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(D) The torque (couple) acting on a bar magnet in a uniform magnetic field is given by $T = MB \sin 	heta$,where $M$ is the magnetic moment,$B$ is th

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$\sin ^{-1}(0.5)$

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(D) The torque (couple) acting on a bar magnet in a uniform magnetic field is given by $T = MB \sin 	heta$,where $M$ is the magnetic moment,$B$ is the magnetic field,and $	heta$ is the angle between the magnet and the field.Initially,the magnet is perpendicular to the field,so $	heta = \frac{\pi}{2} = 90^{\circ}$.The initial torque is $T = MB \sin(90^{\circ}) = MB$.We want the new torque $T'$ to be half of the initial torque,so $T' = \frac{T}{2} = \frac{MB}{2}$.Let the new angle be $	heta'$. Then $T' = MB \sin 	heta' = \frac{MB}{2}$.Dividing both sides by $MB$,we get $\sin 	heta' = \frac{1}{2} = 0.5$.Therefore,the angle $	heta'$ is $\sin^{-1}(0.5) = 30^{\circ}$.

$A$ bar magnet is held perpendicular to a uniform magnetic field. The couple acting on the magnet is to be halved by rotating it. Through what angle should it be rotated?

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