Update 2022-03-02 15:47

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Jean-Sébastien
2022-03-02 15:47:54 +01:00
parent ac1e628013
commit 21bf9fdba5
194 changed files with 1653 additions and 1216 deletions
+57 -39
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@@ -1,7 +1,7 @@
<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-03-01 Tue 08:14 -->
<!-- 2022-03-02 Wed 15:45 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -1631,7 +1631,7 @@ Start from zero current, integrate in time:
W = \frac{1}{2} L I^2
\label{Gr(7.29)}
\]
Nicer way (generalizable to surface and volume currents): from (\ref{Gr(7.25)}), flux through loop is \(\Phi = L I\). But
Nicer way (generalizable to surface and volume currents): from <a href="./emd_Fl_i.html#PLI">PLI</a>, flux through loop is \(\Phi = L I\). But
\[
\Phi = \int_{\cal S} {\bf B} \cdot d{\bf a} = \int_{\cal S} ({\boldsymbol \nabla} \times {\bf A}) \cdot d{\bf a}
= \oint_{\cal P} {\bf A} \cdot d{\bf l},
@@ -1647,16 +1647,16 @@ W = \frac{1}{2} I \oint {\bf A} \cdot d{\bf l} = \frac{1}{2} \oint ({\bf A} \cdo
\]
Generalization to volume currents:
</p>
<div class="eqlabel" id="org9fbfb9d">
<div class="eqlabel" id="org87af1ba">
<p>
<a id="W_intAJ"></a><a href="./emd_Fl_e.html#W_intAJ"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<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"/>
<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"/>
</svg></a>
</p>
<div class="alteqlabels" id="org6b3d2a5">
<div class="alteqlabels" id="org95d3f8b">
<ul class="org-ul">
<li>gr (7.31)</li>
<li>Gr (7.31)</li>
</ul>
</div>
@@ -1664,31 +1664,22 @@ Generalization to volume currents:
</div>
<p>
\[
W = \frac{1}{2} \int_{\cal V} ({\bf A} \cdot {\bf J}) d\tau
W = \frac{1}{2} \int_{\cal V} d\tau ~({\bf A} \cdot {\bf J})
\tag{W_intAJ}\label{W_intAJ}
\]
Even better: use Ampère, \({\boldsymbol \nabla} \times {\bf B} = \mu_0 {\bf J}\):
\[
W = \frac{1}{2\mu_0} \int_{\cal V} {\bf A} \cdot ({\boldsymbol \nabla} \times {\bf B}) d\tau
W = \frac{1}{2\mu_0} \int_{\cal V} d\tau ~{\bf A} \cdot ({\boldsymbol \nabla} \times {\bf B})
\label{Gr(7.32)}
\]
Integrate by parts using product rule 6:
\[
{\boldsymbol ∇} ⋅ ({\bf A} × {\bf B}) = {\bf B} ({\boldsymbol ∇} × {\bf A})
</p>
<ul class="org-ul">
<li>{\bf A} ⋅ ({\boldsymbol ∇} × {\bf B}),</li>
</ul>
<p>
{\boldsymbol \nabla} \cdot ({\bf A} \times {\bf B}) = {\bf B} \cdot ({\boldsymbol \nabla} \times {\bf A}) - {\bf A} \cdot ({\boldsymbol \nabla} \times {\bf B}),
\]
so
\[
{\bf A} ({\boldsymbol ∇} × {\bf B}) = {\bf B} ⋅ {\bf B}
</p>
<ul class="org-ul">
<li>{\boldsymbol ∇} ⋅ ({\bf A} × {\bf B}).</li>
</ul>
<p>
{\bf A} \cdot ({\boldsymbol \nabla} \times {\bf B}) =
{\bf B} \cdot {\bf B} - {\boldsymbol \nabla} \cdot ({\bf A} \times {\bf B}).
\]
Then,
\[
@@ -1698,12 +1689,27 @@ W = \frac{1}{2\mu_0} \left[ \int_{\cal V} d\tau B^2 - \int_{\cal V} d\tau {\bold
\]
We can integrate over all space: after neglecting boundary terms (assuming fields fall to zero at infinity), we are left with
</p>
<div class="core div" id="org2f4a453">
<div class="core div" id="org247bd37">
<div class="eqlabel" id="orgdf4c394">
<p>
<a id="W_intBsq"></a><a href="./emd_Fl_e.html#W_intBsq"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<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"/>
<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"/>
</svg></a>
</p>
<div class="alteqlabels" id="orge069b86">
<ul class="org-ul">
<li>Gr (7.34)</li>
</ul>
</div>
</div>
<p>
\[
W_{mag} = \frac{1}{2\mu_0} \int d\tau B^2
\label{Gr(7.34)}
\]
W_{mag} = \frac{1}{2\mu_0} \int d\tau B^2
\tag{W_intBsq}\label{W_intBsq}
\]
</p>
</div>
@@ -1713,26 +1719,38 @@ We can integrate over all space: after neglecting boundary terms (assuming fiel
Summary: energy in electric and magnetic fields:
</p>
\begin{align}
W_{elec} = \frac{1}{2} \int d\tau V\rho = \frac{\varepsilon_0}{2} \int d\tau E^2, \hspace{2cm}
\mbox{(2.43 and 2.45)}, \\
W_{mag} = \frac{1}{2} \int d\tau ({\bf A} \cdot {\bf J}) = \frac{1}{2\mu_0} \int d\tau B^2,
\hspace{2cm} \mbox{(7.31 and 7.34)}
W_{elec} &amp;= \frac{1}{2} \int d\tau ~V\rho &amp;= \frac{\varepsilon_0}{2} \int d\tau ~E^2, \\
W_{mag} &amp;= \frac{1}{2} \int d\tau ~({\bf A} \cdot {\bf J}) &amp;= \frac{1}{2\mu_0} \int d\tau ~B^2,
\end{align}
<div class="example div" id="orgeb4514a">
<p>
\paragraph{Example 7.13:} coaxial cable (inner cylinder radius \(a\), outer \(b\)) carries current \(I\).
Find energy stored in section of length \(l\).
\paragraph{Solution:} from Ampère,
which are equations <a href="./ems_es_e.html#W_vcd">W_vcd</a>, <a href="./ems_es_e.html#W_intEsq">W_intEsq</a>, <a href="./emd_Fl_e.html#W_intAJ">W_intAJ</a> and <a href="./emd_Fl_e.html#W_intBsq">W_intBsq</a>.
</p>
<div class="example div" id="org03fb4e8">
<p>
<b>Example: energy in coaxial cable</b>
</p>
<p>
Consider a coaxial cable with inner cylinder radius \(a\), outer \(b\),
carrying current \(I\).
</p>
<p>
<b>Task</b>: find the energy stored in a section of length \(l\).
</p>
<p>
<b>Solution</b>: from Ampère,
\[
{\bf B} = \frac{\mu_0 I}{2\pi s} \hat{\boldsymbol \varphi}, \hspace{1cm} a &lt; s &lt; b, \hspace{1cm}
{\bf B} = 0, \hspace{1cm} s &lt; a ~\mbox{or}~ s &gt; b.
\]
{\bf B} = \frac{\mu_0 I}{2\pi s} \hat{\boldsymbol \varphi}, \hspace{1cm} a &lt; s &lt; b, \hspace{1cm}
{\bf B} = 0, \hspace{1cm} s &lt; a ~\mbox{or}~ s &gt; b.
\]
Energy is thus
\[
W_{mag} = \frac{1}{2\mu_0} \int_0^{2\pi} d\varphi \int_0^l dz \int_a^b s ds \left(\frac{\mu_0 I}{2\pi s}\right)^2
= \frac{\mu_0 I^2 l}{4\pi} \ln \frac{b}{a}.
\]
W_{mag} = \frac{1}{2\mu_0} \int_0^{2\pi} d\varphi \int_0^l dz \int_a^b s ds \left(\frac{\mu_0 I}{2\pi s}\right)^2
= \frac{\mu_0 I^2 l}{4\pi} \ln \frac{b}{a}.
\]
</p>
</div>
@@ -1758,7 +1776,7 @@ 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: 2022-03-01 Tue 08:14</p>
<p class="date">Created: 2022-03-02 Wed 15:45</p>
<p class="validation"></p>
</div>