Michael Tope wrote:
I am a bit behind on ARRL Handbooks, Richard, but from
what you describe, this is the same figure that appears in my
1992 edition (chapter 38, figure 2). In any case, what is
shown in the figure agrees with my understanding of "broken",
although admittedly when I made my previous post, I was
thinking of the case where the shield is broken on the side of
the loop opposite the feedpoint. For the purposes of this
discussion, however, it doesn't matter whether the break is
at the top (opposite the feed) or at the bottom (adjacent
to the feed). In either case, current induced on the inside
of the shield by current flowing on the center conductor loop
has a continuous back to ground via the outside surface of the
shield. IOW, the gap doesn't suppress the eddy current, rather
it forces it to flow on the outside surface of the shield, thereby
causing the loop to radiate.

Absolutely nothing, neither electic nor magnetic, couplesthrough the
wall of a conductor more than several skin depths thick. This isn't
anything that can be debated, it is simply how it works. It is very
easy to demonstrate, it takes only a few minutes and a minimum of test
equipment.

It is something very basic in physics and underlies how coaxial cables
and things with shields of all types work.

The gap is the feedpoint no matter where the gap is placed. The
radiation and coupling of any time-varying field, magnetic or electric,
occurs on a frequency where the shield is more than a few skin depths
thick comes by the gap.

This is such a very basic thing it is important everyone understand it.


I think the handbook has it right.


Yes, I agree it does. If you connect the shield at both ends, the
loop can't radiate because the eddy current caused by current
flowing on the inner conductor loop will confined to the inside of
the shield.

Absolutely. When the gap is closed there is no potential difference
across the gap the outside of the shield is not connected to the inside
of the shield via the potential developed across the gap. The outer
wall is not coupled to the inner wall, the feedpoint is shorted.

When the gap is opened, the outside of the shield IS the antenna. Not
the inside or anything inside the inside.

Likewise, eddy currents induced on the outside of the
shield by EM waves passing the antenna will be confined to the
outside of the shield if there is no gap (reciprocity holds - the
antenna won't receive with no gap).

Again true. This is a very basic thing we must understand if we are to
understand how shields, walls, or conductors of any kind or form work
with HF currents, voltages, or fields of any type.

There isn't any way to change this effect.

73 Tom