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The magnetic moment for a point particle is shown to be related  to the angular momentum \(\ell\) and is given by \begin{equation} \vec{m} = \frac{q}{2M}\vec{\ell} \end{equation}


It will be shown that for point charges it is easy to see that the magnetic moment is related to the angular momentum.
Recall that the magnetic moment for a current distribution is  given by
\begin{equation} \vec m = \frac{1}{2} \int \vec r \times \vec j \, d^3 r \end{equation} Note that for a point particle the current density is \begin{equation} \vec{j}(\vec{r})=q_k \vec{v}_k\delta(\vec{r}-\vec{r}_k). \end{equation}
Here the following notation  has been used for \(k^\text{th}\) charged particle.\\ \(q_k=\)charge; \(M_k=\)mass; \(\vec{v}_k=\)velocity; \(\vec{p}_k =\) momentum; \(\vec{\ell}_k\) angular momentum. Therefore, the magnetic moment for system of point particles is given by \begin{equation} \vec{m} = \sum_k \frac{q_k}{2M_k}\big\{ \vec{r}_k\times \vec{p}_k\big\} = \sum_k \frac{q_k}{2M_k}\vec{\ell}_k. \end{equation}

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