Answer to Question #123012 in Classical Mechanics for taylor

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

As per the given question,

$e=1.6\times 10^{-19}Coulomb$

The radius of the circular orbit (r)=50pm

Speed of electron $(v)=2.2 \times 10^6 m/sec$

magnetic dipole moment of an electron =?

Magnetic dipole moment $(m)=iA=\frac{e}{T}\pi r^2$

$=\frac{e}{\frac{2\pi r}{v}}\pi r^2 =\frac{erv}{2}$

$=\frac{1.6\times 10^{-19}\times50\times 10^{-12}\times 2.2 \times 10^6}{2} \times \frac{Coulomb\times m\times m}{sec}$

$=8.8\times 10^{-24}\frac{Coulomb}{sec}\times m^2$

$=8.8\times 10^{-24}Am^2$ (because we know that $Ampere=\frac{Coulomb}{sec}$ )