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 part of the coil. The cross-product can be written as \(d\vec{l}\times \vec{r} = |d\vec{l}|r\sin\theta\). Thus, 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 vectors of radius and length. 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 as follows:

\[  \vec{B} = \frac{\mu_0 I}{4 \pi r^2} \oint_C |d\vec{l}|  \]

This contour integral gives \(2\pi r\). Therefore,

\[  \vec{B} = \frac{\mu_0 I}{4 \pi r^2}2 \pi r = \frac{\mu_0 I}{2 r}  \]