En/na Roy Lewallen ha escrit:
EA3FYA - Toni wrote:

I know that a coax cable does not radiate (if common mode currents
properly suppressed) because both conductors are apparently "in the
same place" (wouldn't know how to express it in more technical terms).

Here's why it doesn't radiate: In a coaxial cable with a solid shield,
the differential mode current is entirely inside the shield. Current and
fields penetrate only a very small distance from the inner surface of
the shield, and no significant amount ever makes it through to the
outside. This is assuming that the shield is at least several skin
depths thick, which is a good assumption at HF and above.

When you say "Current and fields penetrate only a very small
distance...", I agree for the current part, but I'm not so sure for the
fields part:

As I understand it you can not "stop a field" in no way, though you can
certainly nullify it with an identical but opposite field.

Then the question is whether the two fields (the one from the current
flowing in the shield + the one from the current flowing in the inner
conductor) nullify at all points in the immediate vicinity of the
shield. I certainly believe it but would like to understand why this is so.

I guess the mathematical proof would involve assuming the braid is an
infinite number of conductors equally spaced around the center
conductor, each having it's infinitesimal share of the shield current,
and integrating all of their fields at the point of interest (Would
probably be able to do so back when I was at university but now it is
too strong math for me). Would this be a good approximation of the problem?

--
Toni

Toni wrote:
En/na Roy Lewallen ha escrit:
EA3FYA - Toni wrote:

I know that a coax cable does not radiate (if common mode currents
properly suppressed) because both conductors are apparently "in the
same place" (wouldn't know how to express it in more technical terms).

Here's why it doesn't radiate: In a coaxial cable with a solid shield,
the differential mode current is entirely inside the shield. Current
and fields penetrate only a very small distance from the inner surface
of the shield, and no significant amount ever makes it through to the
outside. This is assuming that the shield is at least several skin
depths thick, which is a good assumption at HF and above.

When you say "Current and fields penetrate only a very small
distance...", I agree for the current part, but I'm not so sure for the
fields part:

As I understand it you can not "stop a field" in no way, though you can
certainly nullify it with an identical but opposite field.

You bet you can stop a field. It can be stopped either by reflection,
absorption, or a combination of the two. Inside an anechoic chamber,
absorbing materials stop internal fields to prevent reflections. A
screen room or metallic shield reflects external fields.

Then the question is whether the two fields (the one from the current
flowing in the shield + the one from the current flowing in the inner
conductor) nullify at all points in the immediate vicinity of the
shield. I certainly believe it but would like to understand why this is so.

Indeed they do. Look up Ampere's Law. If you draw a boundary through the
middle of the shield or outside the shield, you'll find that the sum of
currents within that boundary is zero. According to the law, that means
that no net field penetrates the boundary. Because of the physical
symmetry, no net field means no field at all.

I guess the mathematical proof would involve assuming the braid is an
infinite number of conductors equally spaced around the center
conductor, each having it's infinitesimal share of the shield current,
and integrating all of their fields at the point of interest (Would
probably be able to do so back when I was at university but now it is
too strong math for me). Would this be a good approximation of the problem?

No, it's not that complicated, but a path or surface integration is
required to use Ampere's law.

Roy Lewallen, W7EL