Assuming the atom in the ground state, the expression for the magnetic field at a point (nucleus) in hydrogen atom due to circular motion of electron is
\(\left[\mu_{0}=\right.\) permeability of free space, \(\epsilon_{0}=\) permittivity of free space, \(\quad \mathrm{m}=\) mass of
electron, \(\mathrm{e}=\) electronic charge, \(\mathrm{h}=\) Planck's constant \(]\)

Magnetic field at the centre of a circular coil is given by
\(\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \mathrm{r}}\) ....(1)
In the case of electron revolving around the nucleus, the current I can be written as
\(I\) =ef where \(f\) is the frequency of revolution.
\(\mathrm{f}=\frac{\mathrm{V}}{2 \pi \mathrm{r}} \quad \therefore \mathrm{I}=\frac{\mathrm{eV}}{2 \pi \mathrm{r}}\)
But for electron in ground state \(V=\frac{e^{2}}{2 \epsilon_{0} h}\)
\(\therefore \mathrm{I}=\frac{\mathrm{e}^{3}}{4 \pi \epsilon_{\mathrm{o}} \mathrm{hr}}\)
Putting this value in \(\mathrm{Eq}(1)\) :
\(\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{e}^{3}}{8 \pi \epsilon_{\mathrm{o}} \mathrm{hr}^{2}}\) ...(2)
But \(r=\frac{h^{2} \epsilon_{0}}{\pi m e^{2}}\)
Putting this value of \(\mathrm{r}\) in Eq. (2)
we get : \(B=\frac{\mu_{0} e^{7} \pi m^{2}}{8 \epsilon_{0}^{3} h^{5}}\)