An electron moves with a constant speed v along a circle of radius r.(a) find the equivalent current through a point on its path.(b) Find the magnetic moment of the circulating electron.(c ) Find the ratio of the magnetic moment to the angular momentum of the electron.

Explanation

(a) Consider a point P on the path of the
electron. In one revolution of the electron, a charge e
crosses the point P. As the frequency of revolution is
`vl(2pir),` the charge crossing P in unit time, i.e., the
electric current is
`i= (ev)/(2pir).`
(b) The area A enclosed by this circular current is `pir^2`
so that the magnetic moment of the current is
`mu =iA= ((ev)/(2pir)) (pir^2) = (evr)/(2)`
in a direction pependicular to the loop.
(c) The angular momentum of the electron is l = mur.
Its direction is opposite to that of the magnetic moment. Thus,
` (mu)/(l) =(-evr)/(2mvr) = (-e)/(2m).`