Talk:PlanetPhysics/Quarter Loop Example of Biot Savart Law

Original TeX Content from PlanetPhysics Archive
%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: quarter loop example of Biot-Savart law %%% Primary Category Code: 41.20.Gz %%% Filename: QuarterLoopExampleOfBiotSavartLaw.tex %%% Version: 3 %%% Owner: bloftin %%% Author(s): bloftin %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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\begin{document}

A simple example of the \htmladdnormallink{Biot-Savart law}{http://planetphysics.us/encyclopedia/BiotSavartLaw.html} 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{\bf l} is perpendicular to $\hat$, so the Biot-Savart law simplifies from the definition of the \htmladdnormallink{cross product}{http://planetphysics.us/encyclopedia/VectorProduct.html} $$ d{\bf B} = -\frac{\mu_0}{4 \pi}\frac{I dl \, sin (\pi/2)}{r^2} \hat $$

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

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

which simply yields

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

\end{document}