Read Full Step-by-Step Solution →

Question #45

Practice Question

Concept Applied

$\mathcal{E}_{\text{peak}} = N A \omega B_0 = 50 \times 0.1 \times 100 \times 0.02 = 10\,\text{V}$...

Read Full Step-by-Step Solution →

Question #46

Practice Question

A.A) 1.57 V
B.B) 3.14 V
C.C) 5 V
D.D) 10 V

Concept Applied

The maximum emf induced in the coil is given by ε = NABω, where N is the number of turns, A is the area of the coil, B is the magnetic field, and ω is...

Read Full Step-by-Step Solution →

Question #47

Practice Question

A.10 V
B.20 V
C.30 V
D.40 V

Concept Applied

The emf induced in the coil is given by the equation ε = nABω sin(ωt), where n is the number of turns, A is the area of the coil, B is the magnetic fi...

Read Full Step-by-Step Solution →

Question #48

Practice Question

A.A) 10 V
B.B) 20 V
C.C) 50 V
D.D) 100 V

Concept Applied

The maximum emf induced in the coil is given by the equation ε = NABω, where N is the number of turns, A is the area of each turn, B is the magnetic f...

Read Full Step-by-Step Solution →

Question #49

Practice Question

A.A. 0.1 V
B.B. 0.2 V
C.C. 0.5 V
D.D. 1 V

Concept Applied

The induced emf in the coil can be calculated using Faraday's law of electromagnetic induction....

Read Full Step-by-Step Solution →

Question #50

Practice Question

A.A) 0.5 V
B.B) 1 V
C.C) 0.1 V
D.D) 0.05 V

Concept Applied

The induced emf is calculated using the formula ε = N(dΦ/dt), where ε is the induced emf, N is the number of turns, and dΦ/dt is the rate of change of...

Read Full Step-by-Step Solution →
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