Magnetic field in the hydrogen atom with Bohr theory

Question:
Use Bohr theory. Find the magnetic field by electron's exerting at the nucleus of the hydrogen atom. The radius of hydrogen is assumed to be 0.529 $\times$ 10$^{-10}$ m.

Answer:
We can assume that the electron moves around the nucleus as making a circular current. According to the Biot-Savart Law, we have the magnetic field of a circular current as
\[
B_z = \frac{\mu_0 I}{4 \pi}\int^{2\pi r}_{0}\frac{dl}{r^2}=\frac{\mu_0 I}{2r}
\]
Now, we need to obtain the current. The electric current is defined by $I=\frac{Q}{T}$. The charge of an electron or a proton is 1.61 $\times$ 10$^{-19}$ C. In order to get the current, we have to find the time period. That is
\[
T=\frac{2\pi r}{v}
\]
where $v$ is the tangential velocity of the electron. So the current is expressed by
\[
I = \frac{Q}{\frac{2\pi r}{v}}
\]
In terms of the classical theory, the centripetal force is equal to the Coulomb force. Then, solve for the velocity.
\begin{eqnarray}
\frac{mv^2}{r}&=&\frac{ke^2}{r^2} \\
v&=&\sqrt{\frac{ke^2}{mr}}  \\
  &=& \sqrt{\frac{8.99 \times 10^9 (1.61 \times 10^{-19})^2}{9.11 \times 10^{-31}\cdot 0.529 \times 10^{-10}}}  \\
  &=& 2.20 \times 10^6 \ \mathrm{m/s}
\end{eqnarray}
Therefore, we calculate the current.
\[
I=\frac{Q}{\frac{2\pi r}{v}}=\frac{1.61 \times 10^{-19}}{\frac{2\pi \times 0.529\times 10^{-10}}{2.20\times 10^6}}=1.07\times 10^{-3} \ \mathrm{A}
\]
Then, we can find the magnetic field.
\[
B=\frac{\mu_0 I}{2r}=\frac{4\pi \times 10^{-7} \times1.07\times 10^{-3}}{2 \times 0.529\times 10^{-10}}= 12.7 \ \mathrm{T}
\]

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