A short bar magnet placed with its axis at 30° with an external field of 800G, experiences a torque of 0.016 Nm. Work done in moving it from its most stable to most unstable position is:

0.064 J

The correct answer is Option (4) → 0.064 J

The torque on a magnetic dipole in uniform magnetic field is -

$τ=MB\sin θ$

$⇒M=\frac{τ}{B\sin θ}=\frac{0.016}{800×10^{-4}×\sin 30°}$

$=0.4Am^2$

The potential energy of a magnetic dipole is -

$U=-MB\cos θ$

$W=U_{final}-U_{initial}$

$=-MB\cos 180°-(-MB\cos 0°)$

$=2MB$

$=2×0.4×0.08$

$=0.064J$