5.3 A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10–2 J. What is the magnitude of the magnetic moment of the magnet?

Magnetic field strength, B=0.25T
Torque on the bar magnet,
$T=4.5×{10}^{-2}J$
Angle between the bar magnet and the external magnetic field$\mathrm{\theta }=30°$
Torque is related to the magnetic moment (M) as:
$\begin{array}{rl}\therefore M& =\frac{T}{B\mathrm{sin}\theta }\\ & =\frac{4.5×{10}^{-2}}{0.25×\mathrm{sin}{30}^{\circ }}=0.36{\mathrm{JT}}^{-1}\end{array}$
Hence, the magnetic moment of the magnet is
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