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Electromagnetic Induction

Electromagnetic InductionNEET Physics · Class 12 · NCERT Chapter 6

16 NEET previous-year questions on Electromagnetic Induction, each with the correct answer and a step-by-step solution. Sourced directly from official NEET papers across every booklet code.

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All (16)
Electromagnetic Induction (16)

A

Zero

B

Bv π r 2 /2 and P is at higher potential

C

π rBv and R is at higher potential

D

2rBv and R is at higher potential

Solution

The potential difference across the semicircular ring is given by . The length of the semicircular ring is , so . Since the motion is downward and the magnetic field is horizontal, the potential at R is higher due to the direction of the induced current. Therefore, the potential difference is and R is at higher potential, but the correct option is (d) and R is at higher potential, considering the full length of the semicircle.

A

0.15 Nm

B

0.20 Nm

C

0.24 Nm

D

0.12 Nm www.vedantu.com 39

Solution

τ⃗ = M⃗⃗⃗ × B⃗⃗ =MBsin60o =Ni ABsin60o 50 ×2×0.12 ×0.1 ×0.2 × √3 2 =12√3 ×10−2 Nm=0.20748 Nm

A

Cells

B

Potential gradients

C

A condition of no current flow through the galvanometer

D

A combination of cells, galvanometer and resistances

Solution

Reading of potentiometer is accurate because during taking reading it does not draw any current from the circuit.

A

2 mA

B

0.2 A

C

2 A

D

0 ampere

Solution

Sol – +ε C R3R2L1 R1 L2 At t = 0, no current flows through R 1 and R3 ∴ + –ε i R2 2 i R ε= = 18 9 = 2 A Note : Not correctly framed but the best option out of given is (3).

A

1·389 H

B

138·88 H

C

0·138 H

D

13·89 H ACHLA/AA/Page 5 SPACE FOR ROUGH WORK English

Solution

The magnetic potential energy stored in an inductor is given by . Substituting the values, , so option (d) is correct.

A

68 cm

B

6.8 cm

C

113.9 cm

D

88 cm

Solution

Δl1 = Δl2 l1∝1 Δθ = l2∝2Δθ 88 × 1.7 × 10−5 = l2 × 2.2 × 10−5 Δl2 = 88×1.7/2.2 = 68 cm

A

12.2 nm

B

12.2 × 10−13 m

C

12.2 × 10−12 m

D

12.2 × 10−4 nm

Solution

λ = ℎ √2𝑚𝐾= 6.63×10−34 √2×9.1×10−31×104×1.6×10−19 = 12.2 ×10−12 m

A

metals

B

insulators only

C

semiconductors only

D

insulators and semiconductors

Solution

Metals have a positive temperature coefficient of resistance, while insulators and semiconductors have a negative temperature coefficient of resistance. This is due to the increase in carrier concentration in insulators and semiconductors with temperature, so option (d) is correct.

A

1 2 R R

B

2 1 R R

C

2 1 2 R R

D

2 2 1 R R

Solution

The mutual inductance between two coaxial circular loops of radii and (with ) is directly proportional to . This is derived from the formula for mutual inductance of two coaxial loops, so option (b) is correct.

A

3 Ia 2 and 3 Ia 2

B

3 Ia 2 and Ia 2

C

3 Ia 2 and 4 Ia 2

D

4 Ia 2 and 3 Ia 2

Solution

For the equilateral triangle, the number of turns and the area . The magnetic dipole moment is . For the square, and , so . Therefore, the correct option is (b).

A

Magnetic flux

B

Self inductance

C

Magnetic permeability

D

Electric permittivity

Solution

Dimensional formula of magnetic permeability is [MLT–2A–2] - 6 - NEET (UG)-2022 (Code-Q1)

A

2 weber

B

0.5 weber

C

1 weber

D

Zero weber

Solution

Magnetic flux () ·φ= ρρ B BA and ρρ BA are in same direction, therefore φB = B.A = 0.5 × 12 = 0.5 Wb

A

8 μJ

B

4 μJ

C

4 mJ

D

8 mJ

Solution

The magnetic energy stored in an inductor is given by . Substituting the values, , so option (a) is correct.

A

AB and DC

B

BA and CD

C

AB and CD

D

BA and DC

Solution

North of magnet is moving away from solenoid 1 so end B of solenoid 1 is South and as south of magnet is approaching solenoid 2 so end C of solenoid 2 is South.

A

1 : 2

B

2 : 1

C

1 : 1

D

1 : 4

Solution

According to transformer ratio, 2 :1SS PP VN VN ==

A

Zero at all places

B

Constant between the plates and zero outside the plates

C

Non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates

D

Zero between the plates and non-zero outside

Solution

Let the surface charge density be

Given

It means displacement current is constant.

This system will act like a cylindrical wire.

The graph of magnetic field (B) vs r is

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