(i) Given the electric field (in cartesian components)

$$\mathbf{E}(\mathbf{r}, t)=\left(0, x / t^{2}, 0\right)$$

use the Maxwell equation

$$\nabla \times \mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}$$

to find $\mathbf{B}$ subject to the boundary condition that $|\mathbf{B}| \rightarrow 0$ as $t \rightarrow \infty$.

Let $S$ be the planar rectangular surface in the $x y$-plane with corners at

$$(0,0,0), \quad(L, 0,0), \quad(L, a, 0), \quad(0, a, 0)$$

where $a$ is a constant and $L=L(t)$ is some function of time. The magnetic flux through $S$ is given by the surface integral

$$\Phi=\int_{S} \mathbf{B} \cdot d \mathbf{S}$$

Compute $\Phi$ as a function of $t$.

Let $\mathcal{C}$ be the closed rectangular curve that bounds the surface $S$, taken anticlockwise in the $x y$-plane, and let $\mathbf{v}$ be its velocity (which depends, in this case, on the segment of $\mathcal{C}$ being considered). Compute the line integral

$$\oint_{\mathcal{C}}(\mathbf{E}+\mathbf{v} \times \mathbf{B}) \cdot d \mathbf{r}$$

Hence verify that

$$\oint_{\mathcal{C}}(\mathbf{E}+\mathbf{v} \times \mathbf{B}) \cdot d \mathbf{r}=-\frac{d \Phi}{d t}$$

(ii) A surface $S$ is bounded by a time-dependent closed curve $\mathcal{C}(t)$ such that in time $\delta t$ it sweeps out a volume $\delta V$. By considering the volume integral

$$\int_{\delta V} \nabla \cdot \mathbf{B} d \tau$$

and using the divergence theorem, show that the Maxwell equation $\nabla \cdot \mathbf{B}=0$ implies that

$$\frac{d \Phi}{d t}=\int_{S} \frac{\partial \mathbf{B}}{\partial t} \cdot d \mathbf{S}-\oint_{\mathcal{C}}(\mathbf{v} \times \mathbf{B}) \cdot d \mathbf{r}$$

where $\Phi$ is the magnetic flux through $S$ as given in Part (i). Hence show, using (1) and Stokes' theorem, that (2) is a consequence of Maxwell's equations.