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Electric field density at the surface of a current carring wire

  1. Mar 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Calculate the magnetic and electric energy densities at the surface of a 3.0 mm diameter copper wire carrying a 15-A current



    2. Relevant equations
    uB=.5[tex]\frac{B2}{\mu0}[/tex]
    uE=.5[tex]\epsilon0[/tex]E2
    R=[tex]\rho[/tex](L/A)
    B=([tex]\mu0[/tex]I)/(2[tex]\pi[/tex]r)
    [tex]\rho[/tex]=1.68 x 10^-8 ohm-meters
    3. The attempt at a solution
    Okay, so finding the magnetic energy density isn't too difficult. My problem is with the electric energy density. I can use the area of the wire and the fact that it's copper to find the resistance and then use ohm's law to find the voltage. but then I get in this bind. E=V/d, but at the surface of the wire, d=0 so you get V/0 which kind of implies infinity and this agrees with my thoughts anyway. However, I feel like this doesn't really make any sense in terms of an electric energy density. Does some one see where the reasoning is going wrong and how I can make it right?
     
  2. jcsd
  3. Mar 5, 2009 #2
    whoa....okay the equations got screwed up there. Hope you can understand them...B^2 is obviously the one in magnetic energy density, mu sub zero, E^2. Sorry about that folks.
     
  4. Mar 5, 2009 #3
    I've just realized that the same problem comes up with my magnetic field. So. Basically I have no idea what I'm doing and am in need of desperate help.
     
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