A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.24 T experiences a torque of magnitude 0.048 N m. The magnetic moment of the magnet is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $0.4\, A m^2$
Torque on a magnetic dipole: $\tau = M B \sin \theta$
Given: $\tau = 0.048\ \text{N·m}$, $B = 0.24\ \text{T}$, $\theta = 30^\circ$
Magnetic moment:
$M = \frac{\tau}{B \sin \theta} = \frac{0.048}{0.24 \cdot \sin 30^\circ} = \frac{0.048}{0.24 \cdot 0.5} = \frac{0.048}{0.12} = 0.4\ \text{A·m}^2$
Final Answer: $M = 0.4\ \text{A·m}^2$