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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?
Solution
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 = `"T"/("B" sin θ)`
= `(4.5 xx 10^-2)/(0.25 xx sin 30°)`
= 0.36 J T−1
Hence, the magnetic moment of the magnet is 0.36 J T−1.
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