Question:

Use Bohr theory. Find the magnetic field due to electron’s revolution about 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.

\[  \frac{mv^2}{r} = \frac{ke^2}{r^2} \]

\[  v = \sqrt{\frac{ke^2}{mr}}  \]

\[  = \sqrt{\frac{8.99 \times 10^9 (1.61 \times 10^{-19})^2}{9.11 \times 10^{-31}\cdot 0.529 \times 10^{-10}}}  \]

\[  =  2.20 \times 10^6 \ \mathrm{m/s}  \]

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.

\[  B=\frac{\mu_0 I}{2r}=\frac{4\pi \times 10^{-7} \times1.07\times 10^{-3}}{2 \times 0.529\times 10^{-10}}= 12.7 \ \mathrm{T}  \]