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Small transmitting aka "magnetic" loop antennas are often constructed out of copper or aluminum pipe material. Antennas don't need to be solid due to the skin effect. Does the skin effect take place on both surfaces of a pipe conductor, outside and inside, or only the outside? (I don't understand the underlying physics)

Where I am going with this is: if instead of a pipe, I use a broad sheet metal piece which is to have the same "active surface" area that corresponds to a wouldbe pipe that a magloop calculator spat out as efficient enough: It does not have an inside - it's solid, though thin. I guess given that it is thicker than a few skin depths at a given frequency, do both sides of the sheet "count" as active surface for the RF current in a resonant loop, and if the inside surfaces, OTOH, of a pipe do not contribute, would that mean that a sheet metal piece with, let's call it "top", surface equals that of the outer surface of a half pipe, and the "bottom" surface of the sheet likewise equals the other half pipe's outer surface, one could get away with half the material when using the sheet metal vs. the pipe, yielding the same efficiency?

-- EDIT: sketch of the question: Left are a loop of pipe material and its crossection below, right are a loop of sheet material with its crossection below. enter image description here

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  • \$\begingroup\$ I think you mean an antenna like this one -- but I'm also pretty sure that the one answer so far doesn't speak to that design. You need to edit your question for clarity. I suggest that you drop a picture of the antenna design you're talking about (the page I cite here has an excellent one), cite the picture properly (by linking to the page it comes from, for instance), and see if any answers change. \$\endgroup\$ Commented 13 hours ago
  • \$\begingroup\$ No, that's an RX-only loop with "special tricks" used for various purposes. I mean a naked one-turn loop without any other shenanigans, a variable HV capacitor between both ends of the conductor, and a smaller coupling loop at the other end of the circle, where the RF power is fed to. \$\endgroup\$ Commented 12 hours ago
  • \$\begingroup\$ @TimWescott OK, when one thinks something up, it always seems obvious enough what's meant - to oneself! :D I added a drawing. \$\endgroup\$ Commented 12 hours ago
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    \$\begingroup\$ I think you need more clarification, because Andy and the other Tim are giving you answers as if you're asking about the difference in conduction inside the tube and outside the tube, not the surface of the tube toward the center of the antenna and the surface of the tube toward the outside of the antenna. That's what I thought you meant. I'm now 99.44% sure you're talking about inside the antenna, not the tube. However, usually when you tell an RF guy "inside" and "outside" and there's a tube involved, that's what they're going to think. \$\endgroup\$ Commented 10 hours ago
  • \$\begingroup\$ @user1847129 - Where did the photo come from? To comply with the site referencing rule, details of the original source of any copied/adapted material must be provided by you, next to each copied/adapted item. If the original source is online, please edit the question & add the webpage (or PDF/video timestamp) name & link (e.g. website name + webpage title + URL). (It doesn't apply here, but if the source is offline (e.g. printed book / private intranet) then add source details "to the best of your ability" e.g. title, authors, page, edition etc.). TY \$\endgroup\$ Commented 8 hours ago

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You're asking about currents toward the inside of the antenna (not the inside surface of the tube), and toward the outside of the antenna (not the outside surface of the tube). Those will be substantially equal, until the width of your strap started getting up to a substantial fraction (probably 1/10, but I'm not that much of an antenna guy) of the diameter of the loop.

So yes, you could approximate that with a flat sheet of metal. I think you'd be disappointed in how well it held up to wind or any motion of its support*, but electrically it'd work.

I'm guessing that if you just used a strip of metal twice the diameter of the tubing then you'd get roughly the same electrical performance out of it. Other shapes, like "U" shaped channel, or "T" shape, or "V" shape angle would behave similarly electrically and be much stronger -- but then you're getting into the same difficulties of fabricating the thing that you would with tube, and they'd be weaker, so you may as well use tube.

* "Wibbly-warbly" is not a good technical term -- but it'd be wibbly-warbly.

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  • \$\begingroup\$ "currents toward the inside of the antenna" - you mean the whole structure, like an imagined center point of the circle? No, not at all. That's not what a drawing of cross-sections illustrates - or I'm really confused now what you mean by that ;) I have thought about the disadvantes of wind, for people who would expose this to wind. That's not my scenario, though. \$\endgroup\$ Commented 9 hours ago
  • \$\begingroup\$ Then you need to clarify your question even more. Try sketching the whole electrically active part of the thing you're planning on constructing. \$\endgroup\$ Commented 9 hours ago
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This question was answered before the OP edited his answer with loop images. I can give no guarantees on whether my answer below is relevant to the new information that is disclosed in that edit. The circular pipe was not disclosed; I assumed a straight pipe (as I suspect anyone would when giving an answer to the original body of the post). I don't do shifting goalposts.

Does the skin effect take place on both surfaces of a pipe conductor, outside and inside, or only the outside? (I don't understand the underlying physics)

A tube of current does not produce a magnetic field on the inside thus, skin effect (and the majority of current) exists at the outer surface.

It's a similar story when you analyse a length of coaxial cable; the magnetic field external to the cable caused by forward current in the inner conductor is completely cancelled by the return current flowing back through the shield (aka tube). Thus there is no net external magnetic field (as we know and love about coaxial cable). But, there is an internal magnetic field between inner conductor and shield however, that field is purely due to the forward current carried by the inner conductor.

This gives rise to the fact that for an ideal length of coaxial cable, the shield (aka tube) possesses zero inductance. It can only have zero inductance if there is no net magnetic field associated with it. And, of course that means there is no internal magnetic field produced by the shield.

A self-made QuickField simulation might help you believe this: -

enter image description here

The above image is for coaxial cable but, the inner conductor is not being used to pass current. The field inside the tube/shield is effectively zero despite the shield passing 1000 amps. This means there is no skin effect on the inside of a tube carrying current. It's a physics thing really but, is a really important aspect of how coaxial cable works.

For a sheet carrying current, skin effect occurs on both sides of the sheet.

would that mean that a sheet metal piece with, let's call it "top", surface equals that of the outer surface of a half pipe, and the "bottom" surface of the sheet likewise equals the other half pipe's outer surface, one could get away with half the material when using the sheet metal vs. the pipe, yielding the same efficiency?

Whether you could get away with it depends on your application. The beauty about the tube/pipe is that it produces an equal magnetic field in all directions along the length of the pipe. This may indeed be superior in some applications compared to a sheet antenna.

Also, the inductance of a sheet will be smaller hence, the overall magnetic field produced for a given current will be less than that of a tube (because it's inductance is higher). This is down to what inductance means: the amount of magnetic field produced (per turn) for a given current (per amp).

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  • \$\begingroup\$ Andy, I think your model doesn't match what amateur radio enthusiasts call a "magnetic antenna" (and I think the question needs clarification). The "magnetic antenna" uses a multi-turn loop inside of pipe, which is interrupted at the antenna feed point, and grounded. According to the literature, this has the effect of shielding the antenna from the electric field, while leaving it as sensitive to the magnetic field as if the wire loop were unshielded. I don't see how you're modeling that here. \$\endgroup\$ Commented 13 hours ago
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    \$\begingroup\$ No, not necessarily multi-turn at all. "Magnetic" refers to a property ascribed to a loop with a circumference of <= 1/10 lambda (a rule of thumb, I guess), of supposedly predominantly radiating magnetic component in the near field (a few wavelengths) and also receiving, Making it less prone to be influenced by metal stuff in the vicinity vs. mostly electrical antennas. Some even say it's less prone to receive electrical disturbances nearby. That's the lore anyway, partly disputed by some. The supposedly more technical term is "electrically small transmitting loop" \$\endgroup\$ Commented 12 hours ago
  • \$\begingroup\$ Also, it does not need to be shielded (which your description suggests), which is only a variant of it, but not the default. Most people simply use copper tubing to form a single turn loop, and a high voltage variable capacitor at the gap where the 2 ends meet. I could try to make a drawing some time later, if that helps. \$\endgroup\$ Commented 12 hours ago
  • \$\begingroup\$ @user1847129 - Hi, If you intended your comments to be seen by TimWescott then I'm not sure he will have received a notification, as you didn't include @TimWescott in the text. Therefore he may be unaware of them. || So if you intended to respond to his comment, I suggest you post one more comment e.g. @TimWescott Please see my replies to your comments above. to trigger a notification. || On the other hand, the answer's author Andy aka will have received a notification for your comments, but I doubt he was the intended recipient so he may ignore them. Can you clarify the addressee? TY \$\endgroup\$ Commented 10 hours ago
  • \$\begingroup\$ @SamGibson you're right, but he also commented on the OP, where I mentioned it also. \$\endgroup\$ Commented 9 hours ago
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Current flows on the surface, in response to surrounding magnetic field.

It's actually more specific than that: current flow and magnetic field are equivalent (Ampere's law).

In the case of the loop antenna, current flows on the outside, due to exterior magnetic field.

There is no current flow, and no magnetic field, within the tube. Since these fields are equivalent, both are necessary.

(Note the implicit assumption that frequency is high enough that skin effect is dominant, so that inner and outer surfaces have largely independent currents. Put another way, the tube is an effective shield at the chosen frequency.)

There must be something else to carry such a current inside the tube. (Or, frequency is high enough it's a waveguide, but to be perfectly specific: there must be an inner conductor to support a TEM00 field mode, which has the current and magnetic field orientation as you'd expect). If we add one, we get a coaxial cable geometry. See: https://en.wikipedia.org/wiki/Skin_effect#Coaxial_cable

Can you cheat it? Can skin effect be used as a (frequency) filter, or something? Not quite.

Consider the case of current flowing on a tube, then you tie a wire to the inside of that tube (where, presumably, no current flows at high frequency, but could at low frequency). Can you measure current on the wire? Yes! -- But, ah, realize what you have to do, first: you can't just connect to that wire through solid metal. You had to open a hole in the tube, to pass the wire through; and now the topology is different. Now current can flow in through that hole, along its surface, current comes inside, up the wire and back out!

You'd need some sort of 4-dimensional tunneling to get the wire out, without changing the tube topology. And we simply can't do that (at least not yet, and certainly not affordably :) ).

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