Derivation of the gyromagnetic ratio of electron | Quantum Science Philippines
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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}

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