A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torque needed to maintain the needle in this position will be: (a) √(3) W (b) W (c) √(3)/2W (d) 2W

Answer: (a). W = MB(1-cos 60°) = MB/2 ⇒ MB = 2W τ = MBsin 60° = 2W · √(3)/2 = √(3)W So the answer is (a) √(3) W.

A magnetic needle lying parallel to a magnetic field requires W units… | JEE Main 2003 (AIEEE)

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Field of a Magnet & Magnet in a Uniform Field

A magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^\circ.$ The torque needed to maintain the needle in this position will be:

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Answer: (a)

$W = MB(1-\cos 60^\circ) = \frac{MB}{2} \Rightarrow MB = 2W$
$\tau = MB\sin 60^\circ = 2W \cdot \frac{\sqrt{3}}{2} = \sqrt{3}W$
So the answer is (a) $\sqrt{3}\,W$ .

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