A travelling field
Now for the new consequences. They come from putting together all of
Maxwell’s equations. First, let’s see what would happen in a
circumstance which we pick to be particularly simple. By assuming that
all the quantities vary only in one coordinate, we will have a
one-dimensional problem. The situation is shown in
Fig. 18–3. We have a sheet of charge located on the
$yz$-plane. The sheet is first at rest, then instantaneously given a
velocity $u$ in the $y$-direction, and kept moving with this constant
velocity. You might worry about having such an “infinite”
acceleration, but it doesn’t really matter; just imagine that the
velocity is brought to $u$ very quickly. So we have suddenly a surface
current $J$ ($J$ is the current per unit width in the $z$-direction). To
keep the problem simple, we suppose that there is also a stationary
sheet of charge of opposite sign superposed on the $yz$-plane, so that
there are no electrostatic effects. Also, although in the figure we show
only what is happening in a finite region, we imagine that the sheet
extends to infinity in $\pm y$ and $\pm z$. In other words, we have a
situation where there is no current, and then suddenly there is a
uniform sheet of current. What will happen? Well, when there is a sheet of current in the plus $y$-direction,
there is, as we know, a magnetic field generated which will be in the
minus $z$-direction for $x>0$ and in the opposite direction for
$x<0$. We could find the magnitude of $\FLPB$ by using the fact that
the line integral of the magnetic field will be equal to the current
over $\epsO c^2$. We would get that $B=J/2\epsO c^2$ (since the
current $I$ in a strip of width $w$ is $Jw$ and the line integral
of $\FLPB$ is $2Bw$). This gives us the field next to the sheet—for small $x$—but since
we are imagining an infinite sheet, we would expect the same argument
to give the magnetic field farther out for larger values of $x$.
However, that would mean that the moment we turn on the current,
the magnetic field is suddenly changed from zero to a finite value
everywhere. But wait!  If the magnetic field is suddenly changed, it
will produce tremendous electrical effects. (If it changes in
any way, there are electrical effects.) So because we moved the
sheet of charge, we make a changing magnetic field, and therefore
electric fields must be generated. If there are electric fields
generated, they had to start from zero and change to something
else. There will be some $\ddpl{\FLPE}{t}$ that will make a
contribution, together with the current $J$, to the production of the
magnetic field. So through the various equations there is a big
intermixing, and we have to try to solve for all the fields at once. By looking at the Maxwell equations alone, it is not easy to see
directly how to get the solution. So we will first show you what the
answer is and then verify that it does indeed satisfy the
equations. The answer is the following: The field $\FLPB$ that we
computed is, in fact, generated right next to the current sheet (for
small $x$). It must be so, because if we make a tiny loop around the
sheet, there is no room for any electric flux to go through it. But
the field $\FLPB$ out farther—for larger $x$—is, at first,
zero. It stays zero for awhile, and then suddenly turns on. In short,
we turn on the current and the magnetic field immediately next to it
turns on to a constant value $\FLPB$; then the turning on of $\FLPB$
spreads out from the source region. After a certain time, there is a
uniform magnetic field everywhere out to some value $x$, and then zero
beyond. Because of the symmetry, it spreads in both the plus and minus
$x$-directions. The $\FLPE$-field does the same thing. Before $t=0$ (when we turn on
the current), the field is zero everywhere. Then after the time $t$,
both $\FLPE$ and $\FLPB$ are uniform out to the distance $x=vt$, and
zero beyond. The fields make their way forward like a tidal wave, with
a front moving at a uniform velocity which turns out to be $c$, but
for a while we will just call it $v$. A graph of the magnitude of
$\FLPE$ or $\FLPB$ versus $x$, as they appear at the time $t$, is
shown in Fig. 18–4(a). Looking again at
Fig. 18–3, at the time $t$, the region between $x=\pm vt$
is “filled” with the fields, but they have not yet reached beyond. We
emphasize again that we are assuming that the current sheet and,
therefore the fields $\FLPE$ and $\FLPB$, extend infinitely far in both
the $y$- and $z$-directions. (We cannot draw an infinite sheet, so we
have shown only what happens in a finite area.) We want now to analyze quantitatively what is happening. To do that,
we want to look at two cross-sectional views, a top view looking down
along the $y$-axis, as shown in Fig. 18–5, and a side
view looking back along the $z$-axis, as shown in
Fig. 18–6. Suppose we start with the side view. We see the
charged sheet moving up; the magnetic field points into the page
for $+x$, and out of the page for $-x$, and the electric field is downward
everywhere—out to $x=\pm vt$. Let’s see if these fields are consistent with Maxwell’s
equations. Let’s first draw one of those loops that we use to
calculate a line integral, say the rectangle $\Gamma_2$ shown in
Fig. 18–6. You notice that one side of the rectangle is
in the region where there are fields, but one side is in the region
the fields have still not reached. There is some magnetic flux through
this loop. If it is changing, there should be an emf around it. If the
wavefront is moving, we will have a changing magnetic flux, because
the area in which $\FLPB$ exists is progressively increasing at the
velocity $v$. The flux inside $\Gamma_2$ is $B$ times the part of the
area inside $\Gamma_2$ which has a magnetic field. The rate of change
of the flux, since the magnitude of $\FLPB$ is constant, is the
magnitude times the rate of change of the area. The rate of change of
the area is easy. If the width of the rectangle $\Gamma_2$ is $L$, the
area in which $\FLPB$ exists changes by $Lv\,\Delta t$ in the
time $\Delta t$. (See Fig. 18–6.) The rate of change of flux
is then $BLv$. According to Faraday’s law, this should equal minus the
line integral of $\FLPE$ around $\Gamma_2$, which is just $EL$. We
have the equation
\begin{equation}
\label{Eq:II:18:10}
E=vB.
\end{equation}

So if the ratio of $E$ to $B$ is $v$, the fields we have assumed will
satisfy Faraday’s equation. But that is not the only equation; we have the other equation relating
$\FLPE$ and $\FLPB$:
\begin{equation}
\label{Eq:II:18:11}
c^2\FLPcurl{\FLPB}=\frac{\FLPj}{\epsO}+\ddp{\FLPE}{t}.
\end{equation}

To apply this equation, we look at the top view in
Fig. 18–5. We have seen that this equation will give us the
value of $B$ next to the current sheet. Also, for any loop drawn outside
the sheet but behind the wavefront, there is no curl of $\FLPB$ nor
any $\FLPj$ or changing $\FLPE$, so the equation is correct there. Now let’s
look at what happens for the curve $\Gamma_1$ that intersects the
wavefront, as shown in Fig. 18–5. Here there are no
currents, so Eq. (18.11) can be written—in integral
form—as
\begin{equation}
\label{Eq:II:18:12}
c^2\oint_{\Gamma_1}\FLPB\cdot d\FLPs=\ddt{}{t}\!\!
\underset{\text{inside $\Gamma_1$}}{\int}
\!\FLPE\cdot\FLPn\,da.
\end{equation}

The line integral of $\FLPB$ is just $B$ times $L$. The rate of change
of the flux of $\FLPE$ is due only to the advancing wavefront. The
area inside $\Gamma_1$, where $\FLPE$ is not zero, is increasing at
the rate $vL$. The right-hand side of Eq. (18.12) is
then $vLE$. That equation becomes
\begin{equation}
\label{Eq:II:18:13}
c^2B=Ev.
\end{equation}
 We have a solution in which we have a constant $\FLPB$ and a constant $\FLPE$
behind the front, both at right angles to the direction in which the front is
moving and at right angles to each other. Maxwell’s equations specify the ratio
of $E$ to $B$. From Eqs. (18.10) and (18.13),
\begin{equation*}
E=vB,\quad\text{and}\quad E=\frac{c^2}{v}\,B.
\end{equation*}

But one moment! We have found two different conditions on the
ratio $E/B$. Can such a field as we describe really exist? There is,
of course, only one velocity $v$ for which both of these equations can
hold, namely $v=c$. The wavefront must travel with the
velocity $c$. We have an example in which the electrical influence from a
current propagates at a certain finite velocity $c$. Now let’s ask what happens if we suddenly stop the motion of the
charged sheet after it has been on for a short time $T$. We can see
what will happen by the principle of superposition. We had a current
that was zero and then was suddenly turned on. We know the solution
for that case. Now we are going to add another set of fields. We take
another charged sheet and suddenly start it moving, in the opposite
direction with the same speed, only at the time $T$ after we started
the first current. The total current of the two added together is
first zero, then on for a time $T$, then off again—because the two
currents cancel. We have a square “pulse” of current.
 The new negative current produces the same fields as the positive one,
only with all the signs reversed and, of course, delayed in time
by $T$. A wavefront again travels out at the velocity $c$. At the
time $t$ it has reached the distance $x=\pm c(t-T)$, as shown in
Fig. 18–4(b). So we have two “blocks” of field marching
out at the speed $c$, as in parts (a) and (b) of Fig. 18–4.
The combined fields are as shown in part (c) of the figure. The fields
are zero for $x>ct$, they are constant (with the values we found above)
between $x=c(t-T)$ and $x=ct$, and again zero for $x<c(t - T)$. In short, we have a little piece of field—a block of
thickness $cT$—which has left the current sheet and is travelling through
space all by itself. The fields have “taken off”; they are
propagating freely through space, no longer connected in any way with
the source. The caterpillar has turned into a butterfly! How can this bundle of electric and magnetic fields maintain itself? The
answer is: by the combined effects of the Faraday law,
$\FLPcurl{\FLPE}=-\ddpl{\FLPB}{t}$, and the new term of
Maxwell,
$c^2\FLPcurl{\FLPB}=\ddpl{\FLPE}{t}$. They cannot help maintaining
themselves. Suppose the magnetic field were to disappear. There would be
a changing magnetic field which would produce an electric field. If this
electric field tries to go away, the changing electric field would
create a magnetic field back again. So by a perpetual interplay—by the
swishing back and forth from one field to the other—they must go on
forever. It is impossible for them to disappear.1 They maintain
themselves in a kind of a dance—one making the other, the second
making the first—propagating onward through space.