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Suppose a current flows in a straight cylindrical wire so that an electric field $\textbf{E}$ is maintained in the wire. Will there be an electric field just outside the wire..?

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3 Answers 3

When a steady current is flowing through the wire, the overall wire is charge neutral. In the presence of an external electric field $\textbf{E}$ the electrons are simply moving from one end of the wire to the other. The electron density at end of the wire from which the electrons are leaving gets replenished by (say) the battery. An equal flux of electrons is obviously coming out of the other end of the wire and going into the battery. In each region of the wire the free (or conducting) electron density matches that of the positive ion cores. As a result, the wire is charge neutral. Therefore there should not be any electric field outside the wire (Gauss's law).

Also, because you said that "$\textbf{E}$ is maintained," the magnetic field produced by the wire (Biot-Savart law) will be time independent. Therefore any electric field outside the wire, due to induction, will also not be possible (Lenz's law).

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This is wrong. the integral of E from one point to another is the same along any path, requiring an E field continuously extending outside the wire when a current is flowing. –  Ron Maimon Oct 5 '12 at 7:32

From your setup, it sounds like you have an E field directed along the wire that is driving the current (i.e. the wire has some finite conductivity). If that is the case, then just outside the wire there must also be an E field, because of Faraday's Law. The curl of E must be finite, and if you had a discontinuity of the tangential E field at the surface of the wire then the curl would be infinite.

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I believe it is Faraday's law here rather than Ampere's law. –  Alfred Centauri Oct 4 '12 at 18:08
    
You are right, Faraday's Law. –  user1631 Oct 4 '12 at 20:59

The voltage difference in a steady state current is independent of path, and deforming the path out of the wire, you can see that the electric field must be continuous. The electric field not only extends outside the wire, in conjunction with the magnetic field surrounding the wire, it is carrying the bulk of the momentum of the current.

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Ron Maimon, isn't there another problem: what if you take your path very far from the wire, I assume the integral $\int E.ds$ is nonzero on that portion of path which is parallel to wire then you will find that the electric filed must be independent of distance from wire to give the same voltage and it seems nonphysical! Am I right? –  richard Apr 23 '13 at 16:36
    
@richard: That's not a paradox. The field bend, so that you get a nonzero integral along the parts of the path relatively close to the wire, which will give the voltage difference between each point and infinity. –  Ron Maimon Apr 23 '13 at 18:46
    
why should field bend in a symmetric condition (long wire) ? actually I'm confused about this! because there are few texts talking about it which are not quite consistent. For example in Feynman lectures the electric field is assumed parallel to wire! –  richard Apr 23 '13 at 19:39
    
@Richard, if it doesn't bend, E can't fall off with distance (imagine two opposite infinite plates of charge sourcing the current). The wire isn't infinite. –  Ron Maimon Apr 24 '13 at 14:21

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