Question

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 \( 4.5\times 10^{-2}\; J \). What is the magnitude of the magnetic moment of the magnet?

Answer 1

Magnetic field strength, B = 0.25 T

Torque on the bar magnet, \( T = 4.5\times 10^{-2}\; J \)

The angle between the bar magnet and the external magnetic field, \( \theta = 30° \)

Torque is related to magnetic moment (M) as:

\( T = MB sin\theta \)

\( ∴ M = \frac{T}{B sin\theta } = \frac{4.5\times 10^{-2}}{0.25\times sin 30°} = 0.36\; J\; T^{-1} \)

Hence, the magnetic moment of the magnet is \( 0.36\; J\; T^{-1} \)

The magnetic moment of the magnet is \( 0.36\; J\; T^{-1} \)