The fine structure of an atomic spectrum results from the magnetic field "seen" by an orbiting electron. In this question you will make a semiclassical estimate of the $B$ field seen by a $2 p$ electron in hydrogen. The $B$ field at the center of a circular current loop, $i$, of radius $r$ is known to be $B=\mu_{0} i / 2 r$.
(a) Treating the electron and proton as classical particles in circular orbits (each as seen by the other), show that the $B$ field seen by the electron is
$$
B=\frac{\mu_{0}}{4 \pi} \frac{e L}{m_{\mathrm{e}} r^{3}}
$$
where $L$ is the electron's orbital angular momentum (L $=m_{\mathrm{e}} v r$ for a circular orbit). Remember that the current produced by the orbiting proton is $i=e v / 2 \pi r$, where $v$ is the speed of the proton as seen by the electron (or vice versa). (b) For a rough estimate, you can give $L$ and $r$ their values for the $n=2$ orbit of the Bohr model, $L=2 \hbar$ and $r=4 a_{\mathrm{B}}$. Show that this gives $B \approx 0.39 \mathrm{~T}$ and hence that the separation, $2 \mu_{\mathrm{B}} B$, of the two $2 p$ levels is about $4.5 \times 10^{-5} \mathrm{eV}$.
It should be clear that this semi-classical calculation is only a rough estimate. You have used the Bohr values for $L$ and $r .$ If, for example, you had used the quantum value $L=\sqrt{2} \hbar$, this would have changed your answer by a factor of $\sqrt{2} .$ There is another very important reason that the argument used here is only roughly correct: The electron's rest frame is noninertial (since it is accelerated) and a careful analysis by the British physicist L. H. Thomas showed that the energy separation calculated here should include an additional factor of $\frac{1}{2}$. That our answer, $4.5 \times 10^{-5} \mathrm{eV}$, is correct to two significant figures is just a lucky accident.