# Maximum $B$-field at the surface of a permanent magnet

I have the dipole formula for a permanent spherical magnet in coordinate-free form.

$$\vec{B}=\frac{\mu_{0}}{4\pi}\frac{3(\vec{m}\cdot\hat{r})\hat{r}-\vec{m}}{r^{3}}$$

I am trying to find the maximum $$B$$-field at the surface of the magnet, so I know I have to "maximize" the dot product which turns out to be just $$mr*cos(\theta)$$. Is $$\theta$$ just 0 degrees in this case since the magnetic field would be strongest at the north or south poles? Then the magnetic field would simply be $$\vec{B}=\frac{\mu_{0}}{4\pi}\frac{3mr\hat{r}-\vec{m}}{r^{3}}$$, and if I have values for the radius of the magnet, can I put that in for r and the magnetization as m? Unit vector $$\hat{r}$$ should not matter since it can be taken in any direction, although, for north or south pole it would be best to have that be $$B_{z}$$?

Any help would be appreciated, I'm still new to magnetism and just feeling it out for now. I was told by a friend that the max. field strength would be at 45 degrees but this made no sense to me.

• You can't use the magnetic moment field near the magnet. Commented Apr 29, 2022 at 13:58

If you stay at the surface of the magnet, your $$r=R$$ the radius of the magnet. If you want to maximize $$B$$, you need to be careful, because you have a second term that isn’t aligned so you could have potential compensations.
At position $$\hat r$$ at azimutal angle $$\theta$$ from $$m$$, you have (projecting on $$\hat r$$ and its orthogonal direction) $$B^2= B_0^2 (4\cos^2\theta+\sin^2 \theta)=B_0^2 (3\cos^2\theta+1)$$ with $$B_0=\frac{\mu_0m}{4\pi R^3}$$. (Notice that the relevant quantity is not the magnetic moment or the size but the magnetization).
Luckily for you, the conclusion is still the same with the optimum angle at $$\theta=0,\pi$$ and maximum value $$B=2B_0$$.