A magnetic needle lying parallel to magnetic field requires a work W to turn it by 60∘. The external torque required to maintain needle in this position is

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A magnetic needle lying parallel to magnetic field requires a work W to turn it by $60^{\circ} .$ The external torque required to maintain needle in this position is

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$\frac{\sqrt{3}}{2} W$

$\sqrt{3} W$

answer is D.

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$W = M B \cos 60^{0} = \frac{M B}{2}$

$τ = M B \sin 60^{0} = \frac{\sqrt{3}}{2} M B = \sqrt{3} W$

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