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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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build/emsm_msm_H_A.html
+8 -8
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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>
@@ -1611,7 +1611,7 @@ Ampère's law:
\]
so we can define
</p>
<div class="core div" id="org020f4e9">
<div class="core div" id="orgb7bb08b">
<p>
\[
{\bf H} \equiv \frac{1}{\mu_0} {\bf B} - {\bf M}
@@ -1623,7 +1623,7 @@ so we can define
<p>
and rewrite Ampère's law as
</p>
<div class="core div" id="orgf520f44">
<div class="core div" id="orgf499c02">
<p>
\[
{\boldsymbol \nabla} \times {\bf H} = {\bf J}_f
@@ -1635,7 +1635,7 @@ and rewrite Ampère's law as
<p>
or in integral form,
</p>
<div class="core div" id="org5405d94">
<div class="core div" id="org2211577">
<p>
\[
\oint {\bf H} \cdot d{\bf l} = I_{f_{enc}}
@@ -1649,7 +1649,7 @@ or in integral form,
rewrite Ampère's law in terms of free currents alone. Bound currents come along for the ride.
</p>
<div class="example div" id="org5a3e208">
<div class="example div" id="orgf303933">
<p>
\paragraph{Example 6.2:} long copper rod radius \({\bf R}\) carries uniformly distributed free current \(I\).
Find \({\bf H}\) inside and outside rod.
@@ -1658,12 +1658,12 @@ Bound currents antiparallel to \(I\) in bulk and parallel at surface. All curre
so \({\bf B}, {\bf M}, {\bf H}\) are circumferential. Apply integral form of Ampère's law with
radius \(s &lt; R\): \(H (2\pi s) = I_{f_{enc}} = I \frac{\pi s^2}{\pi R^2}\) so
\[
{\bf H} = \frac{I s}{2\pi R^2} \hat{\boldsymbol \phi}, \hspace{5mm} s \leq R
{\bf H} = \frac{I s}{2\pi R^2} \hat{\boldsymbol \varphi}, \hspace{5mm} s \leq R
\label{Gr(6.21)}
\]
Outside,
\[
{\bf H} = \frac{I}{2\pi s} \hat{\boldsymbol \phi}, \hspace{5mm} s \geq R.
{\bf H} = \frac{I}{2\pi s} \hat{\boldsymbol \varphi}, \hspace{5mm} s \geq R.
\label{Gr(6.22)}
\]
There, \({\bf M} = 0\) so \({\bf B} = \mu_0 {\bf H}\).
@@ -1768,7 +1768,7 @@ 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>