Suppose that the current density $\mathbf{J}(\mathbf{r})$ is constant in time but the charge density $\rho(\mathbf{r}, t)$ is not.

(i) Show that $\rho$ is a linear function of time:

$$\rho(\mathbf{r}, t)=\rho(\mathbf{r}, 0)+\dot{\rho}(\mathbf{r}, 0) t$$

where $\dot{\rho}(\mathbf{r}, 0)$ is the time derivative of $\rho$ at time $t=0$.

(ii) The magnetic induction due to a current density $\mathbf{J}(\mathbf{r})$ can be written as

$$\mathbf{B}(\mathbf{r})=\frac{\mu_{0}}{4 \pi} \int \frac{\mathbf{J}\left(\mathbf{r}^{\prime}\right) \times\left(\mathbf{r}-\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^{3}} d V^{\prime}$$

Show that this can also be written as

$$\mathbf{B}(\mathbf{r})=\frac{\mu_{0}}{4 \pi} \nabla \times \int \frac{\mathbf{J}\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|} d V^{\prime}$$

(iii) Assuming that $\mathbf{J}$ vanishes at infinity, show that the curl of the magnetic field in (1) can be written as

$$\nabla \times \mathbf{B}(\mathbf{r})=\mu_{0} \mathbf{J}(\mathbf{r})+\frac{\mu_{0}}{4 \pi} \nabla \int \frac{\nabla^{\prime} \cdot \mathbf{J}\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|} d V^{\prime}$$

[You may find useful the identities $\nabla \times(\nabla \times \mathbf{A})=\nabla(\nabla \cdot \mathbf{A})-\nabla^{2} \mathbf{A}$ and also $\left.\nabla^{2}\left(1 /\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\right)=-4 \pi \delta\left(\mathbf{r}-\mathbf{r}^{\prime}\right) .\right]$

(iv) Show that the second term on the right hand side of (2) can be expressed in terms of the time derivative of the electric field in such a way that $\mathbf{B}$ itself obeys Ampère's law with Maxwell's displacement current term, i.e. $\nabla \times \mathbf{B}=\mu_{0} \mathbf{J}+\mu_{0} \epsilon_{0} \partial \mathbf{E} / \partial t$.