Formula
Torque on magnetic dipole
τ = m × B. Magnitude τ = m B sin θ. PE U = -m·B = -m B cos θ. Stable equilibrium when m parallel B.
-- NCERT Class 12 Physics, Ch. 5, p. 141Torque Magnetic Dipole
8 MCQs9-step worked example
Source: NCERT Magnetic Effects of Current and MagnetismOfficial key: NTA-verifiedLast updated: 23 Sep 2026
Torque Magnetic Dipole, explained for NEET
You have already met τ = NIAB sin θ for a coil. This topic looks like that statement with different letters, and the resemblance is what costs marks. Here the dipole is a bar magnet: there is no N, I or A to multiply. The moment m, in A·m², is handed to you, and τ = mB sin θ — in vector form, τ = m × B. The angle is still the one between m and B.
What is new is the energy. NCERT Class 12 Physics, Chapter 5, page 141 sets the torque beside the potential energy of the dipole, U = −m·B = −mB cos θ. Two consequences follow. The corpus records this as a low-frequency topic — roughly one question every two to three years, weighted to recall and single-step application.
A dipole has two equilibrium orientations, not one. At θ = 0° the torque is zero and U = −mB, the minimum: stable. At θ = 180° the torque is also zero but U = +mB, the maximum: unstable, and the smallest nudge turns it further away. Zero torque alone never tells you which of the two you are in — the energy does.
The zero of this potential energy sits at θ = 90°, not at θ = 0°. That is built into U = −m·B, and sign slips there are a common loss. Work done against the field in turning from θ₁ to θ₂ is W = mB(cos θ₁ − cos θ₂). Write the initial angle first.
Displace the magnet slightly from stable equilibrium and it oscillates: T = 2π√(I/mB), where I is the moment of inertia about the axis of oscillation, not a current. This is how a uniform field's magnitude is measured in the laboratory.
Watch out: sin θ governs the torque, cos θ governs the energy. Reaching for the wrong one turns a solved problem into a negative mark.
Can you answer these Torque Magnetic Dipole MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
For a magnetic dipole of moment m placed in a uniform magnetic field B, the magnetic potential energy of the dipole is given by
Show answer and why every option is right or wrong
Answer: C. NCERT Class 12 Physics, Chapter 5, page 141 states the dipole's potential energy in a uniform field as U = −m·B = −mB cos θ, minimum at θ = 0° and maximum at θ = 180°.
Why A is wrong: A is wrong because dropping the minus sign would make the aligned orientation (θ = 0°) the maximum-energy state, whereas alignment is the stable minimum.
Why B is wrong: B is wrong because m × B is the torque, a vector quantity; potential energy is a scalar, so the product here must be the dot product.
Why D is wrong: D is wrong because sin θ belongs to the torque expression τ = mB sin θ; the energy varies as cos θ, and −mB sin θ would wrongly give zero energy at θ = 0°.
MCQ 2Easy RecallPractice
A bar magnet is free to rotate in a uniform magnetic field. It is in stable equilibrium when
Show answer and why every option is right or wrong
Answer: A. At θ = 0° the torque is zero and the potential energy U = −mB is a minimum, which is the condition for stable equilibrium (NCERT Class 12 Physics, Chapter 5, page 141).
Why B is wrong: B is wrong because θ = 180° gives U = +mB, an energy maximum; the torque is zero there, but any small displacement produces a torque that turns the magnet further away — unstable equilibrium.
Why C is wrong: C is wrong because θ = 90° is where the torque is greatest, not zero, so the magnet cannot rest in that orientation at all.
Why D is wrong: D is wrong because although a uniform field exerts no net force, it exerts a torque at every orientation except 0° and 180°; absence of net force does not mean absence of a preferred orientation.
MCQ 3Easy RecallPractice
A bar magnet of moment m, displaced slightly from alignment with a uniform field B, oscillates with period T = 2π√(I/mB). In this expression, I denotes
Show answer and why every option is right or wrong
Answer: D. The relation comes from equating the restoring torque mB sin θ to I times the angular acceleration, so I is the magnet's moment of inertia about the axis of oscillation (NCERT Class 12 Physics, Chapter 5, in the section on the dipole in a uniform field beginning on page 141).
Why A is wrong: A is wrong because no current appears in this relation; a bar magnet's magnetic character enters only through its moment m.
Why B is wrong: B is wrong because intensity of magnetisation is the moment per unit volume and does not appear in the period expression at all.
Why C is wrong: C is wrong because impulse has units of N·s and cannot combine with m and B to yield a time.
MCQ 4Direct ApplicationPractice
A bar magnet of magnetic moment 0.60 A·m² is placed in a uniform magnetic field of 0.25 T, with its axis at 60° to the field direction. The magnitude of the torque on the magnet is
Show answer and why every option is right or wrong
Answer: B. τ = mB sin θ = 0.60 × 0.25 × sin 60° = 0.15 × 0.866 = 1.3 × 10⁻¹ N·m, using the torque relation on NCERT Class 12 Physics, Chapter 5, page 141.
Why A is wrong: A is wrong because 7.5 × 10⁻² N·m uses cos 60° = 0.500 in place of sin 60°; cosine belongs to the potential energy, sine to the torque.
Why C is wrong: C is wrong because 1.5 × 10⁻¹ N·m is the product mB alone — the maximum torque, which occurs only at θ = 90°.
Why D is wrong: D is wrong because 2.6 × 10⁻¹ N·m carries a factor of 2 from the axial dipole-field expression into the torque, where no such factor exists.
MCQ 5Direct ApplicationPractice
A magnetic dipole of moment 0.40 A·m² is held in a uniform field of 0.50 T with its moment at 120° to the field. Its magnetic potential energy is
Show answer and why every option is right or wrong
Answer: D. U = −mB cos θ = −(0.40 × 0.50) × cos 120° = −0.20 × (−0.500) = +1.0 × 10⁻¹ J; past 90° the cosine is negative, so U is positive (NCERT Class 12 Physics, Chapter 5, page 141).
Why A is wrong: A is wrong because −1.0 × 10⁻¹ J is the value at θ = 60°; at 120° the cosine itself is negative and the two minus signs leave U positive.
Why B is wrong: B is wrong because it substitutes sin 120° = 0.866 for cos 120°; the energy varies as the cosine, not the sine.
Why C is wrong: C is wrong because it also uses sin 120° in place of cos 120°, and correcting the sign afterwards does not repair the wrong trigonometric function.
MCQ 6Direct ApplicationPractice
A bar magnet of moment 2.0 A·m² rests aligned with a uniform field of 0.15 T. The work that must be done against the field to turn it into the antiparallel position is
Show answer and why every option is right or wrong
Answer: A. W = mB(cos θ₁ − cos θ₂) = mB(cos 0° − cos 180°) = 2mB = 2 × 2.0 × 0.15 = 6.0 × 10⁻¹ J, following the energy expression on NCERT Class 12 Physics, Chapter 5, page 141.
Why B is wrong: B is wrong because 3.0 × 10⁻¹ J is mB, the work needed to go from θ = 0° to θ = 90°, not all the way to 180°.
Why C is wrong: C is wrong because a uniform field exerts no net force but does exert a torque; work must be done against that torque, so the answer cannot be zero.
Why D is wrong: D is wrong because the negative sign comes from writing the final angle first; W = mB(cos θ₁ − cos θ₂) takes the initial angle first, and the work done against the field here is positive.
MCQ 7CalculationPractice
A bar magnet experiences a torque of 3.0 × 10⁻² N·m when its axis makes 30° with a uniform field of magnitude 0.20 T. The work that must be done against the field to rotate it from that position until its axis is perpendicular to the field is
Show answer and why every option is right or wrong
Answer: C. From τ = mB sin θ, m = 3.0 × 10⁻²/(0.20 × 0.500) = 0.30 A·m²; then W = mB(cos 30° − cos 90°) = 0.30 × 0.20 × 0.866 = 5.2 × 10⁻² J (NCERT Class 12 Physics, Chapter 5, page 141).
Why A is wrong: A is wrong because 3.0 × 10⁻² is the given torque in N·m; torque and work are different quantities and the number cannot be carried straight across.
Why B is wrong: B is wrong because 6.0 × 10⁻² J is mB, the work from θ = 0° to 90°; starting at 30° brings in the factor cos 30°.
Why D is wrong: D is wrong because 1.2 × 10⁻¹ J is 2mB, the work for a complete 0° to 180° rotation, not for 30° to 90°.
MCQ 8CalculationPractice
A bar magnet suspended in a uniform horizontal field oscillates with period T about its stable position. It is replaced by a second magnet having twice the moment of inertia and twice the magnetic moment, and the field magnitude is halved. The new period of oscillation is
Show answer and why every option is right or wrong
Answer: B. T = 2π√(I/mB); the product mB becomes (2m)(B/2) = mB, unchanged, while I doubles, so T′ = 2π√(2I/mB) = √2 T (NCERT Class 12 Physics, Chapter 5, section on the dipole in a uniform field, page 141).
Why A is wrong: A is wrong because T/2 would require mB to become four times larger at fixed I, whereas here the product mB is unchanged.
Why C is wrong: C is wrong because 2T would need the moment of inertia to be four times larger; it is only doubled.
Why D is wrong: D is wrong because T/√2 reverses the effect of the moment of inertia — a larger I lengthens the period, it does not shorten it.
Free NEET study resources
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
How do you solve a Torque Magnetic Dipole question? A worked example
- 1
Given.
A bar magnet of magnetic moment m = 0.50 A·m² is suspended in a uniform magnetic field of magnitude B = 0.40 T. In its equilibrium-displaced position it experiences a torque of magnitude τ = 1.0 × 10⁻¹ N·m.
- 2
Required.
The angle θ between the magnet's axis and the field, and the magnetic potential energy of the magnet in that position.
- 3
Concept.
A dipole in a uniform field feels a torque set by the sine of the angle between m and B, and stores a potential energy set by the cosine of the same angle. Given the torque, the angle is recovered from the sine relation and then fed into the energy relation — the two use different trigonometric functions of the same θ.
- 4
Formula.
τ = mB sin θ and U = −mB cos θ (NCERT Class 12 Physics, Chapter 5, page 141).
- 5
Substitution.
sin θ = τ/(mB) = (1.0 × 10⁻¹)/(0.50 × 0.40), then U = −(0.50 × 0.40) cos θ.
- 6
Calculation.
mB = 0.50 × 0.40 = 0.20 A·m²·T. sin θ = 0.10/0.20 = 0.500, so θ = 30°. Then U = −0.20 × cos 30° = −0.20 × 0.8660 = −0.1732 J. The value cos 30° = √3/2 is a mathematical constant and contributes no significant figures; the two significant figures in the answer come from τ, m and B.
- 7
Final answer.
θ = 30° and U = −1.7 × 10⁻¹ J.
- 8
Common trap.
sin θ = 0.500 has two solutions in the range 0° to 180°: θ = 30° and θ = 150°. The torque magnitude alone cannot distinguish them — at 150° the same torque acts but U = +1.7 × 10⁻¹ J, a positive energy on the unstable side of the field. If a question supplies only a torque, check whether it also tells you which side of the field the magnet lies on. The second trap is arithmetic: substituting sin 30° into the energy expression gives −1.0 × 10⁻¹ J, the right magnitude for the wrong reason and the wrong number here.
- 9
Similar NEET-style question.
A magnet of moment 1.2 A·m² is held with its axis perpendicular to a uniform field of 0.25 T and then released. How much work does the field do on the magnet as it turns to its stable position? (W = U_initial − U_final = 0 − (−mB) = 3.0 × 10⁻¹ J.)
What to remember before solving Torque Magnetic Dipole questions
More in Magnetic Effects of Current and Magnetism: 3 exam traps and mistakes · 11 formulas · 5 question patterns from its other lessons.
Torque Magnetic Dipole questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
All 17 past-paper questions from Magnetic Effects of Current and Magnetism →
Sources
NCERT refs: Class 12 Physics Chapter 5, p.141
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →