Potential Energy Of A Magnetic Dipole

Question

A magnetic dipole is rotated from its stable equilibrium position to a position where it points exactly opposite to a uniform magnetic field; the work done on the dipole equals

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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 $U = -mB\cos	heta$, where $	heta$ is the angle between the moment and the field. Stable equilibrium occurs at $	heta = 0^{\circ}$, where $U = -mB$ (minimum energy). The position pointing opposite to the field is $	heta = 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 $U = -pE\cos	heta$, 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 $mB$.

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Potential Energy Of A Magnetic Dipole

Select the correct option:

🔒 Solution Hidden from View

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Solution

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