A short bar magnet placed with its axis at 30º 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 magnetic moment of the magnet?

Magnetic field strength, B = 0.25 T

Torque on the bar magnet, T = 4.5 x 10^-2 J

Angle between the bar magnet and the external magnetic field, 0 = 30° Torque is related to magnetic moment (M) as: T = MB sin 0

Therefore , M = T/Bsin 0

= (4.5 x 10^-2 )/(0.25 x sin 30º)  =0.36 J T⁻¹

Hence, the magnetic moment of the magnet is 0.36 IT^-1.