For a static current density $\mathbf{J}(\mathbf{x})$ show that we may choose the vector potential $\mathbf{A}(\mathbf{x})$ so that

$$-\nabla^{2} \mathbf{A}=\mu_{0} \mathbf{J} .$$

For a loop $L$, centred at the origin, carrying a current $I$ show that

$$\mathbf{A}(\mathbf{x})=\frac{\mu_{0} I}{4 \pi} \oint_{L} \frac{1}{|\mathbf{x}-\mathbf{r}|} \mathrm{d} \mathbf{r} \sim-\frac{\mu_{0} I}{4 \pi} \frac{1}{|\mathbf{x}|^{3}} \oint_{L} \frac{1}{2} \mathbf{x} \times(\mathbf{r} \times \mathrm{d} \mathbf{r}) \quad \text { as } \quad|\mathbf{x}| \rightarrow \infty$$

[You may assume

$$-\nabla^{2} \frac{1}{4 \pi|\mathbf{x}|}=\delta^{3}(\mathbf{x})$$

and for fixed vectors $\mathbf{a}, \mathbf{b}$

$$\left.\oint_{L} \mathbf{a} \cdot \mathrm{d} \mathbf{r}=0, \quad \oint_{L}(\mathbf{a} \cdot \mathbf{r} \mathbf{b} \cdot \mathrm{d} \mathbf{r}+\mathbf{b} \cdot \mathbf{r} \mathbf{a} \cdot \mathrm{d} \mathbf{r})=0 .\right]$$