The vector potential $A^{\mu}$ is determined by a current density distribution $j^{\mu}$ in the gauge $\partial_{\mu} A^{\mu}=0$ by

$$\square A^{\mu}=-\mu_{0} j^{\mu}, \quad \square=-\frac{\partial^{2}}{\partial t^{2}}+\nabla^{2},$$

in units where $c=1$.

Describe how to justify the result

$$A^{\mu}(\mathbf{x}, t)=\frac{\mu_{0}}{4 \pi} \int d^{3} x^{\prime} \frac{j^{\mu}\left(\mathbf{x}^{\prime}, t^{\prime}\right)}{\left|\mathbf{x}-\mathbf{x}^{\prime}\right|}, \quad t^{\prime}=t-\left|\mathbf{x}-\mathbf{x}^{\prime}\right|$$

A plane square loop of thin wire, edge lengths $l$, has its centre at the origin and lies in the $(x, y)$ plane. For $t<0$, no current is flowing in the loop, but at $t=0$ a constant current $I$ is turned on.

Find the vector potential at the point $(0,0, z)$ as a function of time due to a single edge of the loop.

What is the electric field due to the entire loop at $(0,0, z)$ as a function of time? Give a careful justification of your answer.