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Faraday and Lenz laws; induced emf/current and eddy currents: NEET MCQs with answers

90 questions on this topic of Electromagnetic Induction and Alternating Currents (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. The magnetic flux through each turn of a 50-turn coil changes by 20 mWb in 1 s. The average induced emf is:

  1. 1000 V
  2. 0.02 V
  3. 1 V
  4. 2 V
Show answer and explanation

Answer: (C) 1 V

By Faraday's law the average emf is |e| = N ΔΦ/Δt = 50 × 20 × 10⁻³/1 = 1 V.

medium

2. A square loop of wire of side a lies in the xy-plane in a region with a time-dependent magnetic field B→ = B₀ (1 - αt) k̂, where B₀ and α are positive constants. The magnitude and direction of the induced electric field along the perimeter of the loop at t < 1/α (viewed from the positive z-axis looking down) are:

  1. Zero, because induced fields are non-conservative
  2. Induced emf is B₀ a² α with clockwise induced current
  3. Induced emf is B₀ a² α with counter-clockwise induced current
  4. Induced emf is B₀ a α² with clockwise induced current
Show answer and explanation

Answer: (C) Induced emf is B₀ a² α with counter-clockwise induced current

The magnetic flux through the loop is Φ = B A = B₀(1 - αt)a². As time progresses, the upward flux decreases because dΦ/dt = -α B₀ a². By Lenz's law, the induced current must produce upward magnetic field to oppose the decrease, which corresponds to a counter-clockwise direction as viewed from the positive z-axis. The magnitude of induced emf is |dΦ/dt| = B₀ a² α.

hard

3. A copper disc of radius 20 cm is rotated at 1200 rpm about its central axis perpendicular to a uniform magnetic field of 0.2 T parallel to the axis of rotation. The induced electromotive force developed between the center and the rim of the disc is:

  1. 0.25 V
  2. 1.01 V
  3. 0.50 V
  4. 0.125 V
Show answer and explanation

Answer: (C) 0.50 V

The induced emf between the center and the rim of a rotating disc is given by e = (1/2) B ω R². Here, R = 0.2 m, B = 0.2 T, and angular frequency ω = 2π(1200 / 60) = 40π rad/s. Substituting these values: e = 0.5 × 0.2 × 40π × (0.2)² = 0.5 × 0.2 × 40π × 0.04 = 0.16π ≈ 0.503 V ≈ 0.50 V.

medium

4. A conducting circular loop of area A and resistance R is placed in a magnetic field directed perpendicular to the plane of the loop. If the magnetic field changes from B₁ to B₂ in a time interval Δt, the total heat dissipated in the loop during this time, assuming the field changes at a constant rate, is:

  1. A(B₂ - B₁) / (R Δt)
  2. A²(B₂ - B₁)² / (R Δt²)
  3. A²(B₂ - B₁)² / (R Δt)
  4. A²(B₂ - B₁)² Δt / R
Show answer and explanation

Answer: (C) A²(B₂ - B₁)² / (R Δt)

Since the field changes at a constant rate, dB/dt = (B₂ - B₁) / Δt. The induced emf is constant: e = A(B₂ - B₁) / Δt. The power dissipated as heat is P = e² / R = A²(B₂ - B₁)² / (R Δt²). Total heat dissipated in time Δt is H = P · Δt = A²(B₂ - B₁)² / (R Δt).

easy

5. The magnetic flux through each turn of a 200-turn coil changes by 5 mWb in 0.1 s. The average induced emf is:

  1. 10000 V
  2. 10 V
  3. 0.1 V
  4. 0.05 V
Show answer and explanation

Answer: (B) 10 V

By Faraday's law the average emf is |e| = N ΔΦ/Δt = 200 × 5 × 10⁻³/0.1 = 10 V.

hard

6. A thin conducting ring of radius a and resistance R falls vertically under gravity through a non-uniform horizontal magnetic field whose vertical gradient is constant, dB/dz = β. If the horizontal component of the magnetic field is negligible and the magnetic flux through the ring at height z is Φ = π a² β z, what is the terminal velocity attained by the ring? (Mass of the ring is m and acceleration due to gravity is g).

  1. m g R / (π² a⁴ β²)
  2. m g R / (2 π² a⁴ β²)
  3. m g R / (π a² β)
  4. 2 m g R / (π² a⁴ β²)
Show answer and explanation

Answer: (A) m g R / (π² a⁴ β²)

The induced emf in the ring as it falls with velocity v is e = |dΦ/dt| = (dΦ/dz)(dz/dt) = π a² β v. The induced current is I = e / R = (π a² β v) / R. Rate of electrical energy dissipation is I² R = (π² a⁴ β² v²) / R. At terminal velocity, this power equals the rate of loss of gravitational potential energy, P = m g v. Equating the two gives v = m g R / (π² a⁴ β²).

easy

7. The magnetic flux through each turn of a 100-turn coil changes by 2 mWb in 0.5 s. The average induced emf is:

  1. 0.004 V
  2. 0.4 V
  3. 400 V
  4. 0.1 V
Show answer and explanation

Answer: (B) 0.4 V

By Faraday's law the average emf is |e| = N ΔΦ/Δt = 100 × 2 × 10⁻³/0.5 = 0.4 V.

easy

8. The magnetic flux through each turn of a 200-turn coil changes by 10 mWb in 0.5 s. The average induced emf is:

  1. 1 V
  2. 4000 V
  3. 4 V
  4. 0.02 V
Show answer and explanation

Answer: (C) 4 V

By Faraday's law the average emf is |e| = N ΔΦ/Δt = 200 × 10 × 10⁻³/0.5 = 4 V.

medium

9. Given below are two statements: Statement I: Eddy currents are produced in bulk pieces of conductors when magnetic flux linked with them changes. Statement II: Eddy currents are always undesirable because they only cause unwanted dissipation of electrical energy into heat. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct as changing magnetic flux in bulk conductors induces circulating eddy currents. Statement II is incorrect because eddy currents are useful in practical applications such as induction furnaces, electromagnetic damping in galvanometers, and magnetic braking in electric trains.

easy

10. The magnetic flux through each turn of a 100-turn coil changes by 2 mWb in 0.1 s. The average induced emf is:

  1. 4 V
  2. 0.02 V
  3. 2000 V
  4. 2 V
Show answer and explanation

Answer: (D) 2 V

By Faraday's law the average emf is |e| = N ΔΦ/Δt = 100 × 2 × 10⁻³/0.1 = 2 V.

easy

11. A cylindrical bar magnet is dropped vertically along the central axis of a long, vertically suspended copper tube. Neglecting air resistance, the acceleration of the falling magnet is:

  1. continuously equal to g throughout the fall
  2. greater than g due to induced magnetic attraction
  3. initially g, then decreases and eventually becomes zero
  4. initially zero and gradually increases to g
Show answer and explanation

Answer: (C) initially g, then decreases and eventually becomes zero

As the magnet falls, changing magnetic flux induces eddy currents in the copper tube. By Lenz's law, these currents produce an upward retarding magnetic force on the magnet. When this upward force balances the weight of the magnet (mg), the net acceleration becomes zero and the magnet attains a terminal velocity.

medium

12. A circular loop of radius r and electrical resistance R is placed in a uniform magnetic field B perpendicular to the plane of the loop. If the radius of the loop shrinks uniformly with time such that r(t) = r₀ - vt (where r₀ and v are positive constants), the magnitude of the induced current in the loop as a function of time t is:

  1. (π B v / R) (r₀ - vt)
  2. (π B v / R) (r₀ - vt)²
  3. (2π B v² / R) t
  4. (2π B v / R) (r₀ - vt)
Show answer and explanation

Answer: (D) (2π B v / R) (r₀ - vt)

The magnetic flux through the loop is Φ = B A = B π r² = B π (r₀ - vt)². By Faraday's law, the magnitude of induced emf is ε = |dΦ/dt| = B π · 2(r₀ - vt) · v = 2π B v (r₀ - vt). The induced current is I = ε / R = (2π B v / R)(r₀ - vt).

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