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 × 10^–2 J. What is the magnitude of magnetic moment of the magnet?

Answer 1

Given: 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°

Formula: T = MB sin θ

∴ M = `T/(B sin θ)`

= `(4.5 xx 10^-2)/(0.25 xx sin 30°)`

= `(4.5 xx 10^-2)/(0.25 xx 1/2)`

= `(4.5 xx 10^-2)/(0.25 xx 0.5)`

= `(4.5 xx 10^-2)/0.125`

= 360 × 10⁻² 

= 0.36 J T⁻¹

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