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Torque on current loop and moving-coil galvanometer: NEET MCQs with answers

50 questions on this topic of Magnetic Effects of Current and Magnetism (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. A circular coil of radius 10 cm consisting of 100 closely wound turns carries a steady current of 0.5 A. The magnitude of the magnetic dipole moment associated with this coil is:

  1. 0.785 A m²
  2. 0.393 A m²
  3. 3.14 A m²
  4. 1.57 A m²
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Answer: (D) 1.57 A m²

The area of one turn is A = π r² = 3.14 × (0.10)² = 3.14 × 10⁻² m². The total magnetic dipole moment is M = N I A = 100 × 0.5 × (3.14 × 10⁻²) = 1.57 A m².

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2. Given below are two statements: Statement I: A moving-coil galvanometer can be used to measure alternating current (AC) directly because the deflection is directly proportional to the current. Statement II: The torque produced on the coil in a radial magnetic field alternates in sign every half-cycle of an AC of 50 Hz, resulting in a net average deflection of zero. In 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.
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Answer: (C) Statement I is incorrect but Statement II is correct.

A moving-coil galvanometer cannot measure AC directly because for 50 Hz AC, the instantaneous torque τ = NIAB reverses 100 times per second. Due to the coil's rotational inertia, it cannot follow such rapid reversals and remains at zero. Thus Statement I is false and Statement II is true.

hard

3. Two galvanometers P and Q have coils of equal area and are suspended in identical magnetic fields with identical suspension fibers. Coil P has 30 turns and resistance 12 Ω, while coil Q has 60 turns and resistance 30 Ω. The ratio of the voltage sensitivity of P to that of Q is:

  1. 5 : 4
  2. 4 : 5
  3. 1 : 1
  4. 2 : 1
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Answer: (A) 5 : 4

Voltage sensitivity is defined as V_s = θ / V = (N A B) / (k R). Since A, B, and k are identical for both, V_s is proportional to N / R. For galvanometer P: (N / R)_P = 30 / 12 = 2.5. For galvanometer Q: (N / R)_Q = 60 / 30 = 2.0. The ratio is (V_s)_P / (V_s)_Q = 2.5 / 2.0 = 5 / 4.

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4. A square coil of side a carries a current I. It is placed in a uniform magnetic field B such that its magnetic dipole moment vector m makes an angle of 60° with the direction of B. The work done in slowly rotating the coil to an angle of 90° with B is:

  1. I a² B
  2. (√3/2) I a² B
  3. (1/2) I a² B
  4. 2 I a² B
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Answer: (C) (1/2) I a² B

The magnetic moment of the square coil is m = I a². The work done by an external agent is W = -m B (cos θ₂ - cos θ₁) = m B (cos 60° - cos 90°) = m B (1/2 - 0) = (1/2) I a² B.

easy

5. A current-carrying planar loop is placed in a uniform magnetic field. Which of the following statements is always correct regarding the net magnetic force and net torque experienced by the loop?

  1. The net force is always non-zero, but the net torque may be zero.
  2. The net force is always zero, while the net torque depends on the orientation of the loop.
  3. Both the net force and net torque are always zero regardless of orientation.
  4. Both the net force and net torque are non-zero for any orientation.
Show answer and explanation

Answer: (B) The net force is always zero, while the net torque depends on the orientation of the loop.

For any closed planar or non-planar loop carrying a steady current in a uniform magnetic field, the vector sum of forces on all elements vanishes (∮ dL = 0, so F_net = 0). However, the net torque τ = M × B depends on the angle between the magnetic dipole moment M and the field B.

hard

6. A rigid wire loop carrying current I consists of two semicircular arcs of radii R₁ and R₂ (with R₂ > R₁) in the x-y plane centered at the origin, joined by straight radial segments along the x-axis. The entire loop is placed in a uniform magnetic field B directed along the +y-axis. What is the magnitude of the net torque acting on this loop?

  1. Zero
  2. I · π · (R₂² + R₁²) · B
  3. (1/2) · I · π · (R₂² + R₁²) · B
  4. (1/2) · I · π · (R₂² - R₁²) · B
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Answer: (D) (1/2) · I · π · (R₂² - R₁²) · B

The magnetic moment of a planar loop is M = I · A n, where A is the enclosed area. The enclosed area between the two semicircles is A = (1/2) π (R₂² - R₁²). The unit normal to the x-y plane is along the z-axis (k). Since the magnetic field is along the +y-axis (B = B j), the angle between M and B is 90°. Hence, the magnitude of the torque is τ = |M × B| = M · B = (1/2) · I · π · (R₂² - R₁²) · B.

easy

7. A microammeter has a resistance of 100 Ω and a full-scale range of 50 μA. It is used as a voltmeter by connecting a resistance of 19.9 kΩ in series with it. The maximum voltage that can be measured by this voltmeter is:

  1. 0.5 V
  2. 5.0 V
  3. 2.0 V
  4. 1.0 V
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Answer: (D) 1.0 V

Total resistance of the voltmeter is R_total = G + R = 100 Ω + 19900 Ω = 20000 Ω = 2 × 10⁴ Ω. The maximum measurable voltage is V = I_g × R_total = (50 × 10⁻⁶ A) × (2 × 10⁴ Ω) = 1.0 V.

easy

8. A galvanometer with a coil resistance of 25 Ω produces a full-scale deflection when a current of 10 mA flows through it. A shunt resistor of 0.25 Ω is connected across its terminals to form an ammeter. What fraction of the total measured current flows through the galvanometer coil?

  1. 1/100
  2. 1/10
  3. 1/99
  4. 1/101
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Answer: (D) 1/101

The current through the galvanometer is given by Ig = I × [S / (G + S)]. Substituting the given values, Ig / I = 0.25 / (25 + 0.25) = 0.25 / 25.25 = 1 / 101. Thus, 1/101 of the total circuit current passes through the coil.

medium

9. A circular loop and a square loop are constructed from identical lengths of uniform conducting wire. When identical currents are passed through both loops in the same plane and placed in a uniform magnetic field parallel to their planes, the ratio of maximum torque on the circular loop to that on the square loop is:

  1. π : 4
  2. 2 : π
  3. 1 : 1
  4. 4 : π
Show answer and explanation

Answer: (D) 4 : π

Let the length of the wire be L. For the circle, 2π r = L, so A_c = π (L / 2π)² = L² / (4π). For the square, 4 a = L, so A_s = (L / 4)² = L² / 16. Maximum torque is τ_max = I A B. The ratio τ_c / τ_s = A_c / A_s = (L² / 4π) / (L² / 16) = 16 / (4π) = 4 / π.

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10. A square loop of side 5 cm carrying a current of 2 A is placed in a uniform magnetic field of 0.4 T. If the plane of the loop makes an angle of 30° with the magnetic field lines, what is the magnitude of the torque acting on the loop?

  1. 1.0 × 10⁻³ N m
  2. 0.86 × 10⁻³ N m
  3. 2.0 × 10⁻³ N m
  4. 1.73 × 10⁻³ N m
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Answer: (D) 1.73 × 10⁻³ N m

The area of the square loop is A = (0.05 m)² = 2.5 × 10⁻³ m². The magnetic moment is M = I A = 2 × 2.5 × 10⁻³ = 5 × 10⁻³ A m². The angle between the magnetic moment (normal to the loop) and the field B is θ = 90° - 30° = 60°. Therefore, the torque is τ = M B sin 60° = (5 × 10⁻³) × 0.4 × (√3/2) ≈ 1.73 × 10⁻³ N m.

easy

11. A galvanometer coil has an area of 2.0 × 10⁻⁴ m² and 80 turns. It is suspended in a uniform radial magnetic field of 0.25 T. If the torsional spring constant of the suspension is 1.0 × 10⁻⁶ N m/rad, the current required to produce a deflection of 0.2 rad is:

  1. 100 μA
  2. 25 μA
  3. 50 μA
  4. 75 μA
Show answer and explanation

Answer: (C) 50 μA

In equilibrium for a radial magnetic field, N I A B = k θ. Therefore, I = (k θ) / (N A B) = (1.0 × 10⁻⁶ × 0.2) / (80 × 2.0 × 10⁻⁴ × 0.25) = (0.2 × 10⁻⁶) / (4.0 × 10⁻³) = 0.05 × 10⁻³ A = 50 μA.

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12. A square loop of side 0.2 m carries a current of 1.5 A in the x-y plane, oriented such that its magnetic moment points in the +z-direction. A uniform magnetic field B = (0.3 i + 0.4 k) T is applied across the region. The torque vector acting on the loop is:

  1. (-0.024 i) N m
  2. (-0.018 j) N m
  3. (0.024 j) N m
  4. (0.018 j) N m
Show answer and explanation

Answer: (D) (0.018 j) N m

The area of the loop is A = (0.2)² = 0.04 m². The magnetic moment vector is M = I · A k = 1.5 × 0.04 k = 0.06 k A m². The magnetic field is B = 0.3 i + 0.4 k T. The torque is τ = M × B = (0.06 k) × (0.3 i + 0.4 k) = 0.06 × 0.3 (k × i) + 0 = 0.018 j N m.

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