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Vector potential of a current carrying circular loop

  1. Jul 21, 2014 #1

    ShayanJ

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    Gold Member

    Consider a circular loop of radius R that carrys a uniform current I. We know(by Biot-Savart law) that the magnetic field it produces on its axis is given by [itex] \vec{B}=\frac{\mu_0 I R^2 \hat z }{2(z^2+R^2)^\frac 3 2} [/itex].
    But let's calculate its vector potential:
    [itex]
    \vec{A}=\frac{\mu_0}{4\pi} \int \frac{I \vec{dl}}{|\vec{r}-\vec{r}'|}=\frac{\mu_0}{4\pi} \int_0^{2\pi} \frac{I R d\varphi \hat \varphi}{\sqrt{R^2+z^2}}=\frac{\mu_0 I R}{4\pi \sqrt{R^2+z^2}}\int_0^{2\pi} \hat \varphi d\varphi \\
    [/itex]
    But we have [itex] \hat \varphi=-\sin\varphi \hat x +\cos\varphi \hat y [/itex] and so [itex] \int_0 ^{2\pi} \hat\varphi d\varphi=0 [/itex] which gives [itex] \vec{A}=0 [/itex] and so [itex] \vec{B}=0 [/itex] !!!
    What's wrong?
    Thanks
     
  2. jcsd
  3. Jul 21, 2014 #2

    Dale

    Staff: Mentor

    What is the relationship between A and B? Given that relationship, does A = 0 at a point imply B = 0 at that point?
     
  4. Jul 21, 2014 #3

    ShayanJ

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    [itex] \vec{B}=\vec\nabla \times \vec A [/itex] which gives B=0 for A=0. Gauge freedom can't help too(By definition).
     
    Last edited: Jul 21, 2014
  5. Jul 21, 2014 #4

    Dale

    Staff: Mentor

    No, it doesn't.

    Remember, you have evaluated A only along a line. Is that enough information to calculate its curl?
     
  6. Jul 21, 2014 #5

    ShayanJ

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    Oh God....yeah man!!!
    Sorry...thanks
     
  7. Jul 21, 2014 #6
    Well if i understand correctly you find the potential only along the z-axis. For a point outside the z-axis [itex]|\vec{r}-\vec{r'}|[/itex] will depend on [itex]\varphi[/itex] as well so the integral wouldnt be always zero.
     
  8. Jul 22, 2014 #7

    vanhees71

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    2016 Award

    Yep, and unfortunately you are led to elliptic integrals :-(.
     
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