Past exam of the physics 2 paper of the University of Oxford / 2022 / 3

Assume the electrical conductor is nonmagnetic, with magnetic permeability ${\mu }_0$, and neglect end effects. Rotational symmetry and translational symmetry make the magnetic flux density azimuthal and constant along a coaxial circle. The magnetostatic Ampère law therefore givesUniform current density is $j=I/(\pi a^2)$, so $I_{\mathrm{enc}}(r)=j\pi r^2=I{r}^2/a^2$. ThusThe direction follows the right-hand rule around the electric current. For a uniform linear magnetic material, replace ${\mu }_0$ by its magnetic permeability.

Write the axial current density as $j(r)=Kr$. Normalize it using the total electric current:The enclosed electric current at radius $r$ is $I_{\mathrm{enc}}(r)=2\pi K{r}^3/3=I(r/a{)}^3$. The magnetostatic Ampère law now yieldsBoth profiles join the exterior magnetic flux density ${\mu }_{0}I/(2\pi a)$ at the surface, but the second vanishes quadratically at the axis because less electric current is concentrated near it.