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<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-02-17 Thu 08:42 -->
<!-- 2022-02-21 Mon 10:33 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -602,11 +602,11 @@ Table of contents
</summary>
<ul>
<li>
<a href="./ems_ms_lf_pc.html#ems_ms_lf_pc">Point Charge</a><span class="headline-id">ems.ms.lf.pc</span>
<a href="./ems_ms_lf_pc.html#ems_ms_lf_pc">Point Charges</a><span class="headline-id">ems.ms.lf.pc</span>
</li>
<li>
<a href="./ems_ms_lf_c.html#ems_ms_lf_c">Currents</a><span class="headline-id">ems.ms.lf.c</span>
<a href="./ems_ms_lf_sc.html#ems_ms_lf_sc">Steady Currents</a><span class="headline-id">ems.ms.lf.sc</span>
</li>
@@ -614,21 +614,12 @@ Table of contents
</details>
</li>
<li>
<a href="./ems_ms_ce.html#ems_ms_ce">Charge Conservation and the Continuity Equation</a><span class="headline-id">ems.ms.ce</span>
<details>
<summary>
</li>
<li>
<a href="./ems_ms_BS.html#ems_ms_BS">Steady Currents: the Biot-Savart Law</a><span class="headline-id">ems.ms.BS</span>
</summary>
<ul>
<li>
<a href="./ems_ms_BS_sc.html#ems_ms_BS_sc">The Magnetic Field issuing from a Steady Current</a><span class="headline-id">ems.ms.BS.sc</span>
</li>
</ul>
</details>
</li>
<li>
@@ -640,11 +631,15 @@ Table of contents
</summary>
<ul>
<li>
<a href="./ems_ms_dcB_sc.html#ems_ms_dcB_sc">Straight-line Currents</a><span class="headline-id">ems.ms.dcB.sc</span>
<a href="./ems_ms_dcB_iw.html#ems_ms_dcB_iw">Simplistic case: infinite wire</a><span class="headline-id">ems.ms.dcB.iw</span>
</li>
<li>
<a href="./ems_ms_dcB_BS.html#ems_ms_dcB_BS">Divergence and Curl of \({\bf B}\) from Biot-Savart</a><span class="headline-id">ems.ms.dcB.BS</span>
<a href="./ems_ms_dcB_d.html#ems_ms_dcB_d">Divergence of \({\bf B}\) from Biot-Savart</a><span class="headline-id">ems.ms.dcB.d</span>
</li>
<li>
<a href="./ems_ms_dcB_c.html#ems_ms_dcB_c">Curl of \({\bf B}\) from Biot-Savart; Ampère's Law</a><span class="headline-id">ems.ms.dcB.c</span>
</li>
@@ -661,6 +656,10 @@ Table of contents
</summary>
<ul>
<li>
<a href="./ems_ms_vp_A.html#ems_ms_vp_A">Definition; Gauge Choices</a><span class="headline-id">ems.ms.vp.A</span>
</li>
<li>
<a href="./ems_ms_vp_mbc.html#ems_ms_vp_mbc">Magnetic Boundary Conditions</a><span class="headline-id">ems.ms.vp.mbc</span>
</li>
@@ -698,10 +697,6 @@ Table of contents
</summary>
<ul>
<li>
<a href="./emsm_esm_s.html#emsm_esm_s">A proper definition of "statics"</a><span class="headline-id">emsm.esm.s</span>
</li>
<li>
<details>
<summary>
@@ -1435,7 +1430,7 @@ Table of contents
</li>
<li>
<a href="./c_m_dc_pr.html#c_m_dc_pr">Product Rules</a><span class="headline-id">c.m.dc.pr</span>
<a href="./c_m_dc_pr.html#c_m_dc_pr">Product arguments</a><span class="headline-id">c.m.dc.pr</span>
</li>
<li>
@@ -1598,150 +1593,233 @@ Table of contents
</svg></a><span class="headline-id">ems.ms.vp.me</span></h5>
<div class="outline-text-5" id="text-ems_ms_vp_me">
<p>
Remember our expansion for the electrostatic field <a href="./ems_ca_me_a.html#1or_Leg">1or_Leg</a>
Remember our expansion for the inverse distance (which we made good use of in the
multipole expansion of the electrostatic field), equation <a href="./ems_ca_me_a.html#1or_Leg">1or_Leg</a>
\[
\frac{1}{|{\bf r} - {\bf r}'|} = \frac{1}{[r^2 + (r')^2 - 2r r' \cos \theta']^{1/2}}
= \frac{1}{r} \sum_{l=0}^{\infty} \left(\frac{r'}{r}\right)^l P_l (\cos \theta').
\label{Gr(5.77)}
\frac{1}{|{\bf r} - {\bf r}'|}
= \frac{1}{r} \sum_{l=0}^{\infty} \left(\frac{r'}{r}\right)^l P_l (\hat{\bf r} \cdot \hat{\bf r}').
\]
The expansion for the vector potential for a current loop carrying current \(I\) over path \({\cal P}\)
can thus be written
\[
{\bf A} ({\bf r}) = \frac{\mu_0 I}{4\pi} \sum_{l=0}^{\infty} \frac{1}{r^{l+1}} \oint_{\cal P} d{\bf l}'
(r')^l P_l (\cos \theta')
\label{Gr(5.78)}
\]
or (in my notations)
\[
{\bf A} ({\bf r}) = \frac{\mu_0 I}{4\pi} \sum_{l=0}^{\infty} \frac{1}{|{\bf r}|^{l+1}}
\oint_{\cal P} d{\bf l}_s |{\bf r}_s|^l P_l (\hat{\bf r} \cdot \hat{\bf r}_s) \rho({\bf r}_s),
\hspace{1cm} |{\bf r}| &gt; |{\bf r}_s| ~\forall~ {\bf r}_s \in {\cal P}.
\]
Explicitly, the first few terms are
\[
{\bf A}({\bf r}) = \frac{\mu_0 I}{4\pi} \left[ \frac{1}{r} \oint_{\cal P} d{\bf l}'
Explicitly, the first few terms are (letting \(\hat{\bf r} \cdot \hat{\bf r}_s = \cos \theta\))
</p>
<ul class="org-ul">
<li>\frac{1}{r^2} \oint d{\bf l}' r' cos θ'</li>
<li>\frac{1}{r^3} \oint d{\bf l}' (r')^2 \left( \frac{3}{2} cos^2 θ' - \frac{1}{2} \right) + … \right]</li>
</ul>
<div class="eqlabel" id="orge0dc13a">
<p>
\label{Gr(5.79)}
\]
Again, these are known as the {\bf monopole}, {\bf dipole}, {\bf quadrupole} terms.
<a id="A_Leg"></a><a href="./ems_ms_vp_me.html#A_Leg"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
</svg></a>
</p>
<div class="alteqlabels" id="org4f1a878">
<ul class="org-ul">
<li>Gr (5.80)</li>
</ul>
</div>
</div>
\begin{align}
{\bf A}({\bf r}) = \frac{\mu_0 I}{4\pi} \left[ \frac{1}{r} \oint_{\cal P} d{\bf l}_s
+ \frac{1}{r^2} \oint d{\bf l}_s r_s \cos \theta
+ \frac{1}{r^3} \oint d{\bf l}_s (r_s)^2 \left( \frac{3}{2} \cos^2 \theta - \frac{1}{2} \right) + ... \right]
\tag{A_Leg}\label{A_Leg}
\end{align}
<p>
Again, these are known as the magnetic <b>monopole</b>, <b>dipole</b>, <b>quadrupole</b> terms.
</p>
<p>
{\bf Note:} {\bf the magnetic monopole term always vanishes.} This is simply because the total vector
displacement on a closed loop is zero, \(\oint d{\bf l}' = 0\), or in other words: there are no magnetic
monopoles (also from Maxwell's equation \({\boldsymbol \nabla} \cdot {\bf B} = 0\)).
Note: <b>the magnetic monopole term always vanishes</b>. This is simply because the total vector
displacement on a closed loop is zero, \(\oint d{\bf l}_s = 0\), or in other words: there are no magnetic
monopoles (also from Maxwell's equation <a href="./ems_ms_dcB_d.html#divB0">divB0</a> \({\boldsymbol \nabla} \cdot {\bf B} = 0\)).
</p>
<p>
The dominant term is thus the dipole,
</p>
<div class="eqlabel" id="org662943d">
<p>
<a id="A_di1"></a><a href="./ems_ms_vp_me.html#A_di1"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
</svg></a>
</p>
<div class="alteqlabels" id="orgf466457">
<ul class="org-ul">
<li>Gr (5.85)</li>
</ul>
</div>
</div>
<p>
\[
{\bf A}_{di} ({\bf r}) = \frac{\mu_0 I}{4\pi} \frac{1}{r^2} \oint d{\bf l}' (\hat{\bf r} \cdot {\bf r}')
\label{Gr(5.81)}
{\bf A}_{di} ({\bf r}) = \frac{\mu_0 I}{4\pi} \frac{1}{r^2} \oint d{\bf l}_s (\hat{\bf r} \cdot {\bf r}_s)
\tag{A_di1}\label{A_di1}
\]
Using equation \ref{Gr(1.108)} from Problem 1.61,
Using the identity (see Gr Problem 1.62),
</p>
<div class="eqlabel" id="orgd42c168">
<p>
<a id="intcr"></a><a href="./ems_ms_vp_me.html#intcr"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
</svg></a>
</p>
<div class="alteqlabels" id="org892de03">
<ul class="org-ul">
<li>Gr (1.108)</li>
</ul>
</div>
</div>
<p>
\[
\oint_{\cal P} ({\bf c} \cdot {\bf r}) d{\bf l} = {\bf a} \times {\bf c}, \hspace{1cm}
\oint_{\cal P} d{\bf l} ({\bf c} \cdot {\bf r}) = {\bf a} \times {\bf c}, \hspace{1cm}
{\bf a} = \int_{\cal S} d{\bf a}
\label{Gr(1.108)}
\tag{intcr}\label{intcr}
\]
with \({\bf c} = \hat {\bf r}\),
</p>
<div class="info div" id="org6b178f6">
<div class="info div" id="org8f1a4cb">
<p>
\paragraph{Parenthesis: Problem 1.61 (e)} Start from Stokes' theorem,
<b>Parenthesis: derivation of</b> <a href="./ems_ms_vp_me.html#intcr">intcr</a>
</p>
<p>
Start from Stokes' theorem,
\[
\int_{\cal S} {\boldsymbol \nabla} \times {\bf V} \cdot d{\bf a} = \oint_{\cal P} {\bf V} \cdot d{\bf l}
\]
Let \({\bf V} = {\bf c} T\), where \({\bf c}\) is constant.
\int_{\cal S} d{\bf a} \cdot {\boldsymbol \nabla} \times {\bf v} = \oint_{\cal P} d{\bf l} \cdot {\bf v}
\]
Let \({\bf v} = {\bf c} T\), where \({\bf c}\) is constant.
On the left-hand side:
\[
LHS = \int_{\cal S} T ({\boldsymbol \nabla} \times c) \cdot d{\bf a} - \int_{\cal S} {\bf c} \times ({\boldsymbol \nabla} T) \cdot d{\bf a}
\]
The first term is zero since \({\bf c}\) is constant. For the second term: use vector identity nr 1,
\(({\bf c} \times ({\boldsymbol \nabla} T)) \cdot d{\bf a} = {\bf c} \cdot ({\boldsymbol \nabla}T \times d{\bf a})\). The second
term thus becomes
LHS = \int_{\cal S} d{\bf a} \cdot (T ({\boldsymbol \nabla} \times c)) - \int_{\cal S} d{\bf a} \cdot ({\bf c} \times ({\boldsymbol \nabla} T))
\]
The first term is zero since \({\bf c}\) is constant. For the second term: use the triple product vector identity
\(d{\bf a} \cdot ({\bf c} \times ({\boldsymbol \nabla} T)) = {\bf c} \cdot ({\boldsymbol \nabla}T \times d{\bf a})\). The second term thus becomes
\[
</p>
<ul class="org-ul">
<li>∫_{\cal S} {\bf c} × ({\boldsymbol ∇} T) ⋅ d{\bf a} = -{\bf c} ⋅ ∫_{\cal S} {\boldsymbol ∇}T × d{\bf a}</li>
</ul>
<p>
-\int_{\cal S} d{\bf a} \cdot ({\bf c} \times ({\boldsymbol \nabla} T)) = -{\bf c} \cdot \int_{\cal S} {\boldsymbol \nabla}T \times d{\bf a}
\]
Treating the right-hand side of the original equation now,
\[
RHS = {\bf c} \cdot \oint_{\cal P} T d{\bf l}
\]
RHS = {\bf c} \cdot \oint_{\cal P} T d{\bf l}
\]
so we get (since this is valid for any \({\bf c}\))
\[
\int_{\cal S} {\boldsymbol \nabla} T \times d{\bf a} = -\oint_{\cal P} T d{\bf l}.
\]
Now put \(T = {\bf c} \cdot {\bf r}\) in this: conclusion is \ref{Gr(1.108)}.
\int_{\cal S} {\boldsymbol \nabla} T \times d{\bf a} = -\oint_{\cal P} T d{\bf l}.
\]
Now put \(T = {\bf c} \cdot {\bf r}\) in this, yielding <a href="./ems_ms_vp_me.html#intcr">intcr</a>.
</p>
</div>
<p>
Back to our problem:
</p>
<p>
\[
\oint d{\bf l}' (\hat{\bf r} \cdot {\bf r}') = -\hat{\bf r} \times \int_{\cal S} d{\bf a}'
\label{Gr(5.82)}
\oint d{\bf l}_s (\hat{\bf r} \cdot {\bf r}_s) = -\hat{\bf r} \times \int_{\cal S} d{\bf a}_s
\]
and defining the
</p>
<div class="core div" id="org3564ba4">
<div class="core div" id="org8827bc7">
<p>
<b>magnetic dipole moment</b>
\[
{\bf m} \equiv I \int_{\cal S} d{\bf a} = I {\bf a}
\label{Gr(5.84)}
\]
</p>
<div class="eqlabel" id="org172b482">
<p>
<a id="magdim"></a><a href="./ems_ms_vp_me.html#magdim"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
</svg></a>
</p>
<div class="alteqlabels" id="org2dff5c7">
<ul class="org-ul">
<li>Gr (5.86)</li>
</ul>
</div>
</div>
<p>
\[
{\bf m} \equiv I \int_{\cal S} d{\bf a} = I {\bf a}
\tag{magdim}\label{magdim}
\]
</p>
</div>
<p>
we obtain the convenient expression for the
</p>
<div class="main div" id="orge3c27a3">
<div class="main div" id="orgbade3a2">
<p>
<b>dipole term of the vector potential</b>
</p>
<div class="eqlabel" id="orgcea620a">
<p>
<a id="A_di"></a><a href="./ems_ms_vp_me.html#A_di"><svg xmlns="http://www.w3.org/2000/svg" width="16" height="16" fill="currentColor" class="bi bi-link" viewBox="0 0 16 16">
<path d="M6.354 5.5H4a3 3 0 0 0 0 6h3a3 3 0 0 0 2.83-4H9c-.086 0-.17.01-.25.031A2 2 0 0 1 7 10.5H4a2 2 0 1 1 0-4h1.535c.218-.376.495-.714.82-1z"/>
<path d="M9 5.5a3 3 0 0 0-2.83 4h1.098A2 2 0 0 1 9 6.5h3a2 2 0 1 1 0 4h-1.535a4.02 4.02 0 0 1-.82 1H12a3 3 0 1 0 0-6H9z"/>
</svg></a>
</p>
<div class="alteqlabels" id="org58d35ca">
<ul class="org-ul">
<li>Gr (5.85)</li>
</ul>
</div>
</div>
<p>
{\bf dipole term of the vector potential}
\[
{\bf A}_{di} ({\bf r}) = \frac{\mu_0}{4\pi} \frac{{\bf m} \times \hat{\bf r}}{r^2}
\label{Gr(5.83)}
\]
{\bf A}_{di} ({\bf r}) = \frac{\mu_0}{4\pi} \frac{{\bf m} \times \hat{\bf r}}{r^2}
\tag{A_di}\label{A_di}
\]
</p>
</div>
<div class="example div" id="org18d7423">
<div class="example div" id="org95095e4">
<p>
\paragraph{Example 5.13:} find magnetic dipole moment of bookend shape of Gr. Fig. 5.52.
All sides have length \(w\) and carry current \(I\).
\paragraph{Solution:} combine two loops, use \ref{Gr(5.83)}
<b>Example: bookend shape</b>
</p>
<p>
<b>Task</b>: find the magnetic dipole moment of a bookend shape current-carrying loop
(c.f. Gr Fig. 5.52). All sides have length \(w\) and carry current \(I\).
</p>
<p>
<b>Solution</b>: combine two loops, use <a href="./ems_ms_vp_me.html#magdim">magdim</a>
\[
{\bf m} = I w^2 \hat{\bf y} + I w^2 \hat{\bf z}
\]
{\bf m} = I w^2 \hat{\bf y} + I w^2 \hat{\bf z}
\]
</p>
</div>
<p>
{\bf Note:} {\it the magnetic dipole moment is independent of the choice of origin.}
<b>Note</b>: the magnetic dipole moment is independent of the choice of origin.
</p>
<p>
{\bf Note:} does there exist a {\it pure magnetic dipole} ? Well, yes, but it's an
<b>Note</b>: does there exist a <i>pure magnetic dipole</i>? Well, yes, but it's an
infinitely small loop carrying an infinitely large current, so that the dipole term
is finite.
</p>
<p>
In practice: dipole approximation often good enough when far away from source on a
In practice: the dipole approximation often good enough when far away from source on a
scale of the source's current loops.
</p>
</div>
@@ -1765,7 +1843,7 @@ 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: 2022-02-17 Thu 08:42</p>
<p class="date">Created: 2022-02-21 Mon 10:33</p>
<p class="validation"></p>
</div>