Magnetic Dipole in Uniform Magnetic Field – Torque on a magnetic dipole in a uniform magnetic field: NEET MCQs with Answers

14 free questions on Magnetic Dipole in Uniform Magnetic Field – Torque on a magnetic dipole in a uniform magnetic field for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Torque on a magnetic dipole in a uniform magnetic field.

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Question 1

Previous-year question
A magnetic needle of magnetic moment 6.7 × 10⁻² A m² and moment of inertia 7.5 × 10⁻⁶ kg m² is performing simple harmonic oscillations in a magnetic field of 0.01 T. The time taken for 10 complete oscillations is [NEET 2017]
  1. 6.98 s
  2. 8.76 s
  3. 6.65 s
  4. 8.89 s
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Answer: (C) 6.65 s
Explanation: T = 2π√(I/MB) = 2π√(7.5 × 10⁻⁶/(6.7 × 10⁻² × 0.01)) ≈ 0.665 s, so ten oscillations take about 6.65 s.

Question 2

Previous-year question
A bar magnet of magnetic moment M is placed in a magnetic field of induction B. The torque exerted on it is [AIPMT 1999]
  1. M · B
  2. −M · B
  3. M × B
  4. −M × B
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Answer: (C) M × B
Explanation: The torque on a magnetic dipole in an external field is the vector product τ = M × B, of magnitude MB sin θ, where θ is the angle between M and B. A dot product of M and B is a scalar and gives the potential energy U = −M · B, not a torque.

Question 3

Previous-year question
A bar magnet having a magnetic moment of 2 × 10⁴ J T⁻¹ is free to rotate in a horizontal plane. A horizontal magnetic field B = 6 × 10⁻⁴ T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60° from the field is [AIPMT 2009]
  1. 0.6 J
  2. 12 J
  3. 6 J
  4. 2 J
Show answer
Answer: (C) 6 J
Explanation: W = MB(cos θ₁ − cos θ₂) with θ₁ = 0° and θ₂ = 60°, so W = MB(1 − ½) = MB/2 = (2 × 10⁴ × 6 × 10⁻⁴)/2 = 12/2 = 6 J.

Question 4

Previous-year question
A short bar magnet of magnetic moment 0.4 J T⁻¹ is placed in a uniform magnetic field of 0.16 T. The magnet is in stable equilibrium. The potential energy is [AIPMT 2011]
  1. 0.064 J
  2. −0.064 J
  3. zero
  4. −0.082 J
Show answer
Answer: (B) −0.064 J
Explanation: The potential energy of a magnetic dipole is U = −M · B = −MB cos θ. Stable equilibrium corresponds to θ = 0°, so U = −MB = −(0.4)(0.16) = −0.064 J. The negative sign indicates the minimum-energy configuration.

Question 5

Previous-year question
A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60°. The torque needed to maintain the needle in this position will be [AIPMT 2012]
  1. 2√3 J
  2. 3 J
  3. √3 J
  4. 3/2 J
Show answer
Answer: (B) 3 J
Explanation: From W = MB(1 − cos 60°) = MB/2 = √3, we get MB = 2√3. The torque required is τ = MB sin 60° = 2√3 × (√3/2) = 3 J (strictly, 3 N m).

Question 6

Previous-year question
A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in the equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in this new position is [NEET 2016]
  1. W/√3
  2. √3 W
  3. √3 W/2
  4. 2W/√3
Show answer
Answer: (B) √3 W
Explanation: The work done in rotating from 0° to 60° is W = MB(cos 0° − cos 60°) = MB(1 − ½) = MB/2, so MB = 2W. The torque in the new position is τ = MB sin 60° = 2W × (√3/2) = √3 W.

Question 7

Previous-year question
A rectangular coil of 250 turns, of length 2.1 cm and width 1.25 cm, carries a current of 85 μA and is placed in a magnetic field of strength 0.85 T. The work done in rotating the coil through 180° against the torque is [NEET 2017]
  1. 9.1 μJ
  2. 4.55 μJ
  3. 2.3 μJ
  4. 1.5 μJ
Show answer
Answer: (A) 9.1 μJ
Explanation: M = NIA = 250 × 85 × 10⁻⁶ × (2.1 × 10⁻² × 1.25 × 10⁻²) = 5.58 × 10⁻⁶ A m². Rotating through 180° needs W = 2MB = 2 × 5.58 × 10⁻⁶ × 0.85 ≈ 9.5 × 10⁻⁶ J, for which the nearest listed value, and the official key, is 9.1 μJ.
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Question 8

A bar magnet of magnetic moment 2 A m² is held with its moment at 120° to a uniform magnetic field of 0.3 T. Its magnetic potential energy and the magnitude of the torque on it are:
  1. −0.52 J and 0.3 N m
  2. −0.3 J and 0.52 N m
  3. +0.52 J and 0.3 N m
  4. +0.3 J and 0.52 N m
Show answer
Answer: (D) +0.3 J and 0.52 N m
Explanation: The potential energy is U = −mB cos θ = −0.6 × cos 120° = +0.3 J. It is positive because the magnet is closer to the unstable orientation than to the stable one. The torque is τ = mB sin θ = 0.6 × sin 120° ≈ 0.52 N m.

Question 9

A small bar magnet has magnetic moment m = 20 ĵ A m² and is placed in a uniform magnetic field B = (0.3 î + 0.4 ĵ) T. What are the torque on it and its potential energy, taking the energy as zero when m is perpendicular to B?
  1. Torque −6 k̂ N m and energy +8 J
  2. Torque −6 k̂ N m and energy −8 J
  3. Torque −8 k̂ N m and energy −6 J
  4. Torque +6 k̂ N m and energy −8 J
Show answer
Answer: (B) Torque −6 k̂ N m and energy −8 J
Explanation: The torque is m × B = 20 ĵ × (0.3 î + 0.4 ĵ) = 6 (ĵ × î) = −6 k̂ N m, because ĵ × ĵ = 0. The energy is U = −m·B = −(20)(0.4) = −8 J. Only the component of B along m contributes to the energy, and only the perpendicular component contributes to the torque.

Question 10

A bar magnet free to rotate is at rest along a uniform magnetic field. Turning it through 60° from this position needs 5 J of work. How much further work is needed to turn it from 60° to 180° from the field direction?
  1. 10 J
  2. 20 J
  3. 15 J
  4. 5 J
Show answer
Answer: (C) 15 J
Explanation: Work to turn a dipole from θ₁ to θ₂ is MB(cos θ₁ − cos θ₂). The first turn gives MB(1 − 1/2) = 5 J, so MB = 10 J. The further turn needs MB(cos 60° − cos 180°) = 10 × (1/2 + 1) = 15 J.

Question 11

Slowly turning a bar magnet from its position of stable equilibrium in a uniform magnetic field to a position perpendicular to the field requires work W. How much further work is needed to turn it slowly from there to its position of unstable equilibrium?
  1. W
  2. 2W
  3. W/2
  4. Zero, because the torque is zero at unstable equilibrium
Show answer
Answer: (A) W
Explanation: The work done in turning a magnet from θ₁ to θ₂ is mB(cos θ₁ − cos θ₂). From 0° to 90° this equals mB, so W = mB. From 90° to 180° it is also mB, which equals W. The torque is zero only at the final position, not along the way.

Question 12

A bar magnet of magnetic dipole moment 2.5 A m² rests with its centre on a frictionless pivot in a uniform magnetic field of 0.4 T. It is held at rest with its axis at 30° to the field by a force applied at right angles to the axis, at a point 20 cm from the pivot. The force is:
  1. 1.25 N
  2. 2.5 N
  3. 5.0 N
  4. 4.33 N
Show answer
Answer: (B) 2.5 N
Explanation: For equilibrium, the torque of the force must balance the magnetic torque, so F × d = mB sin θ. The magnetic torque is 2.5 × 0.4 × sin 30° = 0.5 N m. Hence F = 0.5 / 0.20 = 2.5 N.

Question 13

A small magnetised needle floats on a cork in a bowl of water. In a uniform magnetic field it only turns until it lines up with the field. Near one pole of a strong bar magnet it turns and also drifts through the water. What is the best explanation of the drift?
  1. A uniform field exerts no force on either pole of the needle
  2. In a non-uniform field the forces on the two poles of the needle are unequal, so a net force remains
  3. The bar magnet gives the needle a net electric charge
  4. The bar magnet's field reduces the needle's magnetic moment to zero
Show answer
Answer: (B) In a non-uniform field the forces on the two poles of the needle are unequal, so a net force remains
Explanation: In a uniform field the two poles experience equal and opposite forces, which form a couple, so there is a torque but no net force. Near a bar magnet the field differs at the two ends of the needle. The forces on the poles then do not cancel, leaving a net force along with the torque.

Question 14

A bar magnet of magnetic moment M is held perpendicular to a uniform magnetic field B. It is then turned towards the field direction until the torque on it is half of its initial value. How much work does the magnetic field do on the magnet during this turn?
  1. MB
  2. (√3/2)MB
  3. MB/2
  4. (1 − √3/2)MB
Show answer
Answer: (B) (√3/2)MB
Explanation: The torque is MB sin θ, so half the maximum means sin θ = 1/2 and the magnet is now at θ = 30° to the field. The potential energy is U = −MB cos θ, which is zero at 90° and −(√3/2)MB at 30°. The work done by the field equals the fall in potential energy, (√3/2)MB.

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