My early thought was that, at the very least, Poynting Vectors deserve a video but Transmission Lines also deserve a video.
Here's a problem I recall from Jackson which might (or might not) provide a missing link:
Given two parallel conductors of arbitrary shape, prove that the product of capacitance-per-unit-length and inductance-per-unit-length is a constant (i.e., independent of profile and separation).
I remember finding a suitable answer, but I can't for the life of me remember how.
I always use this video to explain how Time Domain Reflectometry finds a broken cable in a wall. Not everyone is familiar with the idea that a cable going to nowhere can have voltage and current.
Watch his vid again. Transmission line mentioned. He even use the issue with putting cable across the ocean having issue as a possible example explain his thought experiment. Could it be you used to watch his vids in full and remember and not so much for this one?
> * But they have a high parasitic capacitance because the outside conductor has a large surface area.
> So if capacitance is causing too much propagation delay, you can use twisted pair.*
The capacitance is not really "parasitic", it is the circuit element representation of the electric field which is propagating. The dynamic electric field gives rise to a magnetic field, which is the inductive element in the expression 1/sqrt(LC) that you listed.
Twisted pair is no different: There is a capacitive element and inductive element per unit length. As it turns out, the capacitance varies as the log of the distance between the wires, so you can twist the wires and not change the capacitance very much. There used to be lots of "twinax", which kept the two wires at a fixed distance, but the separation plastic was mostly a PITA. (The pics I can find on google don't show the old flat plastic twinax that folks used to use. You can still find that stuff hanging off old television antennas.)
Oh please do. With something like an audio cable at least there's one end that is purely sticking out and one end that is purely sticking in. But with a lot of other cables there's a mix of inny bits and outy bits and it took me a long time to work out which bit I'm supposed to pay attention to to make the analogy work (the answer is to pay attention to any visible bits that carry a single/power. If this is obvious to you I suspect it was just because of how someone explained it to you, but if you're just brillianter than me then great!).
There's another part to this story I remember hearing, that there is a way to measure of impedance? capacitance? in this powerline to detect where the coil is. anyone now how that works?
I understand your point, but could you give more information about the original question? I think the video posted by OP is incredible in terms of analysis (CT scan!) but it would great to follow the thread through "normal" cables without electronics?
Coiling is only relevant to magnetic fields. These cables generally have the same amount of current passing in opposite directions through different wires, so net zero.
There are a handful of videos where other pen-testers show the process from start to finish, and the space needed to get the wire through is not as much as you think.
A wire doesn't look like a point source from far away, unlike a dipole. Cut off a segment of that wire (so it's not infinite-length) and it won't be 1/r at large distances.
Getting the wire through isn't the problem, it's controlling it afterward. Not surprisingly, the 90 minute video you linked didn't find time to show that part.
interesting that you mention this. I can't find any photographs of the line. Some depictions show two wires on poles. Others show what appear to depict a pole with a single conductor wire at the top.
Here's a problem I recall from Jackson which might (or might not) provide a missing link:
Given two parallel conductors of arbitrary shape, prove that the product of capacitance-per-unit-length and inductance-per-unit-length is a constant (i.e., independent of profile and separation).
I remember finding a suitable answer, but I can't for the life of me remember how.
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