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?

(N/A) The potential energy of the configuration arises due to the interaction of one dipole (say,$Q$) in the magnetic field produced by the other $(P)$. The magnetic field $\vec{B}_P$ due to dipole $P$ is given by:
$1$. On the axial line: $\vec{B}_P = \frac{\mu_0}{4\pi} \frac{2\vec{m}_P}{r^3}$
$2$. On the equatorial line: $\vec{B}_P = -\frac{\mu_0}{4\pi} \frac{\vec{m}_P}{r^3}$
Equilibrium occurs when the torque $\vec{\tau} = \vec{m}_Q \times \vec{B}_P = 0$,which means $\vec{m}_Q$ must be parallel or anti-parallel to $\vec{B}_P$.
$(a)$ In configurations $PQ_1$ and $PQ_2$,the magnetic moment $\vec{m}_Q$ is neither parallel nor anti-parallel to the field $\vec{B}_P$ at those points. Thus,there is a non-zero torque,and the system is not in equilibrium.
$(b)$ Equilibrium is stable when $\vec{m}_Q$ is parallel to $\vec{B}_P$ (potential energy $U = -\vec{m}_Q \cdot \vec{B}_P$ is minimum) and unstable when $\vec{m}_Q$ is anti-parallel to $\vec{B}_P$ (potential energy $U$ is maximum).
$(i)$ Stable equilibrium: $PQ_3$ and $PQ_6$.
$(ii)$ Unstable equilibrium: $PQ_4$ and $PQ_5$.
$(c)$ The potential energy $U = -\vec{m}_Q \cdot \vec{B}_P$. The lowest potential energy corresponds to the configuration where $\vec{m}_Q$ and $\vec{B}_P$ are parallel and the magnitude of $\vec{B}_P$ is maximum. Since the axial field is twice the equatorial field,$PQ_6$ (on the axis) provides the lowest potential energy.