1. Magnetic potential energy of a bar magnet

   For a magnetic dipole of magnetic moment $m$ placed in a uniform magnetic field $B$, the potential energy is $U = -mB\cos\theta$ where $\theta$ is the angle between $\vec m$ and $\vec B$.

2. Initial position: parallel

   In the parallel position, $\theta = 0^\circ$ So, $U_i = -mB\cos 0^\circ = -mB$

3. Final position: antiparallel

   In the antiparallel position, $\theta = 180^\circ$ So, $U_f = -mB\cos 180^\circ = +mB$

4. Work required to rotate the magnet

   The external work done in rotating the magnet slowly from parallel to antiparallel position is equal to the increase in potential energy: $W = U_f - U_i$ $W = mB - (-mB) = 2mB$

5. Substitute values

   Given: $m = 5.0\,\mathrm{A\,m^2}, \quad B = 0.4\,\mathrm{T}$

   Therefore, $W = 2 \times 5.0 \times 0.4 = 4.0\,\mathrm{J}$

6. Option check

   • A: zero $\to$ incorrect
   • B: $1\,\mathrm{J}$ $\to$ incorrect
   • C: $2\,\mathrm{J}$ $\to$ incorrect
   • D: $4\,\mathrm{J}$ $\to$ correct

Therefore, the required work done is $4\,\mathrm{J}$.