# Pre-Quantum Electrodynamics
Author: Jean-Sébastien Caux

##### 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 \ref{Gr(5.83)} (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:

\[ {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' ~\frac{{\bf M} ({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \label{Gr(6.11)} \]

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 \({\boldsymbol \nabla}' \frac{1}{|{\bf r} - {\bf r}'|} = \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3}\), \[ {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' {\bf M} ({\bf r}') \times \left( {\boldsymbol \nabla}' \frac{1}{|{\bf r} - {\bf r}'|} \right), \] we can further integrate by parts and use product rule: \({\boldsymbol \nabla} \times (f {\bf A}) = f ( {\boldsymbol \nabla} \times {\bf A}) - {\bf A} \times ({\boldsymbol \nabla} f)\), we get \[ {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \left\{ ∫_{\cal V} dτ' \frac{{\boldsymbol ∇}' × {\bf M} ({\bf r}')}{|{\bf r} - {\bf r}'|}

- ∫_{\cal V} dτ' {\boldsymbol ∇}' × \left( \frac{{\bf M} ({\bf r}')}{|{\bf r} - {\bf r}'|} \right)

\right\} \] Problem 1.61 b) (p.56): leads to \[ {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} ∫_{\cal V} dτ' \frac{{\boldsymbol ∇}' × {\bf M} ({\bf r}')}{|{\bf r} - {\bf r}'|}

- \frac{\mu_0}{4\pi} \oint_{\cal S} \frac{{\bf M} ({\bf r}') × d{\bf a}'}{|{\bf r} - {\bf r}'|}

\label{Gr(6.12)} \] Reinterpretation: first term: potential from volume current,

\[ {\bf J}_b = {\boldsymbol \nabla} \times {\bf M} \label{Gr(6.13)} \]

second term: potential from surface current,

\[ {\bf K}_b = {\bf M} \times \hat{\bf n} \label{Gr(6.14)} \]

With these definitions,

\[ {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} ∫_{\cal V} dτ' \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)} \]

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

\paragraph{Example 6.1:} find field of uniformly magnetized sphere. \paragraph{Solution:} put z axis along \({\bf M}\). \[ {\bf J}_b = {\boldsymbol \nabla} \times {\bf M} = 0, \hspace{1cm} {\bf K}_b = {\bf M} \times \hat{\bf n} = M \sin \theta \hat{\boldsymbol \phi}. \] Rotating spherical shell of uniform surface charge \(\sigma\): surface current density \[ {\bf K} = \sigma {\bf v} = \sigma \omega R \sin \theta \hat{\boldsymbol \phi}. \] Same if \(\sigma R {\boldsymbol \omega} = {\bf M}\). Refer to Example 5.11,% ({\it not done in class !}), \[ {\bf B} = \frac{2}{3} \mu_0 {\bf M} \label{Gr(6.16)} \] inside sphere, whereas outside: pure dipole field, with \[ {\bf m} = \frac{4}{3} \pi R^3 {\bf M}. \]

Author: Jean-Sébastien Caux

Created: 2022-02-08 Tue 06:55

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