= Solution
{c}
For a consistently oriented surface and boundary, <Ampère's law> is
$$
\oint_\Gamma\mathbf B\cdot d\boldsymbol\ell
=\mu_0 I_{\rm enc}
=\mu_0\int_S\mathbf J\cdot\mathbf n\,dS.
$$
<Stokes 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 $2\pi rB_\theta(r)$. For $r\leq a$, the enclosed <electric current> is $I r^2/a^2$; for $r\geq a$, it is $I$. Therefore
$$
\boxed{\mathbf B(r)=
\begin{cases}
\dfrac{\mu_0 I r}{2\pi a^2}\mathbf e_\theta,&0\leq r\leq a,\\[4pt]
\dfrac{\mu_0 I}{2\pi r}\mathbf e_\theta,&r\geq a.
\end{cases}}
$$
The 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 $\partial_zB_\theta=0$, so the expression is valid at every height; no unmentioned axial dependence remains.
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