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Figure shows a small magnetised needle $P$ placed at a point $O$. The arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle $Q$.
$(a)$ In which configuration the system is not in equilibrium?
$(b)$ In which configuration is the system in $(i)$ stable, and $(ii)$ unstable equilibrium?
$(c)$ Which configuration corresponds to the lowest potential energy among all the configurations shown?
Solution
Potential energy of the configuration arises due to the potential energy of one dipole (say, $Q$ ) in the magnetic field due to other (P). Use the result that the field due to $P$ is given by the expression
$B _{ p }=-\frac{\mu_{0}}{4 \pi} \frac{ m _{ P }}{r^{3}}$
(on the normal bisector)
$B _{ p }=\frac{\mu_{0} 2}{4 \pi} \frac{ m _{ p }}{r^{3}}$ (on the axis)
where $m _{ p }$ is the magnetic moment of the dipole $P$. Equilibrium is stable when $m _{ 9 }$ is parallel to $B _{ p },$ and unstable when it is anti-parallel to $B _{ p }$ For instance for the configuration $Q _{3}$ for which $Q$ is along the perpendicular bisector of the dipole $P$, the magnetic moment of $Q$ is parallel to the magnetic field at the position $3 .$ Hence $Q _{3}$ is stable. Thus,
$(a)$ $PQ _{1}$ and $PQ _{2}$
$(b)$ (i) $PQ _{3}, PQ _{6}$ (stable); (ii) $PQ _{5}, PQ _{4}$ (unstable)
$(c)$ $PQ _{6}$
