What are the fields produced around a current carrying conductor?

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To illustrate this issue using a mechanical example, consider the equation ##F = ma##. If an object of mass ##m## is at rest, then the net force acting on it must be zero. The acceleration is zero as well. If we tried to ascertain the mass from this information we get ##m = F/a = 0/0##, which is indeterminate. The mass is non-zero, non-infinite, but force (summation) and acceleration are both zero. With any "##y=mx##" type of equation this is applicable. Does this help?

Your posts helped me. Thank you.
 
  • #27
anorlunda
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I hop this is not too off topic.

A conductor with some finite resistivity will have a slight E-field as well as a B field. A superconductor carrying a steady current will have only a B field.

On the same day I read the above quote, I listened to Leonard Susskind's lecture 6 on Special Relativity and Fields (available on YouTube and iTunes). He said that electric and magnetic fields are frame dependent. What I see as electric/magnetic, another frame will see as magnetic/electric. E becomes B and B becomes E.

DaleSpam, is your statement about superconductors frame dependent?
 
  • #28
Dale
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He said that electric and magnetic fields are frame dependent. What I see as electric/magnetic, another frame will see as magnetic/electric. E becomes B and B becomes E.

DaleSpam, is your statement about superconductors frame dependent?
Yes, all statements about E, B, and J are inherently frame dependent for the reason mentioned by Susskind.
 
  • #29
Dale
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Read the article from Arvix.org: http://arxiv.org/pdf/1202.1851v1.pdf
As the authors explicitly state, they are deliberately going against the established mainstream theory. Time and experiments will decide, but my comments above are (I believe) consistent with the consensus theory and therefore not consistent with challenging theories.
 
  • #30
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The E field for a current carrying material is σE=J\sigma E=J. The source of this is Ohm's law: http://en.wikipedia.org/wiki/Ohm's_law. Ohm's law is also covered in every textbook that covers EM physics or circuit analysis. This E field is small for ordinary conductors (large σ\sigma) and 0 for superconductors (σ=∞\sigma = \infty).

My actual question is regarding the fields produced "around" the conductor.

The E in your equation is the electric field set up across the ends of the conductor (If I am not wrong).

Sorry for not noticing this point before.
 
  • #31
Dale
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The E is the field within the superconductor, not just at the ends, but at any point within.

For the fields outside the conductor you would need to solve Maxwell's equations. I believe that there would also be no E-field outside of a superconducting wire, unless imposed externally, but I cannot justify that at the moment.
 
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  • #32
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For the fields outside the conductor you would need to solve Maxwell's equations. I believe that there would also be no E-field outside of a superconducting wire, unless imposed externally, but I cannot justify that at the moment.

Anyway, thank you for the info on the fields within the current carrying conductors.

I look forward if anyone wants to reply for the actual question. I quote it once again:

If we consider a current carrying conductor, every instant an electron enters the conductor, another electron will be leaving the conductor. Thus, the current carrying conductor will not be charged (i.e, it would not have any net positive or negative charge). Remember dipole has zero net charge, but it does have electric field around it. So, if net charge is zero, it doesn't mean there is no electric field.

It is important to notice that, if we assume only electrons to be moving, and kernels (positive nuclei) to be static, magnetic field will be produced only due to electrons.

Does it mean that electric field and magnetic field exists around the current carrying conductor?
Or
Does it mean that only magnetic field exists around the current carrying conductor?

The question is simple, but I have found varied answers until now.
 
  • #35
Dale
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Yes, I was intrigued by some of the results, particularly the Poynting vector. It seems that in a very reasonable sense power doesn't flow through a wire.
 

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