Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Advanced Physics Homework Help
Short Aerial: Find the scalar potential (retarded sources)
Reply to thread
Message
[QUOTE="Poirot, post: 5477647, member: 587460"] [h2]Homework Statement [/h2] Determine the vector potential due to a short wire running from (−L/2, 0, 0) to (L/2, 0, 0) carrying a current I = I[SUB]0[/SUB]cos(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. [h2]Homework Equations[/h2] V([U]x,[/U] t) = 1/4πε[SUB]0[/SUB] ∫d[SUP]3[/SUP][U]x'[/U] ρ([U]x'[/U] , t - |[U]x'[/U] - [U]x|[/U]/c) / |[U]x'[/U] - [U]x| A[/U]([U]x,[/U] t) = μ[SUB]0[/SUB]/4π ∫d[SUP]3[/SUP][U]x'[/U] [U]j [/U]([U]x'[/U] , t - |[U]x'[/U] - [U]x|[/U]/c) / |[U]x'[/U] - [U]x|[/U] [U][/U] Where A is the vector potential and V is the scalar potential. [h2]The Attempt at a Solution[/h2] I managed to solve the first part for the vector potential by using the fact that r>>L and therefore |[U]x'[/U] - [U]x|[/U] ≈ r and using that [U]j[/U] = I/c [B]x[/B](hat) (in the x direction, and I let C= the cross sectional area of the wire) this gave [U]A[/U]([U]x,[/U] t) = (μ[SUB]0[/SUB]L I[SUB]0[/SUB] 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! [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Advanced Physics Homework Help
Short Aerial: Find the scalar potential (retarded sources)
Back
Top