Page 131 - Electromagnetics

the creation of a collimated beam of charge, as occurs in an electron tube where a series
of permanent magnets can be used to create a beam of steady current.
More typically, steady currents are created using wire conductors to guide the moving
charge. When an external force, such as the electric field created by a battery, is applied
to an uncharged conductor, the free electrons will begin to move through the positive
lattice, forming a current. Each electron moves only a short distance before colliding with
the positive lattice, and if the wire is bent into a loop the resulting macroscopic current
will be steady in the sense that the temporally and spatially averaged microscopic current
will obey ∇· J = 0. We note from the examples above that any charges attempting to
leave the surface of the wire are drawn back by the electrostatic force produced by the
resulting imbalance in electrical charge. For conductors, the “drift” velocity associated
with the moving electrons is proportional to the applied field:

u d =−µ e E
2
−3
where µ e is the electron mobility. The mobility of copper (3.2 × 10 m /V · s)is such
that an applied field of 1 V/m results in a drift velocity of only a third of a centimeter
per second.

Integral properties of a steady current. Steady currents obey several useful inte-
gral properties. To develop these properties we need an integral identity. Let f (r) and
g(r) be scalar functions, continuous and with continuous derivatives in a volume region
V . Let J represent a steady current field of finite extent, completely contained within
V . We begin by using (B.42)to expand
∇· ( fgJ) = fg(∇· J) + J ·∇( fg).

Noting that ∇· J = 0 and using (B.41), we get
∇· ( fgJ) = ( f J) ·∇g + (gJ) ·∇ f.
Now let us integrate over V and employ the divergence theorem:

( fg)J · dS = [( f J) ·∇g + (gJ) ·∇ f ] dV.
S V
Since J is contained entirely within S,wemust have ˆ n · J = 0 everywhere on S. Hence

[( f J) ·∇g + (gJ) ·∇ f ] dV = 0. (3.25)
V
We can obtain a useful relation by letting f = 1 and g = x i in (3.25), where (x, y, z) =
(x 1 , x 2 , x 3 ). This gives

J i (r) dV = 0, (3.26)
V
where J 1 = J x and so on. Hence the volume integral of any rectangular component of J
is zero. Similarly, letting f = g = x i we find that

x i J i (r) dV = 0. (3.27)
V
With f = x i and g = x j we obtain

x i J j (r) + x j J i (r) dV = 0. (3.28)
V

© 2001 by CRC Press LLC