Question

An electron in the ground state of hydrogen atom in revolving in anticlockwise direction in a circular orbit of radius R (see fig).
The atom is placed in a uniform magnetic induction $\overrightarrow{B}$ such that the normal to the plane of electron's orbit makes an angle of ${30}^o$ with the magnetic induction. If the torque experienced by the orbiting electron is $\tau =\frac{ehB}{X\pi m}$. Find X?

Answer 1

$\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}$
$|\overrightarrow{\tau }|=MBsin\theta$
where, $\theta$ is the angle between $\overrightarrow{M}$ and $\overrightarrow{B}\theta ={30}^o$
$\therefore \tau =\left(\frac{eh}{4\pi m}\right)sin{30}^o=\frac{ehB}{8\pi m}$
The direction of $\overrightarrow{\tau }$ is perpendicular to both $\overrightarrow{M}and\overrightarrow{B}$