Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 2 / 6D / b / Solution

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.