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Jean-Sébastien
2022-03-02 15:47:54 +01:00
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@@ -1,7 +1,7 @@
<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2022-03-01 Tue 08:14 -->
<!-- 2022-03-02 Wed 15:45 -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Pre-Quantum Electrodynamics</title>
@@ -1629,7 +1629,7 @@ is (using fact that \({\bf B}_1\) is proportional to \(I_1\))
\Phi_2 = \int {\bf B}_1 \cdot d{\bf a}_2 \Longrightarrow
\Phi_2 = M_{21} I_1
\]
where \(M_{21}\) is the {\bf mutual inductance} of the two loops.
where \(M_{21}\) is the <b>mutual inductance</b> of the two loops.
</p>
<p>
@@ -1638,7 +1638,7 @@ Useful formula:
\Phi_2 = \int {\bf B}_1 \cdot d{\bf a}_2 = \int ({\boldsymbol \nabla} \times {\bf A}_1) \cdot d{\bf a}_2
= \oint {\bf A}_1 \cdot d{\bf l}_2
\]
But from (\ref{Gr(5.63)}),
But from <a href="./ems_ms_vp_A.html#A_CoulG">A_CoulG</a>,
\[
{\bf A}_1 ({\bf r}) = \frac{\mu_0 I_1}{4\pi} \oint_{{\cal P}_1} \frac{d{\bf l}_1}{|{\bf r} - {\bf r}_1|}
\]
@@ -1647,44 +1647,89 @@ so
\Phi_2 = \frac{\mu_0 I_1}{4\pi} \oint_{{\cal P}_2} d{\bf l}_2 \cdot
\left(\oint_{{\cal P}_1} \frac{d{\bf l}_1 }{|{\bf r}_2 - {\bf r}_1|}\right)
\]
and we can write the mutual inductance as the {\bf Neumann formula},
and we can write the mutual inductance as the <b>Neumann formula</b>,
</p>
<div class="eqlabel" id="org197212f">
<p>
<a id="Newmann_M"></a><a href="./emd_Fl_i.html#Newmann_M"><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="org4709e6a">
<ul class="org-ul">
<li>Gr (7.22)</li>
</ul>
</div>
</div>
<p>
\[
M_{21} = \frac{\mu_0}{4\pi} \oint_{{\cal P}_1} \oint_{{\cal P}_2} \frac{d{\bf l}_1 \cdot d{\bf l}_2}
{|{\bf r}_1 - {\bf r}_2|}
\label{Gr(7.22)}
\tag{Neumann_M}\label{Neumann_M}
\]
Two things:
first, \(M_{21}\) is purely geometrical. Second,
</p>
<div class="eqlabel" id="orgb952a44">
<p>
<a id="Msym"></a><a href="./emd_Fl_i.html#Msym"><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="org53f3037">
<ul class="org-ul">
<li>Gr (7.23)</li>
</ul>
</div>
</div>
<p>
\[
M_{12} = M_{21}
\label{Gr(7.23)}
\tag{Msym}\label{Msym}
\]
</p>
<div class="example div" id="org34dd058">
<div class="example div" id="org6347791">
<p>
\paragraph{Example 7.10:}
short solenoid (length \(l\), radius \(a\), \(n_1\) turns per unit length) lies concentrically inside
<b>Example: solenoid in solenoid</b>
</p>
<p>
Consider a short solenoid (length \(l\), radius \(a\), \(n_1\) turns per unit length)
which lies concentrically inside
a very long solenoid (radius \(b\), \(n_2\) turns per unit length). Current \(I\) in short solenoid.
What is flux through long solenoid ?
\paragraph{Solution:} complicated to calculate \({\bf B}_1\). Use mutual inductance, starting from
</p>
<p>
<b>Task</b>: compute the flux through the long solenoid.
</p>
<p>
<b>Solution</b>: it's complicated to calculate \({\bf B}_1\).
Use mutual inductance, starting from
the reverse situation: current \(I\) on outer solenoid, calculate flux through inner one.
Field of outer solenoid: from (\ref{Gr(5.57)}),
Field of outer solenoid: from <a href="./ems_ms_dcB_c.html#Amp_int">Amp_int</a>,
\[
B = \mu_0 n_2 I
\]
B = \mu_0 n_2 I
\]
so flux through a single loop of inner solenoid is
\[
B \pi a^2 = \mu_0 n_2 I \pi a^2.
\]
B \pi a^2 = \mu_0 n_2 I \pi a^2.
\]
For \(n_1 l\) turns in total, total flux through inner solenoid is
\[
\Phi = \mu_0 \pi a^2 n_1 n_2 l I.
\]
\Phi = \mu_0 \pi a^2 n_1 n_2 l I.
\]
Same as flux through outer solenoid if inner one has current \(I\). Mutual inductance is here
\[
M = \mu_0 \pi a^2 n_1 n_2 l.
\]
M = \mu_0 \pi a^2 n_1 n_2 l.
\]
</p>
</div>
@@ -1697,62 +1742,101 @@ What if we vary current in loop 1? Flux in 2 will vary. Induces EMF in loop 2:
\label{Gr(7.24)}
\]
Changing current also induces EMF in the source loop itself:
</p>
<div class="eqlabel" id="org667fec5">
<p>
<a id="PLI"></a><a href="./emd_Fl_i.html#PLI"><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="org28a1a02">
<ul class="org-ul">
<li>Gr (7.25)</li>
</ul>
</div>
</div>
<p>
\[
\Phi = L I
\label{Gr(7.25)}
\tag{PLI}\label{PLI}
\]
where \(L\) is the {\bf self-inductance} (or inductance) of the loop. Depends only on
where \(L\) is the <b>self-inductance</b> (or inductance) of the loop. Depends only on
geometry. Changing current induces EMF of
\[
{\cal E} = -L \frac{dI}{dt}
\label{Gr(7.26)}
\]
Inductance: measured in {\bf henries} (\(H\)). \(H = V s/A\).
Inductance: measured in <b>henries</b> (\(H\)). \(H = V s/A\).
</p>
<div class="example div" id="org7acd05e">
<div class="example div" id="org00e72a3">
<p>
\paragraph{Example 7.11:} find self-inductance of toroidal coil with
<b>Example: self-inductance of toroidal coil</b>
</p>
<p>
Consider a toroidal coil with
rectangular cross-section (inner radius \(a\), outer radius \(b\), height \(h\))
which carries total of \(N\) turns.
\paragraph{Solution:} magnetic field inside toroid is (\ref{Gr(5.58)})
</p>
<p>
<b>Task</b>: find its self-inductance
</p>
<p>
<b>Solution</b>: magnetic field inside toroid is <a href="./ems_ms_dcB_c.html#Btor">Btor</a>
\[
B = \frac{\mu_0 NI}{2\pi s}
\]
B = \frac{\mu_0 NI}{2\pi s}
\]
Flux through single turn:
\[
\int {\bf B} \cdot d{\bf a} = \frac{\mu_0 N I}{2\pi} h \int_a^b \frac{ds}{s}
= \frac{\mu_0 N I h}{2\pi} \ln \frac{b}{a}.
\]
\int {\bf B} \cdot d{\bf a} = \frac{\mu_0 N I}{2\pi} h \int_a^b \frac{ds}{s}
= \frac{\mu_0 N I h}{2\pi} \ln \frac{b}{a}.
\]
Total flux: \(N\) times this, so self-inductance is
\[
L = \frac{\mu_0 N^2 h}{2\pi} \ln \frac{b}{a}
\label{Gr(7.27)}
\]
L = \frac{\mu_0 N^2 h}{2\pi} \ln \frac{b}{a}
\label{Gr(7.27)}
\]
</p>
</div>
<p>
Inductance (like capacitance) is intrinsically positive. Use Lenz law. Think of {\bf back EMF}.
Inductance (like capacitance) is intrinsically positive. Use Lenz law.
Think of <i>back EMF</i>.
</p>
<div class="example div" id="org9f29434">
<div class="example div" id="org958ea6c">
<p>
\paragraph{Example 7.12:} circuit with inductance \(L\), resistor \(R\) and battery \({\cal E}_0\).
What is the current ?
\paragraph{Solution:}
<b>Example: circuit</b>
</p>
<p>
Consider a circuit with inductance \(L\), resistor \(R\) and battery \({\cal E}_0\).
</p>
<p>
<b>Task</b>: find the current
</p>
<p>
<b>Solution</b>:
Ohm's law:
\[
{\cal E}_0 - L \frac{dI}{dt} = IR \Longrightarrow I(t) = \frac{{\cal E}_0}{R} + k e^{-(R/L)t}.
\]
{\cal E}_0 - L \frac{dI}{dt} = IR \Longrightarrow I(t) = \frac{{\cal E}_0}{R} + k e^{-(R/L)t}.
\]
If initial condition: \(I(0) = 0\), then
\[
I(t) = \frac{{\cal E}_0}{R} \left[ 1 - e^{-(R/L)t} \right]
\label{Gr(7.28)}
\]
where \(\tau \equiv L/R\) is the {\bf time constant} of the circuit.
I(t) = \frac{{\cal E}_0}{R} \left[ 1 - e^{-(R/L)t} \right]
\label{Gr(7.28)}
\]
where \(\tau \equiv L/R\) is the <b>time constant</b> of the circuit.
</p>
</div>
@@ -1777,7 +1861,7 @@ 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: 2022-03-01 Tue 08:14</p>
<p class="date">Created: 2022-03-02 Wed 15:45</p>
<p class="validation"></p>
</div>