A magnetic dipole lying parallel to a magnetic field requires ' W ' units of work to turn it through 30°. The torque required to keep this needle in this position is [sin 30° = cos 60° = 0.5, sin 60° = cos 30° = √(3)/2, sin 90° = cos 0° = 1] (a) √(3)W (b) 2/√(3)W (c) (W)/((√(3) - 1)) (d) (W)/((2 - √(3)))

Answer: (d). Work from 0° to 30°: W = MB(1 - cos 30°) = MB(2 - √(3))/(2), so MB = (2W)/(2 - √(3)). Torque at 30°: τ = MBsin 30° = MB/2 τ = (W)/(2 - √(3)) So the answer is (d).

A magnetic dipole lying parallel to a magnetic field requires ' W '… | MHT CET 2026 (19 May, Shift 1)

Q2MHT CET 2026 · 19 MAY · SHIFT 1Medium

Field of a Magnet & Magnet in a Uniform Field

A magnetic dipole lying parallel to a magnetic field requires ' $W$ ' units of work to turn it through $30^\circ$ . The torque required to keep this needle in this position is

$\left[\sin 30^\circ = \cos 60^\circ = 0.5, \sin 60^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2}, \sin 90^\circ = \cos 0^\circ = 1\right]$

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Answer: (d)

Work from $0^\circ$ to $30^\circ$ : $W = MB(1 - \cos 30^\circ) = MB\dfrac{2 - \sqrt{3}}{2}$ , so $MB = \dfrac{2W}{2 - \sqrt{3}}$ .
Torque at $30^\circ$ : $\tau = MB\sin 30^\circ = \dfrac{MB}{2}$
$\tau = \dfrac{W}{2 - \sqrt{3}}$
So the answer is (d).

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Field of a Magnet & Magnet in a Uniform Field question from MHT CET 2026 (19 May, Shift 1)

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