A short bar magnet is placed in a uniform external magnetic field of 0.25 T such that its axis makes an angle 30° with respect to the direction of the field. It experiences a torque of magnitude 0.04 J. The magnitude of the magnetic moment of the magnet is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.32 JT−1
A magnet of magnetic moment m in a uniform field B experiences a torque that depends on the angle between its axis and the field.
\( \displaystyle \tau = mB\sin\theta \)
\( \displaystyle m = \frac{\tau}{B\sin\theta} \)
\( \displaystyle m = \frac{0.04\,\mathrm{J}}{0.25\,\mathrm{T} \times \sin 30^\circ} \)
\( \displaystyle m = \frac{0.04}{0.125}\,\mathrm{J\,T^{-1}} \)
\( \displaystyle m = 0.32\,\mathrm{J\,T^{-1}} \)
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 – …
