An electron of mass m in the ground state of hydrogen atoms is revolving in a circular orbit of radius R. The atom is place in a uniform magnetic induction B, such that the magnetic moment $\mu$ of the atom makes 5 times the angle with B as that the angular momentum L of the electron with the external magnetic induction B. Torque experienced by the orbiting electron is

If angle of L with B is assumed  $\theta$, angle between  $\mu$ and B will be $5\text{\hspace{0.17em}θ}$
As L and $\mu$ are opposite, therefore 
$\text{θ}+5\text{\hspace{0.17em}θ}={180}^0$ 
$\Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}θ}={30}^0\mathrm{.....}\left(i\right)$ 
Gyromagnetic ratio of a current carrying loop or atom is given by $\frac{\mu }{L}=\frac{q}{2m}$
$\Rightarrow \mu =\frac{q}{2m}\times L$

So, for electron in first orbit of Bohr’s atom
 $\mu =\frac{e}{2m}\times \frac{h}{2\pi }=\frac{eh}{4\pi m}\mathrm{.......}\left(ii\right)$
As, magnetic torque is given by  $\tau =\mu B\sin \theta =\frac{eh}{4\pi m}B\sin {30}^0$
[Using Eqs. (i) and (ii)]
$=\frac{ehB}{8\pi m}$