Update 2022-02-14 20:42

This commit is contained in:
Jean-Sébastien
2022-02-14 20:42:37 +01:00
parent 4cfe8cef59
commit 09a8ba5fb6
204 changed files with 1968 additions and 1206 deletions
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@@ -1,7 +1,7 @@
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<!-- 2022-02-13 Sun 21:20 -->
<!-- 2022-02-14 Mon 20:35 -->
<meta charset="utf-8">
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<title>Pre-Quantum Electrodynamics</title>
@@ -1666,7 +1666,7 @@ so we get
Substituting this in \ref{Gr(8.6)} and using the divergence theorem,
we obtain
</p>
<div class="main div" id="org1ad9e6b">
<div class="main div" id="org9d063a8">
<p>
{\bf Poynting's theorem}
\[
@@ -1691,7 +1691,7 @@ energy is carried by EM fields out of \({\cal V}\) across its boundary surface.
<p>
Energy per unit time, per unit area carried by EM fields:
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<p>
{\bf Poynting vector}
\[
@@ -1704,7 +1704,7 @@ Energy per unit time, per unit area carried by EM fields:
<p>
We can thus express Poynting's theorem more compactly:
</p>
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<p>
{\bf Poynting's theorem}
\[
@@ -1717,7 +1717,7 @@ We can thus express Poynting's theorem more compactly:
<p>
where we have defined the total
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<p>
{\bf Energy in electromagnetic fields}
\[
@@ -1740,7 +1740,7 @@ Then,
\]
so we get the
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<p>
{\bf Poynting theorem (differential form)}
\[
@@ -1757,7 +1757,7 @@ and has a similar for to the continuity equation
<div class="example div" id="orgf36d646">
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<p>
\paragraph{Example 8.1} Current in a wire: Joule heating. Energy per unit time delivered to wire: from Poynting.
Assuming that the field is uniform, the electric field parallel to the wire is
@@ -1767,11 +1767,11 @@ Assuming that the field is uniform, the electric field parallel to the wire is
where \(V\) is the potential difference between the ends ald \(L\) is the length. Magnetic field is circumferential:
wire of radius \(a\),
\[
{\boldsymbol B} = \frac{\mu_0 I}{2\pi a} \hat{\boldsymbol \phi}
{\boldsymbol B} = \frac{\mu_0 I}{2\pi a} \hat{\boldsymbol \varphi}
\]
Poynting:
\[
{\boldsymbol S} = \frac{1}{\mu_0} \frac{V}{L} \frac{\mu_0 I}{2\pi a} \hat{\boldsymbol x} \times \hat{\boldsymbol \phi} = -\frac{VI}{2\pi a L} \hat{\boldsymbol s}
{\boldsymbol S} = \frac{1}{\mu_0} \frac{V}{L} \frac{\mu_0 I}{2\pi a} \hat{\boldsymbol x} \times \hat{\boldsymbol \varphi} = -\frac{VI}{2\pi a L} \hat{\boldsymbol s}
\]
and points radially inwards. Energy per unit time passing surface of wire:
\[
@@ -1802,7 +1802,7 @@ 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>