Assume the box is non-conducting. There would be a current difference
between the two terminals unless you shrunk the length of antenna it
contains to a vanishingly small length. In the context of the antenna,
even a few feet of an antenna is a significant fraction of a wavelength.
Put the box around a few feet of a power transmission line carrying 60
Hz current, and you won't be able to measure any difference between the
current going in and coming out. In that situation, a few feet isn't a
significant part of a wavelength.

If you use a conducting box, you'll end up with current on the outside
of the box that gets into the problem. Its value depends on coupling to
the wire inside and to the antenna outside, so it gets stickier than I
want to deal with.

Roy Lewallen, W7EL

Jim Kelley wrote:

Roy Lewallen wrote:

Read again the fourth sentence of the posting you quoted.

Roy Lewallen, W7EL

Jim Kelley wrote:

Roy Lewallen wrote:


If that
two-terminal box contains an inductor, then the current out has to equal
the current in -- that's the only way the sum of currents at the two
terminals can sum to zero.


What if you draw a two terminal black box around the middle few feet of
a 1/4 wave vertical? What makes the current become the same at both
ends?

73, Jim AC6XG


Okay. I read it. I'll try asking the question another way.

What if you draw a two terminal black box around the middle few feet of
a 1/4 wave vertical? What makes the sum of the currents at both ends
become equal to zero?

73, Jim AC6XG