Page 208 - Engineering Electromagnetics, 8th Edition

190 ENGINEERING ELECTROMAGNETICS

Let us again find the magnetic field intensity produced by an infinitely long
filament carrying a current I. The filament lies on the z axis in free space (as in
Figure 7.3), and the current flows in the direction given by a z . Symmetry inspection
comes first, showing that there is no variation with z or φ.Nextwe determine which
components of H are present by using the Biot-Savart law. Without specifically using
the cross product, we may say that the direction of dH is perpendicular to the plane
conaining dL and R and therefore is in the direction of a φ . Hence the only component
of H is H φ , and it is a function only of ρ.
We therefore choose a path, to any section of which H is either perpendicular
or tangential, and along which H is constant. The first requirement (perpendicularity
or tangency) allows us to replace the dot product of Amp`ere’s circuital law with the
product of the scalar magnitudes, except along that portion of the path where H is
normal to the path and the dot product is zero; the second requirement (constancy)
then permits us to remove the magnetic field intensity from the integral sign. The
integration required is usually trivial and consists of finding the length of that portion
of the path to which H is parallel.
In our example, the path must be a circle of radius ρ, and Amp`ere’s circuital law
becomes

2π 2π
H · dL = H φ ρdφ = H φ ρ dφ = H φ 2πρ = I
0 0

or

I
H φ =
2πρ

as before.
As a second example of the application of Amp`ere’s circuital law, consider an
infinitely long coaxial transmission line carrying a uniformly distributed total current
I in the center conductor and −I in the outer conductor. The line is shown in Fig-
ure 7.8a. Symmetry shows that H is not a function of φ or z.In order to determine the
components present, we may use the results of the previous example by considering
the solid conductors as being composed of a large number of filaments. No filament
has a z component of H. Furthermore, the H ρ component at φ = 0 , produced by one
◦
filament located at ρ = ρ 1 , φ = φ 1 ,is canceled by the H ρ component produced by a
symmetrically located filament at ρ = ρ 1 , φ =−φ 1 . This symmetry is illustrated by
Figure 7.8b.Again we find only an H φ component which varies with ρ.
A circular path of radius ρ, where ρ is larger than the radius of the inner conduc-
tor but less than the inner radius of the outer conductor, then leads immediately to

I
H φ = (a <ρ < b)
2πρ