Magnetic field arising due to electron's orbital motion around the nucleus My book writes -

The magnetic field in a reference frame fixed with the electron, arising from the orbital motion of the electron with velocity $\vec{v}$ in the electric field $\vec{E}$ (due to the nucleus) is given by
\begin{equation}
\vec{B}=\frac{1}{c^2}(\vec{E}\times\vec{v})
\end{equation}

My doubts are the following:
• Is this $\vec{E}$ the old Coulomb field, given by $\frac{1}{4\pi\epsilon_0}\frac{Ze}{r^2}$ ($Z$ being the atomic number of nucleus)?
• Why the factor $\frac{1}{c^2}$? Is there any relativistic situation involved?
• And lastly, how do we arrive at this equation?
 A: The answer to the first 2 questions is yes.
The "derivation" comes from the EM fields of a point charge moving at constant speed, see derivation here.
However, you can also get to the final expression "classically". Consider the proton orbiting the electron as a circular loop of current, which will generate a magnetic field according to the usual formula $$ B = \frac{\mu_0 I}{2r}.$$
The current $I$ is $dq/dt$ where $q$ is the nucleus' charge $Ze$ and $t$ is the period of oscillation $2\pi r / v$ ($v$ the velocity), which gives an effective magnetic field strength: $$ B = \frac{\mu_0 Z e v}{4\pi r^2} =  \frac{1}{\epsilon_0 c^2}\frac{Z e v}{4\pi r^2}, $$ where is the second step I used $\mu_0 \epsilon_0 = 1/c^2$.

Finally, if I remember correctly, there should be a overall factor of $1/2$ at the end shoved somewhere in the fine structure. This is due to Thomas precession, an actual and classically-irreproducible relativistic effect that arises from changing between rotating frames of reference. See derivation here.
