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Magnetic Vector Potential

  1. Apr 9, 2009 #1

    Matterwave

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    1. The problem statement, all variables and given/known data
    Find the vector potential [tex]\vec{A}(x,0,0)[/tex] (i.e. on the x-axis) for a current loop of radius a, carrying a current I in the [tex]\phi[/tex] direction.

    2. Relevant equations
    [tex]\vec{A} = \frac{\mu_0}{4\pi}\int_{V'}{\frac{\vec{J}dV'}{R}}[/tex]
    Where R is the distance from the source point to the field point. The coordinates are set up so that all primes denote the source and all non-primed denote the field point. (x means the x position of the field point and x' means the x position of the source point)


    3. The attempt at a solution
    Ok, so the current flows in the [tex]\phi[/tex] direction so my vector potential should also be in the phi direction. i.e.
    [tex]\vec{A}=A_\phi \hat\phi[/tex]

    I'm on the x-axis so
    [tex]R=\sqrt{(x-x')^2+y'^2}=\sqrt{(x-acos(\phi'))^2+a^2sin^2(\phi')}[/tex]

    Expanding out that expression and using the trig identity
    [tex]sin^2(x)+cos^2(x)=1[/tex]

    I get:
    [tex]R = \sqrt{x^2-2axcos(\phi')+a^2}[/tex]

    So, I only have to integrate over [tex]\phi[/tex] because the radius of the loop is constant. So my integral is set up in this manner:

    [tex]A_\phi (x,0,0) = \frac{\mu_0}{4\pi}\int\limits_{0}^{2\pi}\frac{Iad\phi'}{\sqrt{x^2-2xacos(\phi')+a^2}}[/tex]

    Where the a appears as is customary when integrating the angle in polar coordinates. I play around with it a bit more to try to get it into a form where I can use elliptic integrals:

    [tex]\Rightarrow A_\phi (x,0,0) = \frac{\mu_0Ia}{4\pi}\int\limits_{0}^{2\pi}\frac{d\phi}{\sqrt{(x+a)^2-2xa-2xacos(\phi')}}[/tex]
    [tex] = \frac{\mu_0Ia}{4\pi}\int\limits_{0}^{2\pi}\frac{d\phi}{\sqrt{(x+a)^2-2xa(1+cos(\phi'))}}[/tex]
    [tex] = \frac{\mu_0Ia}{4\pi}\int\limits_{0}^{2\pi}\frac{d\phi}{\sqrt{(x+a)^2-4xacos^2(\frac{\phi'}{2})}}[/tex]
    [tex] = \frac{\mu_0Ia}{4\pi(x+a)}\int\limits_{0}^{2\pi}\frac{d\phi}{\sqrt{1 - \frac{4xa}{(x+a)^2}cos^2(\frac{\phi}{2})}}[/tex]

    I now make the substitutions:
    [tex]k^2=\frac{4xa}{(x+a)^2}[/tex]
    [tex]t^2=cos^2(\frac{\phi'}{2})[/tex]

    I find dt in terms of [tex]d\phi'[/tex] and do the usual thing and get finally:

    [tex]A_\phi (x,0,0) = \frac{\mu_0Ia}{2\pi(x+a)}\int\limits_{-1}^{1}\frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}[/tex]

    Noticing that the integrand is even, I can therefore write:


    [tex]A_\phi (x,0,0) = \frac{\mu_0Ia}{\pi(x+a)}\int\limits_{0}^{1}\frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}[/tex]

    My integral is an elliptical integral of the first kind K(k) so my final answer is:


    [tex]A_\phi (x,0,0) = \frac{\mu_0Ia}{\pi(x+a)}K(k)[/tex]

    But herein lies the problem. That's the wrong answer. According to my professor, my integral should consist of not only elliptical integrals of the first kind, but also the second kind E(k). And the general solution should be:

    [tex]A_\phi (x,0,0) = \frac{\mu_0Ia}{\pi(x+a)} (\frac{2E(k)-(2-k^2)K(k)}{k^2})[/tex]

    Which is significantly more complicated than my solution. For the life of me, I can't figure out where I went wrong...someone help please?
     
  2. jcsd
  3. Apr 10, 2009 #2
    Dont forget that the components of the infinitesimal vector are :[tex]d\vec r = (-a\sin\phi d\phi,a\cos\phi d\phi)[/tex]

    So you missed \cos\phi in you integral and it correctly should be:

    [tex]A_{\phi}(x,0,0)=\frac{\mu_0 I }{4\pi}\int_0^{2\pi}\frac{a\cos\phi'\,d\phi'}{\sqrt{...}}[/tex]
     
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