Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 1 / 16B (Electromagnetism)

Charge conservation gives the continuity equation ${\partial }_t\rho +\nabla \cdot \mathbf{J}=0$. Since $\mathbf{J}$ is independent of time, so is ${\partial }_t\rho =-\nabla \cdot \mathbf{J}$. Integrating at each fixed position gives

Let $s=|\mathbf{r}-{\mathbf{r}}^{\prime }|$ and introduce the magnetic vector potential $\mathbf{A}(\mathbf{r})={\mu }_0(4\pi {)}^{-1}\int \mathbf{J}({\mathbf{r}}^{\prime })/sd{V}^{\prime }$. Differentiation is with respect to $\mathbf{r}$, so $\mathbf{J}({\mathbf{r}}^{\prime })$ is constant under it. HenceTaking the curl through the integral reproduces the Biot-Savart law, proving $\boxed{\mathbf{B}=\nabla \times \mathbf{A}}$ and the requested representation.

Use the identity $\nabla \times \mathbf{B}=\nabla (\nabla \cdot \mathbf{A})-{\nabla }^2\mathbf{A}$. The Laplacian identity for the Coulomb kernel gives, component by component,Also $\nabla (s^{-1})=-{\nabla }^{\prime }(s^{-1})$, so integration by parts yieldsHere the surface integral at infinity vanishes under the assumed localization of the current. As usual, this assumption means sufficient decay for these integrals and the boundary limit, rather than merely pointwise convergence to zero. Combining the two terms gives

For the charge-generated electric field, use the electric potentialThe vector potential is time independent, so its usual additional contribution $-{\partial }_t\mathbf{A}$ is zero. The continuity equation now givesThus the second term in part (iii) is exactly the displacement current contribution ${\mu }_0{\epsilon }_0{\partial }_t\mathbf{E}$, andThis is the Ampère-Maxwell equation. The constructed field also has zero curl, consistent with the time-independent magnetic field and Faraday's law. A static addition to the electric field does not change this conclusion; the integral construction selects the fields sourced by the given densities rather than an independently added radiation field.