A bar magnet when suspended horizontally and perpendicular to earth's field experiences a torque of \( 3\times10^{-4}\,\mathrm{Nm} \). The Magnetic moment of the magnet would be: (take horizontal component of Earth's magnetic field at the place \( 0.4\times10^{-4}\,\mathrm{T} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 7.5\,\mathrm{JT^{-1}} \)
The torque on a magnetic dipole of moment \( m \) in a uniform field \( B \) is \( \tau = mB\sin\theta \). The magnet is perpendicular to the field, so \( \theta = 90^\circ \) and \( \sin\theta = 1 \).
\( \displaystyle m = \frac{\tau}{B} = \frac{3\times10^{-4}\,\mathrm{N\,m}}{0.4\times10^{-4}\,\mathrm{T}} = 7.5\,\mathrm{J\,T^{-1}} \)
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 – …
