NEC-LIST: induced voltages

Grant,
 
Your question is not so dumb! I think a question is never dumb, per
se. This is an interesting problem, and I gave it some thought. My
conclusions are as follows:
 
After considering your "wire" as a half-wave dipole, since your
resistor is in the center, I took another approach, since I wanted to
be independent of wavelength or frequency. I considered the wire as an
effective receiving area multiplied by the power density impinging on
the wire. In order to obtain the voltage, I used the radiation
resistance of a Hertz dipole. This resulted in the formula:
 
   V = S * A * R

with S = Power density ( Watts/square meter )
     A = effective area of the wire ( square meter )
     R = radiation resistance
     V = Volt
 
exact : V = ( 0.0625 * E^2 * l^2 )^1/2
 
with E = Volt/m
     l = Length of wire in meter

The above formula is derived from :
 
V = {( E^2/120*pi)*(1.5*L^2/4*pi)*[20*pi^2(l/L)^2]}^1/2
 
with L = Lambda ( wavelength )
     l = length of wire

Under these assumptions ( wavelength is eliminated ), I found the
following results:
 
E = 1 V/m, l = 1 m, V = 0.25 volt
E = 2 V/m, l = 1 m, V = 0.50 volt
E = 1 V/m, l = 2 m, V = 0.50 volt
E = 2 V/m, l = 2 m, V = 1.00 volt
E = 4 V/m, l = 1 m, V = 1.00 volt
E = 1 V/m, l = 4 m, V = 1.00 volt
 
etc.
 
Result : the voltage can never equal the field strength !
 
Comment: This assumes that you have a matched situation. If you load
with different imedances, you must consider this in the formula.
 
Max Schmitt
Received on Thu Jun 25 1998 - 10:05:47 EDT