An electron of charge e moves in a circular orbit of radius r around a nucleus. The magnetic field due to orbital motion of the electron at the site of the nucleus is B. The angular velocity $\omega$ of the electron is

An electron moving in a circular orbit is equivalent to a current carrying loop. As explained above, the current is

$I = v e =\frac{e}{T}$

where T is the time period of the motion of the electron around the nucleus. If v is the speed of the electron,

$$\begin{gathered} T=\frac{2\pi }{\mathrm{\omega }} \\ \therefore I=\frac{e\omega }{2\pi } \end{gathered}$$

Now, the magnetic field at the centre of the loop is

$$\begin{gathered} B=\frac{{\mu }_{0}I}{2r}=\frac{{\mu }_{0}e\omega }{4\pi r} \\ \Rightarrow \omega =\frac{4\pi r B}{{\mu }_{0}e} \end{gathered}$$