Potential Energy Of A Magnetic Dipole

According to NCERT Class 12, Chapter 5 (Magnetism and Matter), the potential energy of a magnetic dipole of moment $m$ in a field $B$ is $U=-mB\cos \theta$, where $\theta$ is the angle between the moment and the field. Stable equilibrium occurs at $\theta =0^{\circ }$, where $U=-mB$ (minimum energy). The position pointing opposite to the field is $\theta =180^{\circ }$, where $U=+mB$ (maximum energy). The work done equals the change in potential energy: $W=U_{180^{\circ }}-U_{0^{\circ }}=(+mB)-(-mB)=2mB$. The value $mB$ corresponds only to rotating from $0^{\circ }$ to $90^{\circ }$. The 'zero' option would only be true for a full $360^{\circ }$ rotation returning to start. The value $mB/2$ has no basis in the energy formula. Plausibility check: the anti-parallel position is the least stable, highest-energy configuration, so it logically requires the largest work input, consistent with the maximum value $2mB$. This mirrors the electric-dipole case $U=-pE\cos \theta$, where flipping a dipole through $180^{\circ }$ likewise costs $2pE$, and the symmetry of the two results reinforces that the total energy swing between the fully aligned and fully reversed states is twice the alignment energy $mB$.