The speed of light
We have a wave which leaves the material source and goes outward at
the velocity $c$, which is the speed of light. But let’s go back a
moment. From a historical point of view, it wasn’t known that the
coefficient $c$ in Maxwell’s equations was also the speed of light
propagation. There was just a constant in the equations. We have
called it $c$ from the beginning, because we knew what it would turn
out to be. We didn’t think it would be sensible to make you learn the
formulas with a different constant and then go back to substitute $c$
wherever it belonged. From the point of view of electricity and
magnetism, however, we just start out with two constants, $\epsO$
and $c^2$, that appear in the equations of electrostatics and
magnetostatics:
\begin{equation}
\label{Eq:II:18:14}
\FLPdiv{\FLPE} =\frac{\rho}{\epsO}
\end{equation}

and
\begin{equation}
\label{Eq:II:18:15}
\FLPcurl{\FLPB} =\frac{\FLPj}{\epsO c^2}.
\end{equation}

If we take any arbitrary definition of a unit of charge, we can
determine experimentally the constant $\epsO$ required in
Eq. (18.14)—say by measuring the force between two unit
charges at rest, using Coulomb’s law. We must also determine
experimentally the constant $\epsO c^2$ that appears in
Eq. (18.15), which we can do, say, by measuring the force
between two unit currents. (A unit current means one unit of charge per
second.) The ratio of these two experimental constants is $c^2$—just
another “electromagnetic constant.” Notice now that this constant $c^2$ is the same no matter what we
choose for our unit of charge. If we put twice as much
“charge”—say twice as many proton charges—in our “unit” of
charge, $\epsO$ would need to be one-fourth as large. When we pass two
of these “unit” currents through two wires, there will be twice as
much “charge” per second in each wire, so the force between two
wires is four times larger. The constant $\epsO c^2$ must be reduced
by one-fourth. But the ratio $\epsO c^2/\epsO$ is unchanged. So just by experiments with charges and currents we find a
number $c^2$ which turns out to be the square of the velocity of propagation
of electromagnetic influences. From static measurements—by measuring
the forces between two unit charges and between two unit currents—we
find that $c=3.00\times10^8$ meters/sec. When
Maxwell first made this
calculation with his equations, he said that bundles of electric and
magnetic fields should be propagated at this speed. He also remarked
on the mysterious coincidence that this was the same as the speed of
light. “We can scarcely avoid the inference,” said
Maxwell, “that
light consists in the transverse undulations of the same medium which
is the cause of electric and magnetic phenomena.” Maxwell had made one of the
great unifications of physics. Before his
time, there was light, and there was electricity and magnetism. The
latter two had been unified by the experimental work of
Faraday, Oersted, and Ampère. Then, all of a
sudden, light was no longer “something else,” but was only electricity
and magnetism in this new form—little pieces of electric and magnetic
fields which propagate through space on their own. We have called your attention to some characteristics of this special
solution, which turn out to be true, however, for any
electromagnetic wave: that the magnetic field is perpendicular to the
direction of motion of the wavefront; that the electric field is
likewise perpendicular to the direction of motion of the wavefront;
and that the two vectors $\FLPE$ and $\FLPB$ are perpendicular to each
other. Furthermore, the magnitude of the electric field $E$ is equal
to $c$ times the magnitude of the magnetic field $B$. These three
facts—that the two fields are transverse to the direction of
propagation, that $\FLPB$ is perpendicular to $\FLPE$, and that
$E=cB$—are generally true for any electromagnetic wave. Our special
case is a good one—it shows all the main features of electromagnetic
waves.