Roy Lewallen , W7EL wrote:
"Do you have a reference which defines field strength in terms of
voltage induced in a wire?"

Here is a reference from a professional source. B. Whitfield Griffith,
Jr. was Director of Advanced Development at General Dynamics Corporation
at Garland, Texas when his book, "Radio-Electronicc Transmission
Fundamentals" was published by Mc Graw-Hill in 1972. On page 322,
Griffith writes:
"The strength of an electromagnetic wave is generally measured in terms
of the intensity of the electric field; this is expressed in volts per
meter, or millivolts or microvolts per meter, as the conditions may
indicate. This value may be understood as being the numbeer of volts
which would be induced in a piece of wire one meter long placed in the
field parallel to the electric lines of force; the induction of voltage
would result from the movement of the magnetic flux across the wire.

Griffith agrees with Terman.

Best regards, Richard Harrison, KB5WZI

"Richard Harrison" bravely wrote to "All" (15 Dec 05 17:18:25)
--- on the heady topic of " Antenna reception theory"

RH From: (Richard Harrison)
RH Xref: core-easynews rec.radio.amateur.antenna:221371

RH Roy Lewallen , W7EL wrote:
RH "Do you have a reference which defines field strength in terms of
RH voltage induced in a wire?"

RH Here is a reference from a professional source. B. Whitfield Griffith,
RH Jr. was Director of Advanced Development at General Dynamics
RH Corporation at Garland, Texas when his book, "Radio-Electronicc
RH Transmission Fundamentals" was published by Mc Graw-Hill in 1972. On
RH page 322, Griffith writes:
RH "The strength of an electromagnetic wave ; this is expressed in
RH volts per meter, or millivolts or microvolts per meter, as the
RH conditions may indicate.

So far so good, however things go awry from here on...


RH This value may be understood as being the
RH numbeer of volts which would be induced in a piece of wire one meter
RH long placed in the field parallel to the electric lines of force; the
RH induction of voltage would result from the movement of the magnetic
RH flux across the wire.

RH Griffith agrees with Terman.

He clearly contradicts himself in my opinion. He says it "is generally
*measured* in terms of the intensity of the electric field" then goes
on to talk about induction by magnetic flux. Further, he is not saying
that "it is" but that it "*may* be understood as". He is making an
example of one way to proceed. He is not saying they are the same
thing, only similar. There is a difference not a matter of syntax.

This reminds me of a contradiction I find in Einstein's equivalence
principle. He states accelerating at 1G is the same as being pulled by
the Earth's gravity at 1G. In my opinion, they are similar but they
are not quite the same thing simply because in the former there is
motion and for the latter there is not.

A*s*i*m*o*v

Asimov wrote:
"He clearly contradicts himself in my opinion."

Griffith echoes Terman.

Terman writes on page 2 of his 1955 edition:
"The strength of a radio wave is measured in terms of the voltage stress
produced by the electric field of the wave and it is usually expressed
in microvolts per meter."

Terman does not literally mean that it is the electric field which is
measured. In his handbook Terman says that a magnetic loop antenna is
usually the field strenngth meter`s antenna. Terman meant that field
strength is quoted in volts per meter. It is irrelevant which field is
actually measured as the two fields are locked in magnitude by the
impedance of space, 377 ohms. If you know one, you know the other.

Terman also wrote on page 2:
"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
of the wave induces in a conductor 1 m long when sweeping across this
conductor with the velocity of light."

The electric and magnetic fields serve to recreate each other in their
flight from their source. Either could be measured to get the strength
of the wave. Neither field directly produces volts in a wire.

As Terman says, it is the wave sweeping the wire at the velocity of
light, the d(phi)/dt that generates the volts.

Best regards, Richard Harrison, KB5WZI

Asimov wrote:

This reminds me of a contradiction I find in Einstein's equivalence
principle. He states accelerating at 1G is the same as being pulled by
the Earth's gravity at 1G. In my opinion, they are similar but they
are not quite the same thing simply because in the former there is
motion and for the latter there is not.

A contradiction which might only appear if you were to confuse little g
and big G. Eistein was referring to the former, which is the
acceleration a falling body experiences as a result of the force of
gravity on Earth - a change of velocity of 9.8 meters per second in one
second. Clearly motion is involved. Big G is the universal
gravitational constant, which is in units of Newton meters squared per
kilogram squared. Clearly, this relates to a force in Newtons which
varies as the square of mass and the inverse square of distance,
according to Newton's law of gravitation. Motion is only involved if
the force is not met with an equal and opposing force - as is obviously
the case when standing on the ground.

ac6xg

Roy Lewallen wrote:

Richard Harrison wrote:

. . .
I agree the voltage induced in 1/2-meter of wire properly placed within
a 1V/m uniform field is 0.5 volt, not 1 volt. The induced voltage in a
wire within a uniform field sweeping the wire rises uniformly along the
wire. It can be assumed to be the summation of tiny increments of
voltage all along the wire. The voltages of the too-short dipole halves
add just as two cells in some flashlights add. Their vectors are head to
tail. But, current will be limited by radiation and loss resistances of
the wires. It will also be limited by reactance in the wires.
Open-circuit, 0.5 V + 0.5 V = 1V.


There are two incorrect statements here.

First, the voltage induced in the wire doesn't rise uniformly along the
wire. It's sinusoidal, even for a very short wire.

Interesting. Assuming a plane wave sweeping broadside, with the field
being the same at every point along the wire, one might be inclined to
argue that the voltage induced on a wire should be the same at every
point along a finite length. The "rise" in voltage as the field sweeps
past would be with respect to time, rather than with respect to
position. Sort of explains why a radio receiver works just fine with
only one wire attached when you think about it. ;-)

73, ac6xg

On Fri, 16 Dec 2005 13:56:12 -0800, Jim Kelley
wrote:

one might be inclined to argue that the voltage induced on a wire should be the same at every
point along a finite length

One might, if it were an exceptionally short piece of wire. Otherwise
it must exhibit some greater reactance than nearly 0 and support a
difference of potential in proportion to the current through it. There
is also the radiation (non-0) resistance to consider (same observation
of a potential difference there too).

Presumably, this all hinges on what is meant, in practical terms, for
"finite length."

The "rise" in voltage as the field sweeps
past would be with respect to time,
With respect to what time? The time for the wave to sweep past? What
frequency? The full wave, or its peak? or its average (0)?

rather than with respect to position.

Position of what?

If we could generate a step-wave, and that wave impinges upon the wire
broadside, and we were to arbitrarily assign the frequency of 3 MHz
(so that the wire length of 1 Meter is one percent long - a short
antenna by any definition); then the first degree would immediately
demand a response in 1nS. 1nS is a very appreciable dT for 1 Meter's
worth of inductance.

Still and all, 1nS is hardly the slice of the wave an antenna wire
sees, unless it is 1nS thick - or 33cm. This comes close to a tower's
size, but receivers work with far thinner stuff and the dV/dT
certainly pushes the voltage drop across 1 meter of that wire up
higher.

73's
Richard Clark, KB7QHC

Jim Kelley wrote:
Sort of explains why a radio receiver works just fine with
only one wire attached when you think about it. ;-)

It doesn't work fine when you don't think about it? :-)
--
73, Cecil http://www.qsl.net/w5dxp

Richard Clark wrote:

On Fri, 16 Dec 2005 13:56:12 -0800, Jim Kelley
wrote:


one might be inclined to argue that the voltage induced on a wire should be the same at every
point along a finite length


One might, if it were an exceptionally short piece of wire. Otherwise
it must exhibit some greater reactance than nearly 0 and support a
difference of potential in proportion to the current through it. There
is also the radiation (non-0) resistance to consider (same observation
of a potential difference there too).

You seem to be arguing that the induced potential in the wire is due to
IR or IZ. I don't think that's true. According to JC Maxwell, the
magnitude of the induced voltage and current should depend on the EM
field and on the in situ permitivity and permeability of the wire. As
long as those things are constant along the length of that wire, there
is no reason to expect anything but uniform voltage and current along
the length of the wire. As Roy said, an antenna does not behave the
same on receive as it does on transmit apparently. A receive antenna
does not appear to behave like a transmission line.

Presumably, this all hinges on what is meant, in practical terms, for
"finite length."

Finite here just means small compared to any curvature of the wave
front. Were the wire long enough compared to the distance to the
source, one couldn't expect the fields to be uniform along the length of
the wire.

The "rise" in voltage as the field sweeps
past would be with respect to time,

With respect to what time? The time for the wave to sweep past? What
frequency? The full wave, or its peak? or its average (0)?

It just means voltage as a function of time - like the waveform induced
by the fields of a passing electromagnetic wave for example.

rather than with respect to position.


Position of what?

It means position along the wire - as in the topic of the conversation
you've joined.

I apologize for any lack of clarity on my part.

73, ac6xg

This thread has presented a clear illustration of the danger of quoting
from a book, even an authoritative one, without fully understanding the
context.

Before I continue, let me stipulate that the conductors being talked
about here are all electrically short (much shorter than a wavelength),
and are placed parallel to the E field. The following statements won't
be generally true if both those conditions aren't met.

I've been saying that the voltage at the center of an open circuited 1 m
dipole immersed in a 1 V/m field is 0.5 volts. Richard has quoted some
references which say that the voltage induced in a 1 m wire immersed in
a 1 V/m field is 1 volt, and implying that it follows that the voltage
at the center of an open circuited dipole of that length is 1 volt. It
isn't, and I'll try to explain why.

The voltage drop, or emf, across a very short conductor in a field of E
volts/meter is E*L volts, where L is the length of the very short
conductor in meters. This is undoubtedly the reason for the statements
like the ones Richard has quoted. If we look at the emf generated across
each tiny part of a one meter long dipole by a 1 V/m field and add them
up, we'll find that they total 1 volt. But what's the voltage across the
dipole, from end to end? The answer to that is it's just about anything
we want it to be. Any time we try to measure the voltage between two
points in space when there's a time-varying H field present, the answer
we get depends on the path we take between the two points. Crudely but
not entirely accurately put, it depends on how we arrange the voltmeter
leads. We can, however, measure the voltage at the dipole center, across
a gap that's arbitrarily small.

So what's that voltage? I can't think of any line of reasoning which
would deduce that it's equal to the total emf along the wire (1 volt for
our example). You can casually open various texts and find the answer to
that -- it's either 0.5 volts, as I've said, or 1 volt, as Richard has
said. To understand the reason for the apparent contradiction requires
digging more deeply into the texts.

Most of the texts I have analyze field-conductor interactions with a
special kind of dipole which has a uniform current distribution -- that
is, the current is the same amplitude all along the dipole's length.
There are a couple of ways you can physically make a dipole like this.
One is to begin with a conventional (but short) dipole and add large end
hats, like a two-ended top loaded vertical. A number of authors (e.g.,
Kraus) show a diagram of such a dipole. Another way to get a
distribution like this is to stipulate that the dipole really be only a
tiny segment of a longer wire. The short conductor with uniform current
is often called a "Hertzian dipole", sometimes an "elemental dipole",
sometimes just a short or infinitesimal dipole, or by Terman, a
"doublet". As Balanis (_Antenna Theory - Analysis and Design_, p. 109)
says, "Although a constant current distribution is not realizable, it is
a mathematical quantity that is used to represent actual current
distributions of antennas that have been incremented into many small
lengths." The voltage at the center of a one meter antenna of this type
in a 1 V/m field is 1 volt.

But this fictitious dipole, used for conceptual and computational
convenience, isn't a real dipole. A real dipole -- that is, just a
single, straight wire with a source or load at its center and no end
hats -- doesn't have a uniform current distribution. Instead, the
current is greatest in the center, dropping to zero at the ends. When
transmitting, the current on a short dipole drops nearly linearly from
the center to the ends. When receiving, the current distribution is
nearly sinusoidal. The net result of this non-uniform current is that
the voltage at the center is less than it is for a uniform-current
dipole -- exactly half as much, actually. Conceptually, it's because the
current near the ends contributes less to the voltage at the center. I'm
going to wave my hands over the significance of the different receiving
and transmitting current distributions, except to say that reciprocity
is still satisfied in all ways, including the transmitting and receiving
impedances being the same. To add confusion, the gain, directivity, and
effective apertures of both types of dipole are the same -- 1.5, 1.5,
and 3 * lambda^2 / (8 * pi) respectively. This means that you can
extract the same amount of power from an impinging wave with either type
of antenna, provided that you terminate each in the complex conjugate of
its transmitting feedpoint impedance. The radiation resistance of the
uniform-current dipole is 4 times that of the conventional dipole.
(Remember, these are all electrically short.)

There's a common term for the relationship between the field strength
and the length of a conductor, called the "effective height" or
"effective length". The voltage at the center of a dipole in a field of
E volts/m is simply E * the effective length. The concept is valid for
any length conductor, not just short ones. The effective length of a
uniform-current dipole is equal to the wire length. The effective length
of a short conventional dipole is 0.5 times the wire length. The
effective length for receiving is the same as the effective length for
transmitting -- in transmitting, it relates the strength of the field
produced to the *voltage* -- not power -- applied across the feedpoint.
If you apply 0.5 volts to a standard dipole and 1.0 volts to a
uniform-current dipole, the power applied to each will be the same
because of the 1:4 ratio of radiation resistance, and the generated
fields will be the same. This is consistent with the antenna gains being
the same.

As I mentioned, most text authors use a uniform-current dipole for
analysis. One which directly derives the voltage of a standard short
dipole is King, Mimno, and Wing, _Transmission Lines, Antennas, and Wave
Guides_. Many others, including Kraus, _Antennas_ (2nd Ed. p. 41),
derive the effective length for a short conventional dipole as 0.5 * the
physical length, from which the open circuit voltage due to an impinging
field can easily be determined.

As a last note on a point of contention, electric field strength is
usually defined not by the voltage induced in a conductor but from the
force between charges using the Lorenz force law. The unit of electric
field strength is found to be newtons/coulomb, which is the same as
volts/meter. Among the texts using this definition are Kraus
(_Electromagnetics_), Terman (_Radio Engineering_), Ida, Majid, and Ramo
et al.

Roy Lewallen, W7EL

Jim Kelley wrote:

Interesting. Assuming a plane wave sweeping broadside, with the field
being the same at every point along the wire, one might be inclined to
argue that the voltage induced on a wire should be the same at every
point along a finite length. The "rise" in voltage as the field sweeps
past would be with respect to time, rather than with respect to
position. Sort of explains why a radio receiver works just fine with
only one wire attached when you think about it. ;-)


The Earth's (static) electric field is about 120 V/m at ground level on
a stormless day. Better not walk too fast. Jogging might be fatal!

Roy Lewallen, W7EL