1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Short Aerial: Find the scalar potential (retarded sources)

  1. May 19, 2016 #1
    1. The problem statement, all variables and given/known data
    Determine the vector potential due to a short wire running from (−L/2, 0, 0) to (L/2, 0, 0) carrying a current I = I0cos(ωt) (consider only points at distance r >> L from the origin). (You may neglect the effect of the return circuit.) Now determine the corresponding scalar potential for large r.

    2. Relevant equations
    V(x, t) = 1/4πε0 ∫d3x' ρ(x' , t - |x' - x|/c) / |x' - x|

    (x, t) = μ0/4π ∫d3x' j (x' , t - |x' - x|/c) / |x' - x|

    Where A is the vector potential and V is the scalar potential.
    3. The attempt at a solution
    I managed to solve the first part for the vector potential by using the fact that r>>L and therefore |x' - x| ≈ r
    and using that j = I/c x(hat) (in the x direction, and I let C= the cross sectional area of the wire)
    this gave A(x, t) = (μ0L I0 cos(ω(t-r/c)))/4πr

    When trying to determine the vector potential I ran into an issue, I'm not entirely sure what ρ is explicitly. I had a guess that is was ρ = Q/LC, and then using the fact that I=dQ/dt and integrating to find Q, but this was a stab in the dark and I have no idea if it's right.

    Any help would be greatly appreciated, thank you!
  2. jcsd
  3. May 19, 2016 #2


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    The problem seems to suggest that you can derive V from A. There is a gauge condition for the potentials that must be satisfied in order for your relevant equations to be valid. This condition can be used to relate V to A.
  4. May 19, 2016 #3
    That seems to make much more sense since ρ was not specified. Thank you!
  5. May 21, 2016 #4

    rude man

    User Avatar
    Homework Helper
    Gold Member

    You didn't relate ρ to your equation for A so I don't know what you mean by it. It's usually used to mean charge density, not applicable here.

    The vector potential A is
    A = (1/4π)(j/r)dV
    where j is current density (a vector),
    dV is a differential of volume within the volume of current, and
    r is distance from dV to the point of observation (not a vector).
    note I use A in the sense of H = ∇ x A).
    In this case I assume r >> L also implies r >> R, the radius of the wire.

    So I suggest orienting your short wire in say the z direction, middle at origin, then doing the simple integration over the length of the wire of j/r.

    The problem suggests you compute A first, then V. Actually, the magnetic scalar potential is of academic interest only since, when current flows, the mag potential cannot be used to compute B or H since ∇ x H is not then zero (Maxwell: ∇ x H = j). So I don't know why you were called upon to compute it to begin with.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted