According to Bohr's theory, the time averaged magnetic field at the centre (i.e., nucleus) of a hydrogen atom due to the motion of electrons in the $$n^{th}$$ orbit is proportional to: ($$n$$ = principal quantum number)

We begin with the standard result from magnetostatics for a circular current loop. The magnetic field at the centre of a loop of radius $$r$$ carrying a steady current $$I$$ is given by the Biot-Savart law in the form

$$B=\frac{\mu_0 I}{2r}.$$

Inside a hydrogen atom the single electron moves around the nucleus in a circular orbit, so the orbiting electron behaves like a tiny current loop. The current associated with the motion of one charge $$e$$ is obtained from the definition $$I=\dfrac{\text{charge}}{\text{time period}}$$:

$$I=\frac{e}{T}.$$

The time period $$T$$ is related to the speed $$v$$ and the circumference $$2\pi r$$ of the orbit by $$T=\dfrac{2\pi r}{v}$$. Substituting this into the current gives

$$I=\frac{e}{\dfrac{2\pi r}{v}}=\frac{e v}{2\pi r}.$$

Putting this expression for $$I$$ back into the magnetic-field formula we get

$$B=\frac{\mu_0}{2r}\left(\frac{e v}{2\pi r}\right)=\frac{\mu_0 e v}{4\pi r^{2}}.$$

So, for the electron in the $$n^{\text{th}}$$ Bohr orbit, the magnetic field depends on the electron’s speed $$v_n$$ and the radius $$r_n$$ through

$$B_n\propto\frac{v_n}{r_n^{2}}.$$

To express $$v_n$$ and $$r_n$$ in terms of the principal quantum number $$n$$ we now invoke the two fundamental relations of the Bohr model.

First, the quantisation of angular momentum: $$m_e v_n r_n = n\hbar.$$ Solving for the radius we have

$$r_n=\frac{n\hbar}{m_e v_n}. \quad -(1)$$

Second, the condition that the electrostatic force provides the centripetal force: $$\dfrac{m_e v_n^{2}}{r_n}=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{e^{2}}{r_n^{2}}.$$ Rearranging, this gives

$$v_n^{2}=\frac{1}{4\pi\varepsilon_0}\,\frac{e^{2}}{m_e r_n}. \quad -(2)$$

We eliminate $$r_n$$ by substituting the expression from (1) into (2). From (1) we have $$r_n=\dfrac{n\hbar}{m_e v_n}$$, so putting this into (2) yields

$$v_n^{2}=\frac{1}{4\pi\varepsilon_0}\,\frac{e^{2}}{m_e}\left(\frac{m_e v_n}{n\hbar}\right)=\frac{1}{4\pi\varepsilon_0}\,\frac{e^{2} v_n}{n\hbar}.$$

Cancel one $$v_n$$ factor from both sides, giving

$$v_n=\frac{1}{4\pi\varepsilon_0}\,\frac{e^{2}}{n\hbar}.$$

This result shows explicitly that

$$v_n\propto n^{-1}. \quad -(3)$$

Now we find the radius. Substituting (3) back into (1):

$$r_n=\frac{n\hbar}{m_e}\left(\frac{n\hbar\,4\pi\varepsilon_0}{e^{2}}\right)=\left(\frac{4\pi\varepsilon_0\hbar^{2}}{m_e e^{2}}\right)n^{2}.$$

All the bracketed quantities are constants, so

$$r_n\propto n^{2}. \quad -(4)$$

We now substitute the dependences (3) and (4) into the earlier proportionality $$B_n\propto\dfrac{v_n}{r_n^{2}}$$:

$$B_n\propto\frac{n^{-1}}{(n^{2})^{2}}=\frac{n^{-1}}{n^{4}}=n^{-5}.$$

Thus the time-averaged magnetic field at the nucleus is proportional to $$n^{-5}$$.

Hence, the correct answer is Option D.