On Feb 22, 3:43 pm, "Richard Fry" wrote:
Other things equal, the far field varies by the square root of the power
change.
If 1 kW into a given antenna system produces 300 mV/m at 1 km, then 0.1 W
into that same system produces SQRT(0.1W / 1000W) x 300 mV/m at 1 km,
which is 3 mV/m.

So in terms of the 20W test, this value at 1km would be 42.4 mV/m? If
so, how would I translate this value to another distance (e.g. 0.4 km
away)? I'm assuming that it would be greater than 42.4mV/m.

A good AM receiver in the absence of natural/man-made interference can
produce useful (but not highly "competitive") performance in fields of less
than 100 µV/m.

Would a typically AM receiver be able to pick up signals 50 uV/m in
field strength? I'm assuming that it wouldn't since the components in
the AM receiver have noise levels probably close to this, huh?