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.


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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.