Update 2022-03-01 08:15
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Jean-Sébastien
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<!DOCTYPE html>
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<html lang="en">
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<head>
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<!-- 2022-02-21 Mon 20:41 -->
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<!-- 2022-03-01 Tue 08:14 -->
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<meta charset="utf-8">
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<meta name="viewport" content="width=device-width, initial-scale=1">
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<title>Pre-Quantum Electrodynamics</title>
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@@ -1624,7 +1624,7 @@ Table of contents
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<p>
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Suppose we have a piece of material with known magnetization \({\bf M}\).
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What is the field produced by this object?
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For a single dipole: refer to \ref{Gr(5.83)} (vector potential of single dipole):
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For a single dipole: refer to <a href="./ems_ms_vp_me.html#A_di">A_di</a> (vector potential of single dipole):
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\[
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{\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \frac{{\bf m} \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3}
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\label{Gr(6.10)}
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@@ -1632,118 +1632,136 @@ For a single dipole: refer to \ref{Gr(5.83)} (vector potential of single dipole
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For a chunk of material with local magnetization \({\bf M} ({\bf r})\),
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by the principle of superposition we thus have:
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</p>
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<div class="main div" id="org293c8c1">
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<div class="main div" id="orga8df7e7">
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<div class="eqlabel" id="org8873f67">
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<p>
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<a id="A_M"></a><a href="./emsm_msm_fmo_bc.html#A_M"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
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<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
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<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
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</svg></a>
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</p>
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<div class="alteqlabels" id="org2aafb5d">
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<ul class="org-ul">
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<li>Gr (6.11)</li>
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</ul>
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</div>
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</div>
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<p>
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\[
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{\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' ~\frac{{\bf M} ({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3}
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\label{Gr(6.11)}
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\]
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{\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}
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\tag{A_M}\label{A_M}
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\]
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</p>
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</div>
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<p>
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In principle, this is all that is needed.
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As in electric case however, a more illuminating version of this equation can be given
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by using some simple identities:
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As in electric case however, a more illuminating version of this equation can be given by using some simple identities:
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using \({\boldsymbol \nabla}' \frac{1}{|{\bf r} - {\bf r}'|} = \frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3}\),
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\[
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</p>
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\begin{equation*}
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{\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' {\bf M} ({\bf r}') \times \left( {\boldsymbol \nabla}'
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\frac{1}{|{\bf r} - {\bf r}'|} \right),
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\]
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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
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\[
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\end{equation*}
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<p>
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we can further integrate by parts and use product rule <a href="./c_m_dc_pr.html#curl_prod">curl_prod</a>
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\({\bf A} \times ({\boldsymbol \nabla} f)
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= f ( {\boldsymbol \nabla} \times {\bf A}) - {\boldsymbol \nabla} \times (f {\bf A})\),
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we get
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</p>
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\begin{equation*}
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{\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \left\{
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∫_{\cal V} dτ' \frac{{\boldsymbol ∇}' × {\bf M} ({\bf r}')}{|{\bf r} - {\bf r}'|}
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</p>
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<ul class="org-ul">
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<li>∫_{\cal V} dτ' {\boldsymbol ∇}' × \left( \frac{{\bf M} ({\bf r}')}{|{\bf r} - {\bf r}'|} \right)</li>
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</ul>
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<p>
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\int_{\cal V} d\tau' \frac{{\boldsymbol \nabla}' \times {\bf M} ({\bf r}')}{|{\bf r} - {\bf r}'|}
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- \int_{\cal V} d\tau' {\boldsymbol \nabla}' \times \left( \frac{{\bf M} ({\bf r}')}{|{\bf r} - {\bf r}'|} \right)
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\right\}
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\]
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Problem 1.61 b) (p.56): leads to
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\[
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{\bf A} ({\bf r}) = \frac{\mu_0}{4\pi}
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∫_{\cal V} dτ' \frac{{\boldsymbol ∇}' × {\bf M} ({\bf r}')}{|{\bf r} - {\bf r}'|}
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</p>
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<ul class="org-ul">
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<li>\frac{\mu_0}{4\pi} \oint_{\cal S} \frac{{\bf M} ({\bf r}') × d{\bf a}'}{|{\bf r} - {\bf r}'|}</li>
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</ul>
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\end{equation*}
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<p>
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Using the identity
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</p>
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\begin{equation*}
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\int_{\cal V} d\tau \nabla \times {\bf v} = \oint_{\cal S} d{\bf a} \times {\bf v}
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\end{equation*}
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<p>
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(this can be derived from the divergence theorem, with \({\bf v} \times {\bf c}\) as argument)
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leads to
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</p>
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\begin{equation*}
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{\bf A} ({\bf r}) = \frac{\mu_0}{4\pi}
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\int_{\cal V} d\tau' \frac{{\boldsymbol \nabla}' \times {\bf M} ({\bf r}')}{|{\bf r} - {\bf r}'|}
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+ \frac{\mu_0}{4\pi} \oint_{\cal S} \frac{{\bf M} ({\bf r}') \times d{\bf a}'}{|{\bf r} - {\bf r}'|}
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\label{Gr(6.12)}
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\]
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\end{equation*}
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<p>
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Reinterpretation: first term: potential from volume current,
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</p>
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<div class="main div" id="org5be5663">
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<div class="main div" id="org2d1d26c">
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<div class="eqlabel" id="org021e10f">
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<p>
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<a id="JbgradM"></a><a href="./emsm_msm_fmo_bc.html#JbgradM"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
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<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
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<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
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</svg></a>
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</p>
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<div class="alteqlabels" id="org6840413">
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<ul class="org-ul">
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<li>Gr (6.13)</li>
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</ul>
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</div>
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</div>
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<p>
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\[
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{\bf J}_b = {\boldsymbol \nabla} \times {\bf M}
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\label{Gr(6.13)}
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\]
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{\bf J}_b = {\boldsymbol \nabla} \times {\bf M}
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\tag{JbgradM}\label{JbgradM}
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\]
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</p>
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</div>
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<p>
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second term: potential from surface current,
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</p>
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<div class="main div" id="orgccad168">
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<div class="main div" id="org2ed4e80">
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<div class="eqlabel" id="org25ca796">
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<p>
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<a id="KbM"></a><a href="./emsm_msm_fmo_bc.html#KbM"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
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<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
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<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
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</svg></a>
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</p>
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<div class="alteqlabels" id="orgdc7bf79">
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<ul class="org-ul">
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<li>Gr (6.14)</li>
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</ul>
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</div>
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</div>
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<p>
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\[
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{\bf K}_b = {\bf M} \times \hat{\bf n}
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\label{Gr(6.14)}
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\]
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{\bf K}_b = {\bf M} \times \hat{\bf n}
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\tag{KbM}\label{KbM}
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\]
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</p>
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</div>
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<p>
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With these definitions,
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</p>
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<div class="main div" id="org4379c6a">
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<p>
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\[
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{\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} ∫_{\cal V} dτ' \frac{{\bf J}_b ({\bf r}')}{|{\bf r} - {\bf r}'|}
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</p>
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<ul class="org-ul">
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<li>\frac{\mu_0}{4\pi} \oint_{\cal S} da' \frac{{\bf K}_b ({\bf r}')}{|{\bf r} - {\bf r}'|}</li>
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</ul>
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<p>
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\label{Gr(6.15)}
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\]
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</p>
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<div class="main div" id="orgf0db421">
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\begin{equation}
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{\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' \frac{{\bf J}_b ({\bf r}')}{|{\bf r} - {\bf r}'|}
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+ \frac{\mu_0}{4\pi} \oint_{\cal S} da' \frac{{\bf K}_b ({\bf r}')}{|{\bf r} - {\bf r}'|}
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\label{Gr(6.15)}
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\end{equation}
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</div>
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<p>
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so the field produced by the material is the same as that produced by {\bf bound currents}
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in the volume and surface of the material.
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so the field produced by the material is the same as that produced by
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<b>bound currents</b> in the volume and surface of the material.
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</p>
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<div class="example div" id="orge4aa513">
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<p>
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\paragraph{Example 6.1:} find field of uniformly magnetized sphere.
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\paragraph{Solution:} put z axis along \({\bf M}\).
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\[
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{\bf J}_b = {\boldsymbol \nabla} \times {\bf M} = 0,
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\hspace{1cm}
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{\bf K}_b = {\bf M} \times \hat{\bf n} = M \sin \theta \hat{\boldsymbol \varphi}.
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\]
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Rotating spherical shell of uniform surface charge \(\sigma\): surface current density
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\[
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{\bf K} = \sigma {\bf v} = \sigma \omega R \sin \theta \hat{\boldsymbol \varphi}.
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\]
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Same if \(\sigma R {\boldsymbol \omega} = {\bf M}\). Refer to Example 5.11,% ({\it not done in class !}),
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\[
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{\bf B} = \frac{2}{3} \mu_0 {\bf M}
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\label{Gr(6.16)}
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\]
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inside sphere, whereas outside: pure dipole field, with
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\[
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{\bf m} = \frac{4}{3} \pi R^3 {\bf M}.
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\]
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</p>
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</div>
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</div>
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</div>
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@@ -1765,7 +1783,7 @@ target="_blank">Creative Commons Attribution 4.0 International License</a>.
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</div>
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<div id="postamble" class="status">
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<p class="author">Author: Jean-Sébastien Caux</p>
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<p class="date">Created: 2022-02-21 Mon 20:41</p>
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<p class="date">Created: 2022-03-01 Tue 08:14</p>
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<p class="validation"></p>
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</div>
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