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# A short bar magnet placed with its axis at 30º with a uniform external magnetic field of 0.25 T

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?

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

$\therefore M=\frac{T}{B \sin \theta}$

$=\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}=0.36 \mathrm{~J} \mathrm{~T}^{-1}$

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