Richard Harrison wrote:
. . .
Strength of an electromagnetic wave is usually measured and quoted in
terms of its electric field in volts per meter. This is the number of
volts which would be induced in a one-meter length of wire placed in the
field parallel to the electric lines of force. . .

If the wavelength is 1 m, the voltage induced in the center of an
open-circuited 1 m diple by a 1 V/m field is 0.5 volt, not 1 volt.

Roy Lewallen, W7EL

Roy,

What is the voltage measured between the bottom-end of a 1 metre
vertical antenna and a perfect ground when the field strength is 1 V/m
and the wavelength is 1 m.
----
Reg.

Reg Edwards wrote:
Roy,

What is the voltage measured between the bottom-end of a 1 metre
vertical antenna and a perfect ground when the field strength is 1 V/m
and the wavelength is 1 m.

1 volt.

Roy Lewallen, W7EL

"Roy Lewallen" wrote in message
...
Reg Edwards wrote:
Roy,

What is the voltage measured between the bottom-end of a 1 metre
vertical antenna and a perfect ground when the field strength is 1
V/m
and the wavelength is 1 m.

1 volt.

Roy Lewallen, W7EL
========================================
Thank you Roy.

I don't doubt that your answer conforms to the learned text books on
the subject. But I am suspicious the text books may be wrong.

I will do some calculations related to radiation resistance and power
available to a matched receiver. If I think my suspicions are correct
then I will come back to you.
----
Reg.

Reg Edwards wrote:
Thank you Roy.

I don't doubt that your answer conforms to the learned text books on
the subject. But I am suspicious the text books may be wrong.

I will do some calculations related to radiation resistance and power
available to a matched receiver. If I think my suspicions are correct
then I will come back to you.
----
Reg.

To tell the truth, I got the result for a wire over ground from an NEC-2
model, after first checking to make sure I got the theoretical 0.5 volt
for the center of a dipole in free space. (NEC-2 has provision for
applying a plane wave to the model.)

In the process of confirming the 0.5 volt value, I found an error in a
popular text, Balanis, _Antenna Theory, Analysis and Design_. On p. 61,
he incorrectly states that the current along a short dipole "can be
assumed to be constant", which isn't true, and from that concludes that
the "induced voltage" would be 1 volt when the dipole feedpoint is short
circuited. How he defines "induced voltage" with a shorted feepoint
isn't clear, but the uniform current assumption he used to get it is
incorrect.

Kraus, in _Antennas_, and others get it right, and modeling confirms it.

Roy Lewallen, W7EL

Roy, you have anticipated my thoughts on the subject.

Nevertheless, I will do some calculations.

The proof of the pudding lies in the type of computer programs whose
input data does not depend on unreliabe human imagination about
antenna gain, mirror images and reflections from the ground.

I have no access to the learned text books or computer programs.
----
Reg, G4FGQ

Reg Edwards wrote:
Roy, you have anticipated my thoughts on the subject.

Nevertheless, I will do some calculations.

You should be able to reason the wire-over-ground case as follows:

Imagine a plane wave of 2 V/m intensity striking a 2 m long open
circuited dipole in free space. The open circuit voltage should be 4
times as great as it would be for a 1 V/m wave striking a 1 m long
dipole. If you bisect the system with a ground plane, you have half the
dipole and half the field above ground -- that's a 1 m wire and 1 V/m
field. And half the original dipole's voltage appears between the bottom
end of the wire and ground. So the resulting voltage is twice what it
would be at the center of a 1 m dipole in free space.

The proof of the pudding lies in the type of computer programs whose
input data does not depend on unreliabe human imagination about
antenna gain, mirror images and reflections from the ground.

I have no idea what you're talking about there, but I'm sure that
whatever it is, it must not apply to the programs you write.

I have no access to the learned text books or computer programs.

Sure you do, as does anyone with access to this newsgroup. Texts are
readily available by web order for the price of a very few bottles of
mediocre wine. Some have even been scanned and posted on the web. And
NEC-2 is free and can be downloaded from the web. But some people just
can't deal with any idea they didn't come up with on their own -- we
call it the NIH (Not Invented Here) syndrome. But each to his own.

Enjoy tonight's Balanis. You can save the Kraus for a special occasion.

Roy Lewallen, W7EL

Roy Lewallen wrote:
But some people just
can't deal with any idea they didn't come up with on their own ...

You mean like Ptotal = P1 + P2 + 2*SQRT(P1*P2)cos(a) ? :-)
The first time I saw that equation was in Dr. Best's QEX article.
--
73, Cecil http://www.qsl.net/w5dxp

Roy Lewallen wrote:
In the process of confirming the 0.5 volt value, I found an error in a
popular text, Balanis, _Antenna Theory, Analysis and Design_. On p. 61,
he incorrectly states that the current along a short dipole "can be
assumed to be constant", which isn't true, ...

You must have the 1st edition. In the second edition Balanis says on page
133 that the current on an infinitesimal dipole can be considered to be
constant. On page 143 he says the current along a small dipole is
triangular shaped.
--
73, Cecil http://www.qsl.net/w5dxp

Richard,

Terman said no such thing, and your interpretation is clearly in error.

Magnetic fields cannot impart ANY energy to charges, such as electrons
in a wire. This is because the force from a magnetic field on a charge
is always perpendicular to the motion of the charge. No work can be done
by the magnetic field, and the energy of the electrons does not change.
Only electric fields can provide energy to an electron.

Fortunately, Faraday's Law saves the day. Changing magnetic flux is
inextricably intertwined with electromotive force. Terman's comment on
page 2 of the 1955 edition simply points out the operation of Faraday's
Law. (Yes, I have this volume of Terman.)

Your conclusion statement is completely reversed. The magnetic field
does nothing to induce current in the antenna, while the electric field
does everything.

Again, however, the laws of physics save the day. Maxwell's equations
link electric and magnetic fields in such a manner that the magnetic
field you favor creates just enough electric field to drive the
electrons in the wire.

As has been stated many times in this newsgroup, it is not possible to
filter out one field component or the other. As long as there is some
time dependence, i.e., other than purely static fields, both the
electric and magnetic fields coexist.

73,
Gene
W4SZ

Richard Harrison wrote:
Roy Lewallen wrote:
"You can find the explanation for why this is in any electromagnetic
text."

I found it in Terman.

As we all know, we place correctly polarized dipoles, for example,
parallel to the wavefront for maximum response. Terman confirms the
electric field in this instance induces no energy in the antenna. It all
comes from the magnetic field.

If antenna current flows, no matter where it comes from, loss resistance
causes a voltge drop. That`s why the wire needs to be perfect. The
electric field produces no voltage in the antenna because the wavefront
has the same voltage across its entire surface. That`s because it all
left the same point at the same time. So, a wire parallel to the front
has no difference of potential induced by the wavefront`s electric
field. It all must come from the mgnetic field.

On page 2 of his 1955 edition, Terman says:
"The strength of the wave measured in terms of microvolts per meter of
stress in space is also exactly the same voltage that the MAGNETIC FLUX
(my emphasis) of the wave induces in a conductor 1 m long when sweeping
across this conductor with the velocity of light."

From the above, it is seen that the electric field is not effective in
inducing current in a receiving antenna parallel to a wavefront. All the
energy intercepted by the antenna is induced by the magnetic field.

Best regards, Richard Harrison, KB5WZI