A short bar magnet placed with its axis at \( 30^\circ \) 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) → \( 0.4\,\mathrm{A\,m^2} \)
The torque on a magnetic dipole in a uniform field is \( \tau = mB\sin\theta \), where \( \theta \) is the angle between the magnet's axis and the field.
\( \displaystyle m = \frac{\tau}{B\sin\theta} \)
\( \displaystyle m = \frac{0.048\,\mathrm{N\,m}}{0.24\,\mathrm{T} \times \sin 30^\circ} = \frac{0.048}{0.24 \times 0.5} = 0.4\,\mathrm{A\,m^2} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
