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Vector potential due to sudden burst of current in a wire

  1. Oct 26, 2013 #1
    There is a wire oriented in the z direction with no current in it for t<0. At t=0, there is a burst of current: I(t) = qδ(t). What is the vector potential [itex]\vec{A}[/itex]? ([itex]\vec{B}[/itex]=∇×[itex]\vec{A}[/itex])

    My solution:
    let point P be located distance s from the wire, z be length along wire, and Ω be distance from point on wire to point P. Then:

    [itex]\vec{A}[/itex](s,t) = [itex]\frac{μq\hat{z}}{4\pi}[/itex][itex]\int^{∞}_{-∞}\frac{δ(t-Ω/c)}{Ω}dz[/itex] = [itex]\frac{μq\hat{z}}{4\pi}[/itex] [itex]\frac{1}{ct}[/itex]

    t-Ω/c is the retarded time, thus δ(t) becomes δ(t-Ω/c), and I substituded Ω = √(s2+z2) to get the integral in terms of z. The limits are from -Inf to +Inf because for a given time t, the only contribution to the potential is from z=±√( (ct)2-s2 ). Thus I integrated the expression by plugging in this z value.

    The answer is wrong however, and despite perusing the solution I cannot figure it out. The solution rewrites the integral in terms of Ω instead of z, but i dont understand why we cannot just integrate wrt z. Help is much appreciated!
     
  2. jcsd
  3. Oct 26, 2013 #2

    TSny

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    Note that the argument of the delta function is some function of ##z##. So, if you are going to use ##z## as the integration variable, you will need to use property (7) given at this link: http://mathworld.wolfram.com/DeltaFunction.html
     
    Last edited: Oct 26, 2013
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