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Physics LibreTexts

5.5: Magnetic Moments

Consider a particle of electric charge $ q$ and speed $ v$ performing a circular orbit of radius $ r$ in the $ x$ -$ y$ plane. The charge is equivalent to a current loop of radius $ r$ in the $ x$ -$ y$ plane carrying current $ I=q\,v/2\pi\, r$ . The magnetic moment $ \mu$ of the loop is of magnitude $ \pi\, r^2\, I$ and is directed along the $ z$ -axis. Thus, we can write


$\displaystyle \mbox{\boldmath$\mu$}$$\displaystyle = \frac{q}{2}\, {\bf x} \times {\bf v},$ (458)



where $ {\bf x}$ and $ {\bf v}$ are the vector position and velocity of the particle, respectively. However, we know that $ {\bf p} = {\bf v} /m$ , where $ {\bf p}$ is the vector momentum of the particle, and $ m$ is its mass. We also know that $ {\bf L} = {\bf x}\times{\bf p}$ , where $ {\bf L}$ is the orbital angular momentum. It follows that


$\displaystyle \mbox{\boldmath$\mu$}$$\displaystyle = \frac{q}{2\,m} \,{\bf L}.$ (459)



Using the usual analogy between classical and quantum mechanics, we expect the above relation to also hold between the quantum mechanical operators, $ \mu$ and $ {\bf L}$ , which represent magnetic moment and orbital angular momentum, respectively. This is indeed found to the the case.

Spin angular momentum also gives rise to a contribution to the magnetic moment of a charged particle. In fact, relativistic quantum mechanics predicts that a charged particle possessing spin must also possess a corresponding magnetic moment (this was first demonstrated by Dirac--see Chapter 11). We can write


$\displaystyle \mbox{\boldmath$\mu$}$$\displaystyle = \frac{q}{2\,m} \left({\bf L} + g \,{\bf S}\right),$ (460)

where \(g\) is called the gyromagnetic ratio. For an electron this ratio is found to be

$\displaystyle g_e = 2\left( 1 + \frac{1}{2\pi} \frac{e^2}{4\pi \,\epsilon_0\,\hbar \,c} \right).$ (461)


The factor 2 is correctly predicted by Dirac's relativistic theory of the electron (see Chapter 11). The small correction $ 1/(2\pi\, 137)$ , derived originally by Schwinger, is due to quantum field effects. We shall ignore this correction in the following, so


$\displaystyle \mbox{\boldmath$\mu$}$$\displaystyle \simeq - \frac{e}{2\,m_e} \left({\bf L} + 2 \,{\bf S}\right)$ (462)


for an electron (here, \(e> 0\)).