Is this the correct?
No.
If not, how to handle this?
Simply convert the E and H values from dB to \$V/m\$ and \$A/m\$ respectively. Although the notation you've used isn't conventional, it's pretty clear what it means. The E field is 100 dB referenced to \$1mV / m\$ and the H field is 80dB referenced to \$1mA / m\$
Now, when converting a voltage or current to dB, you take \$20 \ log\$ of the value divided by the reference unit.
So, in the case of the E field, to convert from dB, you divide 100 by 20 to get the exponent of 10 and then multiply that by the reference unit:
\$100 dB (\frac{mV}{m}) \rightarrow (10^{\frac{100}{20}}) \frac{mV}{m} = 100\frac{V}{m} \$
Similarly, for the H field:
\$80 dB (\frac{mA}{m}) \rightarrow (10^{\frac{80}{20}}) \frac{mA}{m} = 10\frac{A}{m} \$
Now that you have E & H in familiar units, you proceed as usual to find the characteristic impedance.
A more sophisticated approach would be to recall that division becomes subtraction of logs. You do have to be careful of reference units though. The difference of the logs is \$20 dB (\frac{mV}{mA})\$ which gives the same result as the procedure above: \$10 \Omega\$
I'll assume dBA is for amperes and not the 'A weighted' dB sound pressure level measurement.
Zero dBA will therefore mean 1A RMS and 6dBA will mean 2A RMS. Strictly speaking 2A is closer to 6.021dBA but everyone accepts that if you double the current (or voltage) it increases by 6dB
0dBV is 1V RMS and by the reasons above 2V is 6dBV
RMS values are always implied unless otherwise stated.
Calculating: if your meter read 13.5dBA, divide by 20 to get 0.675 and take the inverse log (base 10) to get 4.732A
Or dBA = 20 log(I)
EDIT - if your meter read 13.5dBV it actually measures 4.732V and if your clamp has a gain of 100mV per amp, the current flowing will be 47.32A. All values RMS.
Best Answer
Sure, take 10 * log of the ratio between the measured power, and the full scale power.
Since power goes as the square of voltage, you can remove the square root operation.
That gives you:
$$ dBFs = 10 \log({(I^{2}+Q^{2}) / (2^{11}-1)^2 }) $$
You can of course pull the denominator out of the expression, take its log, and convert it to a constant value to subtract from the log of the numerator.
Defining the maximum range of the inputs can get tricky; the range in two's complement form would be from -2048 to +2047 (though other representations are possible). Considering +/- 2047 the maximum is tempting, though some might say +/- 2047.5. And there's even a school of thought which offsets zero by .5. Rounding errors can have some very interesting effects after multiple DSP operations.
Also, it is tempting to think of the maximum as I=2047 Q=2047, however this is a vector which can only occur at the 45-degree phases - you could see it in an impulse, but not in an undistorted signal. Normally, you would want to adjust your gain to stay within the maximimum vector rotatable to any phase, ie, I or Q = 2047 and the other zero, or their combined magnitude = 2047, so that is what should be considered full scale.