Magnetic Field at a Point Along the Axial Line of the Magnetic Dipole (Bar Magnet)

Consider a bar magnet NS of magnetic length $2l$ and pole strength $q_m$, centred at $O$. To find the field at a point $C$ on its extended axis, at distance $r$ from $O$, imagine placing a unit north pole ($q_{mC}=1$ A m) at $C$ and adding the fields due to the real north and south poles individually.

The north pole, at distance $(r-l)$ from $C$, produces $B_N = \dfrac{\mu_0}{4\pi}\dfrac{q_m}{(r-l)^2}$ pointing away from N (i.e. along the axis, away from the magnet); the south pole, at distance $(r+l)$, produces $B_S = \dfrac{\mu_0}{4\pi}\dfrac{q_m}{(r+l)^2}$ pointing toward S. Since $C$ is closer to N than to S, $B_N > B_S$ and the two partially cancel, leaving a net field along the axis pointing from S to N:

$B_{axial} = \frac{\mu_0}{4\pi}\,q_m\left[\frac{1}{(r-l)^2} - \frac{1}{(r+l)^2}\right] = \frac{\mu_0}{4\pi}\,\frac{2q_m(2l)\,r}{(r^2-l^2)^2} = \frac{\mu_0}{4\pi}\,\frac{2p_m\,r}{(r^2-l^2)^2}$ …