Vector potential created by a magnetic moment

Question

I want to proove that the vector potential created by a magnetic moment $\vec{m}$ is $$\vec{A}(\vec{r}) = \frac{\mu_{0}}{4\pi} \frac{\vec{m} \times \vec{u}}{r^2}$$ by using the formula $$\vec{A} = \frac{\mu_{0}}{4{\pi}} \int \frac {\vec{j} (\vec{r}')} {\lvert \vec{r} - \vec{r}' \rvert} \mathrm{d} \vec{r}'.$$

My idea was to consider a loop of current $ I = \frac{m}{\pi r^2} $ with r the radius of the loop. I then have: $$ \vec{j}(\vec{r}') = \frac{I}{2\pi r} \vec{u}_{\theta} = \frac {m}{2\pi^2 r^2} \vec{u}_{\theta}$$

By using the formula $$\vec{A} = \frac{\mu_{0}}{4{\pi}} \frac {m}{2\pi^2 r^2} \int_{0}^{2\pi} \frac {1} {\lvert \vec{r} - \vec{r}' \rvert} \mathrm{d} \vec{r}',$$ but I don't know how to continue.

Answer 1

Let's say the point at which you're trying to find the vector potential lies at $(r,\phi,\theta)$ and that any point on your current loop lies at $(R,\phi',0)$ where R is the radius of the loop.

You should use $$A=\frac{\mu_0}{4\pi}\int \frac{I}{r}dl'=\frac{\mu_0}{4\pi}\int_{0}^{2\pi} \frac{IRd\phi'}{(r^2+R^2-2Rr(cos(\theta)cos(\phi-\phi')))}$$

This becomes a nasty elliptic integral. From what I've heard, it turns into no man's land from here on out.