Derivation of the gyromagnetic ratio of electron | Quantum Science Philippines

## Derivation of the gyromagnetic ratio of electron

By: Argielou A. Flores, MSU-IIT

Derive the gyromagnetic ratio, $\mu$, of electron $e^-$

Solution:

The magnitude of the magnetic moment of a current loop,

$\mu = IA$

where $I$ is the current, $A$ is the area enclosed

For an electron with frequency $f$ , in a circular orbit of radius $r$, it has a current

$I = -ef$

and area of,

$A = \pi r^2$

Substituting these, we have

$\mu = -ef\pi r^2$

Its velocity $v =2\pi rf$ and for its angular momentum $L$,

$\begin{array}{lll} L&= &mvr\\& = &2m\pi r^2f\end{array}$

$r^2 = \frac{1}{2\pi mf} L$

Therefore, the magnitude of magnetic moment,

$\begin{array}{lll} \mu& = &-ef \pi ( \frac{1}{2\pi mf} L )\\& = &\frac{-e}{2m}L\end{array}$

and since

$\mu = \alpha L$

From this we can finally say that our gyromagnetic ratio is,

$\alpha = \frac{-e}{2m}$