Find an expression for the magnetic dipole moment and magnetic field induction at the way of a Boh's hypothetical hydrogen atom in the $n^{th}$ orbit of the electron in terms of universal constants.

Magnetic moment: According to Bohr's hypothesis, angular momentum in n^th orbit is $L=n\left(\frac{h}{2\mathrm{\pi }}\right)$

Further, $\frac{M}{L}=\frac{q}{2m}$ or $\frac{e}{2m}$

$\therefore$Magnetic moment, $M=\left(\frac{e}{2m}\right)\left(L\right)=\left(\frac{e}{2m}\right)\left(n\frac{h}{2\mathrm{\pi }}\right)=\left(\frac{neh}{4\mathrm{\pi m}}\right)$

Magnetic field induction $\frac{m{v}^2}{r}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(e\right)\left(e\right)}{r^2}$ …..(i)

From bohr's hypothesis: $mvr=\frac{nh}{2\mathrm{\pi }}$ …..(ii)

Solving these two equations, we find:

$v=\frac{e^2}{2{\epsilon }_{0}nh}$ and $r=\frac{{\epsilon }_0{n}^2{h}^2}{{\mathrm{\pi me}}^2}$

Now, magnetic field induction at centre $B=\frac{{\mu }_{0}i}{2r}$

Here,

 $$\begin{gathered} i=qf=\left(e\right)\left(\frac{v}{2\mathrm{\pi }r}\right) \\ \therefore B=\left(\frac{{\mu }_0}{2r}\right)\left(\frac{ev}{2\mathrm{\pi r}}\right)=\frac{{\mu }_{0}e}{4\mathrm{\pi }}\left(\frac{v}{r^2}\right) \\ =\left(\frac{{\mu }_{0}e}{4\mathrm{\pi }}\right)\left(\frac{e^2}{2{\epsilon }_{0}nh}\right)\left(\frac{{\mathrm{\pi }}^2{\mathrm{m}}^2{\mathrm{e}}^4}{{\epsilon }_0^2{n}^4{h}^4}\right) \\ =\frac{{\mu }_0{\mathrm{\pi m}}^2{\mathrm{e}}^7}{8{\epsilon }_0^3{h}^5{n}^5} \end{gathered}$$