I went to the library and took a look in Johnson and Jasik's "Antenna
Engineering Handbook," where a good explanation of the shielded loop antenna
is provided. In brief, what happens is that the incident magnetic field
induces a current around the _outside_ of the shield and -- due to the slit
in the shield -- the current is then forced to flow "around the edge" and
onto the _inside_ of the shield through whatever impedance it sees at the
slit. If you're using a piece of coax where the center conductor has been
shorted to the shield at the location of the slit, the impedance looking
into the slit is the load impedance 'transformed' back the length of the
transmission line (i.e., assuming a matched load, it would be the same as
the transmission line impedance). If you're using a coil of insulated wire
inside, e.g., a copper pipe, the slit impedance is pretty much a short
circuit (well, a small inductance, actually -- the self inductance of the
inside loop) so all the current induced on the outside is 'mirrored' into
the inside. This then couples into the coil of insulated wire just as if
the shield weren't there. Some small loss occurs due to the loop inductance
and the finite conductivity of the shield. (I've found web sites with
measured results showing that shielded loops do have readily measurable loss
over unshielded loops.)

An interesting result of the above is that the polarity of the voltage
coming out of a shielded loop should be opposite that of what's coming out
of an unshielded loop. (Because in the case of the shielded loop, as the
current flows over the edge of the slit it's reversing direction.) It would
be fun to try this experimentally... (I'm thinking something like building
another coil and discharging a capacitor through it to induce spikes in the
antenna.)

The explanation on web sites about the shield acting something like an
electric dipole doesn't really fly, IMO.

---Joel

On Thu, 11 Nov 2004 09:24:07 -0800, "Joel Kolstad"
wrote:

An interesting result of the above is that the polarity of the voltage
coming out of a shielded loop should be opposite that of what's coming out
of an unshielded loop. (Because in the case of the shielded loop, as the
current flows over the edge of the slit it's reversing direction.) It would
be fun to try this experimentally...

Hi Joel,

Build two loop antennas: one shielded, one un-shielded. Combine them
and see if the cancel (less the differential you've observed as loss
at various web sites (which bears reporting here)).

73's
Richard Clark, KB7QHC

Joel Kolstad wrote:
"An Interesting result of the above is that the polarity of the voltage
coming out of a shielded loop should be opposite that of what`s coming
out of an unshielded loop."

Yes, a phase reversal is produced by the change of current direction
between the outside and inside surfaces of the loop shield. But carry
your logic a bit further. There is a second phase reversal provided by
the inductive coupling between the inner surface of the loop shield and
the wire loop inside the shield.

The wire loop is the secondary of a transformer. Induced and counter emf
are 180-degrees out of phase with the primary or source emf. For
graphics see Fig. 5-14 on page 79 of Robert l. Shrader`s "Electronic
Communication".

Best regards, Richard Harrison, KB5WZI

"Richard Harrison" wrote in message
...
Yes, a phase reversal is produced by the change of current direction
between the outside and inside surfaces of the loop shield. But carry
your logic a bit further. There is a second phase reversal provided by
the inductive coupling between the inner surface of the loop shield and
the wire loop inside the shield.

Sure, of course... but there's also a (single) phase reversal in the
unshielded loop. Hence the overall difference between the shielded and
unshielded loop is still a single phase reversal. Or do you disagree?

---Joel

Joel Kolstad wrote:
"Hence the overall difference between the shielded and unshielded loop
is still a single phase reversal."

Let`s reason together, as Lyndon Johnson used to say. An unshielded wire
loop and/or a same size shield-pipe loop`s outer surface when placed in
the same RF field and having the same attitudes in that field will have
currents which attempt to flow in the same direction. Be it clockwise or
counter-clockwise, it will be the same direction on the outside of both
loops, shielded or unshielded.

As Joel noted, current flowing inside of the shielded loop has its
direction (sense) reversed by the configuration of the shield pipe. As
current flows from outside to inside the loop around the butt ends of
the pipe at the shield gap, there is a phase reversal.

Current flowing on the outside of the shield toward the gap is flowing
away from the gap once it has entered the shield. Also current flowing
on the outside of the shield away from the gap, while it is on the
inside of the shield it is flowing toward the gap. This is a 180-degree
phase reversal between current on the outside of the loop as compared
with that on the inside of the loop, This results just from current
flowing around the butt ends of the shield pipe.

The current inside the loop on the surface of its shield is the primary
current in a two-winding transformer.Current induced into the wire loop
contained by the shield is secondary current in this transformer.

Transformer secondary current is in the same phase as self-induced
current, that is it is 180-degrees out of phase with the primary
current. Two phase reversals in the shielded loop place its output in
the same phase as the unshielded loop.

Best regards, Richard Harrison, KB5WZI

Hi Richard,

Ah, I see now that you're correct; thanks. I was forgetting that current
'reversal' on the exterior of the shield needed to oppose the incident
field. Oops.

---Joel

On Thu, 11 Nov 2004 13:52:37 -0800, "Joel Kolstad"
wrote:

"Richard Harrison" wrote in message
...
Yes, a phase reversal is produced by the change of current direction
between the outside and inside surfaces of the loop shield. But carry
your logic a bit further. There is a second phase reversal provided by
the inductive coupling between the inner surface of the loop shield and
the wire loop inside the shield.

Sure, of course... but there's also a (single) phase reversal in the
unshielded loop. Hence the overall difference between the shielded and
unshielded loop is still a single phase reversal. Or do you disagree?

---Joel

Disagree.
w.
--
On the Internet nobody knows that I am a dog.

Joel, think of it this way... Consider the "shielded loop" in which
the gap is at the top, and the left side is a piece of coax. The
outer of the coax is just the loop "shield" on the left side. The
coax center conductor connects to the right side of the loop "shield"
at the top, across the gap. So the current in the coax center
conductor is just the same as in the shield right side. In other
words, for a loop which is very small compared with a wavelength (at
least), the current in the center conductor is the same as the current
on the outside of the left half of the shield, and in phase with it.
In other words, it's the same as in an unshielded loop. I don't see
any phase reversal there.

Or consider it this way: the loop encloses essentially the same
time-varying magnetic field, whether you look at the loop formed by
the "shield" or the loop formed by the wires inside the "shield."

Cheers,
Tom

"Joel Kolstad" wrote in message ...
"Richard Harrison" wrote in message
...
Yes, a phase reversal is produced by the change of current direction
between the outside and inside surfaces of the loop shield. But carry
your logic a bit further. There is a second phase reversal provided by
the inductive coupling between the inner surface of the loop shield and
the wire loop inside the shield.

Sure, of course... but there's also a (single) phase reversal in the
unshielded loop. Hence the overall difference between the shielded and
unshielded loop is still a single phase reversal. Or do you disagree?

---Joel