PlanetPhysics/Quarter Loop Example of Biot Savart Law

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 $$\hat$$, so the Biot-Savart law simplifies from the definition of the cross product $$ d{\mathbf B} = -\frac{\mu_0}{4 \pi}\frac{I dl \, sin (\pi/2)}{r^2} \hat $$

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

$$dl = r d\theta$$

plugging this in and setting up the integral gives

$$ {\mathbf B} = - \hat \frac{\mu_0 I}{4 \pi r} \int_0^{\frac{\pi}{2}} \, d \theta $$

which simply yields

$$ {\mathbf B} = -\frac{\mu_0 I \hat}{8 r} $$