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Update 2022-02-09 22:41

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Jean-Sébastien
2022-02-09 22:41:42 +01:00
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commit f3c2446d19
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build/ems_ms_dcB_BS.html
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@@ -1,7 +1,7 @@
<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-02-09 Wed 07:31 -->
<!-- 2022-02-09 Wed 22:40 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -408,17 +408,13 @@ Table of contents
<li>
<a href="./ems_es_ep_fp.html#ems_es_ep_fp">Field in terms of the potential</a><span class="headline-id">ems.es.ep.fp</span>
</li>
<li>
<a href="./ems_es_ep_c.html#ems_es_ep_c">Comments on the Electrostatic Potential</a><span class="headline-id">ems.es.ep.c</span>
</li>
<li>
<a href="./ems_es_ep_ex.html#ems_es_ep_ex">Example calculations for the potential</a><span class="headline-id">ems.es.ep.ex</span>
</li>
<li>
<a href="./ems_es_ep_PL.html#ems_es_ep_PL">The Poisson Equation and the Laplace Equation</a><span class="headline-id">ems.es.ep.PL</span>
<a href="./ems_es_ep_PL.html#ems_es_ep_PL">Poisson's and Laplace's Equations</a><span class="headline-id">ems.es.ep.PL</span>
</li>
<li>
@@ -430,29 +426,8 @@ Table of contents
</details>
</li>
<li>
<details>
<summary>
<a href="./ems_es_e.html#ems_es_e">Electrostatic Energy from the Potential</a><span class="headline-id">ems.es.e</span>
</summary>
<ul>
<li>
<a href="./ems_es_e_pcd.html#ems_es_e_pcd">The Energy of a Point Charge Distribution</a><span class="headline-id">ems.es.e.pcd</span>
</li>
<li>
<a href="./ems_es_e_ccd.html#ems_es_e_ccd">The Energy of a Continuous Charge Distribution</a><span class="headline-id">ems.es.e.ccd</span>
</li>
<li>
<a href="./ems_es_e_c.html#ems_es_e_c">Comments on Electrostatic Energy</a><span class="headline-id">ems.es.e.c</span>
</li>
</ul>
</details>
</li>
<li>
@@ -1633,7 +1608,7 @@ Apply the divergence:
\left(\frac{{\bf J}({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \right)
\label{Gr(5.46)}
\]
Note that we can write (using \ref{Gr(1.101)}) \(\frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} = -{\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|}\). Substituting this in and using product rule number 6,
Note that we can write (using <a href="./c_m_dd_3d.html#div1or">div1or</a>) \(\frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} = -{\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|}\). Substituting this in and using product rule number 6,
\[
{\boldsymbol \nabla} \cdot ({\bf A} \times {\bf B}) = {\bf B} \cdot ({\boldsymbol \nabla} \times {\bf A}) - {\bf A} \cdot ({\boldsymbol \nabla} \times {\bf B}),
\]
@@ -1650,7 +1625,7 @@ we get
But \({\bf J}\) depends only on \({\bf r}'\) so \({\boldsymbol \nabla} \times {\bf J} ({\bf r}') = 0\), and since
the curl of a gradient always vanishes, we obtain
</p>
<div class="core div" id="orgd962857">
<div class="core div" id="orgf9f177f">
<p>
\[
{\boldsymbol \nabla} \cdot {\bf B} = 0
@@ -1718,7 +1693,7 @@ at infinity), and in the third step we have used the assumption of steady-state
<p>
We thus obtain in total
</p>
<div class="core div" id="orgcfe305c">
<div class="core div" id="orgc0230aa">
<p>
<b>Ampère's law</b>
\[
@@ -1735,7 +1710,7 @@ We thus obtain in total
\]
so
</p>
<div class="core div" id="org5d8a7de">
<div class="core div" id="org344d23c">
<p>
\[
\oint_{\cal P} {\bf B} \cdot d{\bf l} = \mu_0 I_{enc} \hspace{2cm}
@@ -1757,7 +1732,7 @@ Sign ambiguity: resolved by right-hand rule as usual.
Ampère's law in magnetostatics takes a parallel role to Gauss's law in electrostatics.
</p>
<div class="example div" id="orgdb60bd0">
<div class="example div" id="orgab648a7">
<p>
\paragraph{Example 5.7:} same as Example 5.5, but now with Ampère.
\paragraph{Solution:} by symmetry, \({\bf B}\) is circumferential and can only depend on \(s\). Then,
@@ -1769,7 +1744,7 @@ choosing an amperian loop at a fixed radius \(s\), we get
</div>
<div class="example div" id="orgd1801de">
<div class="example div" id="org8965f93">
<p>
\paragraph{Example 5.8:} uniform surface current \({\bf K} = K \hat{\bf x}\) flowing in \(xy\) plane.
\paragraph{Solution:} Biot-Savart: \({\bf B}\) must be perpendicular to \({\bf K}\). Intuition:
@@ -1786,7 +1761,7 @@ and along \(\hat{\bf y}\) for \(z &lt; 0\). Amperian loop of width \(l\) punchi
</div>
<div class="example div" id="org808fe1f">
<div class="example div" id="org2825c5d">
<p>
\paragraph{Example 5.9:} solenoid along \(\hat{\bf z}\), wire carrying current \(I\) doing \(n\) turns per unit length on cylinder of radius \(R\).
\paragraph{Solution:} by symmetry, \({\bf B}\) must be along axis of solenoid. Outside: infinitely far away, \({\bf B}\) must vanish.
@@ -1807,7 +1782,7 @@ Amperian loop of length \(l\), half-inside and half-outside:
i) infinite straight lines, ii) infinite planes, iii) infinite solenoids, iv) toroids.
</p>
<div class="example div" id="org7db2d3a">
<div class="example div" id="org535691e">
<p>
\paragraph{Example 5.10:} toroidal coil (no matter the shape, as long as it is rotationally symmetric).
\paragraph{Solution:} magnetic field is circumferential everywhere. Outside coil, field again zero.
@@ -1838,7 +1813,7 @@ 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: 2022-02-09 Wed 07:31</p>
<p class="date">Created: 2022-02-09 Wed 22:40</p>
<p class="validation"></p>
</div>