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NEET Electromagnetic Induction and Alternating Currents MCQs with answers

324 questions on Electromagnetic Induction and Alternating Currents for NEET 2027, 1 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

mediumNEET 2019

1. Lenz's law is consequence of:

  1. Conservation of energy
  2. Conservation of charge
  3. Conservation of momentum
  4. Newton's 3rd law
Show answer and explanation

Answer: (A) Conservation of energy

Induced current opposes change to conserve energy.

Quick tip: Lenz = Energy conservation in EM form.

NCERT Class 12, Chapter 6, page 208

easy

2. The dimensions of self-inductance L in terms of fundamental dimensions [M], [L], [T], and [A] are:

  1. [M L² T⁻² A⁻²]
  2. [M L² T⁻¹ A⁻²]
  3. [M L T⁻² A⁻¹]
  4. [M L² T⁻² A⁻¹]
Show answer and explanation

Answer: (A) [M L² T⁻² A⁻²]

Energy stored in an inductor is given by U = (1/2) L I². Therefore, [L] = [U] / [I²] = [M L² T⁻²] / [A²] = [M L² T⁻² A⁻²].

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3. A closely wound circular coil of 500 turns has a mean radius of 10 cm. When a current of 4 A flows through it, the total magnetic flux linked with all the turns of the coil is 0.8 Wb. The self-inductance of the coil is:

  1. 0.05 H
  2. 0.4 H
  3. 1.6 H
  4. 0.2 H
Show answer and explanation

Answer: (D) 0.2 H

Total flux linkage NΦ is given as 0.8 Wb. By definition of self-inductance, NΦ = L I. Therefore, L = (NΦ) / I = 0.8 Wb / 4 A = 0.2 H.

hard

4. In an alternating current circuit, the instantaneous voltage is V = 200 sin(100πt) V and the instantaneous current is I = 5 sin(100πt - π/3) A. The wattful current (in-phase component) and wattless current (quadrature component) of the circuit are respectively:

  1. 1.25 A and 2.17 A
  2. 2.5 A and 4.33 A
  3. 1.77 A and 3.06 A
  4. 3.54 A and 6.12 A
Show answer and explanation

Answer: (C) 1.77 A and 3.06 A

The rms current is I_rms = I₀ / √2 = 5 / √2 ≈ 3.536 A. The phase difference is ϕ = π/3. The wattful component is I_p = I_rms cos ϕ = (5 / √2) × (1/2) = 2.5 / 1.414 ≈ 1.77 A. The wattless component is I_w = I_rms sin ϕ = (5 / √2) × (√3 / 2) ≈ 3.06 A.

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5. A step-down transformer having an efficiency of 90% is connected to a 220 V line and supplies power to a 22 V, 44 W bulb. The current drawn by the primary coil of the transformer is:

  1. 0.20 A
  2. 0.22 A
  3. 0.18 A
  4. 0.25 A
Show answer and explanation

Answer: (B) 0.22 A

The output power is P_out = 44 W. Efficiency η = P_out / P_in = 0.90, which implies P_in = 44 / 0.90 = 440 / 9 W. Since P_in = V_p × I_p, we have 220 × I_p = 440 / 9, giving I_p = 2 / 9 ≈ 0.22 A.

easy

6. In a series LCR circuit connected to an AC source of fixed voltage amplitude and variable frequency, the current amplitude becomes 1/√2 of its maximum value at frequencies 480 Hz and 520 Hz. The sharpness of resonance, represented by the quality factor Q, is:

  1. 25.0
  2. 12.5
  3. 50.0
  4. 6.25
Show answer and explanation

Answer: (B) 12.5

The half-power frequencies are f₁ = 480 Hz and f₂ = 520 Hz. The resonant frequency is f₀ = (f₁ + f₂)/2 = 500 Hz, and the bandwidth is Δf = f₂ - f₁ = 520 - 480 = 40 Hz. The quality factor Q is given by Q = f₀ / Δf = 500 / 40 = 12.5.

hard

7. A coil has an inductance of 1/(2π) H and negligible resistance. When an AC voltage of frequency 50 Hz is connected across it, a current of 2 A flows. If a second identical coil is connected in parallel with the first across the same supply, the total current drawn from the supply will be:

  1. 1 A
  2. 2 A
  3. 8 A
  4. 4 A
Show answer and explanation

Answer: (D) 4 A

The reactance of each coil is X_L = 2πfL = 2π(50)(1/(2π)) = 50 Ω. When two identical inductors are connected in parallel, the equivalent inductive reactance is halved to X_eq = 25 Ω. Since the supply voltage is V = I × X_L = 2 × 50 = 100 V, the new total current is I' = 100 / 25 = 4 A.

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8. An alternating voltage V = 141.4 sin(100πt) V is applied across a series combination of a 40 Ω resistor and a pure capacitor having reactance 30 Ω. The rms value of current flowing through the circuit is:

  1. 1.41 A
  2. 2.83 A
  3. 2.0 A
  4. 3.54 A
Show answer and explanation

Answer: (C) 2.0 A

The peak voltage is V₀ = 141.4 V ≈ 100√2 V, so V_rms = 100 V. The total impedance is Z = √(R² + X_C²) = √(40² + 30²) = 50 Ω. Hence, the rms current is I_rms = V_rms / Z = 100 / 50 = 2.0 A.

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9. A steady direct current flows through a long straight wire lying along the x-axis in the +x direction. A rectangular wire loop in the xy-plane is moving with a constant velocity v in the +y direction (away from the wire). The direction of the induced current in the loop, as viewed from above the xy-plane, is:

  1. clockwise
  2. alternating periodically
  3. zero because velocity is constant
  4. counter-clockwise
Show answer and explanation

Answer: (D) counter-clockwise

The current along +x produces a magnetic field directed out of the page (+z direction) in the upper half-plane. As the loop moves away along +y, the magnetic field strength decreases, so the outward magnetic flux through the loop decreases. By Lenz's law, the induced current must produce an outward magnetic field to oppose this decrease, which corresponds to a counter-clockwise direction.

easy

10. A 12 V, 60 W lamp is run from the secondary of an ideal step-down transformer connected to a 240 V AC supply. The current drawn from the 240 V supply is:

  1. 0.50 A
  2. 5.0 A
  3. 0.25 A
  4. 2.5 A
Show answer and explanation

Answer: (C) 0.25 A

For an ideal transformer, input power equals output power: P_in = P_out = 60 W. Therefore, V_p × I_p = 60 W, which gives I_p = 60 / 240 = 0.25 A.

easy

11. Two coaxial circular loops of radii r and R (r << R) are kept coplanar with their centres coinciding. If the radius of the outer loop R is doubled while keeping the inner loop unchanged, the mutual inductance between the two loops becomes:

  1. Four times the initial value
  2. Twice the initial value
  3. Half of the initial value
  4. One-fourth of the initial value
Show answer and explanation

Answer: (C) Half of the initial value

The magnetic field produced by current I in the outer loop at the centre is B = μ₀I / (2R). The flux through the inner loop is Φ = B × (πr²) = (μ₀πr² / 2R) I. Thus, mutual inductance M = μ₀πr² / (2R), which is inversely proportional to R. Doubling R halves the mutual inductance.

easy

12. The primary winding of a transformer has 400 turns and is connected to a 240 V AC supply. To obtain an output of 12 V from the secondary winding, how many turns should the secondary coil have, assuming an ideal transformer?

  1. 15 turns
  2. 40 turns
  3. 20 turns
  4. 8000 turns
Show answer and explanation

Answer: (C) 20 turns

For an ideal transformer, V_s / V_p = N_s / N_p. Rearranging gives N_s = N_p × (V_s / V_p) = 400 × (12 / 240) = 400 × (1/20) = 20 turns.

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13. The capacitive reactance of a 50 μF capacitor at 50 Hz is:

  1. 127 Ω
  2. 63.7 Ω
  3. 0.0157 Ω
  4. 400 Ω
Show answer and explanation

Answer: (B) 63.7 Ω

X_C = 1/(2πfC) = 1/(2π × 50 × 50 × 10⁻⁶) = 63.7 Ω.

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14. The inductive reactance of a 200 mH inductor at 50 Hz is:

  1. 10 Ω
  2. 0.0159 Ω
  3. 62800 Ω
  4. 62.8 Ω
Show answer and explanation

Answer: (D) 62.8 Ω

X_L = 2πfL = 2π × 50 × 200 × 10⁻³ = 62.8 Ω.

easy

15. A 1 m conducting rod moves at 10 m/s perpendicular to a uniform 0.1 T magnetic field, with the rod, velocity and field mutually perpendicular. The emf across the rod is:

  1. 0.5 V
  2. 0.01 V
  3. 1 V
  4. 2 V
Show answer and explanation

Answer: (C) 1 V

For a rod moving perpendicular to B, the motional emf is e = Blv = 0.1 × 1 × 10 = 1 V.

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