Page 397 - Engineering Electromagnetics, 8th Edition

CHAPTER 11 The Uniform Plane Wave 379

The two fields are once again perpendicular to each other, perpendicular to the
direction of propagation, and in phase with each other everywhere. Note that when E
is crossed into H, the resultant vector is in the direction of propagation. We shall see
the reason for this when we discuss the Poynting vector.

EXAMPLE 11.3
Let us apply these results to a 1-MHz plane wave propagating in fresh water. At
this frequency, losses in water are negligible, which means that we can assume that
.
= 0. In water, µ r = 1 and at 1 MHz, = 81.

r
Solution. We begin by calculating the phase constant. Using (45) with = 0, we

have
√
6
√ ω 2π × 10 81
r

β = ω µ = ω µ 0 0 = = = 0.19 rad/m
r
c 3.0 × 10 8
Using this result, we can determine the wavelength and phase velocity:
2π 2π
λ = = = 33 m
β .19
ω 2π × 10 6 7
ν p = = = 3.3 × 10 m/s
β .19
The wavelength in air would have been 300 m. Continuing our calculations, we find
the intrinsic impedance using (48) with = 0:

µ η 0 377
η = = = = 42
r 9
If we let the electric field intensity have a maximum amplitude of 0.1 V/m, then

6
E x = 0.1 cos(2π10 t − .19z) V/m
E x −3 6
H y = = (2.4 × 10 ) cos(2π10 t − .19z) A/m
η

D11.3. A 9.375-GHz uniform plane wave is propagating in polyethylene
(see Appendix C). If the amplitude of the electric field intensity is 500 V/m
and the material is assumed to be lossless, find: (a) the phase constant; (b) the
wavelength in the polyethylene; (c) the velocity of propagation; (d) the intrinsic
impedance; (e) the amplitude of the magnetic field intensity.

8
Ans. 295 rad/m; 2.13 cm; 1.99 × 10 m/s; 251
; 1.99 A/m