Consider a frame $S^{\prime}$ moving with velocity v relative to the laboratory frame $S$ where $|\mathbf{v}|^{2} \ll c^{2}$. The electric and magnetic fields in $S$ are $\mathbf{E}$ and $\mathbf{B}$, while those measured in $S^{\prime}$ are $\mathbf{E}^{\prime}$ and $\mathbf{B}^{\prime}$. Given that $\mathbf{B}^{\prime}=\mathbf{B}$, show that

$$\oint_{\Gamma} \mathbf{E}^{\prime} \cdot d \mathbf{l}=\oint_{\Gamma}(\mathbf{E}+\mathbf{v} \wedge \mathbf{B}) \cdot d \mathbf{l},$$

for any closed circuit $\Gamma$ and hence that $\mathbf{E}^{\prime}=\mathbf{E}+\mathbf{v} \wedge \mathbf{B}$.

Now consider a fluid with electrical conductivity $\sigma$ and moving with velocity $\mathbf{v}(\mathbf{r})$. Use Ohm's law in the moving frame to relate the current density $\mathbf{j}$ to the electric field $\mathbf{E}$ in the laboratory frame, and show that if $\mathbf{j}$ remains finite in the limit $\sigma \rightarrow \infty$ then

$$\frac{\partial \mathbf{B}}{\partial t}=\nabla \wedge(\mathbf{v} \wedge \mathbf{B})$$

The magnetic helicity $H$ in a volume $V$ is given by $\int_{V} \mathbf{A} \cdot \mathbf{B} d \tau$ where $\mathbf{A}$ is the vector potential. Show that if the normal components of $\mathbf{v}$ and $\mathbf{B}$ both vanish on the surface bounding $V$ then $d H / d t=0$.