# Topband: QST Jun 06 RX Loop

Tom Rauch w8ji at contesting.com
Sun May 14 22:12:46 EDT 2006

```> One question that some readers of this forum might ask:
> If that is the
> case, how does an inductively coupled network, where two
> resonant inductors
> are separated by a Faraday shield, transfer energy?

We can modify the direct capacitive coupling (change
impedance), but we cannot stop the time-varying electric
field without stoping the time-varying magnetic field. Take
either to zero and the other goes to zero. Nothing goes
through a solid shield. It can't because inside the shield
wall the electric field goes to zero.

> Recall the old B&W HDVL series plug-in coils with
SNIP
> feedline.  But if "essentially nothing goes through that
> shield," how is the
> fundamental-frequency energy from the tank coil
> transferred to the pickup

The same way it does in a loop.The outer shield becomes the
actual radiator. The inner conductor loop couples to the
inside of the shield, the inside of the shield has a voltage
(potential)  difference across the gap, and that potential
causes current to flow on the shield outside. The current on
the shield outside, which has a voltage difference along its
length and current flowing,  is what actually couples to the
tank coil.

It's no different than a larger loop.

http://www.w8ji.com/coaxial%20currents/coaxial_line_and_shielded_wires.htm

> My explanation is that in the case of the link-coupled
> tank circuit, we are
> talking about what is essentially a transformer, where the
> primary and
> secondary coils are magnetically coupled, and the shield
> prevents capacity
> coupling.

It also prevents the magnetic field from passing directly
through.

What you really have is a three layer transformer. The inner
conductor's fields set up a longitudinal voltage that is the
same as the inner conductors voltage. This voltage appears
across the gap. The voltage difference across the gap causes
current to flow on the inside and the outside of the shield.
The current inside the shield flows the opposite direction
of the center conductor current and equals the center
conductor's current. The current on the outside flows the
same direction as the current in the center conductor, but
opposite the current on the inside of the shield. (This can
actually be measured on a bench with a few boxes made from
something like double sided blank PC boards.)

What happens if we short the gap is we force the voltage
across the gap to zero. Now we have zero time-varying
voltage outside the cable shield, and we also have zero time
varying current (assuming a zero ohm short).

Any time we take the time-varying voltage to zero we do the
same for current, and vice versa.

In a "shielded link" the shield does exactly the same thing
as in a larger loop antenna. The outside of the shield

We can measure the loss in Q and increase in heating caused
by current having to flow over the additional length on the
inside length plus the outside length of the shield.

Of course what you say about the antenna is absolutely true.
The electric and magnetic fields in a "radio wave" are
always inseparable. The field impedance very close to the
radiator can be changed as can balance of the system, but
the link behaves just like the loop. It all follows the same
rules.

Even the shields around our transmitters or a screen room
behaves the same way. The screen room doesn't let a
time-varying magnetic field in unless we crack the door open
and allow a voltage across the gap, and then we have both
time-varying voltage and current. The outside of the screen
room becomes the "antenna".

73 Tom

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