A magnetic needle suspended parallel to a magnetic field requires $\sqrt 3\,J $ of work to turn it through $60^o$. The torque needed to maintain the needle in this position will be
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$\mathrm{W}=\mathrm{MB}\left(\cos \theta_{1}-\cos \theta_{2}\right)$
$\sqrt{3}=\mathrm{MB}\left(\cos 0^{\circ}-\cos 60^{\circ}\right)$
$\mathrm{MB}=2 \sqrt{3}$
$\tau=M B \sin \theta=2 \sqrt{3} \times \sin 60^{\circ}=3\, \mathrm{J}$
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