K7ITM wrote:
I'll say I disagree with the wording of an early sentence,
where it says that ground return currents are "greatly attenuated" if
they come through lossy earth. Clearly, the current is not
attenuated; the current is what it is.

There is some confusion between DC circuits and
distributed RF networks. In a DC circuit, the
current is the same throughout the circuit. In
a distributed RF network, the current is usually
*NOT* the same throughout the network. One can
put one amp of current into a buried radial and
measure zero amps from some point outward.

In particular, when dealing with RF EM waves, the
H-field to which the current is proportional in
a transmission line is attenuated by the same
attenuation factor as is the E-field to which
the voltage is proportional. Please reference
the transmission line equations to verify that
fact.

Such is easy to see. If one has a flat Z0=50 ohm
transmission line with 100 watts in and 50 watts
out, there is 1.414 amps in and 1.0 amp out because
the ratio of voltage to current is fixed at 50 ohms.
The traveling-wave current in an EM wave in a
transmission line (or in a radial in lossy earth)
is attenuated exactly as much as the voltage.

And don't feel too ignorant about that fact of physics.
Some of the gurus on this newsgroup make a similiar
mistake about RF EM wave current through a loading coil.
--
73, Cecil http://www.w5dxp.com