A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.16 T experiences a torque of magnitude 0.48 J. The magnetic moment of the bar magnet will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 6 J/T
Given:
Magnetic field: $B = 0.16\ \text{T}$
Torque: $\tau = 0.48\ \text{Nm}$
Angle with field: $\theta = 30^\circ$
Torque on a magnetic dipole: $\tau = M B \sin\theta$
Solving for magnetic moment $M$:
$M = \frac{\tau}{B \sin\theta} = \frac{0.48}{0.16 \cdot \frac{1}{2}} = \frac{0.48}{0.08} = 6\ \text{A·m²}$
∴ Magnetic moment of the bar magnet = 6 A·m²