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

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$60$

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(C) The torque (couple) acting on a bar magnet in a uniform magnetic field is given by $	au = MB \sin 	heta$,where $M$ is the magnetic moment,$B$ is the magnetic field,and $	heta$ is the angle between the magnetic moment and the field.Initially,the magnet is perpendicular to the field,so $	heta_1 = 90^\circ$. Thus,$	au_1 = MB \sin 90^\circ = MB$.We want the new torque $	au_2$ to be half of the initial torque,so $	au_2 = \frac{	au_1}{2} = \frac{MB}{2}$.Using the formula $	au_2 = MB \sin 	heta_2$,we have $\frac{MB}{2} = MB \sin 	heta_2$.This gives $\sin 	heta_2 = \frac{1}{2}$,which means $	heta_2 = 30^\circ$.The angle of rotation is the difference between the initial and final angles: $\Delta 	heta = 90^\circ - 30^\circ = 60^\circ$.

$A$ bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it,then the angle by which it is to be rotated is....$^o$

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