Magnetic Field of a Bar Magnet on Its Equatorial Line

Now consider a point $P$ on the bar magnet's equatorial line -- the perpendicular bisector of the line joining its S and N poles -- at distance $d$ from the centre $O$. By the geometry of a right triangle, $P$ is equidistant from BOTH poles, at distance $\sqrt{d^2+l^2}$ from each, so the two individual pole fields have equal magnitude:

Resolving the two fields. $\vec{B}_N$ points away from the N pole (along the line from $N$ to $P$) and $\vec{B}_S$ points toward the S pole (along the line from $P$ to $S$). Resolving each into a component parallel to the magnetic axis and a component along the equatorial direction: by symmetry, the two equatorial-direction components are equal and opposite and cancel exactly, while the two axial components both point in the SAME direction -- opposite to $\vec{m}$ -- and add. Each axial component has magnitude $B_N\cos\theta$, where $\theta$ is the angle each pole-to-$P$ line makes with the equatorial line, and $\cos\theta = l/\sqrt{d^2+l^2}$:

The short-magnet limit. For $d\gg l$, $(d^2+l^2)^{3/2}\approx d^3$, so

$\boxed{B_{\text{eq}} \approx \frac{\mu_0}{4\pi}\cdot\frac{m}{d^3}}$ …