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Update 2022-02-14 20:42

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Jean-Sébastien
2022-02-14 20:42:37 +01:00
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build/ems_ms_vp.html
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<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-02-13 Sun 21:20 -->
<!-- 2022-02-14 Mon 20:35 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -1601,7 +1601,7 @@ Table of contents
<p>
Since \({\boldsymbol \nabla} \cdot {\bf B} = 0\) in magnetostatics, following Helmholtz's theorem we can write
</p>
<div class="core div" id="orgd9330fe">
<div class="core div" id="org326de0f">
<p>
\[
{\bf B} = {\boldsymbol \nabla} \times {\bf A}
@@ -1623,7 +1623,7 @@ add any curlless function (so gradient of a scalar field) to the vector potentia
without changing the magnetic field. This is called a {\bf gauge choice} in electrodynamics.
For example, we can {\bf always} eliminate the divergence of \({\bf A}\),
</p>
<div class="main div" id="org736c1a7">
<div class="main div" id="orga176de6">
<p>
{\bf Example gauge choice:}
\[
@@ -1654,7 +1654,7 @@ zero at infinity,
<p>
Under this gauge choice, Ampère's law becomes
</p>
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<div class="main div" id="org94528cd">
<p>
\[
{\boldsymbol \nabla}^2 {\bf A} = -\mu_0 {\bf J}
@@ -1667,7 +1667,7 @@ Under this gauge choice, Ampère's law becomes
Note: this is a Poisson equation for each component.
For currents falling off sufficiently rapidly at infinity,
</p>
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<div class="core div" id="orgfd1bf3a">
<p>
\[
{\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \int d\tau' \frac{J({\bf r}')}{|{\bf r} - {\bf r}'|}
@@ -1679,7 +1679,7 @@ For currents falling off sufficiently rapidly at infinity,
<p>
For line and surface currents, <i>(beware Griffiths' <b>horrendous</b> notation)</i>
</p>
<div class="main div" id="orgd89f55e">
<div class="main div" id="org171b37e">
<p>
\[
{\bf A}({\bf r}) = \frac{\mu_0}{4\pi} \int dl' \frac{{\bf I ({\bf r}')}}{|{\bf r} - {\bf r}'|},
@@ -1693,7 +1693,7 @@ For line and surface currents, <i>(beware Griffiths' <b>horrendous</b> notation)
<div class="example div" id="orgc0631de">
<div class="example div" id="orgd02337a">
<p>
\paragraph{Example 5.11:} a spherical shell of radius \(R\), carrying a uniform surface charge
\(\sigma\), is set spinning at angular velocity \(\omega\). Find the vector potential at \({\bf r}\).
@@ -1707,7 +1707,7 @@ the sphere is uniform !
</div>
<div class="example div" id="org0904d42">
<div class="example div" id="org503768c">
<p>
\paragraph{Example 5.12:} find the vector potential of an infinite solenoid with \(n\) turns
pet unit length, radius \(R\) and current \(I\).
@@ -1740,12 +1740,12 @@ Use symmetry: vector potential can only be cicumferential. Using an 'amperian'
\]
so
\[
{\bf A} = \frac{\mu_0 n I}{2} s \hat{\boldsymbol \phi}, \hspace{1cm} s &lt; R.
{\bf A} = \frac{\mu_0 n I}{2} s \hat{\boldsymbol \varphi}, \hspace{1cm} s &lt; R.
\label{Gr(5.70)}
\]
For an 'amperian' loop outside, the flux is always \(\mu_0 n I (\pi R^2)\), so
\[
{\bf A} = \frac{\mu_0 n I}{2} \frac{R^2}{s} \hat{\boldsymbol \phi}, \hspace{1cm} s &gt; R.
{\bf A} = \frac{\mu_0 n I}{2} \frac{R^2}{s} \hat{\boldsymbol \varphi}, \hspace{1cm} s &gt; R.
\label{Gr(5.71)}
\]
\paragraph{Exercise:} check that \({\boldsymbol \nabla} \times {\bf A} = {\bf B}\) and that
@@ -1780,7 +1780,7 @@ target="_blank">Creative Commons Attribution 4.0 International License</a>.
</div>
<div id="postamble" class="status">
<p class="author">Author: Jean-Sébastien Caux</p>
<p class="date">Created: 2022-02-13 Sun 21:20</p>
<p class="date">Created: 2022-02-14 Mon 20:35</p>
<p class="validation"></p>
</div>