Q.Obtain the magnetic field at a point on the equatorial line of a bar magnet.

Step 1. Place a unit north pole at $C$ on the equatorial line (perpendicular bisector), distance $r$ from the centre. Both poles are equidistant, $r'=\sqrt{r^2+l^2}$, so $B_N=B_S=\dfrac{\mu_0}{4\pi}\dfrac{q_m}{r^2+l^2}$.

Step 2. Resolving each into components parallel and perpendicular to the axis: the perpendicular components cancel by symmetry, and the parallel components (both opposite to $\vec p_m$) add.

Step 3. Using $\cos\alpha=l/r'$ for the angle each makes with the axis: $B_{equatorial}=\dfrac{\mu_0}{4\pi}\dfrac{p_m}{(r^2+l^2)^{3/2}}$. …

Superpose the two equal pole fields, cancel perpendicular components by symmetry, then …