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Relativistic electrodynamics

  1. Nov 20, 2006 #1
    If someone could help me with this one that would be great. My prof has not been in his office in quite some time...

    -A wire of cross-sectional area A carries a current I, and has zero net electric charge in frame O.
    (a) Find the magnetic field a distance r from the axis of the wire
    (b) Find the charge density ρ' and current density j' in frame O' moving with velocity v parallel to axis of wire relative to O.
    (c) Using Ampere's law and Gauss' law with the charge and current densities found in (b), calculate the electric and magnetic fields in O' at a point a distance r from the axis of the wire.
    (d) Verify that these fields agree with the ones obtained by transforming from O.

    OK, so here's where I'm at:
    (a) I expect B to be the same as in non-relativistic e.d., μ0*I/2πr.

    (b) This is the tricky part. I don't know how to look at ρ. If simply looking at the ρ+ (charge density for positive charges), then ρ' should be greater than ρ0 by a factor of gamma due to length contraction. But if we are looking at ρ- (negative charges), then it's more complicated since it increases by more than a factor of gamma if v is moving in the same direction as the electrons. But if we just use the invariant, and ρ is the net charge density, then ρ0=0, and we have j^2=(c^2)(ρ'^2)+(j'^2). I'm assuming that j'=I'/A (where A is the cross-sectional area). So then we have an equation relating j to ρ', but how to solve for either?

    (c) Of course, I can't figure this out until I figure out (b), but I think that ε0*E'*A=q', meaning that E'=q'/(ε0*2*pi*r*l)=ρ'/(ε0*2*pi*r). So once I figure out ρ' I should have that part. Then I have B'=μ0*I'/2πr. I' should be less than I by a factor of gamma due to time dilation, I think. So I believe B'=μ0*(I/γ)/2πr.

    Anyways, if anyone gets a chance to help me, thanks so much. It's due tomorrow, but posts after that will still help me understand.

    Ben
     
  2. jcsd
  3. Nov 21, 2006 #2

    OlderDan

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    Is v meant to be an arbitrary velocity, or the drift velocity of the current charge carriers? If it is arbitrary, the drift velocity in O should still come into play. You will need th relativistic velocity addition formula to find the velocity of the drifting charges in O'
     
  4. Nov 21, 2006 #3
    v is the arbitrary linear velocity of frame O' wrt frame O. I thought about using relativistic velocity addition to determine the velocity of the electrons in the O' frame...but how can you figure that out? I suppose that the velocity of the electrons carrying the current should be some function of the current itself (among other things), but we've never talked about anything like that in class....although that's not exactly unusual.
     
  5. Nov 22, 2006 #4

    OlderDan

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    I think you need the drift velocity to do the problem. There has to be a different effective density of the opposite charges in the wire, so different length contractions come into play. The current is related to the charge density and the drift velocity.
     
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