How the new term works
As our first example we consider what happens with a spherically
symmetric radial distribution of current. Suppose we imagine a little
sphere with radioactive material on it. This radioactive material is
squirting out some charged particles. (Or we could imagine a large
block of jello with a small hole in the center into which some charge
had been injected with a hypodermic needle and from which the charge
is slowly leaking out.) In either case we would have a current that is
everywhere radially outward. We will assume that it has the same
magnitude in all directions. Let the total charge inside any radius $r$ be $Q(r)$. If the radial
current density at the same radius is $\FLPj(r)$, then
Eq. (18.2) requires that $Q$ decreases at the rate
\begin{equation}
\label{Eq:II:18:5}
\ddp{Q(r)}{t}=-4\pi r^2j(r).
\end{equation}
 We now ask about the magnetic field produced by the currents in this
situation. Suppose we draw some loop $\Gamma$ on a sphere of radius $r$,
as shown in Fig. 18–1. There is some current
through this loop, so we might expect to find a magnetic field
circulating in the direction shown. But we are already in difficulty. How can the $\FLPB$ have any
particular direction on the sphere? A different choice of $\Gamma$
would allow us to conclude that its direction is exactly opposite to
that shown. So how can there be any circulation of $\FLPB$
around the currents? We are saved by Maxwell’s equation. The circulation of $\FLPB$ depends
not only on the total current through $\Gamma$ but also on the
rate of change with time of the electric flux through it. It
must be that these two parts just cancel. Let’s see if that works out.
 The electric field at the radius $r$ must be $Q(r)/4\pi\epsO r^2$—so
long as the charge is symmetrically distributed, as we assume. It is
radial, and its rate of change is then
\begin{equation}
\label{Eq:II:18:6}
\ddp{E}{t}=\frac{1}{4\pi\epsO r^2}\,\ddp{Q}{t}.
\end{equation}

Comparing this with Eq. (18.5), we see
\begin{equation}
\label{Eq:II:18:7}
\ddp{E}{t}=-\frac{j}{\epsO}.
\end{equation}

In Eq. IV the two source terms cancel and the curl of $\FLPB$ is
always zero. There is no magnetic field in our example. As our second example, we consider the magnetic field of a wire used to
charge a parallel-plate condenser (see Fig. 18–2). If the
charge $Q$ on the plates is changing with time (but not too fast), the
current in the wires is equal to $dQ/dt$. We would expect that this
current will produce a magnetic field that encircles the wire. Surely,
the current close to the plate must produce the normal magnetic
field—it cannot depend on where the current is going. Suppose we take a loop $\Gamma_1$ which is a circle with radius $r$,
as shown in part (a) of the figure. The line integral of the magnetic
field should be equal to the current $I$ divided by $\epsO c^2$. We
have
\begin{equation}
\label{Eq:II:18:8}
2\pi rB=\frac{I}{\epsO c^2}.
\end{equation}

This is what we would get for a steady current, but it is also correct
with Maxwell’s
addition, because if we consider the plane surface $S$
inside the circle, there are no electric fields on it (assuming the
wire to be a very good conductor). The surface integral
of $\ddpl{\FLPE}{t}$ is zero. Suppose, however, that we now slowly move the curve $\Gamma$ downward.
We get always the same result until we draw even with the plates of the
condenser. Then the current $I$ goes to zero. Does the magnetic field
disappear? That would be quite strange. Let’s see what Maxwell’s
equation says for the curve $\Gamma_2$, which is a circle of radius $r$
whose plane passes between the condenser plates
[Fig. 18–2(b)]. The line integral of $\FLPB$
around $\Gamma_2$ is $2\pi rB$. This must equal the time derivative of the flux
of $\FLPE$ through the plane circular surface $S_2$. This flux
of $\FLPE$, we know from Gauss’ law, must be equal to $1/\epsO$ times the
charge $Q$ on one of the condenser plates. We have
\begin{equation}
\label{Eq:II:18:9}
c^2\,2\pi rB=\ddt{}{t}\biggl(\frac{Q}{\epsO}\biggr).
\end{equation}
 That is very convenient. It is the same result we found in
Eq. (18.8). Integrating over the changing electric field
gives the same magnetic field as does integrating over the current in
the wire. Of course, that is just what Maxwell’s equation says. It is
easy to see that this must always be so by applying our same arguments
to the two surfaces $S_1$ and $S_1'$ that are bounded by the same
circle $\Gamma_1$ in Fig. 18–2(b). Through $S_1$ there is the
current $I$, but no electric flux. Through $S_1'$ there is no current,
but an electric flux changing at the rate $I/\epsO$. The same $\FLPB$ is
obtained if we use Eq. IV with either surface. From our discussion so far of
Maxwell’s new term, you may have the
impression that it doesn’t add much—that it just fixes up the
equations to agree with what we already expect. It is true that if we
just consider Eq. IV by itself, nothing particularly new
comes out. The words “by itself” are, however,
all-important. Maxwell’s
small change in Eq. IV, when combined
with the other equations, does indeed produce much that is new and
important. Before we take up these matters, however, we want to speak
more about Table 18–1.