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\begin{document}



\begin{center}

\Huge{Part III}

\Huge{Electrodynamics}

\end{center}


\textsf  
\sffamily   



\tableofcontents

Appendix A: Exercises Electrodynamics (page 500)

Appendix B: Practicum Electrodynamics

Appended: Slides Electrodynamics


\newpage

\section{Introduction}
In this part we discuss the electric charge configurations
and currents that are not constant in time. We start with a study of the
effects of and electromagnetic force acting on charge. This study
allows us to understand time dependent currents in electric circuits 
consisting of combinations of resistors, capacitors and inductors.
 Our main 
quest is the time dependent behavior of electric and  magnetic fields,
leading to an elegant description in term of the Maxwell equations.


\section{Current and force}
\subsection{Why does current flow?}
Current flows because something is pushing the charge carriers (electrons).
 And that something is an electric (or magnetic as we will see later) force. Let's analyze that.
Consider a piece of wire made of copper with a current density $\vec{J}$ as 
illustrated in Fig.~\ref{wire}. 
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=6cm \epsffile{wire.ps}}
\caption{\sl A piece of wire made of copper with a current density $\vec{J}$.}
\label{wire}
\end{center}
\end{figure}
To make the electrons that constitute the current  a force~$\vec{f}$ 
is needed. The force $\vec{f}$ is defined as a force per unit charge.
For normal everyday materials and current densities, the current is
proportional to the force:
\begin{equation}
\vec{J}=\sigma \vec{f}
\end{equation}
with $\sigma$ the conductivity~\footnote{Historically the symbol $\sigma$ is
used for conductivity, while we (and others) also use it for a surface charge
 density. So, don't get confused.} depending on the material. 
Usually, in tables describing characteristics of materials,
 one gives the reciprocal value of this quantity, 
called  the resistivity $\rho=1/\sigma$. 

The above expression looks reasonable at first sight, but when you think
of it you may get confused about the following. When a force acts on the 
electrons in the wire, one would expect that the electrons acquire
more and more speed with time. Consequently, the current would increase with
 time, which is not the case. Why not? Well, there appears to be a
cancellation due to 'collisions' of the free electrons and the
nuclei in the material. This effect, that manifests itself as a 
constant friction,
 is illustrated in Fig.~\ref{jfric}.
  Right after the current starts an equilibrium
between the accelerating force $\vec{f}$ and the de-accelarating 
friction sets in.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfxsize=8cm \epsffile{jfric.ps}}
\caption{\sl Top: The free electron undergoes a force $\vec{f}$. Since it is
 really free its speed increases rapidly with time. Bottom:
Another electron that also undergoes a force $\vec{f}$ 
 is constrained in a material and collides with 
nuclei, such that it effectively senses a frictional force that
cancels $\vec{f}$ eventually obtains a constant velocity.}
\label{jfric}
\end{center}
\end{figure}

\subsection{Ohm's law}
So far, we discussed the relation between force $\vec{f}$ and the current
 density $\vec{J}$,
without bothering about the origin of it. The origin of the force $\vec{f}$ for
our purpose is an 
 electric field caused by any device that establishes a potential
difference, such as a battery, a van~der~Graaff generator or a dynamo.
We know that in general the electric force is given by $\vec{F}=q\vec{E}$. The
force $\vec{f}$ was defined per unit charge, thus we may write $\vec{f}=\vec{E}$.
Now we consider a piece of wire with cross section $A$ and length $l$.
A potential difference $V$ over the wire leads to a current $I$ as shown
in Fig.~\ref{wohm}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=5cm \epsffile{wohm.ps}}
\caption{\sl A piece of wire with length $l$ and cross section $A$. The
 potential difference $V$ over its ends leads to a current $\vec{I}$.}
\label{wohm}
\end{center}
\end{figure}
 According to Ohm's law, $V$ and $I$ are related by:
\begin{equation}
V=I R
\end{equation}
with $R$ the resistance of the wire. This law is the subject of many
physics lectures at high school.  Can we derive this law? Yes, in fact
its derivation is rather straightforward!

For a perfect conductor, the conductivity $\sigma=\infty$ and thus $\vec{E}=\vec{J}/\sigma=0$.
For a real, everyday, conductor the electric field inside may be zero for
stationary charges, but certainly not for the situation when current flows,
like in our case. In our piece of wire the electric field is given by:
\begin{equation}
|\vec{E}|=\frac{V}{l}
\end{equation}
The current density in the wire can be written as:
\begin{equation}
|\vec{J}|=\sigma |\vec{f}|=\sigma |\vec{E}|=\sigma \frac{V}{l}
\end{equation}
For $|\vec{J}|$ we write $I/A$ and obtain:
\begin{equation}
|\vec{J}|=\frac{I}{A}=\sigma \frac{V}{l}
\end{equation}
and thus:
\begin{equation}
V=\frac{I l}{\sigma  A}\equiv I R 
\end{equation}
with $R\equiv \frac{l}{\sigma  A}$. Hence, when we double the length $l$ of the
wire the resistance becomes twice as large. When we double the radius of
the wire the resistance drops by a factor four. The conductivity of copper
is about $\sigma=6 \times 10^7$~($\ohm$m)$^{-1}$. A copper wire of 1 meter length and a cross section of 0.75~mm$^2$ has a resistance of about $R=0.02$~$\ohm$.





\subsection{Electromotive force}
Figure~\ref{emf} shows an electric circuit with a current density
 $\vec{J}$.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=5cm \epsffile{emf.ps}}
\caption{\sl An electric circuit with current density $\vec{J}$.
The electric field and a battery ($\vec{f}_b$) are also indicated. }
\label{emf}
\end{center}
\end{figure}
 We know that a force $\vec{f}$ drives the current through the wire. In the wire
this force is an electric field, an electrostatic field. The field through 
the wire is produced by the $+$ and $-$ side of a battery.
We know from electrostatic theory 
that $\int_{circuit} \vec{E} \dot d\vec{l} = 0 $, and thus
$\int_{wire} \vec{E} \dot d\vec{l} = -\int_{battery} \vec{E} \dot d\vec{l} $.
Hence, the electric field in the battery points opposite to that in the
wire. Anyway, you may ask now: if the field integrals cancel each other there
is no net force, so there can't be a current. Well, we forgot the
force, $f_b$ delivered by the battery itself, which is a chemical reaction that
maintains the electric potential over the $+$ and $-$ side.
The total force that drives a current to a circuit is thus: 
\begin{equation}
\vec{f}=\vec{f}_b + \vec{E}
\end{equation}
Note that $\vec{f}$ is not constant in the circuit:  
$f_b$ is only present in the battery and $\vec{E}$ in the
wire is different from $\vec{E}$ in the battery.
 
The net effect of this force is the line integral over the circuit and is
called the electromotive force: 
\begin{eqnarray}
EMF &=&\int_{circuit} \vec{f} \dot d\vec{l} = \int_{cicuit} (\vec{f}_b + \vec{E}) \dot d\vec{l}\nonumber \\
    &=& \int_{battery} \vec{f}_b \dot d\vec{l}  +
         \int_{cicuit} \vec{E} \dot d\vec{l}  \nonumber \\
    &=&	 \int_{battery} \vec{f}_b \dot d\vec{l}  
\end{eqnarray}
In an ideal battery there is no friction or net force on the charges and 
thus:
\begin{equation}
EMF= \int_{battery} \vec{f}_b \dot d\vec{l} = 
 - \int_{battery} \vec{E} \dot d\vec{l} = V_b \left( =   \int_{wire} \vec{E} \dot d\vec{l} \right)
\end{equation}
Hence, the electromotive force is just the
 potential (difference), $V_b$, of the battery.  


%Is the electric field in the wire as constant as suggested in the figure?
% The answer is yes if the wire has a uniform conductivity.
%If the electric field is not constant the current will not be
% Suppose the electric field is larger in 
%some part of the wire. Then, more electric force is applied to
%the charge carriers and they will pile up. But if 
%charge starts to pile up it reduces the originally large electric field.
%Consequntly, a current is a dynamic but stable current
%
% A 'pile of charge' will reduce
%the originally larger electric field, thus less pile-up. Consequently


\subsection{Induced EMF}
Consider the experiment of a circuit pulled out of an magnetic field with
a speed $\vec{v}$ as illustrated in Fig.~\ref{ind}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=3cm \epsffile{ind.ps}}
\caption{\sl An electric circuit with a light bulb with resistance $R$
is pulled out of a magnetic field $\vec{B}$. The vertical height
of the circuit is $h$. The length of the part of 
the circuit that is in the magnetic field
is labeled $s$. The Lorentz force on the positive charge carriers
is also indicated.}
\label{ind}
\end{center}
\end{figure}
The movement of the circuit out of the magnetic field leads to an $EMF$.
What is the source of the $EMF$? It is the Lorentz force on the (moving) 
charges in the wire. In the figure the Lorentz force on the positive charge
carries is indicated, because that defines the
direction of the current, while in reality the Lorentz force on
the electrons is the relevant driving force.
 The Lorentz force in the horizontal pieces of the circuit point downward,
perpendicular to the wire,
as indicated.
 In the vertical piece of wire in the magnetic field
the charge carriers that constitute a current
 will be pushed in the direction as indicated in the
figure. The force per unit charge is
 $\vec{f}=\vec{v} \times \vec{B}$.
For the $EMF$ follows:
\begin{equation}
EMF=\int_{circuit} \vec{f} \dot d\vec{l} = \int_0^h | \vec{f} |  dl = 
\int_{0}^{h} |\vec{v} \times \vec{B}| dl =
 |\vec{v}| |\vec{B}| h
\end{equation} 
 Note that the horizontal pieces do not contribute to the $EMF$.
The $EMF$ generated by the movement of the wire is called inductance.
A potential difference over the light bulb is generated of size $V_{ind}=EMF$.

For reasons that become clear in a moment we study the flux of
the magnetic field through the circuit. The magnetic flux is given by:
\begin{equation}
\Phi_{B}=\int_{surface} \vec{B} \dot d\vec{o} = B h s
\end{equation}
where $s$ is the part of the circuit that lies in the magnetic field,
which depends on the time when we pull the circuit out of the field.
The time derivative of the magnetic flux is given by:
\begin{equation}
\frac{d\Phi_B}{dt}=\frac{d(Bhs)}{dt}=Bh \frac{ds}{dt}=-Bhv=-V_{ind}
\end{equation}
Hence, we found a relation between the EMF and the change of the
magnetic flux:
\begin{equation}
EMF=-\frac{d\Phi_B}{dt}=-V_{ind}
\end{equation}
This principle to create electromotive force is
exploited by  electric generators (dynamo's).


\subsection{Faraday's law}
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=3cm \epsffile{farad.ps}}
\caption{\sl An electric circuit with a light bulb with resistance $R$
is located in an magnetic field $\vec{B}$. At time $t=0$ the magnetic
field is pulled to the left.}
\label{farad}
\end{center}
\end{figure}
Figure~\ref{farad} shows one of the experiments Faraday conducted in 1831.
In this experiment an electric circuit is first located in a magnetic
field. Then, at time $t=0$, the magnet is pulled away with a speed $\vec{v}$.
The height of the wire loop is $h$ and vertical part of the wire that
is in the area of the magnetic field is $s$. 
An $EMF$ leads to a current and the light bulb  flashes, just like in
the 'experiment' when the circuit was pulled away described in the
 previous section.

We ask the same question now: what is the source of the $EMF$? Well,
 based on relativity principle, this is physics-wise 
exactly the same experiment as we did before.
 So, to calculate the $EMF$ we transform this experiment 
to the previous experiment and conclude that the $EMF$ originates
from the Lorentz force.

Good, but no cigar! Let's forget about the relativity principle and
analyze this experiment based on what we
know already about electrostatic, magnetostatic and electrodynamics.
We do not transform to the previous experiment: physics-wise the
results should be invariant anyway!
In this experiment there is no moving charge, so there is no Lorentz force
as source of the $EMF$. We can write:
\begin{equation}
EMF=\int_{circuit} \vec{f} \dot d\vec{l}
    = \int_{circuit} \vec{E} \dot d\vec{l} \neq 0
\end{equation} 
How can  $\int_{circuit} \vec{E} \dot d\vec{l} \neq 0$? Well, this is
electrodynamics, not electrostatics! So, this time the $EMF$ is electric in
nature.
When we use Stokes Law, we obtain:
\begin{eqnarray}
EMF &=&  \int_{circuit} \vec{E} \dot d\vec{l} \nonumber \\
    &=&  \int_{surface-circuit} (\div \times \vec{E})\dot d\vec{o} 
\end{eqnarray} 
From the previous section we know that
\begin{eqnarray}
EMF &=& -\frac{d\Phi_B}{dt}=
        -\frac{d}{dt} \int_{surface-circuit} \vec{B} \dot d\vec{o} \nonumber \\
    &=&  - \int_{surface-circuit}  
            \frac{\partial \vec{B}}{\partial t} \dot d\vec{o}
\end{eqnarray}
When we combine these results we find Faraday's Law:
\begin{equation}
\div \times \vec{E} =  - \frac{\partial \vec{B}}{\partial t} 
\end{equation}
Hence, a changing magnetic field induces an electric field.

In the first experiment we calculated the Lorentz force and therefore
knew the direction of the current. In the second experiment it is already
much harder to figure out what the direction of the current will be.
And, you could get really lost when you have to predict the
direction of the current in case of a time dependent magnetic field!
Fortunately, there is a handy trick to do this, called Lenz's Law. It
states that the: 

'\sl{induced current attempts to compensate the change of
magnetic flux}'.

 Hence, when the magnetic flux decreases (when the circuit
is pulled out of the field) the induced current generates
a magnetic field in the original field direction 
(think of the circuit as a solenoid with one winding). Use your right hand to deduce that
the current flows counter clock wise in the experiments we just
discussed.

\subsection{Knowledge and skills}
The knowledge and skills
you should have acquired during reading of the previous can be summarized as
follows:
\begin{itemize}
\item
You can derive Ohm's Law $V=I R$ from $\vec{J}=\sigma \vec{f}$.
\item
You can calculate the $EMF$ of an electric circuit.
\item
You understand the following relation:
\begin{equation}
EMF=V_{induced}=-\frac{d\Phi_B}{dt}  \nonumber
\end{equation}
and that a changing magnetic field induces an electric field
according to the relation:
\begin{equation}
\div \times \vec{E} =  - \frac{\partial \vec{B}}{\partial t} \nonumber 
\end{equation}
\item
You know how to predict the induced current in an electric circuit 
using Lenz's Law.
\end{itemize}




\section{Electromagnetic inductance and circuits}
\subsection{Self-inductance}
Figure ~\ref{bloop} depicts an illustration
of  a current loop and its magnetic field.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=3cm \epsffile{bloop.ps}}
\caption{\sl An illustration of a current loop and its magnetic field.}
\label{bloop}
\end{center}
\end{figure}
From Biot-Savart's Law we know that whatever the exact shape
of the current loop, its magnetic field is proportional
to the current: $|\vec{B}| \propto I$. When we consider 
the magnetic flux through the loop, originating from its own
magnetic field we can write:
\begin{eqnarray}
\Phi_B &=& \int_{loop} \vec{B} \dot d\vec{o} \propto I \nonumber \\
       &\equiv& L I 
\end{eqnarray}
with $L$ a factor of proportionality called self-inductance, with unit
 Henry, 1~H=1~Vs/A.
The self-inductance only depends on the geometry
and size of the loop.

When the wire loop is placed in a magnetic field, it requires an
electromotive force to change its current:
\begin{equation}
EMF_{loop}=V_{loop}=-\frac{d\Phi_B}{dt}=L \frac{dI}{dt}
\end{equation}




\subsubsection{Self-inductance of a solenoid}
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=3cm \epsffile{lsol.ps}}
\caption{\sl A solenoid with $N$ windings per meter, radius $R$
 and length $l$. }
\label{lsol}
\end{center}
\end{figure}
Figure~\ref{lsol} shows a solenoid with radius $R$, length $l$ and $N$ windings per meter.
To determine the self-inductance of this solenoid we calculate the flux
through all windings from its own magnetic field. Using Amp\`{e}re's law we know
the field in the solenoid: $\vec{B}=\mu_0 N I \dakje{z}$.
The magnetic flux though a single winding is:
\begin{equation}
\Phi_B^1 = \int_{winding-surface} \vec{B} \dot d\vec{o} 
\end{equation}
The field points in the direction of normal of the surface, so 
\begin{eqnarray}
\Phi_B^1 &=& \int_{winding-surface} |\vec{B}| do \nonumber \\
         &=& \int_{\phi=0}^{2\pi} \int_{r=0}^R \mu_0 N I r dr d\phi \nonumber \\
         &=& \mu_0 N I \pi R^2 
\end{eqnarray}
The number of windings of the solenoid is $Nl$, which leads to a total
 flux through the solenoid of:
\begin{equation}
\Phi_B = N l \Phi_B^1  =  \mu_0 N^2 I \pi R^2 l
\end{equation}
As predicted, this flux is proportional to the current $I$. Hence, 
the self-inductance, $L$, of this solenoid is:
\begin{equation}
L=\Phi_B/I = \mu_0 N^2 l  \pi R^2
\end{equation}

\subsubsection{Self-inductance of a coaxial cable}
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=3cm \epsffile{lcoax.ps}}
\caption{\sl A coaxial cable with inner radius $a$ and outer radius $b$ and
length $l$. In the inner and outer core, but in opposite direction, flows
a current $I$.}
\label{lcoax}
\end{center}
\end{figure}
Figure~\ref{lcoax} shows a coaxial cable with inner radius $a$
and outer radius $b$ and length $l$. A current $I$ flows though the
inner core in one direction and runs in the opposite direction
in the outer core.

Inside the cable in the region $a<r<b$, there is
a  magnetic field in the $\phi$ direction. Using Amp\`{e}re's Law
we determine the magnitude of this field:
\begin{equation}
\int_{loop} \vec{B} \dot d\vec{l}= 
 \int_{0}^{2\pi} B_{\phi} d\phi= B_{\phi} 2\pi r=\mu_0 I
\end{equation}
and thus $B_{\phi}=(\mu_0 I) / (2\pi r)$.

To calculate the magnetic flux in the coaxial cable
we have to determine 'amount' of magnetic field that runs through
the region between $a<r<b$ and length $l$:
\begin{eqnarray}
\Phi_B &=& \int_0^l dl' \int_a^b dr \frac{\mu_0 I}{2\pi r} \nonumber \\
       &=& l \int_a^b dr \frac{\mu_0 I}{2\pi r} \nonumber \\
       &=& l \frac{\mu_0 I}{2\pi} ln(b/a)
\end{eqnarray}
And again, the flux is proportional to $I$ and the self-inductance becomes:
\begin{equation}
L=\frac{\Phi_B}{I}=l \frac{\mu_0}{2\pi} ln(b/a)
\end{equation}


\subsection{Mutual inductance}
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=6cm \epsffile{twoloop.ps}}
\caption{\sl Illustration of two infinitesimal loops (1 and 2) with current
$I$ that 'sense' each other's magnetic field, indicated by the arrows.
The left and right drawing depict two different orientations of the loops. }
\label{twoloop}
\end{center}
\end{figure}
Consider two identical infinitesimal loops (we use square loops) as shown
in Fig.~\ref{twoloop}, both with current $I=I_1=I_2$.
For the flux  through loop 2 from loop 1 we can 
write $\Phi_2^{(1)}=M I_1=M I$. With $M$ a
factor of proportionality, called the mutual inductance.
 Physically the situation  is completely symmetric. 
Hence, for the flux through loop 1, we can write $\Phi_1^{(2)}=M I_2=M I$,
with the same factor $M$. If we change the orientation of the loops, the
factor $M$ changes accordingly, but the symmetry
 $\Phi_1^{(2)}=\Phi_2^{(1)}$
remains.


Is this relation also valid for macroscopic loops of any size and shape?
We consider the infinitesimal loops again.
Suppose we extend the second loop, 2a, with an additional infinitesimal 
loop, 2b, (with the same current) as depicted in Fig.~\ref{threeloop}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=6cm \epsffile{threeloop.ps}}
\caption{\sl Illustration of two loops (1 and 2) with current
$I$ that 'sense' each other's magnetic field, indicated by the arrows.
The second loop consists of two infinitesimal loops, a and b.}
\label{threeloop}
\end{center}
\end{figure}
The extended loop 2=2a+2b now forms a single loop again, but just twice
as large as the original. We can write: 
\begin{equation}
\Phi_1^{(2)}= 
\Phi_1^{(a+b)}=\Phi_1^{(a)} + \Phi_1^{(b)} = M_{a} I + M_{b} I = M_{a+b} I
\end{equation}
All the arguments before allow us to write for
the individual contributions from a and b: 
\begin{eqnarray}
\Phi_1^{(a)}= M_a I =\Phi_a^{(1)} \nonumber \\
\Phi_1^{(b)}= M_b I =\Phi_b^{(1)} \nonumber 
\end{eqnarray}
and thus:
\begin{equation}
\Phi_2^{(1)}=\Phi_{(a+b)}^{(1)}=
\Phi_a^{(1)} + \Phi_b^{(1)} = M_{a} I + M_{b} I = M_{a+b} I
\end{equation}
We could extend the loop 2 again by adding another additional 
infinitesimal loop and would find the same relation again.
In fact, we can make macroscopic loop of any size and shape of a 
collection of infinitesimal loops and thus we can
conclude that
\begin{eqnarray}
\Phi_1^{(2)}= M I_2 \nonumber \\  
\Phi_2^{(1)}= M I_1 
\end{eqnarray}
where $M$ depends on the exact geometry of the configuration.

As an example we consider the configuration with a single current loop $L$,
radius $R_L$,
placed in a long solenoid $S$ with radius $R_S$ and 
N windings per meter as illustrated in
Fig.~\ref{lands}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=4cm \epsffile{lands.ps}}
\caption{\sl Illustration of a single loop $L$ in a long solenoid $S$,
both with current $I$.}
\label{lands}
\end{center}
\end{figure}
 A current $I$ runs through the winding(s) of
 $L$ and $S$. What is the flux through the solenoid $S$
from $L$. You could calculate the dipole field of $L$ and determine
 the flux that $S$ sees by integration. Yes, you could, but you should not.
Instead, you should write down:
$\Phi_L^{(S)}= M I = \Phi_S^{(L)} $. Thus, when you calculate the
flux through $L$ from $S$, $\Phi_L^{(S)}$, you also have $\Phi_S^{(L)}$.
The solenoid generates a magnetic field of $B=B_S=\mu_0 N I$ in the axial
direction. The flux through $L$ is:
\begin{equation}
\Phi_L^{(S)}= B \pi R_L^2 = \mu_0 \pi R_L^2 N I 
\end{equation}
We can read off that $M=\mu_0 \pi R_L^2 N $ and thus the
flux through the solenoid from the single loop is thus:
\begin{equation}
\Phi_S^{(L)}= M I = \mu_0 \pi R_L^2 N I 
\end{equation}



\subsection{Electric circuits}
\label{secRC}
In the previous sections we discussed the relation between the current and
the $EMF$ for a resistor and for an inductor. Another object is the capacitor
which was introduced in the course on electrostatics: $V_C=Q/C$.
 Analogous to the
self-inductance $L$ for an inductor, the capacitance $C$ depends only
on the geometry of the configuration. The $EMF$ to charge a capacitor
equals  $V_C$ and a relation with the current can be easily derived:
\begin{equation}
\frac{dV_C}{dt}=\frac{1}{C} \frac{dQ}{dt}=\frac{1}{C} I
\end{equation}

Now we have 
three objects that we can combine to build electric circuits.
Electric circuits that we consider consist of a battery that provides
the potential difference $V_0$ leading to current through
a closed circuit that consist of, or combinations of, 
resistor(s), capacitor(s) and inductor(s).
The voltage change going around the circuit in the direction
of the current flow should be zero (Kirchoff's Voltage Law). Note
that we wrote voltage $change$, we can have a voltage drop or a voltage rise,
 implying that we should keep track of $+$ and $-$ signs once again.

For our three objects the voltage always drops when we follow the
current $I$ in the positive direction. Remember that when 
we cross a battery (in the direction of positive current) the voltage
increases by the battery potential $V_0$.
We summarize these  properties in Table~\ref{tablelrc}.
 \begin{table}[thb]
\begin{center}
\begin{tabular}{|l|c|c|} \hline
   Object          &   Potential change  &  relation with $I$   \\ \hline
Resistor $R$       &  $-V_R$             &  $V_R= I R$          \\ 
Inductor $L$           &  $-V_L$             &  $V_L=L\frac{dI}{dt}$ \\
Capacitor $C$      &  $-V_C$             &  $\frac{dV_C}{dt}=\frac{1}{C} I$ \\
Battery $V_0 $     &  $ V_0$             &   no resistance  \\ \hline
\end{tabular}
\caption{\sl The voltage change over possible elements in an
electric circuit when we follow the direction 
of the positive current.}
\label{tablelrc}
\end{center}
\end{table}

Consider the electric circuit consisting of a battery with 
potential $V_0$,  
a self-inductance $L$ and a resistor $R$, shown in Fig.~\ref{LR}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=4cm \epsffile{LR.ps}}
\caption{\sl An electric circuit with a battery,
 resistor and a self-inductance.}
\label{LR}
\end{center}
\end{figure}
At time $t=0$ the battery is just hooked on and
 the current through the circuit is zero: $I(0)=0$.
What is the current at any time? To solve this problem, we 'walk' around
starting just before the resistor and add all voltage changes:
We use Table~\ref{tablelrc} and obtain:
\begin{eqnarray}
-V_R - V_L + V_0 = 0 \nonumber \\
V_0 - V_L = V_R \nonumber \\
V_0 - L \frac{dI}{dt} = I(t) R \nonumber \\
\frac{dI}{dt} = -\frac{R}{L} I(t) - \frac{V_0}{L}  
\end{eqnarray}
Physically we can interpret the last equation as the electromotive force
$V_0 - L\frac{dI}{dt}$ 
of a battery and an inductor that establishes the potential $V_R=I R$ to
drive a current through a resistor.

Mathematically, it is
 a  first order differential equation with the 
structure $\frac{df}{dt}=a f + b$.
The general solution is $f(t)=k e^{at} - \frac{b}{a}$, with $a$, $b$ and $k$ 
constants.
Returning to our $LR$ circuit, we write:
\begin{equation}
I(t)=\frac{V_0}{R}(1-e^{-(R/L)t})
\end{equation}
Figure~\ref{ilr} shows the current as function of time. At time intervals
$L/R$, called the time constant, the current increases with
  fractions $1-1/e$.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=3cm \epsffile{ilr.ps}}
\caption{\sl The current of an $LR$ circuit as function of time.}
\label{ilr}
\end{center}
\end{figure}

Another first order circuit is shown in Fig.~\ref{RC}, which consists 
of a resistor, a capacitor and a battery. 
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=4cm \epsffile{RC.ps}}
\caption{\sl An electric circuit with a battery, resistor and a capacitor.}
\label{RC}
\end{center}
\end{figure}
At time $t=0$, the battery is just hooked on,
 the charge on the capacitor $Q_C(0)=0$ (and $V_C(0)=0$) 
and the current $I(0)=0$.
What is the potential over the capacitor at any time? 
We 'walk' around
starting just before the resistor and add all voltage changes:
We use Table~\ref{tablelrc} and obtain:
\begin{eqnarray}
-V_R - V_C + V_0 &=& 0 \nonumber \\
V_0 - V_C &=& V_R \nonumber \\
V_0 - V_C &=& I(t) R \nonumber \\ 
V_0 - V_C &=& RC \frac{dV_C}{dt}   
\end{eqnarray}
with solution: 
\begin{equation}
V_C(t)=V_0(1-e^{-t/(RC)})
\end{equation}
The time constant of this circuit is $RC$ as
illustrated by Fig.~\ref{irc}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=3cm \epsffile{irc.ps}}
\caption{\sl The potential difference over a capacitor in
 an $RC$ circuit as function of time.}
\label{irc}
\end{center}
\end{figure}

\subsection{Energy in electric circuits}
\subsubsection{Energy dissipation in a resistor}
Consider the basic circuit of a battery and a resistor, 
shown in Fig.~\ref{rcircuit} below. 
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=4cm \epsffile{rcircuit.ps}}
\caption{\sl An electric circuit consisting of a battery $V_0$ and
a resistor $R$.}
\label{rcircuit}
\end{center}
\end{figure}
Each second a charge $Q=I$ is pumped through the
 resistor, which requires work $W$.
 Some chemical reaction in the battery provides the
necessary energy. The dissipated energy in the resistor is transformed
into heat. 
 Since the voltage change over the resistor is $V=V_0$, the
energy needed to transport the charge equals $W=V Q$.
 The energy per second $dW/dt$, the power, 
dissipated by resistor is $P=V Q/1$s~$=V I$ with
unit Watt. Using Ohm's Law we obtain the equivalent expressions
 $P=I^2 R$ and $P=V^2/R$.

\subsubsection{Energy stored in a capacitor}
In the course on electrostatics we have derived that the energy 
in a capacitor is given by $W=\frac{1}{2} C V^2$. In this section we
study the energy stored in a capacitor 
in an electric circuit and obviously expect to find the same result.
Consider an $RC$ circuit as discussed in section~\ref{secRC}.
From the calculation in this section we conclude that the
capacitor gets charged (with time constant $RC$).
 For the power required to charge the capacitor we write:
\begin{equation}
P=\frac{dW_C}{dt}=V_C(t) I(t) = \frac{Q_C}{C} \frac{dQ_C}{dt}
\end{equation}
The total energy in a charged capacitor ($Q_C=Q$) is obtained by integration
over time:
\begin{eqnarray}
W&=& \int_{t=0}^{\infty} \frac{dW}{dt} dt =
     \int_{t=0}^{\infty}  \frac{Q_C}{C} \frac{dQ_C}{dt} dt \nonumber \\
 &=&  \int_{t=0}^{\infty}  \frac{Q_C}{C} \frac{dQ_C}{dt} dt \nonumber \\
 &=&  \int_{Q_C=0}^{Q}  \frac{Q_C}{C} dQ_C =
         \frac{1}{2} \frac{Q^2}{C} = \frac{1}{2} C V^2
\end{eqnarray}
which is no surprise.


An alternative expression for the energy in the electric field, also derived
in the Chapter on electrostatics, is:
\begin{equation}
W_C=\frac{\epsilon_0}{2} \int_{volume} E^2 dv
\end{equation}
Let's check this expression for a parallel plate capacitor.
The distance between the plates is $d$ and the plates have a
surface area $A$, leading to a capacity $C=\frac{\epsilon_0 A}{d}$.
In the ideal case, the electric field outside the plates
is zero, while in between the plates the field is given by $|\vec{E}|=V/d$.
For the energy we find
\begin{eqnarray}
W_C &=&\frac{\epsilon_0}{2} \int_{volume} \frac{V^2}{d^2} dv \nonumber \\
  &=&\frac{\epsilon_0}{2} \frac{V^2}{d^2} A d=
     \frac{\epsilon_0 A}{2 d} V^2=\frac{1}{2} C V^2
\end{eqnarray}
as expected.


\subsubsection{Energy stored in a self-inductance}
To determine the energy stored in a self-inductance we can follow the 
same strategy as above.
For the power required to reach a current $I_L=I$ in a self-inductance $L$
 we write:
\begin{equation}
P=\frac{dW_L}{dt}=I_L(t) V_{induced}(t) = I_L(t) L \frac{dI_L}{dt}
\end{equation}
The total energy in the self-inductance is obtained by integration
over time:
\begin{eqnarray}
W_L&=& \int_{t=0}^{\infty} \frac{dW}{dt} dt =
     \int_{t=0}^{\infty}  I_L(t) L \frac{dI_L}{dt} dt \nonumber \\
 &=&  \int_{t=0}^{\infty}  I_L(t) L dI_L \nonumber \\
 &=&     \frac{1}{2} L I^2 \label{eq:ul}
\end{eqnarray}

This formula has the same structure as the expression for a capacitor.
Is there also an alternative expression for energy in a self-inductance
in terms of the magnetic field? Well, we could just try:
\begin{equation}
\label{eq:ub}
W_L=\frac{1}{2 \mu_0} \int_{volume} B^2 dv
\end{equation}
Let's check this expression for a solenoid. 
For a solenoid $L=\mu_0 l N^2 \pi R^2$ and thus (using equation~\ref{eq:ul})
 $W_L=\frac{\mu_0}{2}  l N^2 \pi R^2 I^2$. Now we use the
trial equation~\ref{eq:ub}:
\begin{eqnarray}
W_L&=& \frac{1}{2 \mu_0} \int_{volume} (\mu_0 N I)^2 dv \nonumber \\
   &=& \frac{1}{2 \mu_0} (\mu_0 N I)^2 l\pi R^2  \nonumber \\
   &=& \frac{\mu_0}{2} l N^2 \pi R^2 I^2 
\end{eqnarray}
 which leads to the same result! Equation~\ref{eq:ub} is indeed
the correct expression for the energy of the magnetic field. By the
way, we can also use this expression for the energy to calculate
the self-inductance of an object: $L=2 W_L /I^2$.



\subsection{Knowledge and skills}
The knowledge and skills
you should have acquired during reading of the previous can be summarized as
follows:
\begin{itemize}
\item
You understand the meaning of self-inductance and mutual inductance.
\item
You can calculate the self-inductance of the solenoid and a coaxial cable.
\item
You can calculate the time dependent current in  $RC$ and $LR$ circuits.
\item
You can calculate the energy in a self-inductance and in the magnetic field.
\end{itemize}



\section{Maxwell equations}
In this section we complete the field equations and derive the existence of
electromagnetic waves: light! First we summarize the field equations,
we encountered so far.
\begin{table}[thb]
\begin{center}
\begin{tabular}{|l|l|} \hline
Comment              & Integral           \\ \hline 
Gauss          & $\int_{surface} \vec{E} 
                        \dot d\vec{o}=Q_{enclosed}/\epsilon_0 = 
                        \int_{volume} \frac{\rho}{\epsilon_0} dv$   \\
No magnetic monopoles& $\int_{surface} \vec{B} \dot d\vec{o}=0 $ \\
 Faraday             & $\int_{loop} \vec{E} \dot d\vec{l}=-\frac{d\Phi_B}{dt}$ 
                                                                            \\
Amp\`{e}re           & $\int_{loop} \vec{B} \dot d\vec{l}=\mu_0 I_{enclosed}=
                         \int_{surface} \vec{J} \dot d\vec{o} $
                  \\ \hline \hline
                & Differential \\ \hline
Gauss            &  $\div \dot \vec{E}=\rho/\epsilon_0$ \\
No magnetic monopoles     &  $\div \dot \vec{B}=0$ \\
Faraday             &  $\div \times \vec{E} = -\frac{\partial \vec{B} }{\partial t}$ \\
Amp\`{e}re            &  $\div \times \vec{B} = \mu_0 \vec{J}$ \\ \hline
\end{tabular}
\caption{\sl The (incomplete) field equations based on electrostatics, magnetostatics and Faraday's Law in integral and differential form.
}
\label{sofar}
\end{center}
\end{table}
Look at the structure of these equations in Table~\ref{sofar}.
 There is an asymmetry between
the electric and magnetic field, due to the fact that there exist no
magnetic monopoles. In addition, there is nowhere a term
  $\frac{\partial \vec{E} }{\partial t}$. Did we miss something? How
can we find that term, called the Maxwell term?

\subsection{The Maxwell term from a gedanken experiment}
 At time $t=0$ an empty capacitor is charged in a circuit as depicted
in Fig.~\ref{cmaxwell}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=7cm \epsffile{cmaxwell.ps}}
\caption{\sl An electric circuit with a parallel plate capacitor. 
An Amperian loop is indicated. Also indicated are two surfaces ($a$ and $b$)
that are enclosed by the Amperian loop. Surface $a$ has the 'usual' shape of
a disk. Surface $b$ has the shape of a balloon and is stretched in between
the plates of the capacitor.}
\label{cmaxwell}
\end{center}
\end{figure}
 Around the wire at position $a$ a magnetic field
is generated by the current $I$ such that Amp\`{e}re's law is fulfilled:
\begin{equation}
\int_{loop\,a} \vec{B} \dot d\vec{l} = \mu_0 I
\label{eq:sa}
\end{equation}
In this case, the  loop $a$ spans a surface, a disk,
 penetrated by the the wire, thus the enclosed current is $I$.

Now we modify the surface a little bit and stretch it in between the capacitor plates,
such that it acquires the shape of a balloon $b$.
In fact, no wire is pointing through the surface and the the enclosed
current is zero:
\begin{equation}
\int_{loop\,b} \vec{B} \dot d\vec{l} = 0
\end{equation}
which suggests that the magnetic field has vanished
in conflict with equation~\ref{eq:sa}. Something must be
wrong with Amp\`{e}re's law! How can we fix that? Well, the electric field
between the plates of the capacitors is
 $|\vec{E}|=\frac{\sigma}{\epsilon_0}=\frac{Q(t)}{A \epsilon_0}$. The time
derivative is
 $| \frac{\partial \vec{E} }{\partial t} |= 
    \frac{1}{A \epsilon_0} \frac{dQ}{dt}= \frac{I}{A \epsilon_0}$.
A changing electric field is related to a current, which fixes Amp\`{e}re's Law:
\begin{equation}
\int_{loop} \vec{B} \dot d\vec{l} = \mu_0 I_{enclosed} + 
\mu_0 \epsilon_0 \int_{surface}  \frac{\partial \vec{E} }{\partial t} \dot d\vec{o} 
\end{equation}
and we conclude that a changing electric field induced a
magnetic field!

This calculation may look not so scientific to you or perhaps it even
 looks like a hat-trick. Right, but nevertheless the result correctly
describes the original gedanken experiment!

\subsection{The Maxwell term from a controlled charge explosion}
Another example that yields the Maxwell term is a slowly exploding charge as
illustrated in Fig.~\ref{spuit}.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=5cm \epsffile{spuit.ps}}
\caption{\sl A large amount of charge at the origin slowly 'explodes'. 
 The charge that emerges from the explosion traverses an imaginary sphere.
On the surface of this sphere an
 infinitesimal Amperian loop is also indicated }
\label{spuit}
\end{center}
\end{figure}
A large collection of charge at the origin slowly explodes. In the figure,
an imaginary sphere is also shown. The charge that emerges from the explosion
uniformly traverses the sphere. You could compare this situation with
a radioactive decay. However, in radioactive decays usually, besides
electrons, also photons and  neutrinos are produced. Anyway, charge-wise, the
explosion is comparable to radioactive decays. When we start
with $N_0$ nuclei with charge $q$ and  lifetime $\tau$, 
then the number of nuclei at a given time is $N(t)=N_0 e^{-t/\tau}$ and the total charge is
$Q(t)=q N(t)$. The electricfield of the charged nuclei is given by:
\begin{equation}
\vec{E}=\frac{1}{4\pi \epsilon_0} \frac{q N_0  e^{-t/\tau}}{r^2} \dakje{r}
\label{eq:efcn}
\end{equation}
Each decay produces a charge $q$ which leads to a 'shower of charge' that escapes:
with corresponding current:
\begin{equation}
 I=-\frac{dQ(t)}{dt}= \frac{q N_0}{\tau} e^{-t/\tau}
\end{equation}
For the current density through a spherical surface we find:
\begin{equation}
\vec{J}= \frac{q N_0}{\tau} \frac{e^{-t/\tau}}{4\pi r^2} \dakje{r}
\label{eq:cd}
\end{equation}
where $r$ represents the radius of the surface.

Now we investigate the integral $\int_{loop} \vec{B}(r) \dot d\vec{l} $
of the magnetic field over a small loop on the surface of the sphere, 
also indicated in Fig.~\ref{spuit}.
The dot-product filters out the component of the magnetic field along the
surface of the sphere, $B_{//}$. What is the magnitude of this component?
Well, look at the symmetry of the configuration. The can not be a component
of the magnetic field along the surface, and thus:
\begin{equation}
\int_{loop} \vec{B}(r) \dot d\vec{l} =
       \int_{loop} B_{//}(r) dl = 0
\label{eq:bt}
\end{equation}
According to Amp\`{e}re's law, this loop integral should be
proportional with the enclosed
current: $ \mu_0 I_{enclosed}= \mu_0 \vec{J} \dot d\vec{o} $.
 Is the enclosed current zero as equation~\ref{eq:bt} demands? No, because
there is certainly charge showering through the surface 
enclosed by the loop. So,
Amp\`{e}re's Law fails again. Let's see whether the changing electric field
completes the equation:
\begin{eqnarray}
\int_{loop} \vec{B} \dot d\vec{l} &=& \mu_0 I_{enclosed} + 
\mu_0 \epsilon_0 \int_{surface}  \frac{\partial \vec{E} }{\partial t} \dot d\vec{o} 
 \\ \nonumber
  &=& \mu_0  \int_{surface}  \vec{J} \dot d\vec{o} +
\mu_0 \epsilon_0 \int_{surface}  \frac{\partial \vec{E} }{\partial t} \dot d\vec{o} 
\end{eqnarray}
Now we substitute the expressions we derived for the current density
 (equation~\ref{eq:cd})
and the electric field (equation~\ref{eq:efcn}) and obtain:
\begin{eqnarray}
\int_{loop} \vec{B} \dot d\vec{l}   &=&
  \mu_0  \int_{surface}  N_0 \frac{Q}{\tau} \frac{e^{-t/\tau}}{4\pi r^2}  do +
\mu_0 \epsilon_0 \int_{surface} \frac{N_0 Q}{4\pi \epsilon_0 r^2}
        \frac{\partial e^{-t/\tau} }{\partial t}  do  \nonumber \\
&=&
  \mu_0  \int_{surface}  N_0 \frac{Q}{\tau} \frac{e^{-t/\tau}}{4\pi r^2}  do -
  \mu_0  \int_{surface}  N_0 \frac{Q}{\tau} \frac{e^{-t/\tau}}{4\pi r^2}  do
 \nonumber \\ 
&=& 0
\end{eqnarray}
So, the Maxwell term 
$\mu_0 \epsilon_0 \int_{surface} 
 \frac{\partial \vec{E} }{\partial t} \dot d\vec{o}$, 
 saves the day again!

\subsection{Continuity equation}
In the previous section we used the fact that charge is conserved.
A current through a (closed) surface was the
result of a changing charge in the enclosed volume:
\begin{eqnarray}
I_{surface}  &=& - \frac{dQ_{volume}}{dt} \nonumber \\
\int_{surface} \vec{J} \dot d\vec{o} &=&
           - \frac{d}{dt} \int_{volume}\rho dv  \nonumber \\
\int_{surface} \vec{J} \dot d\vec{o} + \frac{d}{dt} \int_{volume}\rho dv &=&0
\end{eqnarray}
Using Gauss's rule we can write:
\begin{equation}
\int_{volume} \div \dot \vec{J}  dv + \int_{volume}\frac{d\rho}{dt} dv =0
\end{equation}
and thus:
\begin{equation}
\div \dot \vec{J} + \frac{d\rho}{dt} =0
\label{eq:cont}
\end{equation}
This expression is known as the continuity equation. 


Is this consistent with our renewed Amp\`{e}re's Law? We start with:
\begin{equation}
\int_{loop} \vec{B} \dot d\vec{l} = 
\mu_0 \int_{surface} \vec{J} \dot d\vec{o} +
\mu_0 \epsilon_0 \int_{surface}  \frac{\partial \vec{E} }{\partial t} \dot d\vec{o} 
\end{equation}
and use Stoke's Law to write:
\begin{equation}
\int_{surface} \div \times \vec{B} \dot d\vec{o} = 
\mu_0 \int_{surface} \vec{J} \dot d\vec{o} +
\mu_0 \epsilon_0 \int_{surface}  \frac{\partial \vec{E} }{\partial t} \dot d\vec{o} 
\end{equation}
 Note that the surface is arbitrary, thus the integrands must be equal:
\begin{equation}
 \div \times \vec{B}  = 
\mu_0  \vec{J} + \mu_0 \epsilon_0   \frac{\partial \vec{E} }{\partial t} 
\end{equation}
(which by the way is the renewed Amp\`{e}re's Law in differential form!)
Now take the divergence of this equation and realize that the divergence of
a rotation is zero by construction:
\begin{equation}
\div \dot  \div \times \vec{B}  = 0 = 
\mu_0 \div \dot \vec{J} + \mu_0 \epsilon_0 \div \dot  \frac{\partial \vec{E} }{\partial t} 
\end{equation}
or equivalently
\begin{equation}
\div \dot  \div \times \vec{B}  = 0 = 
\mu_0 \div \dot \vec{J} + \mu_0   \frac{\partial \rho }{\partial t} 
\end{equation}
from which we can read off the continuity equation. As a matter of fact
we can conclude that  Amp\`{e}re's Law extended with Maxwell's term
leads to charge conservation.


\subsection{The complete set of Maxwell equations}
The time has come to write down the complete set of field equations which we
derived during our tour through electrostatics, magnetostatics and
electrodynamics leading to Faraday's Law and the Maxwell term. The
equations are know as the Maxwell Equations and are listed in table~\ref{tmax}.
\begin{table}[thb]
\begin{center}
\begin{tabular}{|l|l|} \hline
Comment              & Integral           \\ \hline 
Gauss          & $\int_{surface} \vec{E} 
                        \dot d\vec{o}=Q_{enclosed}/\epsilon_0 = 
                        \int_{volume} \frac{\rho}{\epsilon_0} dv$   \\
No magnetic monopoles& $\int_{surface} \vec{B} \dot d\vec{o}=0 $ \\
 Faraday             & $\int_{loop} \vec{E} \dot d\vec{l}=-\frac{d\Phi_B}{dt}$ 
                                                                            \\
Amp\`{e}re+Maxwell term  &
  $\int_{loop} \vec{B} \dot d\vec{l}=\mu_0 I_{enclosed} + \mu_0 \epsilon_0 \int_{surface}  \frac{\partial \vec{E} }{\partial t} \dot d\vec{o} $
                  \\ \hline \hline
                & Differential \\ \hline
Gauss           &  $\div \dot \vec{E}=\rho/\epsilon_0$ \\
No magnetic monopoles     &  $\div \dot \vec{B}=0$ \\
Faraday             &  $\div \times \vec{E} = -\frac{\partial B}{\partial t}$ \\
Amp\`{e}re+Maxwell term  & 
 $\div \times \vec{B} = \mu_0 \vec{J} + \mu_0 \epsilon_0   \frac{\partial \vec{E} }{\partial t}$ \\ \hline
\end{tabular}
\caption{\sl The complete set of 
 field equations based on electrostatics, magnetostatics and electrodynamics,
called the Maxwell Equations, in integral and differential form.
}
\label{tmax}
\end{center}
\end{table}
The physical behavior of electric and magnetic field are described by these
equations. In the following section we use the Maxwell Equations
to derive the existence of electromagnetic waves of which light is a specific example.


\subsection{Electromagnetic waves}
The  Maxwell equations, based on empirical studies, can now be used to 
further investigate the physics of the electric and magnetic field.
In this section we study the fields in free space, which
 requires some mathematics, but the result will be worth it.

In vacuum, away from electric charges and currents the Maxwell Equations
(in differential form) simplify to:
\begin{eqnarray}
  (a) & &  \div \dot \vec{E}=0   \nonumber \\ 
  (b) & & \div \times \vec{E}=-\frac{\partial \vec{B}}{\partial t} \nonumber \\
  (c) & &  \div \dot \vec{B}=0    \nonumber \\ 
  (d) & &  \div \times \vec{B}=\mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t} \label{eq:vac} 
\end{eqnarray}
Now take the rotation of~\ref{eq:vac}$b$ and write:
\begin{equation}
\div \times \div \times \vec{E} = - \div \times \frac{\partial \vec{B}}{\partial t}
\end{equation}

First, we concentrate on the left hand side. Using basic calculus we obtain:
\begin{equation}
\div \times \div \times \vec{E} = \div (\div \dot \vec{E}) - \div^2 \vec{E}
\end{equation}
Remember that we are in vacuum and use equation~\ref{eq:vac}$a$ to write:
\begin{equation}
\div \times \div \times \vec{E} = - \div^2 \vec{E}
\end{equation}

Now, we proceed with the right hand side:
\begin{equation}
- \div \times \frac{\partial \vec{B}}{\partial t} =
- \frac{\partial \div \times \vec{B}}{\partial t} 
\end{equation}
and use~\ref{eq:vac}$d$ to write:
\begin{equation}
- \div \times \frac{\partial \vec{B}}{\partial t} =
- \mu_0 \epsilon_0 \frac{\partial^2 \vec{E} }{\partial t^2} 
\end{equation}

When the results are combined we obtain:
\begin{equation}
\div^2 \vec{E} = \mu_0 \epsilon_0 \frac{\partial^2 \vec{E} }{\partial t^2} 
\label{egolf}
\end{equation}
Starting with the rotation of~\ref{eq:vac}$d$ you find a similar relation for the magnetic field:
\begin{equation}
\div^2 \vec{B} = \mu_0 \epsilon_0 \frac{\partial^2 \vec{B} }{\partial t^2} 
\label{bgolf}
\end{equation}

Fine, so what? Well, remember the theory of waves. The classical wave equation
for a wave in the $z$ direction with speed $v$ is:
\begin{equation}
\frac{\partial^2 \psi}{\partial t^2} = v^2 \frac{\partial^2 \psi}{\partial z^2}
\end{equation}
with typical solution $A\, cos(k z - \omega t)$ where $\omega^2=k^2 v^2$.
Hence, the equations~\ref{egolf} and~\ref{bgolf} imply the existence of electromagnetic waves
with speed $v=\sqrt{ \frac{1}{\mu_0 \epsilon_0 } } = c$, the speed
of light!

\subsubsection{Monochromatic waves in one dimension}
In this section we study
a typical solution of the electromagnetic wave equation.
To simplify the math, we assume:
\begin{itemize}
\item The electric and magnetic field only depend on $z$ and $t$;
\item We stay in the vacuum which is infinitely large.
\end{itemize}
This leads to the following wave equations:
\begin{eqnarray}
\frac{\partial^2 \vec{E} }{\partial t^2}&=& c^2
         \frac{\partial^2 \vec{E} }{\partial z^2}
\nonumber \\
\frac{\partial^2 \vec{B} }{\partial t^2}&=& c^2
         \frac{\partial^2 \vec{B} }{\partial z^2}
\nonumber 
\end{eqnarray}
with solutions of the form:
\begin{eqnarray}
\vec{E}=\vec{E}^0 cos(k z - \omega t) 
\nonumber \\
\vec{B}= \vec{B}^0 cos(k z - \omega t) 
\end{eqnarray}
with $\omega=k c$. The $constants$ $\vec{E}^0$ and $\vec{B}^0$
can be determined by applying Maxwell Equations in vacuum (again):
\begin{itemize}
\item $\div \dot \vec{E}=0 = -E^0_z k sin(k z -\omega t)$  implies that
 $E^0_z=0$.
\item  $\div \dot \vec{B}=0 = -B^0_z k sin(k z -\omega t)$  implies that
 $B^0_z=0$.
\item 
$\div \times \vec{B} = \frac{1}{c^2}\frac{\partial \vec{E} }{\partial t}$
leads to:
\begin{eqnarray}
+cB^0_y=E_x^0 \nonumber \\
-cB^0_x=E_y^0 
\end{eqnarray}
which can be written as:
\begin{equation}
\vec{B}^0=\frac{1}{c} \dakje{z} \times \vec{E}^0
\end{equation}
The equation
$\div \times \vec{E} = -\frac{\partial \vec{B} }{\partial t}$ implies the
same and adds no information.
\end{itemize}

We summarize our findings by saying that the Maxwell Equations have solutions
that can be interpreted as electromagnetic waves with speed $c$, the speed
of light. These waves are transverse: the
 electric and magnetic fields have no components in the
direction of propagation ($E_z=B_z=0$). In addition the fields are also
mutual transverse and in phase. Figure~\ref{wave} shows an illustration
of an electromagnetic wave.
\begin{figure}[htb]
\begin{center}
\mbox{\epsfysize=8cm \epsffile{wave.ps}}
\caption{\sl A schematic view of an electromagnetic wave. }
\label{wave}
\end{center}
\end{figure}
 
\subsection{Knowledge and skills}
The knowledge and skills
you should have acquired during reading of the previous can be summarized as
follows:
\begin{itemize}
\item
You understand the problem in Amp\`{e}re's Law.
\item
You can fix Amp\`{e}re's Law.
\item
You can write down the Maxwell Equations.
\item
You can derive the continuity equation from these equations.
\item
From the Maxwell Equations in vacuum you can proof the existence
of light and its properties.
\end{itemize}






\end{document}
