Q.A magnetic pole of a bar magnet with pole strength of 100 A m is 20 cm away from the centre of a bar magnet. The bar magnet has pole strength of 200 A m and has a length 5 cm. If the magnetic pole is on the axis of the bar magnet, find the force on the magnetic pole.

For a point far from a bar magnet ($r\gg l$), the field can be found by analogy with the electric dipole. On the AXIS, at distance $r$ from the centre, $B_{axis}=\dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}$, directed along $\vec m$; on the perpendicular bisector (the EQUATOR), $B_{eq}=\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}$, directed opposite to $\vec m$. At the same distance $r$ from the centre, the axial field is always exactly twice the equatorial field, $B_{axis}=2B_{eq}$. Nearer the magnet, the exact axial expression is $B=\dfrac{\mu_0}{4\pi}\dfrac{2mr}{(r^2-l^2)^2}$, which reduces to the simple $1/r^3$ dipole law only once $r\gg l$.

At a general point $P$, neither on the axis nor the equator, the moment $\vec m$ is resolved (about the centre) into a component $m\cos\theta$ along the position vector $\vec r$ (treated as an "axial" contribution) and a component $m\sin\theta$ perpendicular to $\vec r$ (treated as an "equatorial" contribution), where $\theta$ is the angle between $\vec r$ and $\vec m$. These give $B_a=\dfrac{\mu_0}{4\pi}\dfrac{2m\cos\theta}{r^3}$ (along $\vec r$) and $B_{eq}=\dfrac{\mu_0}{4\pi}\dfrac{m\sin\theta}{r^3}$ (perpendicular to $\vec r$); since these two contributions are mutually perpendicular, the resultant magnitude is $B=\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}\sqrt{1+3\cos^2\theta}$, making an angle $\alpha$ with $\vec r$ given by $\tan\alpha=\tfrac12\tan\theta$. This general formula correctly reduces to the pure axial result at $\theta=0$ and the pure equatorial result at $\theta=90^\circ$.

All of this comes from a single "electrostatic analogue" trick: every magnetic formula here has a matching electric-dipole formula, with the pole strength $q_m$ playing the role of charge $q$ and the constant $\mu_0/4\pi$ playing the role of $1/4\pi\varepsilon_0$; because the underlying $1/r^2$ (Coulomb-like force between poles) and $1/r^3$ (resulting dipole field) mathematics is identical in both cases, every electric-dipole result carries straight over to magnetism.