A bar magnet of magnetic moment \( m \) is placed in a uniform magnetic field \( B \) at most stable state and then rotated to the most unstable state. Work done in this process will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( +2mB \)
The potential energy of a magnetic dipole in a uniform field is \( U = -mB\cos\theta \). The most stable state is \( \theta = 0^\circ \) (minimum energy) and the most unstable state is \( \theta = 180^\circ \) (maximum energy). The work done in rotating the magnet (by the external agent, slowly, without giving it kinetic energy) equals the increase in potential energy.
\( U_{1} = -mB\cos0^\circ = -mB \)
\( U_{2} = -mB\cos180^\circ = +mB \)
\( W = U_{2} - U_{1} = mB - (-mB) = +2mB \)
Note: The question does not say by whom the work is done. The work done by the magnetic field on the magnet in the same rotation is the negative of the change in potential energy, \( -(U_{2} - U_{1}) = -2mB \). Under that reading Option (2) would be correct. When a question says only “work done” in turning a dipole, it is usually the work done by the external agent, which gives the keyed Option (1).