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(C) Consider a bar magnet of length $2l$ placed in a uniform magnetic field $\overrightarrow{B}$ at an angle $	heta$ with the direction of the field.

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$\overrightarrow{M} 	imes \overrightarrow{B}$

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(C) Consider a bar magnet of length $2l$ placed in a uniform magnetic field $\overrightarrow{B}$ at an angle $	heta$ with the direction of the field.The force acting on each pole of the magnet is $F = mB$, where $m$ is the pole strength.These two equal and opposite forces form a couple that exerts a torque $	au$ on the magnet.The torque is given by: $	au = 	ext{Force} 	imes 	ext{perpendicular distance } (d)$.From the geometry, the perpendicular distance between the forces is $d = 2l \sin 	heta$.Therefore, $	au = (mB) 	imes (2l \sin 	heta) = (m 	imes 2l) B \sin 	heta$.Since the magnetic moment is defined as $M = m 	imes 2l$, we have $	au = MB \sin 	heta$.In vector form, this is expressed as $\overrightarrow{	au} = \overrightarrow{M} 	imes \overrightarrow{B}$.

$A$ bar magnet of magnetic moment $\overrightarrow{M}$ is placed in a magnetic field of induction $\overrightarrow{B}$. The torque exerted on it is

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