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Old August 2nd 15, 06:25 PM posted to rec.radio.amateur.antenna
[email protected] jimp@specsol.spam.sux.com is offline
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First recorded activity by RadioBanter: Jun 2006
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Default "Bal uhn" or "bayl uhn"?

rickman wrote:
On 8/1/2015 8:24 PM, wrote:

There is no current in the shield inner surface, the energy is in the
ELECTROMAGNETIC FIELD between the inner and outer conductors. To be
nit pickingly precise, there is some small current in the inner
surface of the shield and the center wire, but for real coax that
surface current is insignificant.


this is a pretty amazing revelation. So what are the assumptions to
make this true?


https://en.wikipedia.org/wiki/Transm...#Coaxial_cable

Do you think there is significant current in the walls of a wave guide?

If there is no appreciable current flow in the coax, then the resistance
of the wires is of no significance. Funny, when I make a loop antenna
from coax, the Q still seems to be limited by the conductor resistance.
Odd...


Not at all, it is just that you don't understand that in that case it
is not a transmission line.

You can look at coax as a wave
guide if that makes it easier to understand, though the mode is
different than the mode in what is normally called wave guide.

https://en.wikipedia.org/wiki/Transm...#Coaxial_cable


I don't see anything here that says there is no current flow in the coax.


What it says is where the energy IS, not where it is NOT.

If you want to understand the effect of ohmic resistance, read
and understand the differential equations described he

https://en.wikipedia.org/wiki/Transm....27s_equations

At the end of the coaxial structure, the electromagnetic field
becomes a current flow in any conductors connected to the end
of the coax.


Where do the electrons come from that the current consists of? Does the
wire end act as a capacitor?


Are you really asking this?

Where do the electrons come from at the terminals of a receiving antenna?


One of those conductors is always the outside of the shield because
of the physical structure of coax.

The sum of the currents in the outside of the shield plus all other
conductors connected to the outside shield is equal to the sum of the
currents of all the conductors connected to the center wire.


That seems rather obvious.


Then why do you keep asking about it?

--
Jim Pennino