= Solution
In SI units the <Ampère-Maxwell equation> is $\nabla\times\mathbf B=\mu_0\mathbf J+\mu_0\varepsilon_0\partial_t\mathbf E$. Since $\mathbf E=0$, the displacement-current term vanishes and $\nabla\times\mathbf B=\mu_0\mathbf J$. Apply <Stokes theorem> to an oriented surface $S$ with positively oriented boundary $C$:
$$
\oint_C\mathbf B\cdot d\boldsymbol\ell=\int_S(\nabla\times\mathbf B)\cdot\mathbf n\,dS=\mu_0\int_S\mathbf J\cdot\mathbf n\,dS.
$$
Hence the integral form of <Ampère's law> is
$$
\boxed{\oint_C\mathbf B\cdot d\boldsymbol\ell=\mu_0 I_{\rm enclosed}.}
$$
Here $\mu_0$ is vacuum <magnetic permeability>, $\mathbf J$ is <current density>, and positive enclosed <electric current> is defined by the orientation of $S$.
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