Verify Ampere's law for the magnetic field of a point dipole with dipole moment $\vec{M} = M\hat{k}$. Take $C$ as the closed curve running clockwise along the $x$-axis from $x = R$ to $x = a$,then along a circular arc of radius $r$ in the $xz$-plane from $\theta = 0$ to $\theta = \pi$,and finally back to the starting point.

(N/A) The magnetic field of a point dipole $\vec{M} = M\hat{k}$ at a position $\vec{r}$ is given by $\vec{B}(\vec{r}) = \frac{\mu_0}{4\pi} \left[ \frac{3(\vec{M} \cdot \hat{r})\hat{r} - \vec{M}}{r^3} \right]$.
Ampere's law states that $\oint_C \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$.
For a point dipole,there is no physical current enclosed by the loop $C$,so $I_{enclosed} = 0$.
Thus,we must verify that $\oint_C \vec{B} \cdot d\vec{l} = 0$.
Along the $x$-axis,$\vec{r} = x\hat{i}$,$\hat{r} = \hat{i}$,and $\vec{M} \cdot \hat{r} = 0$. So $\vec{B} = \frac{\mu_0}{4\pi} \frac{-\vec{M}}{x^3} = -\frac{\mu_0 M}{4\pi x^3} \hat{k}$.
Since $d\vec{l}$ is along the $x$-axis $(d\vec{l} = dx \hat{i})$,$\vec{B} \cdot d\vec{l} = 0$.
Along the circular arc in the $xz$-plane,$\vec{B} \cdot d\vec{l}$ can be calculated using the magnetic vector potential $\vec{A} = \frac{\mu_0}{4\pi} \frac{\vec{M} \times \vec{r}}{r^3}$.
By Stokes' theorem,$\oint_C \vec{B} \cdot d\vec{l} = \oint_C (\nabla \times \vec{A}) \cdot d\vec{l} = \oint_C \vec{A} \cdot d\vec{l}$.
Since the dipole field is conservative everywhere except at the origin,the line integral over any closed loop not enclosing the origin is zero. Thus,Ampere's law is verified.