Magnetic field at the centre of the hydrogen atom due to motion of electron in \(\mathrm{n}^{\text {th }}\) orbit is proportional to

The radius of the \(\mathrm{n}^{\text {th }}\) Bohr orbit is, \(r_n \propto n^2\).
The angular velocity of the electron, \(\omega_n \propto \frac{1}{n^3}\)
Also, current \(I_n=\frac{q}{T_n}=\frac{q \omega_n}{2 \pi}\)
\( \begin{array}{ll} \therefore & \mathrm{I}_{\mathrm{n}} \propto \omega_{\mathrm{n}} \propto \frac{1}{\mathrm{n}^3} \\ & \text { Now, } \mathrm{B}_{\mathrm{n}}=\frac{\mu_0 \mathrm{I}_{\mathrm{n}}}{2 \mathrm{r}_{\mathrm{n}}} \\ \therefore & \mathrm{B}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}^3} \times \frac{1}{\mathrm{n}^2} \\ \therefore & \mathrm{B}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}^5} \end{array} \)