Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 314 / 1 / b

Put $q(x)=4\pi \alpha (x)/c$. Because every component depends only on $x$, the solenoidal constraint gives $B_x^{\prime }=0$, while the force-free equation givesFor a general nonzero, nonconstant $q$, $B_x=0$. Eliminating $B_y=-B_z^{\prime }/q$ gives the closed equationor equivalently $(B_z^{\prime }/q{)}^{\prime }+q{B}_z=0$. Since $\nabla \alpha ={\alpha }^{\prime }(x){\mathbf{e}}_x$ and $B_x=0$, the constraint $\mathbf{B}\cdot \nabla \alpha =0$ is automatically satisfied.