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Old November 30th 09, 08:10 PM posted to rec.radio.amateur.antenna
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First recorded activity by RadioBanter: Mar 2008
Posts: 1,339
Default Faraday shields and radiation and misinterpretations

I have been reading the groups archives on shield antennas and Faraday
shields and the different auguments regarding how shielding or the
Faraday shield works. Frankly it is a total mess and should be removed
so that hams are not mislead.
Shielding is very simple.
A particle with a electromagnetic field strikes the outside of the
shield.
The magnetic field of same passes thru the shield some might say it is
coupled to the inside of the shield.
The magnetic vector component is out of phase with the electrical
field so it will be just a static particle at rest on the inside but
no inline with the electrical field vector which is now a staic
particle at rest on the outside
We now have a arbitrary boundary as discused by Gauss
For equilibrium all vectors impinging on the boundary must be aligned
such that they cancel.
To accomplish this the inner vector or charge MUST move sideways

THE CHARGE WHEN ACCELERATED CREATES A TIME VARYING CURRENT ALONE
WHILE THE OTHER FIELD VECTORS CANCEL OUT
( I believe that this was the object intended in the cross field
antenna)

As with a applied varying current leaves a xmitter to create
radiation, so must the receiver obtain a time varying current.

Maxwells equations show equations with the electric field, the
magnetic field and a time varying current. When you have a electrical
field or vector of a static particle at rest outside the boundary
opposing the static vector on the inside of the boundary you have
nothing left EXCEPT a time varying current in the closed circuit.
For informative descriptions of how radiation occurs view the QRZ
forum of ( antenna construction and design ) threads (3) on the
double helix
antenna ( see you there)
Somebody some where should re write the above such that a definition
is left for those who follow and remove the garbage which is now in
place
 
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