Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 2 / 6D (Electromagnetism)

In SI units the Ampère-Maxwell equation is $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}+{\mu }_0{\epsilon }_0{\partial }_t\mathbf{E}$. Since $\mathbf{E}=0$, the displacement-current term vanishes and $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}$. Apply Stokes theorem to an oriented surface $S$ with positively oriented boundary $C$:Hence the integral form of Ampère's law isHere ${\mu }_0$ is vacuum magnetic permeability, $\mathbf{J}$ is current density, and positive enclosed electric current is defined by the orientation of $S$.

For the field generated by the infinitely long, axially symmetric shell currents, axisymmetry and the right-hand rule give $\mathbf{B}=B_{\varphi }(r){\mathbf{e}}_{\varphi }$. A circular loop has $\oint \mathbf{B}\cdot d\mathbf{\ell }=2\pi r{B}_{\varphi }(r)$. Its enclosed electric current is zero inside the inner shell, $I$ between the shells and $2I$ outside both. Ampère's law therefore givesThe positive ${\mathbf{e}}_{\varphi }$ direction is counterclockwise viewed from positive $z$. The ideal current sheets make the tangential magnetic field discontinuous at each shell; these formulas specify the three open regions, with no additional imposed background field.

Only the signed enclosed electric current changes. Between the shells it remains $I$, but outside both it is $I-I=0$. The same circular-loop application of Ampère's law givesOpposite currents confine the generated magnetic field to the annulus, as in an ideal infinitely long coaxial cable. The two shell current sheets account for the field jumps.