Question

An electron of charge $e$ moves in a circular orbit of radius $r$ around a nucleus the magnetic field due to orbit motion of the electron at the site of the nucleus is $B$. The angular velocity $\omega$ of the electron is:

• A. $\omega =\frac{{\mu }_{0}eB}{4\pi r}$
• B. $\omega =\frac{{\mu }_{0}eB}{\pi r}$
• C. $\omega =\frac{4\pi rB}{{\mu }_{0}e}$
• D. $\omega =\frac{2\pi rB}{{\mu }_{0}e}$

Answer 1

The correct option is C $\omega =\frac{4\pi rB}{{\mu }_{0}e}$

$v=\omega r$

$I=\frac{\Delta Q}{\Delta t}=\frac{e}{(\frac{2\pi r}{4})}$

$I=\frac{ew}{2\pi }$

$B=(\frac{ew}{2\pi })\frac{{\mu }_o}{2r}$

$\omega =\frac{4\pi rB}{{\mu }_{o}e}$