Update 2022-02-14 20:42
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@@ -1,7 +1,7 @@
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<!DOCTYPE html>
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<html lang="en">
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<head>
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<!-- 2022-02-13 Sun 21:20 -->
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<!-- 2022-02-14 Mon 20:35 -->
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<meta charset="utf-8">
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<meta name="viewport" content="width=device-width, initial-scale=1">
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<title>Pre-Quantum Electrodynamics</title>
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@@ -1657,7 +1657,7 @@ W = \frac{1}{2\mu_0} \left[ \int_{\cal V} d\tau B^2 - \int_{\cal V} d\tau {\bold
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\]
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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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</p>
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<div class="core div" id="org0ecd8ee">
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<div class="core div" id="org2618711">
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<p>
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\[
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W_{mag} = \frac{1}{2\mu_0} \int d\tau B^2
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@@ -1678,18 +1678,18 @@ W_{mag} = \frac{1}{2} \int d\tau ({\bf A} \cdot {\bf J}) = \frac{1}{2\mu_0} \int
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\hspace{2cm} \mbox{(7.31 and 7.34)}
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\end{align}
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<div class="example div" id="org46b2091">
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<div class="example div" id="org9178294">
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<p>
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\paragraph{Example 7.13:} coaxial cable (inner cylinder radius \(a\), outer \(b\)) carries current \(I\).
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Find energy stored in section of length \(l\).
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\paragraph{Solution:} from Ampère,
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\[
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{\bf B} = \frac{\mu_0 I}{2\pi s} \hat{\boldsymbol \phi}, \hspace{1cm} a < s < b, \hspace{1cm}
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{\bf B} = \frac{\mu_0 I}{2\pi s} \hat{\boldsymbol \varphi}, \hspace{1cm} a < s < b, \hspace{1cm}
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{\bf B} = 0, \hspace{1cm} s < a ~\mbox{or}~ s > b.
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\]
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Energy is thus
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\[
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W_{mag} = \frac{1}{2\mu_0} \int_0^{2\pi} d\phi \int_0^l dz \int_a^b s ds \left(\frac{\mu_0 I}{2\pi s}\right)^2
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W_{mag} = \frac{1}{2\mu_0} \int_0^{2\pi} d\varphi \int_0^l dz \int_a^b s ds \left(\frac{\mu_0 I}{2\pi s}\right)^2
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= \frac{\mu_0 I^2 l}{4\pi} \ln \frac{b}{a}.
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\]
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</p>
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@@ -1717,7 +1717,7 @@ target="_blank">Creative Commons Attribution 4.0 International License</a>.
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</div>
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<div id="postamble" class="status">
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<p class="author">Author: Jean-Sébastien Caux</p>
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<p class="date">Created: 2022-02-13 Sun 21:20</p>
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<p class="date">Created: 2022-02-14 Mon 20:35</p>
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<p class="validation"></p>
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</div>
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