 Solution

For a consistently oriented surface and boundary, Ampère's law isStokes theorem gives the left side as ${\int }_S(\nabla \times \mathbf{B})\cdot \mathbf{n}dS$, so the given curl equation proves the law.

On a circle of radius $r$ around the axis, the assumed azimuthal field is tangent and constant in magnitude. Thus the circulation is $2\pi r{B}_{\theta }(r)$. For $r\le a$, the enclosed electric current is $I{r}^2/a^2$; for $r\ge a$, it is $I$. ThereforeThe field is continuous at $r=a$ and vanishes regularly at the axis. Its direction follows the right-hand rule for positive electric current along ${\mathbf{e}}_z$. The radial component of the curl equation also gives ${\partial }_z{B}_{\theta }=0$, so the expression is valid at every height; no unmentioned axial dependence remains.