PlanetPhysics/Quarter Loop Example of Biot Savart Law

A simple example of the [[../BiotSavartLaw/|Biot-Savart law]] is given with a current carrying loop for a quater of a circle as shown in Figure 1.

\begin{figure} \includegraphics[scale=.8]{QuarterCurrentLoop.eps} \vspace{10 pt} \caption{Figure 1: Quarter Current Loop} \end{figure}

The differential line element d{\mathbf l} is perpendicular to $\widehat{𝐫}$, so the Biot-Savart law simplifies from the definition of the [[../VectorProduct/|cross product]] $d𝐁=-\frac{{\mu }_0}{4\pi }\frac{Idlsin(\pi /2)}{r^2}\widehat{𝐤}$

Note that the direction is into the web browser, so we have a $-\widehat{𝐤}$. For the differential dl of a circular arc

$dl=rd\theta$

plugging this in and setting up the integral gives

$𝐁=-\widehat{𝐤}\frac{{\mu }_{0}I}{4\pi r}\underset{0}{\overset{\frac{\pi }{2}}{\int }}d\theta$

which simply yields

$𝐁=-\frac{{\mu }_{0}I\widehat{𝐤}}{8r}$