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Question on derivation of potential of a small current loop

  1. Jun 29, 2013 #1
    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
     
  2. jcsd
  3. Jun 29, 2013 #2

    tiny-tim

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    hi yungman! :smile:

    A is a vector, and so is dl'

    (and think "polygon" :wink:)
     
  4. Jun 29, 2013 #3
    Thanks for the reply

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

    Thanks
     
  5. Jun 29, 2013 #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:
     
  6. Jun 29, 2013 #5
    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.
     
    Last edited: Jun 29, 2013
  7. Jun 29, 2013 #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:)
     
  8. Jun 29, 2013 #7
    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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