ve2pid wrote:

Fb Roy and thanks (to all) for the explanations.

But... returning to your document, I wonder why a balanced antenna
'see' that third wire, but the dummy load does not recognize that
third conductor. Is it because the common-mode current outside of the
shield has something to do with a kind of RF coupling?

Current doesn't flow one way or another because the path is "seen" or
"recognized". It has to obey Kirchoff's Current Law, which it might be
useful to review. It states that the (vector) sum of currents into or
out of any node must equal zero. In this case, the node of interest is
the connection of the dummy load to the coax shield, which is
effectively the connection of three conductors: the conductor to the
dummy load, the inside of the coax shield, and the outside of the coax
shield. If the current on the inside of the shield equals the current to
the dummy load, the current on the outside shield equals zero per
Kerchoff. And this is what happens if the currents in the two dummy load
leads are equal and opposite in phase. Similarly, if the current to the
two halves of a dipole are equal and opposite in phase, the current on
the outside of the coax is zero, and conversely. The whole article is
based on that principle, and explains how it can be accomplished.

Please look at Fig. 1, and the currents going to each side of the
dipole. If the two coax conductors go to opposite sides of a resistor
rather than to two sides of a dipole, those currents have to be equal or
nearly equal, as I explained in my last posting. If they are equal,
there is no extra current available to go along the outside of the coax.

You asked about RF coupling. I explained in my last posting that
coupling to various objects could cause the currents in the two dummy
load wires to be unequal, resulting in current on the outside of the coax.

Sorry, I'm running out of ways to explain this. You might try reading
the other responses if you still don't understand.

Roy Lewallen, W7EL