(a) Formula and Calculation of work done in the two cases (b) Calculation of torque in case (ii) (a) Work done $= m_B (\cos\theta_1 - \cos\theta_2)$ (i) $\;\theta_1=60^{\circ}, \theta_2=90^{\circ}$ $\;\therefore\;\text{ work done }\;=m B (\cos 60^{\circ}-\cos 90^{\circ})$ $\;=m B \left(\;\frac{1}{2}-0\right) =\;\frac{1}{2} m B$ $\;=\;\frac{1}{2} \times 6 \times 0.44\,\mathrm{J}=1.32\,\mathrm{J}$ (ii) $\;\theta_1=60^{\circ}, \theta_2=180^{\circ}$ $\;\therefore\;\text{ work done }\;=m B (\cos 60^{\circ}-\cos 180^{\circ})$ $\;\; =m B \left(\;\frac{1}{2}-(-1)\right) =\;\frac{3}{2} m B$ $\;\; =\;\frac{3}{2} \times 6 \times 0.44\,\mathrm{J}-3.96\,\mathrm{J} \text{end{aligned}}$ [Also accept calculations done through changes in potential energy.] (b) Torque $=\left|{\overset{\rightarrow}{m}} \times \overset{\rightarrow}{B}\right|= m B \sin\theta$ For $\theta=180^{\circ}$, we have Torque $=6 \times 0.44 \sin 180^{\circ}=0$ [If the student straight away writes that the torque is zero since magnetic moment and magnetic field are anti parallel in this orientation, award full]