How Do You Calculate Retarded Vector Potential for a Current-Carrying Loop?

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Homework Help Overview

The discussion revolves around calculating the retarded vector potential for a current-carrying loop, specifically focusing on the integration involved in the process. The original poster presents a problem related to a wire bent into a loop carrying a current that increases linearly with time.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration process, questioning the limits and factors involved in the integrals. The original poster expresses confusion about the derivation of certain integrals and the presence of a factor of 2 in related problems. Others clarify aspects of the integration limits and the reasoning behind using logarithmic functions.

Discussion Status

The conversation is ongoing, with participants sharing insights and personal experiences related to the problem. Some guidance has been offered regarding the integration process, but there is no explicit consensus on the interpretations or methods being discussed.

Contextual Notes

Participants reference a textbook by Griffiths, indicating that the problem is part of a broader study of electromagnetic theory. There is mention of a final exam context, suggesting time constraints and the pressure of academic deadlines.

leonne
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Homework Statement


a wire bent into a loop it looks like [URL]http://www.pitt.edu/~medart/menuglossary/arch.htm[/URL] it carries a current increases linearly with time I=kt calc retarded vector potentaik

Homework Equations


A=[tex]\mu[/tex]o/4[tex]\pi[/tex] [tex]\int[/tex](k(T-r/c))/r) dl

The Attempt at a Solution


I just don't understand one part.
so i have 4 integrals first one is for the small arch second one is the length third bigger arch and last the other length on the other side.
so i have [tex]\int[/tex]dl/r r is radius and so the smaller radius is a and i just have
1/a [tex]\int[/tex]dl I am not sure what the integration is to in the book, they have dl=2a x^ i am not seeing how they got 2a then the straight line dx/x same for here not sure why they integrate it from a to b they got ln b/a

Also for a similar problem but with straight line why did they add a factor of 2?
A=[tex]\mu[/tex]o/4[tex]\pi[/tex] 2z^ [tex]\int[/tex](k(T-r/c))/r) dz
 
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they get ln|b/a| because they are integrating over that straight line segment at distance x away, and the radius changes from a to b . But they did it in Cartesian so they go from -b to -a but the negatives cancel once inside the log. As for the factor of 2 in the other parts i will have to think about it . But on those 2 arcs you would be integrating over the angle correct.
 
to late already had my final lol thanks anyways
 
Ya this was a problem out of Griffiths , I had to do this problem like a month ago.
 
yeah it is was just reviewing for finals
 
I was just working on this two weeks ago in Griffiths. I was going to post but I saw your response. Even after the finals, you should still try to learn it.
 
lol ok so how do u find the integrations?
 

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