Update 2022-02-14 20:42

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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 -->
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<title>Pre-Quantum Electrodynamics</title>
@@ -1657,7 +1657,7 @@ 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
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\[
W_{mag} = \frac{1}{2\mu_0} \int d\tau B^2
@@ -1678,18 +1678,18 @@ W_{mag} = \frac{1}{2} \int d\tau ({\bf A} \cdot {\bf J}) = \frac{1}{2\mu_0} \int
\hspace{2cm} \mbox{(7.31 and 7.34)}
\end{align}
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<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,
\[
{\bf B} = \frac{\mu_0 I}{2\pi s} \hat{\boldsymbol \phi}, \hspace{1cm} a &lt; s &lt; b, \hspace{1cm}
{\bf B} = \frac{\mu_0 I}{2\pi s} \hat{\boldsymbol \varphi}, \hspace{1cm} a &lt; s &lt; b, \hspace{1cm}
{\bf B} = 0, \hspace{1cm} s &lt; a ~\mbox{or}~ s &gt; b.
\]
Energy is thus
\[
W_{mag} = \frac{1}{2\mu_0} \int_0^{2\pi} d\phi \int_0^l dz \int_a^b s ds \left(\frac{\mu_0 I}{2\pi s}\right)^2
W_{mag} = \frac{1}{2\mu_0} \int_0^{2\pi} d\varphi \int_0^l dz \int_a^b s ds \left(\frac{\mu_0 I}{2\pi s}\right)^2
= \frac{\mu_0 I^2 l}{4\pi} \ln \frac{b}{a}.
\]
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@@ -1717,7 +1717,7 @@ target="_blank">Creative Commons Attribution 4.0 International License</a>.
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<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>
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