Page 409 - Engineering Electromagnetics, 8th Edition

CHAPTER 11 The Uniform Plane Wave 391

Solution. We first evaluate the loss tangent, using the given data:
σ 4 2
= = 8.9 × 10 1
6
ω (2π × 10 )(81)(8.85 × 10 −12 )
Seawater is therefore a good conductor at 1 MHz (and at frequencies lower than this).
The skin depth is
1 1
δ = √ = = 0.25 m = 25 cm
−7
6
π f µσ (π × 10 )(4π × 10 )(4)
Now
λ = 2πδ = 1.6m

and
6
6
ν p = ωδ = (2π × 10 )(0.25) = 1.6 × 10 m/sec
In free space, these values would have been λ = 300 m and of course ν = c.
With a 25-cm skin depth, it is obvious that radio frequency communication in
√
seawater is quite impractical. Notice, however, that δ varies as 1/ f ,so that things
will improve at lower frequencies. For example, if we use a frequency of 10 Hz (in
the ELF, or extremely low frequency range), the skin depth is increased over that at
6
1 MHz by a factor of 10 /10, so that
.
δ(10 Hz) = 80 m
.
The corresponding wavelength is λ = 2πδ = 500 m. Frequencies in the ELF range
were used for many years in submarine communications. Signals were transmitted
from gigantic ground-based antennas (required because the free-space wavelength
7
associated with 10 Hz is 3 × 10 m). The signals were then received by submarines,
from which a suspended wire antenna of length shorter than 500 m is sufficient. The
drawback is that signal data rates at ELF are slow enough that a single word can
take several minutes to transmit. Typically, ELF signals would be used to tell the
submarine to initiate emergency procedures, or to come near the surface in order to
receive a more detailed message via satellite.

We next turn our attention to finding the magnetic field, H y , associated with E x .
To do so, we need an expression for the intrinsic impedance of a good conductor. We
begin with Eq. (48), Section 11.2, with = σ/ω,

jωµ
η =
σ + jω
Since σ ω ,wehave

jωµ
η =
σ