A bar magnet of magnetic moment 1 . 5JT^ - 1 is aligned with the...

Potential energy of a magnetic dipole is given as

$U = - MBcos\theta$~where,~$M = Magneticdipole, B = Magneticfield, \theta = Anglebetween\overset{→}{M}and\overset{→}{B}$

$W = {U}_{f} - {U}_{i} ⇒ W = - MBcos{θ}_{f} - \left(- MBcos{θ}_{i}\right) ⇒ W = MB\left(cos{θ}_{i} - cos{θ}_{f}\right) ⇒ W = 1 . 5 \times 0 . 22\left(\cos 0 ° - \cos 180 °\right) ⇒ W = 0 . 66J$

$∵ \overset{→}{τ} = MBsin\theta$

Given~$\theta = 180 °$

$∴ \overset{→}{τ} = MBsin180 ° ⇒ \overset{→}{τ} = 0$