Magnetic field by a circular ring with current flow

Question:
As shown in the figure, there is a circular coil with current flow, $I$. Knowing that the radius is $r$, find the magnetic field at the center of the coil.

Answer:
Use the Biot-Savart Law.
\[
d\vec{B} = \frac{\mu_0 I}{4 \pi}\frac{d\vec{l}\times \vec{r}}{r^3}
\]
where $d\vec{l}$ is the infinitesimal length of the coil. The cross-product can be written as $d\vec{l}\times \vec{r} = |d\vec{l}|r\sin\theta$ The equation becomes
\[
d\vec{B} = \frac{\mu_0 I}{4 \pi}\frac{|d\vec{l}|\sin\theta}{r^2}
\]
$\theta$ is the angle between the radius and length vectors. They are always perpendicular, so $\theta = \frac{\pi}{2}$. Thus, $\sin\frac{\pi}{2}=1$. Now integrate it with the circular contour.
\[
\vec{B} = \oint_C \frac{\mu_0 I}{4 \pi}\frac{|d\vec{l}|}{r^2}
\]
All the constants can be placed outside the integral.
\[
\vec{B} = \frac{\mu_0 I}{4 \pi r^2} \oint_C |d\vec{l}|
\]
The contour integral gives $2\pi r$. Therefore,
\[
\vec{B} = \frac{\mu_0 I}{4 \pi r^2}2 \pi r = \frac{\mu_0 I}{2 r}
\]

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