Re: LF: Untuned and tuned loops

[MarkusVester@aol.com]

Dear Mike and LF Group,

thank you for your inspiring comments. I would like to add in a few remarks; hope they will not be regarded as too theoretical...

>> 1. My strong support for what Jim has said.  The immediate environment always affects all field measurements, not just magnetic field measurements. 

The good side for loop measurements is that the magnetic field is much less influenced by conducting or dielectric surroundings than th
e electric field. It is easy for a tree or stone wall to shunt E to ground, whereas distorting H requires a significant current flow, which can only be supported by a large sheet of metal or a nearby wire loop. A nonresonant short pole has no effect at all. And this is "the secret" why TX loops work so well in forested terrain.

>> 2. Marcus’s formula is correct for free-space and it is in all the books. Wire loss is not an issue for a typical open-circuit loop

The formula connecting U to B is just an embodyment of Maxwell's  
 rot E = -dB
/dt ,
it should very generally give the correct answer for B and (excluding ferromagnetic cores) also H.  It's true even for an electrically loaded loop (see 4. below). However the often-used relationship between the magnetic and electric field (E/H = Zo = 377ohm) is restricted to the undisturbed far field zone.

>> 3. In an ‘environment’ near real ground, buildings etc., the ‘physics’ insists that such a loop gives an open circuit voltage proportional to B = mu * H.  The presence of the environment means that mu is not equal to the free-space value of mu(0).

Yes, but this difference is mostly neglegible. Though our environments may have a lot of dielectric and or conducting objects around, they are hardly magnetic, and µr = 1 is an excellent approximation in almost any situation. At RF, even a steel wall looks like a short circuit (due to the eddy currents) rather than a "magnetic wall", which would have high surface impedance. The exeption of course are nonconducting ferromagnetics, like ferrites.

>> 4. The current in a short-circuit lossless loop is directly proportional to the H field and it does not depend on the area.  This is not in ‘the antenna books’, but it is in some physics books dealing with super-conductivity.  

The short-circuit current does depend on the effective length of the magnetic flux lines. For the schoolbook case of a long cylindrical coil (n turns, arbitrary core material), we have
 U = - n area j omega µ H, L = n^2 µ area / length,
 n I = n U / (j omega L), = - H / length ,
easily memorized by the unit "amps per meter" for H. For a "short" ring coil however, the magnetic length is a function of both the wire and loop diameters.

The original induction rule is still correct in the short circuit case, as the current creates a secondary field such that the total flux and the (averaged) H passing through the loop area is indeed nulled.

>> 5. ...Wh
at the system Q ends up being does not matter for this proof.  Some may have noticed however that it is a matter of considerable (unnecessary) dispute, as reported in the pages of RadCom. However loop (or any antenna) losses do matter, because the cross-section area of an antenna is reduced in proportion to its (in)efficiency.  This point is in most of the antenna books.   After real conductor losses have been minimised and factored out we find practical outdoor Qs of 100 to 350.  In a screened room the loop Q can exceed 600.  

This raises a fundamental issue on the efficiency limit of a small antenna. For a magnetic antenna represented by a cylinder coil in free space, we would get
 
;  Rrad = Zo/(6pi) * area^2 * (2pi/lambda)^4
   X = omega L = 2pi / lambda * c * µo * area / length,
and using Zo/c = µo we get a radiation Q-factor
   Qrad = X / Rrad  =  3/(2pi)^2 * lambda^3 / volume.

For an electrical antenna represented by a plate capacitor, we have
  Rrad = Zo/(6pi) * height^2 * (2pi/lambda)^2
  X = 1/omega/C = height / (omega*epsilon*area) ,
and with 1/Zo/c = epsilon we again get the very same value
  Qrad = 3/(2pi)^2 * lambda^3 / volume.

If we have a lossy antenna confined in a small volume, with a given Q<<Qrad we get a maximum possible efficiency of
 eta = Q/Qrad = 13.2 * Q * volume / lambda^3, 
no matter whether it is an electric or magnetic antenna! The energy storage is necessary to support a reactive dipole field (r^-3) between the physical antenna volume and the radiation zone, beginning at a distance lambda/2pi ("CFA"-promoters will d
isagree here...).

Above earth, the radiation resistances and efficiencies for both types of antenna is doubled, which is plausible if we include the image antenna volume below ground. Somewhat surprisingly, this still holds true for loops above weakly conducting ground, as long as the skin depth (a few 10m at LF) is still small compared to a freespace wavelength.

For a thin wire antenna, the "width" of the effective volume is not given by the wire diameter itself but by the capacitance of the wire. For the rule-of thumb 5 pF/m wire, we would have to use approximately half the height of the wire times its length as the effective cross section. My TX antenna (220pF at 10 m above ground, Q = 200) has an effective volume of 2483m^3 and an efficiency of  0.093%

So much for now, it has become late...
73 and good night

Markus, DF6NM