Past exam of the physics 2 paper of the University of Oxford / 2015 resit / 4 / Solution

The magnetostatic Ampère's law is $\oint \mathbf{B}\cdot d\mathbf{\ell }={\mu }_0{I}_{\mathrm{enc}}$, equivalently $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}$ in vacuum. Let the wire radius be $a$ and take positive electric current along its axis. Cylindrical symmetry makes $\mathbf{B}$ azimuthal and constant in magnitude on a radius-$r$ circular path, so $2\pi rB={\mu }_0{I}_{\mathrm{enc}}$.

Uniform electric current density gives $J=I/(\pi a^2)$. Inside the wire, $I_{\mathrm{enc}}=J\pi r^2=I{r}^2/a^2$; outside, the full electric current is enclosed. ThereforeThe direction is set by the right-hand rule around the electric current. The magnitude starts at zero on the axis, rises linearly to ${\mu }_{0}I/(2\pi a)$ at the surface and then falls as $1/r$. The two expressions are continuous at $r=a$. The long-wire approximation neglects end effects and any high-frequency redistribution of electric current.