Question on derivation of potential of a small current loop

  • Thread starter yungman
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  • #1
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The vector magnetic potential is given
[tex]\vec A=\frac{\mu I}{4\pi}\oint\frac{e^{-j\beta R_1}}{R_1}dl'[/tex]
After a few steps, the equation becomes:
[tex]\vec A=\frac{\mu I}{4\pi}e^{-j \beta R}\left[ (1+j\betaR)\oint\frac{dl'}{R_1}-j\beta\oint dl'\right][/tex]
The Book claim the second integral obviously vanishes!!!

I don't understand this. For a small loop, ##dl'=rd\phi##
[tex]\oint dl'=\int_0^{2\pi}rd\phi=2\pi r[/tex]

That is not zero by any stretch. Can anyone explain this?

thanks
 

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  • #2
tiny-tim
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hi yungman! :smile:

A is a vector, and so is dl'

(and think "polygon" :wink:)
 
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  • #3
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hi yungman! :smile:

A is a vector, and so is dl'

(and think "polygon" :wink:)
Thanks for the reply

So I have to integrate the vector in a closed loop and is zero?

Thanks
 
  • #4
tiny-tim
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yes, "the vector" is just a bit of the loop

add them end-to-end (as in a polygon, like a vector triangle), and they obviously make zero :smile:
 
  • #5
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Thanks for your help Tim. I want to verify:
[tex]\oint d\vec l'=\int_0^{2\pi}\hat \phi rd\phi=2\pi r=\int_0^{2\pi}(-\hat x \sin\phi+\hat y \cos \phi)r d\phi[/tex]

Integrate sine and cosine from 0 to ##2\pi## are both zero. So the integral is zero.
 
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  • #6
tiny-tim
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(i'm ignoring the = 2πr = :wink:)

yeees :smile:

but you're assuming the loop is a circle, and my method works for any loop! o:)
 
  • #7
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Thanks, I understand that it works for non circle. My original question is a circle, that's why.

Thanks for your help.

Alan
 

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