Calculate a vector potential of a simple loop

1. Nov 13, 2014

skrat

1. The problem statement, all variables and given/known data
Calculate the vector potential of a loop with current $I$, raidus $a$. Calculate it for anywhere in space and use approximation where $r>>a$.

2. Relevant equations
$\vec A=\frac{\mu _0I}{4\pi }\oint\frac{d\vec l}{|\vec r-\vec{r(l)}|}$

3. The attempt at a solution
Ok, let's put the origin of my coordinate system in the centre of the loop and let the loop be in xy plane.
than $$d\vec l=dl(-\sin \varphi , \cos \varphi , 0)$$ $$\vec r=r(\sin \vartheta \cos \varphi, \sin \vartheta \sin \varphi , \cos \vartheta)$$ and $$\vec{r(l)}=a(\cos \varphi , \sin \varphi,0)$$.

BUT this is all a complete nonsense because:
$$|\vec r - \vec{r(l)}|=\sqrt{r^2+a^2-2ar\sin \vartheta}$$ and therefore the integral $$\vec A=\frac{\mu _0I}{4\pi }\oint \frac{dl(-\sin \varphi , \cos \varphi , 0)}{\sqrt{r^2+a^2-2ar\sin \vartheta}}$$ considering only cases when $r>>a$ gives me $$\vec A=\frac{\mu _0I}{4\pi r}\oint dl(-\sin \varphi , \cos \varphi , 0)(1+\frac a r \sin \vartheta)$$ but for $dl=ad\varphi$ both integrals are $0$.

Ammm, what am I doing wrong here? :/

2. Nov 16, 2014

skrat

Hmm,

Ok, nevermind. I couldn't sleep last night when it suddenly hit me... This $$\vec r=r(\sin \vartheta \cos \varphi, \sin \vartheta \sin \varphi , \cos \vartheta)$$ is wrong. I should define a new angle $\alpha$ therefore $$\vec r=r(\sin \vartheta \cos \alpha , \sin \vartheta \sin \alpha, \cos \vartheta)$$
This seriously complicates things until you realize that it doesn't matter where your integration starts. You can easily start it at $\alpha$ which sets $\alpha =0$ and this simplifies the problem a lot!

I was able to get the right result. So the problem was only that I said that $\varphi =\alpha$ which is obviously not the case!

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