Magnetic Field due to a Bar Magnet at an Arbitrary Point

The axial and equatorial formulas of the previous section only apply to points lying exactly on the magnet's axis or exactly on its equatorial plane. For a general point $P$ at distance $r$ from the centre $O$ of a bar magnet of moment $\vec m$, where the line $OP$ makes some arbitrary angle $\theta$ with the axis, a more general result is needed.

The trick is to resolve the magnetic moment $\vec m$ (about the centre $O$) into two components relative to the direction of $\vec r=OP$: a component $m\cos\theta$ ALONG $\vec r$, and a component $m\sin\theta$ PERPENDICULAR to $\vec r$. For the component $m\cos\theta$ along $\vec r$, the point $P$ is effectively an axial point of this smaller "sub-dipole," so it contributes a field $B_a=\dfrac{\mu_0}{4\pi}\dfrac{2m\cos\theta}{r^3}$ (Eq. 12.6), directed along $\vec r$ (i.e. along the direction of the $m\cos\theta$ component). For the component $m\sin\theta$ perpendicular to $\vec r$, the point $P$ is effectively an equatorial point of THAT sub-dipole, at the same distance $r$, so it contributes a field $B_{eq}=\dfrac{\mu_0}{4\pi}\dfrac{m\sin\theta}{r^3}$ (Eq. 12.7), directed opposite to the $m\sin\theta$ component (i.e. perpendicular to $\vec r$).

Because these two contributions, $B_a$ (along $\vec r$) and $B_{eq}$ (perpendicular to $\vec r$), are mutually perpendicular, the magnitude of the total resultant field at $P$ is found by Pythagoras: $B=\sqrt{B_a^2+B_{eq}^2}=\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}\sqrt{1+3\cos^2\theta}$ (Eq. 12.8). If $\alpha$ is the angle the resultant field $\vec B$ makes with $\vec r$, then from the two perpendicular components, $\tan\alpha=\dfrac{B_{eq}}{B_a}=\dfrac12\tan\theta$ (Eq. 12.9); working through the geometry further, the angle between the resultant field $\vec B$ and the magnet's own moment $\vec m$ comes out to be $\theta+\alpha$. …