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I'm trying to wrap my head round Skin Depth, we've derived the wave equation from Maxwell's equations, using the conditions that it's a "good conductor" ρ = 0, σ >> ωε and therefore you get that the current density decreases exponentially from its value at the surface.

I'm happy that it pops out the maths however it's an intuitive sense I'm after, given that we've used Maxwell's equations I've sure it'll have something to do with self inductance and I think I've sort of got something with current "strands" being induced in the opposite direction (lenz's law) etc, and then thinking about the force between two strands of current flowing in the opposite direction (they'll repel) but then I've got current flowing two different ways on the surface of my wire and now I'm confused.

Any help would be greatly appreciated, I've looked round quite a lot but if anyone has an explanation or can link me one that isn't a mathematic proof that'd be super

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    \$\begingroup\$ Here's an easy one - look at the cross section of the wire and decide which part looks more like a capacitor, even if its in air. As frequency increases, the impedance on the outside looks more favorable than inside as a result of this - mostly. Kind of. \$\endgroup\$
    – user39962
    Apr 18, 2015 at 20:32
  • \$\begingroup\$ Sorry if I'm being obtuse, what do you mean by which part looks more like a capacitor? Do you mean the fact that charge will build up on the outside (due to radial electric field etc) therefore the outside looks like a capacitor. Because I can kind of understand that, but then we use the fact that the charge is 0 in a conductor to prove the equation? \$\endgroup\$
    – geeeeeeek
    Apr 19, 2015 at 12:57
  • \$\begingroup\$ Its not great but, yes. And its not a very good capacitor, and we're only on the one plate. \$\endgroup\$
    – user39962
    Apr 19, 2015 at 14:05

2 Answers 2

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If you want to understand this, instead of thinking of a transmission line as two wires with special behavior, you need to think of a transmission line as a guiding structure for electromagnetic waves. When we talk about currents and voltages and capacitance and inductance in a transmission line, that's a useful simplification for many cases. But to understand skin effect you should instead think about an electromagnetic wave that is guided by conductive structures. The currents and voltages that we measure on the transmission line are caused by the EM wave travelling along it, not the other way around.

So then, when an EM wave encounters a conductive material, what happens? The EM wave is reflected. It doesn't penetrate deeply into the conductor. But if the conductivity is not infinite, it does penetrate slightly. And up to the depth that it penetrates, it causes currents to flow in the conductor. That is the skin effect.

I'm slightly confused as to the co-ordinates of how you're defining your transmission line.

Normally we call the axial direction (along the length of the transmission line) the z-dimension. The transverse dimensions are x and y.

It's easiest to visualize in a parallel plate waveguide. There, the propagating mode can be seen as a sum of plane waves reflecting back and forth between the two plates. So these plane waves have a k-vector that has components in both the z and x directions. The reflections off the conductors is what causes the currents in the conductors, and the depth of penetration into the conductors is what determines the skin depth.

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  • \$\begingroup\$ Hey, sorry could you expand a bit more? I'm slightly confused as to the co-ordinates of how you're defining your transmission line. If we had a long wire running in the x direction it's my understanding that the wave, though travelling in the x direction, decays in the z direction from the surface to the centre. So are you getting at snell's law kind of reflections in z? \$\endgroup\$
    – geeeeeeek
    Apr 19, 2015 at 13:02
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For an intuitive understanding of the "skin effect," look at the cross section of the wire. Next, think of the center atom as surrounded by a ring of similar atoms, and then repeat this radially. The center atom will be compressed the most (highest density), the next "ring" will be a little less dense, and so on out to the last ring (conductor's surface). By using this "model," it should be easy to see that an electron traveling along the center of the conductor (the densest area), will experience the most "resistance" and an electron traveling along the surface of the conductor will experience the least "resistance" (the least dense area). The result of this, is that there will be a "tendency" for more electrons to travel along the surface of the wire, this is the "skin effect". I hope this "unorthodox" explanation is helpful to you.

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  • \$\begingroup\$ This explanation does not explain why it happens at AC and not DC. \$\endgroup\$
    – Andy aka
    Aug 24, 2015 at 9:11
  • \$\begingroup\$ Furthermore, this explanation is not at all based on the actual physics of what's going on. \$\endgroup\$
    – The Photon
    Mar 4, 2017 at 16:10

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