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Retarded potential

  1. Jan 20, 2009 #1
    1. The problem statement, all variables and given/known data
    The question concerns a square loop in the presence of an infinitely long sinusoidally varying line current.

    The complete problem is here

    2. Relevant equations
    The retarded potential.


    3. The attempt at a solution
    I defined J along the z direction [tex] = I * \delta (\rho') \delta ( \cos (\phi')) / \rho' [/tex]

    Then I find the vector potential
    [tex] A = {\mu_o \over 4 \pi} \int_{-\infty}^{\infty} {J_{ret} \over |x-x'|} d^3x [/tex]

    with the Sin in J having the retarded time of course. But this leaves me with the integral:

    [tex] \int_{-\infty}^{\infty} \sin ( w(t - \sqrt (\rho^2 + (z-z')^2))) \over \sqrt(\rho^2+(z-z')^2)[/tex]

    Did I go wrong? If not then can someone tell me how to get rid of this.
     
  2. jcsd
  3. Jan 20, 2009 #2

    gabbagabbahey

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    Are you sure you are expected to involve the retarded potential for this problem?


    This expression is only valid for finite current distributions. A "long" current carrying wire does not qualify.
     
  4. Jan 20, 2009 #3
    Pretty sure that you have to and for two reasons:
    [1] Wiki says that its for time varying currents and charges without ever mentioning finiteness
    [2] Just saw that Griffiths example 10.2 solves for an infinite line current as well. However, the time dependence over there is not sinusoidal
     
  5. Jan 20, 2009 #4

    gabbagabbahey

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    Sorry, I was thinking of the non-retarded version where the entire line of current simultaneously contributes to the vector potential.:redface:

    Here, the integral should not really go from -infinity to +infinity; you need to consider which section of the wire actually contributes to the potential at a time [tex]t>\frac{\rho}{c}[/tex]. Note that here, unlike in Griffiths' ex. 10.2, you are considering a general field point, not one confined to the xy-plane.
     
  6. Jan 20, 2009 #5
    For a while you had me worried :)

    I can change the limits later of course but I wanted to know how to solve that wretched integral. But I do think that the problem allows me to consider a point in a single plane.

    In any event, it seems that I am on the right track and only need to solve the integral followed a simple differentiation. But the integral...?
     
  7. Jan 20, 2009 #6

    gabbagabbahey

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    Yes, just be sure to state that you are doing so in your final solution.

    EDIT: Actually, the square loop lies in the ρ-z plane, so you need to find the z-dependence of the magnetic field.

    Hmmmm....good question! Mathematica doesn't even want to evaluate this integral analytically. I'll have to think about it.
     
    Last edited: Jan 20, 2009
  8. Jan 20, 2009 #7

    gabbagabbahey

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    You might be able to make things easier for yourself by assuming (and showing some justification for your assumption) that the current varies slow enough to consider [tex]\sin(\omega t_r)\approx \omega t_r[/tex] and then the problem is essentially the same as Griffiths' 10.9(a).

    EDIT: Another idea is to bypass the vector potential altogether and use Jefimenko's equations to find the field. Although I don't think the integral becomes any easier without some approximations.
     
    Last edited: Jan 20, 2009
  9. Jan 20, 2009 #8

    turin

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    From the wording of the problem, and the association with problems from chapter 6 in Jackson, I don't think that you are supposed to make any assumptions that, for instance wt is small. I agree with gabbagabbahey that you should just use Jefimenko, but indeed you will have basically the same issue with the integral. I would suggest a clever change of variables to make the argument of the trig function as simple as possible. To do this properly, you will need to recognize some symmetry properties of the integrand so that you can properly handle the limits.
     
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