Mark, Tom,
let me try to put some numbers to your
question...
Loop resistance: Wire diameter #8 AWG = 3.26 mm, skin
depth at 8.3 kHz ~ 0.66 mm, AC cross section 6.7
mm^2, AC resistance 2.6 mohm/m,
loop perimeter 244m, => wire resistance 0.62
ohm.
Depending on ground
conductivity, there will be additional eddy current losses in the ground,
maybe on the order of an ohm. Plus some capacitor ESR and connection
losses. Assume 2 ohms total resistance, giving i = 10 A from 200
watts.
Loop area: a=3720
m^2, magnetic field at distance r towards the side of the loop:
H = i*a /
(4*pi*r^3) ~ 3000 Am^2 / r^3
Noise background at 9 kHz:
Approximately 3 to 10
uV/m/sqrt(Hz), but quite variable depending on static background, time
of day, noise blanking etc. Assuming 5 uV/m/sqrtHz.
CW would be readable with 9
dB SNR in 50 Hz bandwidth => required signal
Emin = 14 + 9 + 17 dB
uV/m = 40 dBuV/m = 100 uV/m
Assuming a magnetic receive
antenna, this is equivalent to Hmin = Emin / 377 ohm = 0.26 uA/m
Solving for
distance:
rmax = (3000/0.26e-6)^0.33
m = 2.24 km
... Stefan, nice estimate indeed!
In reality, the receive loop
would also need to be horizontal (ie vertical axis), and may thus benefit from
picking up much less of the atmospheric noise. 20 dB lower noise
would extend the range by a factor of 2.15.
Furthermore, G3XBM and others
have demonstrated that in inhabited areas, the field of the TX loop
may be picked up by infrastructure like buried cables or water
conduits, inducing currents which may carry the signal to much further
distances in "utility aided earth mode".
But...
Though a horizontal loop
(vertical axis) may have significant inductive nearfields, it will not
radiate much into the far field because it has the wrong
polarization. The associated horizontal E-fields are simply being shunted
by the ground. To "get out", you'd need to
put the loop in upright orientation (ie horizontal axis). Assuming the
same loop area as before, the radiation resistance (including the loop
"image" below conducting ground) would be
Rrad = 2 * 31171 ohms
* a^2 / lambda^4 = 0.5 microohms
and radiated power
EMRP = i^2 * Rrad = 50
uW
But that would require a much
larger mechanical construction, using either very high supports or a much
lengthier loop.
Employing the existing supports
for an upright loop, the area would be only eg. 610 m^2, giving
Rrad = 0.013
microohms
EMRP = 1.3
uW.
A second option would be an
earth antenna, grounded at the far end. You might feed one corner of your
rectangle against ground, and connect the opposite corner to ground as
well. This would be an earth loop of length 61m*sqrt(2), and height 10 m above
ground plus say 48 m below ground (depending on conductivity) => a= 5000 m^2. But here the ground reflection is already
"included" so there won't be the extra factor of two for the image.
Thus
Rrad = 31171 ohms * a^2 / lambda^4 = 0.45
microohms
