Calculate a vector potential of a simple loop

[skrat]

Homework Statement

Calculate the vector potential of a loop with current ##I##, raidus ##a##. Calculate it for anywhere in space and use approximation where ##r>>a##.

Homework Equations

##\vec A=\frac{\mu _0I}{4\pi }\oint\frac{d\vec l}{|\vec r-\vec{r(l)}|}##

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? :/

 

[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!