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Skin Effect

  1. Jun 11, 2012 #1
    Hey I know the conclusion of skin effect since secondary year of school, but I still didnt get it. I know the reason is because of eddy current.

    I couldnt include the link here as I havent posted 10 yet. For the link, just search "Skin Effect" at Wikipedia.
    First paragraph of the cause in Wikipedia says "The counter EMF is strongest at the center of the conductor, and forces the conducting electrons to the outside of the conductor, as shown in the diagram on the right." I didnt really get it. Why the counter EMF is strongest? Even if it is strongest, why electrons are going to be driven outside? Shouldnt we have to consider about the positive or negetive then decide whether being driven outside or inside?

    Also, the explaination of the third picture on the right "Skin depth is due to the circulating eddy currents (arising from a changing H field) cancelling the current flow in the center of a conductor and reinforcing it in the skin." Well, it is true when I in read increases. But what if I decreases, aren't eddy currents in the picture should circulate the opposite direction. By that case, eddy current actually cancel the current flow in the skin and enforce the current in the center!

    Also, I believe eddy current is larger in the skin as H is larger compared to it in the center.

    I know skin effect is correct. But I do not know why it works that way.

    Please help me get it!

  2. jcsd
  3. Jun 13, 2012 #2


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    Staff: Mentor

    Thread moved to the EE forum for better views. Here is the link you are referring to:


  4. Jun 13, 2012 #3
    You are right, the Wiki page is a little misleading on this point. This is not an electrostatic situation where electrons are being forced to move from the center to the edges. "I" is an AC current, the electrons in the conductor are wiggling back and forth (a relatively small distance). The counter EMF is simply impeding the electrons in the center from wiggling, electrons on the skin are free to wiggle. Because fewer electrons in the cross section of the conductor are participating in this dance, the resistance of the conductor is lowered.

    Draw out the main current "I", the H field, and the eddy current "Iw" as sinusoids, and bear in mind that they are phase shifted due to the derivatives in Maxwell's equations. For example, when "I" decreases but is still directed upward, the H field will be in the opposite direction, but it will be decreasing in magnitude. This leads to eddy current that opposes "I". (Lenz's Law).
  5. Jun 13, 2012 #4
    Eddy currents is a bad description of this phenomenon. Eddies are circular vortex-like motions. This happens to describe the currents that flow when the magnetic field points into the surface. But it doesn't accurately describe the much more common situation of a plane wave striking a metallic surface or the penetration of AC current into a wire- again fields are parallel to, not normal to a surface.

    Having established that, the skin depth is controlled by a dissipative phenomenon depending (for finite conductors) on the resistivity of the metal.

    The penetration is zero for the ideal conductor and as the resistivity increases the penetration increases. As with any dissipative wave propagation the amplitude decays exponentially.
  6. Jun 13, 2012 #5
    Hey, EMI Guy, I can understand "The counter EMF is simply impeding the electrons in the center from wiggling", but why "electrons on the skin are free to wiggle"? No counter EMF on the skin? I think there should be. Is it because electrons cannot wiggle as free as in the centre at all directions? "Because fewer electrons in the cross section of the conductor are participating in this dance, the resistance of the conductor is lowered. " I thought the resistance is increased as the effective cross section is smaller. Am I correct?

    "when "I" decreases but is still directed upward, the H field will be in the opposite direction, but it will be decreasing in magnitude." If "I" dereases but still directed upward, I thought the original H field is still in the same direction but decreasing in magnitude, which causes eddy current inducing H filed at the same direction of original H. Eddy current here is opposite of the one shown in the graph at wiki page. That's what confused me.

    If I am wrong, tell me where. Thanks for your help.
    Last edited: Jun 13, 2012
  7. Jun 13, 2012 #6
    Hey Antiphon, thanks for your help. But I do not really understand your explanation. Do I have to look into "dissipative phenomenon" to understand skin effect?

    I know the conclusion of the skin depth and the factors affecting it. But I need to understand the root cause of skin effect.
  8. Jun 13, 2012 #7
    That search will probably not help you.

    The root cause of the skin effect is that the penetration of fields into conductors takes time. The lower the resistivity of the material, the longer it will take.

    For example, work out what the resistivity of a wire would be that had a skin depth of 1 mm at a frequency of 1 Hz.

    It will be an almost perfect conductor. Now if you had a long 1 meter thick wire made of it then when you turn on a DC current, at first it will flow along the outside. Slowly the DC current will penetrate into the conductor in exactly the same manner as if heat were penetrating into the side of a cold glass cylinder. After around twenty minutes the current inside would begin to approach the level at the surface.

    Now clearly if you apply AC to this wire you can see how the field penetration won't get very far; it would be as if you were alternately heating and cooling the outside of a glass cylinder. If you do it fast enough there will be very little temperature variation going on in the core.

    Does that make sense so far?

    Ok, the thermal analogy is good but not exact.

    In the wire there is in fact induction going on just like the Wikipedia article says but there is also resistive dissipation.

    The correct picture looks like an LR circuit where you have a DC voltage source at the left, an inductor at the top, the resistor on the right.

    If the resistor is zero, the current in the inductor will steadily increase but you will measure no output voltage on the resistor. If there is a non-zero resistance the voltage will ramp up at a rate governed by the RL time constant (almost the skin effect equation!).

    Now the full physically correct circuit analog is an infinite repeating ladder with series inductors and shunt resistors. When you turn on the source, a voltage "front" (measured across the shunt resistors) will "diffuse" through this network.

    -That is unless the resistors are zero ohms and then the current will only flow in the first inductor- the "skin" of the LR ladder!

    Does this physics make sense to you?
    Last edited: Jun 13, 2012
  9. Jun 17, 2012 #8
    Thanks. I kinda understand it now.
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