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Understanding the behavior of E fields inside wires

  1. Oct 11, 2006 #1
    Hello All,

    I'm currently trudging through my physics class which is teaching about electric and magnetic fields, and their interactions. As I have been reading through the textbook, I have been trying to formulate ideas and understand exactly what is happening inside of current carrying wires. I have been trying to visualize what the E fields inside a copper wire do immediately after a voltage potential is applied, and immediately after the charges start to flow.

    To make sure that I understand this right, let’s say we have a length of wire shorted across a battery that applies a potential difference of 1 volt from one end to the other. Assuming that I am interpreting this correctly, there will be an E field setup in the wire with the vector pointing from the negative side of the battery to the positive side; and because of Kerchhoff’s voltage law, we can calculate the amount of current that will flow based on I = V/R with R being the resistance of the wire (assuming we have an ideal battery for simplicity).

    So, after the voltage potential is applied, it is the E field in the wire that pushes the charges through and creates the current right? This resulting current flow produces a magnetic field whose vector, B, is at right angles to the current flow vector (left hand rule for electrons). Ok, this is where I started asking questions.

    I know that magnetic fields will exert a force on moving charges as they move perpendicular to the field. So in the case of the wire, immediately after the charges start to move and the magnetic field is created, do the charges not get forced towards the outside of the wire? So for an instant the wire has a radial E field with the negative end being the outside of the conductor and the positive side being the center? If my assumptions are true, I know, from reading about the Hall Effect, that the resulting E field eventually produces a force that is equal and opposite the force produced by the magnetic field which keeps the charges from moving radially outwards further. So, here are my questions:

    1. Is what I described anywhere close to what is actually happening?

    2. If I made the right assumptions, is there a way to predict how far the charges will move towards the outside of the wire before the E field cancels their movement if I know the B field strength, the amount of current, and the cross section of the wire?

    3. Is there a way to determine how fast this effect takes place (how long it takes the electrons to move from their neutral state to the polarized state?

    4. During the instant that the electrons rush towards the outside of the wire (if this really does happen), is there a radial magnetic field produced by the charges that emanates from the wire?

    Ok, that’s all I have for now. I have more questions but I’ll wait to ask those since I am already going on a lot of assumptions here. Any help/ input/ corrections are greatly appreciated.

    Thanks,
    Jason O
     
  2. jcsd
  3. Oct 11, 2006 #2

    NoTime

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    Science Advisor
    Homework Helper

    It's called the skin effect.
    You should be able to find more info under that heading.

    Generally it is a consideration for high frequency circuits.
    However, the initial turn on counts as high frequency so it sounds like you have it about right.
     
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