Page 531 - Engineering Electromagnetics, 8th Edition

CHAPTER 14 ELECTROMAGNETIC RADIATION AND ANTENNAS 513

√
v = 1/ µ . Since no integration is required for the short filament, we have
µ I[t − R/v] d
A = a z (2)
4π R
Only the z component of A is present, for the current is only in the a z direction. At
any point P at distance R from the origin, the vector potential is retarded by R/v and
so we use
R

I[t − R/v] = I 0 cos ω t − = I 0 cos [ωt − kR] (3)
v
√
where the wavenumber in the lossless medium is k = ω/v = ω µ .In phasor form,
Eq. (3) becomes
I s = I 0 e − jkR (4)
where the current amplitude, I 0 ,is assumed to be real (as it will be throughout this
chapter). Incorporating (4) into (2), we find the phasor retarded potential:

µI 0 d
A s = A zs a z = e − jkR a z (5)
4π R
Using a mixed coordinate system for the moment, we now replace R by the small r
of the spherical coordinate system and then determine which spherical components
are represented by A zs . Using the projections as illustrated in Figure 14.2, we find
A rs = A zs cos θ (6a)

A θs =−A zs sin θ (6b)
and therefore
µI 0 d
A rs = cos θ e − jkr (7a)
4πr

Figure 14.2 The
resolution of A zs at P(r,θ,φ)
into the two spherical
components A rs and A θs .
The sketch is arbitrarily
drawn in the φ = 90 plane.
◦