Frank
If I have two parallel plates seperated by an air space and the plates are
connect to a batttery I have a capacitor with an E field between the plates
and squeezing out the sides. In steady state there's no current flow so I
have no H field.
If I have a coil or a solenoid and connect it to a battery I have a current
flow with a strong H field but I'm not certainabout the E field.
Can this be taken to the next step to answer my original question?
tnx
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73
Hank WD5JFR
"Frank" wrote in message
news:4Pyie.68625$tg1.4151@edtnps84...
"Henry Kolesnik" wrote in message
...
As I understand it a shielded loop (non-magnetic shield) favors the
magnetic field.
Correct.
If I wanted to measure the difference between the 2 fields how would I
measure the electric field?
Any antennna can measure the electric field, you simply have to know the
antenna factor.
Antenna factors can be calculated for various structures; for example the
aperture of a half wave dipole is given by: 0.13*lambda^2. Simple
calculations can then provide the antenna factor, and relate the E field
(in V/m) to the received signal. The electric and magnetic fields are
related by a constant -- the impedance of free space, 377 ohms. i.e. E/H
= 377. In the vicinty of an antenna (the near field) the impedance of free
space becomes a complex number.
A loop shielded with magnetic material would probably reject both fields.
Probably true, but have never experimented with such antennas.
The 3.5 foot loop for my old HP comparator for WWVB is totally
non-magnetic.
Can I generate and transmit each field separately? If so how would I do
it?
No. The E field cannot exist without the H field. See the relationship
above. Some people claim to have invented antennas that seperately
generate E and H fields. Such antennas are known as "EH" and
"Crossed-field", and have largely been rejected by the engineering
comunity as bogus. The designers claim that they do not conform to
Maxwell's Equations, but some other indefinable mathematics.
Regards,
Frank
tnx
--
73
Hank WD5JFR
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