Radius of the circular path of a revolving electron (cahrge=-e) around the nucleous is r. due to this revolution, magnetic field generated at the nucleus is B. What is the angular velocity of the electron ?

Explanation

If the velocity of the electron be `omega` its time period of revolution.
`T=(2pi)/(omega)`
Through any point on the orbit, the amount of charge flowing per second= effective current (I).
So, `I=(e)/(T)=(e omega)/(2pi)`
Magnetic field at the centre of the circular path.
` B=(mu_(0)I)/(2r)=(mu_(0))/(2r)*(e omega)/(2pi)=(mu_(0) e omega)/(4pi r)`
So, `omega=( 4pi r B)/(e mu_(0))`