(i) A rectangular loop of sides a and b carrying current I is placed in a magnetic field B such that its area vector A makes an angle θ with B. With the help of a suitable diagram, show that the torque acting on the loop is given by = m × B, where m (= I A) is the magnetic dipole moment of the loop. (ii) A circular coil of 100 turns and radius ((10)/(√(π)) ) cm carrying current of 5·0 A is suspended vertically in a uniform horizontal magnetic field of 2·0 T. The field makes an angle 30° with the normal to the coil. Calculate : (I) the magnetic dipole moment of the coil, and (II) the magnitude of the counter torque that must be applied to prevent the coil from turning.

PhysicsCBSE Class 12 · Board exam 2026, set 55/5/1 · 5 marks · Long answer

CBSE Class 12 Physics 2026: question 32(a)

Set in the CBSE Class 12 board exam of 2026, paper 55/5/1, section E, question 32(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(a) · 5 marks · Long answerMoving Charges and Magnetism

(i) A rectangular loop of sides a and b carrying current I is placed in a magnetic field $B$ such that its area vector $A$ makes an angle $θ$ with $B$ . With the help of a suitable diagram, show that the torque $τ$ acting on the loop is given by $τ = m \times B$ , where $m$ (= I $A$ ) is the magnetic dipole moment of the loop.
(ii) A circular coil of 100 turns and radius $(\frac{10}{π})$ cm carrying current of 5·0 A is suspended vertically in a uniform horizontal magnetic field of 2·0 T. The field makes an angle 30° with the normal to the coil. Calculate :
(I) the magnetic dipole moment of the coil, and
(II) the magnitude of the counter torque that must be applied to prevent the coil from turning.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 55/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Find the force on each side of the loop; which pair cancels completely and which pair forms a couple?

Hint 2

The moment arm of the couple depends on sin θ; then write τ in terms of m = IA.

Hint 3

For (ii) the radius has √π in it — compute πr² first.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Diagram: rectangular loop PQRS with sides PQ = RS = b (perpendicular to $B$ , along the rotation axis direction) and QR = SP = a, in uniform $B$ ; the normal $\overset{n}{^}$ (area vector) makes angle θ with $B$ .
Forces on sides QR and SP (length a) are equal, opposite and collinear (along the axis), so they cancel and give no torque.
1 mark
Forces on PQ and RS: each of magnitude $F = I b B$ (since these sides are ⟂ $B$ ), equal and opposite but along different lines — they form a couple. Perpendicular distance between their lines of action $= a sin θ$ .
1 mark
Torque: $τ = F a sin θ = I b B a sin θ = I A B sin θ = m B sin θ$ (with $A = ab$ , $m = I A$ ). The torque tends to align $m$ with $B$ , and its direction is that of $m \times B$ , so $τ = m \times B$ .
1 mark
(ii)(I) $A = π r^{2} = π \times \frac{100}{π} \times 1 0^{- 4} = 1.0 \times 1 0^{- 2}$ m².
$m = N I A = 100 \times 5.0 \times 1.0 \times 1 0^{- 2} = 5.0$ A m²
1 mark
(II) $τ = m B sin θ = 5.0 \times 2.0 \times sin 3 0^{\circ} = 5.0$ N m. A counter torque of 5.0 N m must be applied.
1 mark

Answer: (i) τ = IAB sin θ ⇒ τ⃗ = m⃗ × B⃗; (ii) (I) 5.0 A m²; (II) 5.0 N m

Where the marks are

Diagram and forces on the two sides cancel without torque (1 mark)
Forces IbB on other two sides form a couple with arm a sin θ (1 mark)
τ = IAB sin θ = mB sin θ, written as τ⃗ = m⃗ × B⃗ (1 mark)
Area 10⁻² m² and m = 5.0 A m² (1 mark)
Counter torque = mB sin 30° = 5.0 N m (1 mark)

Revise the chapter

Chapter: Moving Charges and Magnetism (notes, formulas, where marks are lost and every past question)
Topic: Torque on a Current Loop and Magnetic Dipole

Practise this chapterBoard-style questions from Moving Charges and Magnetism, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 55/5/1: every question, by section.

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CBSE Class 12 Physics 2026: question 32(a)

Hints

Step-by-step solution

Revise the chapter

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