(a) Show that the magnetic flux passing through a simple, closed curve $C$ can be written as

$$\Phi=\oint_{C} \mathbf{A} \cdot \mathbf{d} \mathbf{x},$$

where $\mathbf{A}$ is the magnetic vector potential. Explain why this integral is independent of the choice of gauge.

(b) Show that the magnetic vector potential due to a static electric current density $\mathbf{J}$, in the Coulomb gauge, satisfies Poisson's equation

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

Hence obtain an expression for the magnetic vector potential due to a static, thin wire, in the form of a simple, closed curve $C$, that carries an electric current $I$. [You may assume that the electric current density of the wire can be written as

$$\mathbf{J}(\mathbf{x})=I \int_{C} \delta^{(3)}\left(\mathbf{x}-\mathbf{x}^{\prime}\right) \mathbf{d} \mathbf{x}^{\prime}$$

where $\delta^{(3)}$ is the three-dimensional Dirac delta function.]

(c) Consider two thin wires, in the form of simple, closed curves $C_{1}$ and $C_{2}$, that carry electric currents $I_{1}$ and $I_{2}$, respectively. Let $\Phi_{i j}$ (where $i, j \in\{1,2\}$ ) be the magnetic flux passing through the curve $C_{i}$ due to the current $I_{j}$ flowing around $C_{j}$. The inductances are defined by $L_{i j}=\Phi_{i j} / I_{j}$. By combining the results of parts (a) and (b), or otherwise, derive Neumann's formula for the mutual inductance,

$$L_{12}=L_{21}=\frac{\mu_{0}}{4 \pi} \oint_{C_{1}} \oint_{C_{2}} \frac{\mathbf{d} \mathbf{x}_{1} \cdot \mathbf{d} \mathbf{x}_{2}}{\left|\mathbf{x}_{1}-\mathbf{x}_{2}\right|} .$$

Suppose that $C_{1}$ is a circular loop of radius $a$, centred at $(0,0,0)$ and lying in the plane $z=0$, and that $C_{2}$ is a different circular loop of radius $b$, centred at $(0,0, c)$ and lying in the plane $z=c$. Show that the mutual inductance of the two loops is

$$\frac{\mu_{0}}{4} \sqrt{a^{2}+b^{2}+c^{2}} f(q)$$

where

$$q=\frac{2 a b}{a^{2}+b^{2}+c^{2}}$$

and the function $f(q)$ is defined, for $0<q<1$, by the integral

$$f(q)=\int_{0}^{2 \pi} \frac{q \cos \theta d \theta}{\sqrt{1-q \cos \theta}}$$