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Short Aerial: Find the scalar potential (retarded sources)

  • Thread starter Poirot
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  • #1
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Homework Statement


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.

Homework Equations


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

A
(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.

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!
 

Answers and Replies

  • #2
TSny
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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.
 
  • #3
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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.
That seems to make much more sense since ρ was not specified. Thank you!
 
  • #4
rude man
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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.
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).
(EDIT:
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.
 

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