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Update 2022-02-14 20:42

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Jean-Sébastien
2022-02-14 20:42:37 +01:00
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commit 09a8ba5fb6
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build/ems_ms_dcB_BS.html
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@@ -1,7 +1,7 @@
<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-02-13 Sun 21:20 -->
<!-- 2022-02-14 Mon 20:35 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -1625,7 +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>
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<div class="core div" id="org633e5ed">
<p>
\[
{\boldsymbol \nabla} \cdot {\bf B} = 0
@@ -1693,7 +1693,7 @@ at infinity), and in the third step we have used the assumption of steady-state
<p>
We thus obtain in total
</p>
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<div class="core div" id="org0636781">
<p>
<b>Ampère's law</b>
\[
@@ -1710,7 +1710,7 @@ We thus obtain in total
\]
so
</p>
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<p>
\[
\oint_{\cal P} {\bf B} \cdot d{\bf l} = \mu_0 I_{enc} \hspace{2cm}
@@ -1732,7 +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="orgf88e8b6">
<div class="example div" id="orgfdfbce3">
<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,
@@ -1744,7 +1744,7 @@ choosing an amperian loop at a fixed radius \(s\), we get
</div>
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<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:
@@ -1761,7 +1761,7 @@ and along \(\hat{\bf y}\) for \(z &lt; 0\). Amperian loop of width \(l\) punchi
</div>
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<div class="example div" id="org75889e8">
<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.
@@ -1782,14 +1782,14 @@ Amperian loop of length \(l\), half-inside and half-outside:
i) infinite straight lines, ii) infinite planes, iii) infinite solenoids, iv) toroids.
</p>
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<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.
Amperian loop half inside, half outside:
\[
B 2\pi s = \mu_0 I_{enc} ~~\Rightarrow~~
{\bf B} = \left\{ \begin{array}{cc} \frac{\mu_0 N I}{2\pi s} \hat{\boldsymbol \phi}, &amp; \mbox{inside coil}, \\
{\bf B} = \left\{ \begin{array}{cc} \frac{\mu_0 N I}{2\pi s} \hat{\boldsymbol \varphi}, &amp; \mbox{inside coil}, \\
0, &amp; \mbox{outside} \end{array} \right.
\label{Gr(5.58)}
\]
@@ -1815,7 +1815,7 @@ 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>