Ampere's law: A toroid carrying current

Question:
A toroid has $N$ turns both sides carrying equal current $I$. The inner and outer radii are $a$ and $b$, respectively. A distance $r$ is supposed to be exact the middle between $a$ and $b$. Find the magnetic field at $r$.

Answer:
From Ampere's law, if inside the contour has a current $I$, we have the relationship:
\[
\oint \vec{B}\cdot dl = \mu_0 I
\]
The radius of the contour is $r$ which encloses the current $N\times I$. Therefore,
\[
B(2\pi r) = \mu_0 NI
\]
Recall that $r=\frac{a+b}{2}$; thus, the magnetic field is
\[
B = \frac{\mu_0 NI}{\pi(a+b)}
\]

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