A bar magnet of magnetic moment $1.5\;J/T$ lies aligned with the direction of a uniform magnetic field of $0.22\;T$ What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment, opposite to the field direction ?

Question

$\begin{array}{1 1} 0 \\ -3.3\;N-m \\ -0.33\;N-m \\ 0.33\;N-m \end{array}$