A simple example of the 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 r ^ {\displaystyle {\hat {\mathbf {r} }}} , so the Biot-Savart law simplifies from the definition of the cross product d B = − μ 0 4 π I d l s i n ( π / 2 ) r 2 k ^ {\displaystyle d{\mathbf {B} }=-{\frac {\mu _{0}}{4\pi }}{\frac {Idl\,sin(\pi /2)}{r^{2}}}{\hat {\mathbf {k} }}} Note that the direction is into the web browser, so we have a − k ^ {\displaystyle -{\hat {\mathbf {k} }}} . For the differential dl of a circular arc d l = r d θ {\displaystyle dl=rd\theta } plugging this in and setting up the integral gives B = − k ^ μ 0 I 4 π r ∫ 0 π 2 d θ {\displaystyle {\mathbf {B} }=-{\hat {\mathbf {k} }}{\frac {\mu _{0}I}{4\pi r}}\int _{0}^{\frac {\pi }{2}}\,d\theta } which simply yields B = − μ 0 I k ^ 8 r {\displaystyle {\mathbf {B} }=-{\frac {\mu _{0}I{\hat {\mathbf {k} }}}{8r}}}
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