A short bar magnet placed with its axis at 30° with an external field of 800 G, experiences a torque of 0.016 Nm. Work done in moving it from its most stable to most unstable position is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.064 J
The torque on a magnetic dipole is \( \tau = mB\sin\theta \), and its potential energy is \( U = -mB\cos\theta \). The stable position is \( \theta = 0^\circ \) and the most unstable position is \( \theta = 180^\circ \).
First convert the field: \( 1\,\mathrm{G} = 10^{-4}\,\mathrm{T} \), so \( B = 800\,\mathrm{G} = 0.08\,\mathrm{T} \).
\( \displaystyle m = \frac{\tau}{B\sin\theta} = \frac{0.016}{0.08\times0.5} = 0.4\,\mathrm{A\,m^{2}} \)
The work done equals the change in potential energy.
\( W = U_{180^\circ} - U_{0^\circ} = (+mB) - (-mB) = 2mB \)
\( W = 2\times0.4\times0.08 = 0.064\,\mathrm{J} \)
Shortcut: Since \( \tau = mB\sin\theta \) and \( W = 2mB \), we get \( W = 2\tau/\sin\theta = 2\times0.016/0.5 = 0.064\,\mathrm{J} \), without finding m or converting gauss to tesla.