(257) |

However, we know from Stokes' theorem that

where is any surface attached to the loop .

Let us evaluate
directly. According to Eq. (254),

where use has been made of ,

But, we have just demonstrated that . This problem is very reminiscent of the difficulty we had earlier with . Recall that for a volume containing a discrete charge , but that at a general point. We got around this problem by saying that is a three-dimensional delta-function whose spike is coincident with the location of the charge. Likewise, we can get around our present difficulty by saying that is a two-dimensional delta-function. A three-dimensional delta-function is a singular (but integrable)

The above equation certainly gives , and everywhere apart from the -axis, in accordance with Eqs. (260)-(262). Suppose that we integrate over a plane surface connected to the loop . The surface element is , so

(265) |

which is in agreement with Eq. (263).

But, why have we gone to so much trouble to prove
something using vector field theory which can be demonstrated
in one line via conventional
analysis [see Eq. (258)]? The answer, of course, is that the vector field
result is easily generalized, whereas the
conventional result is just a special case.
For instance, it is clear that Eq. (266) is true for *any* surface attached to the loop
C, not just a plane surface.
Moreover, suppose that we distort our simple circular loop
so that it is no longer circular or even lies in one plane.
What now is the line integral
of around the loop? This is no longer a simple problem for conventional
analysis, because the magnetic field is not parallel to a line element of the
loop.
However, according to Stokes' theorem,

(267) |

(268) |

Thus, provided the curve circulates the -axis, and, therefore, any surface attached to intersects the -axis, the line integral is equal to . Of course, if does not circulate the -axis then an attached surface does not intersect the -axis and is zero. There is one more proviso. The line integral is for a loop which circulates the -axis in a clockwise direction (looking up the -axis). However, if the loop circulates in an anti-clockwise direction then the integral is . This follows because in the latter case the -component of the surface element is oppositely directed to the current flow at the point where the surface intersects the wire.

Let us now consider wires directed along the -axis, with coordinates
(, ) in the - plane, each carrying a current in the
positive -direction. It is fairly obvious that Eq. (264) generalizes
to

where is the total current enclosed by the curve. Again, if the curve circulates the th wire in a clockwise direction (looking down the direction of current flow) then the wire contributes to the aggregate current . On the other hand, if the curve circulates in an anti-clockwise direction then the wire contributes . Finally, if the curve does not circulate the wire at all then the wire contributes nothing to .

Equation (269) is a field equation describing how a set of -directed
current carrying wires generate a magnetic field. These wires have
zero-thickness, which implies that we are trying to squeeze a finite amount of
current into an infinitesimal region. This
accounts for the delta-functions on the right-hand side of
the equation. Likewise, we obtained delta-functions in Sect. 3.4
because we were dealing with point charges. Let us now generalize to the more
realistic case of diffuse currents. Suppose that the -current flowing through
a small
rectangle in the - plane, centred on coordinates (, ) and of dimensions
and , is
. Here, is termed the current density in
the -direction. Let us integrate
over this rectangle.
The rectangle is assumed to be sufficiently small that
does not vary appreciably across it. According to Eq. (270), this integral is
equal to times the total -current flowing through the rectangle. Thus,

(271) |

Of course, there is nothing special about the -axis. Suppose we have a set of diffuse currents flowing in the -direction. The current flowing through a small rectangle in the - plane, centred on coordinates (, ) and of dimensions and , is given by , where is the current density in the -direction. It is fairly obvious that we can write

(273) |

where is the vector current density. This is the third Maxwell equation. The electric current flowing through a small area located at position is . Suppose that space is filled with particles of charge , number density , and velocity . The charge density is given by . The current density is given by , and is obviously a proper vector field (velocities are proper vectors since they are ultimately derived from displacements).

If we form the line integral of around some general closed curve
, making use of Stokes' theorem and the field equation (274), then we
obtain

The flux of the current density
through is evaluated by integrating
over any surface
attached to . Suppose that we take two different surfaces and
. It is clear that if Ampère's circuital
law is to make any sense then the surface integral
had better equal the integral
. That is, when we work out the flux of the current
though using two different attached surfaces then we had better get
the same answer, otherwise Eq. (275) is wrong (since the left-hand side is clearly independent
of the surface spanning C).
We saw in Sect. 2 that if the integral of a vector field
over some surface attached to a loop depends only on the loop, and is
independent of the surface which spans it, then this implies that
. The flux of the current
density through any loop
is calculated by evaluating the integral
for
any surface which spans the loop. According to
Ampère's circuital law, this integral depends only on
and is completely independent of
(*i.e.*, it is equal to the line integral of around , which depends
on but not on ). This implies that
. In fact, we
can obtain this relation directly from the field equation (274). We know that
the divergence of a curl is automatically zero, so taking the divergence
of Eq. (274), we obtain

(276) |

We have shown that if Ampère's circuital law is to make any sense then we need
. Physically, this implies that the net current flowing
through any closed surface is zero.
Up to now, we have only considered
stationary charges and steady currents. It is clear that if all charges are
stationary and all currents are steady then there can be no net current flowing
through a closed surface , since this would imply a build up of charge in the
volume enclosed by . In other words, as long as we restrict our investigation
to stationary charges, and steady currents, then we expect
,
and Ampère's circuital law makes sense. However, suppose that we now relax this
restriction. Suppose that some of the charges in a volume decide to move
outside . Clearly, there will be a non-zero net flux of electric current through
the bounding surface whilst this is happening. This implies from
Gauss' theorem that
. Under these circumstances
Ampère's circuital law collapses in a heap. We shall see later that we can rescue
Ampère's circuital law by adding an extra term involving a time derivative to the
right-hand side of the field equation (274).
For steady-state situations (*i.e.*,
), this
extra term can be neglected. Thus, the field equation
is, in fact, only two-thirds of Maxwell's third equation: there is a term missing
on the right-hand side.

We have now derived two field equations involving magnetic fields (actually,
we have only derived one
and two-thirds):

We obtained these equations by looking at the fields generated by infinitely long, straight, steady currents. This, of course, is a rather special class of currents. We should now go back and repeat the process for general currents. In fact, if we did this we would find that the above field equations still hold (provided that the currents are steady). Unfortunately, this demonstration is rather messy and extremely tedious. There is a better approach. Let us