Answer to Question #122978 in Electricity and Magnetism for abraham

Question

The orbital speed of electron orbiting around the nucleus in a circular orbit of radius 50Pm is 2.2x 106 m/s, the magnetic dipole moment of an electron is

Answer 1

Given

velocity of an electron is $v=2.2*10^6 m/s$

Radius of circular path is $r=50*10^{-12} m$

The expression for the magnetic dipole of an electron is

$p=\frac{evr}{2}$ $=\frac{(1.60*10^{-19})(2.2*10^6)(50*10^{-12})}{2}=8.80*10^{-24} A.m^2$