# How can the magnetic field of a dipole be exceedingly large close to its singularity?

I am trying to figure out the magnetic field close to a micrometer sized bar magnet in my experiment. The field gets detected 1 micrometer away from the north of the bar magnet. For now, I just take the approximation that the bar magnet is a dipole with a field:

$$$$\vec{B}(\vec{r}) = \frac{\mu_0}{4\pi}\left(\frac{3\vec{r}(\vec{m}\cdot\vec{r})}{r^5}-\frac{\vec{m}}{r^3} \right).$$$$

Where $$\mu_0/4\pi = 10^{-7}$$, $$m$$ is in the order of $$10^{-6}$$ A/m and $$\vec{m}$$ is in the same direction as $$\vec{r}$$. To put this into perspective, the field will approximately be on the order of:

$$$$B(r) \approx 10^{-7}\left(2m/r^3\right) \approx 10^{-7}*10^{-6}/10^{-18} = 10^5 \text{ T},$$$$

which is just unbelievably high. Does anyone know where it goes wrong here?

• Your magnet is not an ideal dipole. May 27, 2018 at 15:42
• That formula doesn't work when you're too close to the magnet, but if you want to use it to get a rough estimate, you should take $r$ to be the distance from the center of the bar magnet, not its surface. Dec 20, 2022 at 6:25

The parameter $\vec r$ in the formula means the distance to the location of the bar magnet, not the distance to its surface. The formula is an approximation for distances much larger than the dimensions of the bar magnet.
There is however a theoretical case for which you could apply the dipole formula to the surface: it is that of a uniformly magnetized sphere. But $$r$$ is indeed the distance to the center of the sphere.
Finally, there may be a problem in your definition. $$m$$ is the total dipole moment in $$A.m^2$$, different from the magnetization $$M$$ in A/m. For an uniform magnetization, $$m = MV$$ with $$V$$ the volume of the magnet.
In order of magnitude, if we stand on the surface of a spherical magnet, we will have $$B\approx\frac{\mu_0}{4\pi}\frac{Mr^3}{r^3}\approx1T$$ independent of the radius of the sphere.