Electric field density at the surface of a current carring wire

  • Thread starter cuttlefish
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  • #1
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Homework Statement


Calculate the magnetic and electric energy densities at the surface of a 3.0 mm diameter copper wire carrying a 15-A current



Homework 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

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?
 

Answers and Replies

  • #2
13
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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.
 
  • #3
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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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