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:
Answer & explanation
Correct answer: option 4
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 -
$U=-MB\cos θ$
$W=U_{final}-U_{initial}$
$=-MB\cos 180°-(-MB\cos 0°)$
$=2MB$
$=2×0.4×0.08$
$=0.064J$