##### Bound Currents
emsm.msm.fmo.bc

Suppose we have a piece of material with known magnetization \({\bf M}\). What is the field produced by this object? For a single dipole: refer to A_di (vector potential of single dipole): \[ {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \frac{{\bf m} \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \label{Gr(6.10)} \] For a chunk of material with local magnetization \({\bf M} ({\bf r})\), by the principle of superposition we thus have:

- Gr (6.11)

\[ {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' ~\frac{{\bf M} ({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \tag{A_M}\label{A_M} \]

In principle, this is all that is needed. As in electric case however, a more illuminating version of this equation can be given by using some simple identities: using \({\boldsymbol \nabla}' \frac{1}{|{\bf r} - {\bf r}'|} = \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3}\),

\begin{equation*} {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' {\bf M} ({\bf r}') \times \left( {\boldsymbol \nabla}' \frac{1}{|{\bf r} - {\bf r}'|} \right), \end{equation*}

we can further integrate by parts and use product rule curl_prod \({\bf A} \times ({\boldsymbol \nabla} f) = f ( {\boldsymbol \nabla} \times {\bf A}) - {\boldsymbol \nabla} \times (f {\bf A})\), we get

\begin{equation*} {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \left\{ \int_{\cal V} d\tau' \frac{{\boldsymbol \nabla}' \times {\bf M} ({\bf r}')}{|{\bf r} - {\bf r}'|} - \int_{\cal V} d\tau' {\boldsymbol \nabla}' \times \left( \frac{{\bf M} ({\bf r}')}{|{\bf r} - {\bf r}'|} \right) \right\} \end{equation*}

Using the identity

\begin{equation*} \int_{\cal V} d\tau \nabla \times {\bf v} = \oint_{\cal S} d{\bf a} \times {\bf v} \end{equation*}

(this can be derived from the divergence theorem, with \({\bf v} \times {\bf c}\) as argument) leads to

\begin{equation*} {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' \frac{{\boldsymbol \nabla}' \times {\bf M} ({\bf r}')}{|{\bf r} - {\bf r}'|} + \frac{\mu_0}{4\pi} \oint_{\cal S} \frac{{\bf M} ({\bf r}') \times d{\bf a}'}{|{\bf r} - {\bf r}'|} \label{Gr(6.12)} \end{equation*}

Reinterpretation: first term: potential from volume current,

- Gr (6.13)

\[ {\bf J}_b = {\boldsymbol \nabla} \times {\bf M} \tag{JbcurlM}\label{JbcurlM} \]

second term: potential from surface current,

- Gr (6.14)

\[ {\bf K}_b = {\bf M} \times \hat{\bf n} \tag{KbM}\label{KbM} \]

With these definitions,

\begin{equation} {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' \frac{{\bf J}_b ({\bf r}')}{|{\bf r} - {\bf r}'|} + \frac{\mu_0}{4\pi} \oint_{\cal S} da' \frac{{\bf K}_b ({\bf r}')}{|{\bf r} - {\bf r}'|} \label{Gr(6.15)} \end{equation}

so the field produced by the material is the same as that produced by bound currents in the volume and surface of the material.

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42