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.

Figure: Figure 1: Quarter Current Loop
\includegraphics[scale=.8]{QuarterCurrentLoop.eps}

The differential line element dl is perpendicular to $ \hat{{\bf r}}$, so the Biot-Savart law simplifies from the definition of the cross product

$\displaystyle d{\bf B} = -\frac{\mu_0}{4 \pi}\frac{I dl   sin (\pi/2)}{r^2} \hat{{\bf k}} $

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

$\displaystyle dl = r d\theta$

plugging this in and setting up the integral gives

$\displaystyle {\bf B} = - \hat{{\bf k}} \frac{\mu_0 I}{4 \pi r} \int_0^{\frac{\pi}{2}}   d \theta $

which simply yields

$\displaystyle {\bf B} = -\frac{\mu_0 I \hat{{\bf k}}}{8 r} $



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