Question

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?

Answer 1

Magnetic field strength, B = 0.25 T
Torque on the bar magnet, T = 4.5 × 10-2 J
Angle between the bar magnet and the external magnetic field, θ = 30°
Torque is related to magnetic moment (M) as:
T = MB sin θ

∴ $M=\frac{T}{B\sin\theta}$

= $\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30\degree}$= 0.36 JT-1

Hence, the magnetic moment of the magnet is 0.36 J T-1 .