JEE Main Electromagnetic Induction Previous Year Questions with Solutions

Electromagnetic Induction is a current that is produced due to the formation of electromotive force (emf) due to the change in magnetic field. The emf produced is directly proportional to the rate of change of current with time. The proportionality constant in the relation between current and emf is called Inductance.

Download Electromagnetic Induction Previous Year Solved Questions PDF

Important Points to Remember:

Electromotive Induction formula

Where,

• e is the induced voltage (in volts)
• N is the number of turns in the coil
• Φ is the magnetic flux (in Weber)
• t is the time (in seconds)

Self Induction

Self Induction is a phenomenon in which change in current flowing through the coil produces electromotive force (emf) in the same coil. When there is a current flow through the coil at any instant, the magnetic flux will be directly proportional to the current passing through the circuit.

The magnetic flux, Φ = LI

Where,

L = Self Inductance

If e is the emf induced in the coil then

e = dΦ/dt

e = d(LI)/dt

Thus, e = L(dI/dt)

Mutual Induction

Mutual Induction is the phenomenon in which electromotive force (emf) is produced in the secondary coil when there is a change in current in the primary coil. The emf (e) produced in the secondary coil is directly proportional to the change in current (dI/dt) in the primary coil.

JEE Main Previous Year Solved Questions on Electromagnetic Induction

Q1: When current in a coil changes from 5 A to 2 A in 0.1 s, an average voltage of 50 V is produced. The self-inductance of the coil is

(a) 0.67 H

(b) 1.67 H

(c) 3 H

(d) 6 H

Solution

I₁ = 5 A,

I₂ = 2 A

ΔI = 2 – 5 = –3 A

Δt = 0.1

s= 50 V

As, ε = -L(ΔI/Δt)

50 = -L(-3/0.1)

50 = 30L

L = 5/3 = 1.67 H

Answer: (b) 1.67 H

Q2: Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A = 10 cm² and length = 20 cm. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is (μ₀ = 4π × 10^–7 T mA^–1)

(a) 2.4π× 10^–4 H

(b) 2.4π × 10^–5 H

(c) 4.8π× 10^–4 H

(d) 4.8π × 10^–5 H

Solution

A = πr₁² = 10 cm², l = 20 cm, N₁ = 300, N₂ = 400

M = μ₀N₁N₂/l

M = (4π x 10^{– 7} x 300 x 400 x 10 x 10^-4) /0.20 = 2.4π × 10^–4 H

Answer: (a) 2.4π× 10^–4 H

Q3: The self inductance of the motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of

(a) 1μF

(b) 2μF

(c) 4μF

(d) 8μF

Solution

For maximum power, Lω = 1/Cω

C = 1/ Lω²

C = 1/[10 x (2π x 50)²] = 1/(10 x 10⁴ x (π)²) = 10^-6 F = 1μF (Taking π² = 10)

Answer: (a) 1μF

Q4: A coil of inductance 300 mH and resistance 2 is connected to a source of voltage 2 V. The current reaches half of its steady state value in

(a) 0.15 s

(b) 0.3 s

(c) 0.05 s

(d) 0.1 s

Solution

During growth of charge in an inductance, I = I₀ (1 – e ^–Rt/L)

or I₀/2 = I₀(1 – e ^–Rt/L)

e^-Rt/L = ½ = 2^-1

or Rt/L = In 2 ⇒t = (L/R)In 2

t =[(300 x 10^-3)/2] x (0.693)

or t = 0.1 sec

Answer: (d) 0.1 s

Q5: In a LCR circuit capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to

(a) 4L

(b) 2L

(c) L/2

(d) L/4

Solution

At resonance, ω = 1/√LC when ω is constant,

(1/ L₁C₁) = (1/ L₂C₂) = (1/ LC) =1/L₂(2C) = 1/2L₂C

L₂ = L/2

Answer: (c) L/2

Q6: Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon

(a) the rates at which currents are changing in the two coils

(b) relative position and orientation of the two coils

(c) the materials of the wires of the coils

(d) the currents in the two coils

Solution

Mutual inductance between two coils depends on the materials of the wires of the coils

Answer: (c) the materials of the wires of the coils

Q7: The power factor of an AC circuit having resistance (R) and inductance (L) connected in series and an angular velocity is

(a) R/ωL

(b) R/(R² + ω²L²)^1/2

(c) ωL/R

(d) R/(R² – ω²L²)^1/2

Solution

Power Factor, cos Φ = 1/(1 + tan²Φ)^½

cos Φ = 1/(1 + (ωL/R)²)^½ (∵ tan Φ = ωL/R)

cos Φ = R/(R² + ω²L²)^1/2

Answer: (b) R/(R² + ω²L²)^1/2

Q8: An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to

(a) 80 H

(b) 0.08 H

(c) 0.044 H

(d) 0.065 H

Solution

Given I = 10 A, V = 80 V

R = V/I = 80/10 = 8 Ω and ω = 50 Hz

I = V/(8² + X_L²)^½

10 = 220/ (64 + X_L²)^½

(64 + X_L²)^½ = 22

Squaring on both sides, we get

64 + X_L² = 484

X_L² = 484 – 64 = 420

X_L = (420)^½ ⇒ 2π x ωL = (420)^½

Series inductor on an arc lamp,

L = (420)^½ /( 2π x 50) = 0.065 H

Answer: (d) 0.065 H

Q9: A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns, giving the output power at 230 V. If the current in the primary of the transformer is 5 A, and its efficiency is 90%, the output current would be

(a) 50 A

(b) 25 A

(c) 45 A

(d) 20 A

Solution

Efficiency η = 0.9 = P_s/P_p

V_s I_s = 0.9 × V_pI_p

I_s = (0.9 x 2300×5)/ 230 = 45A

Answer: (c) 45 A

Q10: A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity ω, the maximum e.m.f. induced in the coil will be

(a) (3/2)nBAω

(b) nBAω

(c) 3nBAω

(d) (½) nBAω

Solution

Emf induced in the coil is given by ε = BAnωsinωt

ε_max = BAωn (The induced emf is maximum when sinωt = 1)

Answer: (b) nBAω

Q11: A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0 m s^–1, at right angles to the horizontal component of the earth’s magnetic field of 0.3 × 10^–4 Wb/m². The value of the induced emf in wire is

(a) 0.3 × 10^–3 V

(b) 2.5 × 10^–3 V

(c) 1.5 × 10^–3 V

(d) 1.1 × 10^–3 V

Solution

The motional emf is given as

= 5 × (0.3 × 10^–4) (10) sin90° = 1.5 × 10^–3 V

Answer: (c) 1.5 × 10^–3 V

Q12: A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then the self inductance of the coil

(a) increases by a factor of 27

(b) decreases by a factor of 3

(c) increases by a factor of 3

(d) decreases by a factor of 9

Answer: (c) increases by a factor of 3

Q13: The self induced emf of a coil is 25 volts.When the current in it is changed at uniform rate from 10 A to 25 A in 1 s, the change in the energy of the inductance is

(a) 637.5 J

(b) 540 J

(c) 437.5 J

(d) 740 J

Solution

ε = 25 V, I₁ = 10 A, I₂ = 25 A, t = 1 s, ΔE =?

ε = L(ΔL/Δt)

25 = L[(25 – 10)/1]

L = 25/15 = (5/3) H

ΔE = ½L (I₂² – I₁² ) =(½) x (5/3) x (25² – 10²) = (⅚) x 525 = 437.5 J

Answer: (c) 437.5 J

Q14: There are two long co-axial solenoids of the same length l. The inner and outer coils have radii r₁ and r₂ and number of turns per unit length n₁ and n₂ respectively. The ratio of mutual inductance to the self inductance of the inner coil is

(a) n₂/n₁

(b) (n₂/n₁)(r₂²/r₁²)

(c) (n₂/n₁)(r₁/r₂)

(d) n₁/n₂

Solution

The mutual inductance of the inner coil

M = μ₀n₁n₂r₁² …(1)

Self inductance (L) of the inner coil is

L = μ₀n₁²r₁² …(2)

Using (1) and (2),

M/L = n₂/n₁

Answer: (a) n₂/n₁

Q15: A solid metal cube of edge length 2 cm is moving in a positive y-direction at a constant speed of 6 m/s. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube perpendicular to the x-axis, is

(a) 2 mV

(b) 6 mV

(c) 1 mV

(d) 12 mV

Solution

Emf developed across the given edges,

ε = Blv = 0.1 × 0.02 × 6 = 12 × 10^–3 V = 12 mV

As all the edges are parallel between the faces perpendicular to the x-axis, hence required potential difference is 12 mV

Answer: (d) 12 mV

Also Read:

Electromagnetic Induction JEE Advanced Previous Year Questions With Solutions

Electromagnetic Induction and Alternating Current JEE Main Questions