Magnetism and Matter

Magnetic moment of the bar magnet:$M = 10^4 J T^{-1}$Magnetic field:$B = 4 \times 10^{-5} T$Initial orientation: $\theta_1 = 0^{\circ}$ (parallel to $B$ )Final orientation: $\theta_2 = 60^{\circ}$Relevant conceptThe potential energy of a magnetic dipole in a magnetic field is$U = -M B \cos \theta$When the magnet is rotated slowly (quasi-statically), the work done by an external agent equals the change in potential energy of the system:$W = U_2 - U_1 = \left[ -M B \cos \theta_2 \right] - \left[ -M B \cos \theta_1 \right]$$W = M B (\cos \theta_1 - \cos \theta_2)$Substitute the values$W = (10^4) (4 \times 10^{-5}) (\cos 0^{\circ} - \cos 60^{\circ})$Compute step by step:$\cos 0^{\circ} = 1, \quad \cos 60^{\circ} = \frac{1}{2}$So,$W = (10^4) (4 \times 10^{-5}) (1-0.5)$$W = (10^4) (4 \times 10^{-5}) (0.5)$$W = (10^4 \times 4 \times 10^{-5} \times 0.5)$$W = (4 \times 10^{-1} \times 0.5) = 0.2 \mathrm{J}$