Solution

Using cylindrical coordinates, we find the additional magnetic field experienced by the electron as

$$\vec B=-\frac{1}{c^2}(\vec v\times\vec E)=-\frac{1}{c^2}\left[v\hat\theta\times\left(\frac{ke}{r^2}\right)\hat r\right]=\left(\frac{kev}{c^2r^2}\right)\hat z.$$

The orbital magnetic moment of the electron is

$$\vec\mu=-\left(\frac{e}{2m}\right)\vec L=-\left(\frac{evr}{2}\right)\hat z.$$

Consequently, the potential energy of the magnetic moment in the magnetic field is

$$U=-\vec\mu\cdot\vec B=\frac{ke^2v^2}{2c^2r}=\frac{k^2e^4}{2mc^2r^2},$$

where the last step is a result of considering electrical attraction between the nucleus and the electron as centripetal force, i.e.

$$\frac{mv^2}{r}=\frac{ke^2}{r^2}.$$

Thus, the force on the magnetic moment is

$$F=-\frac{\partial U}{\partial r}=\frac{k^2e^4}{mc^2r^3}=\frac{2\mu kev}{c^2r^3}.$$