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The above picture shows a rectangular wire carrying alternating current wound across a core with permeability much much greater than of free space(and also assumed is the fact that the width of the conductor is much greater than the skin depth at this frequency). The current is flowing into the screen, denoted by a cross. The graph shows the current density distribution along its cross section. Basically, every transformer core and its nearest winding.

Why is the distribution one sided? Shouldn't skin effect be the same on both sides of the current carrying conductor? Why does the core affect the current distribution in this way?

I've been looking this up for a while now, and no one adequately explains this. When the concept of skin effect(due to eddy currents) was explained to me, it was shown as a completely internal phenomena. Current carrying conductors induce a magnetic field that links with the conductor itself and causes current to flow opposing the cause. Because currents flow in loops, the current in the center is diminished and current on the surface is exaggerated. Now, no one said anything about materials outside the wire here. And how these materials may influence fields inside. If someone could shed some light on that, It'd be much appreciated.

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  • \$\begingroup\$ I also have the same doubt. How the magnetic field inside the wire is getting affected when it is near to the core? If you have got any answer, please let me know. Thanks in advance. \$\endgroup\$
    – aswin
    Jul 29, 2020 at 3:55
  • \$\begingroup\$ In the discussion you mention you are dealing with a single conductor in isolation. If present, conductive materials outside the primary conductor are also influenced by the primary magnetic field, and fields may be induced in the exterior conductor and currents may be induced which overall can cause an interaction between the two conductors. \$\endgroup\$
    – Russell McMahon
    Jul 29, 2020 at 8:57

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It is similar to the proximity effect, but it is not the proximity since there is only one conductor (as you have stated in your long comment thread with Andy aka , but maybe that is too much insisting on semantics since it is basically same phenomena ).

It comes down to Maxwell's equations, but this is the intuition:

By placing a core with high permeability at one side of the conductor your are changing the magnetic field H around the conductor. You can never really separate the inside of the conductor from the outside of the conductor and you have to solve Maxwell's equations for the whole system, by choosing the right boundary conditions i.e. your assumption about the skin effect that it is only an internal effect doesn't quite work because that calculation assumes a certain boundary condition for the surface of the wire.

So the two differential equations have to be solved: $$ \nabla \times \pmb{H} = \pmb{J} \tag{1} $$

and $$ \nabla \times \pmb{E} = - j \omega B \tag{2} $$

Now assuming the core has a permeability that is much higher than the permeability of free space the H field in (1) is very close to zero on the left side of the conductor (where the core is) and very high on the right side of the conductor (where you have free space). Now deriving the whole expression for this setup is very similar to the skin effect, but the boundary conditions are a bit different since left and right surface of the conductor aren't symmetric anymore.

As an aside: Andy aka has it backwards in their answer "there is no magnetic field on the right of the rectangle and hence current finds it much easier to travel on the right edge of the rectangle." This is exactly the wrong way around. The current flow is precisely stronger where the magnetic field H is larger. Which you can actually see in the picture they posted with the proximity effect also and you can see in equation (1). You can convince yourself with the right hand rule that when the current goes in the same direction the H field cancels between the two conductors and no current flows there. When the current flows in opposite direction the H fields adds up between the two conductors and the current crowds towards the stronger H field towards the middle of the conductors.

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