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Transformation between electric and magnetic fields

  1. Aug 27, 2007 #1

    The following is the standard qualitative explanation of the duality between electric field and the magnetic field given in our textbooks:

    Let a long straight wire carry a constant current i. A charge kept at rest near it does not experience any force, implying the absence of electric fields. Now it is set in motion parallel to the wire. A radially outward force is needed to hold it in course, implying the presence of a magnetic field.

    Now if the coordinate frame is transformed to the rest frame of the charge, the charge is static but the force stays, meaning an electric field has appeared.

    What I don't understand (and the textbooks don't explain) is the source of this electric field. According to Gauss's law, only charge can create electric fields. But a current-carrying conductor is electrically neutral. Then what is the origin of this electric field? Does transformation of reference frame create charge? How?

    Quantitatively, the magnetic force (and therefore the electric force) has a magnitude [tex]q v \frac {\mu i}{2 \pi r}[/tex], which implies a linear charge density of [tex]v \frac {i}{c^2}[/tex] on the wire (excuse any mistakes as I was doing the calculation in my head) which is a very small but still non-zero charge density. Where does it appear from? Does this have anything to do with Lorentz transformations?

    Thanks a lot.

    Last edited: Aug 27, 2007
  2. jcsd
  3. Aug 28, 2007 #2


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  4. Aug 28, 2007 #3
    Thanks, that's a nice explanation. So the equivalence of electric and magnetic fields cannot be explained in non-relativistic physics (though non-relativistic electrodynamics is perhaps a contradiction).

  5. Aug 29, 2007 #4
    Does it work if the drift velocity is different from the speed of the charge?
  6. Aug 29, 2007 #5
    Does what work?

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