Talk:PlanetPhysics/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: loop example of Biot-Savart law %%% Primary Category Code: 41.20.Gz %%% Filename: LoopExampleOfBiotSavartLaw.tex %%% Version: 4 %%% 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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Here we will examine two examples of the \htmladdnormallink{Biot-Savart law}{http://planetphysics.us/encyclopedia/BiotSavartLaw.html}, one simple and the other more challenging. To begin we will find the \htmladdnormallink{magnetic field}{http://planetphysics.us/encyclopedia/NeutrinoRestMass.html} at the center of a current carrying loop as shown in figure 1

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

this setup is the same as the quater loop example of Biot-Savart law where we have

$$ d{\bf l} \times \hatTemplate:\bf r = dl \, sin(\pi/2) $$ $$dl = r d\theta$$

giving us a similar integral, except from 0 to 2 $\pi$ and the lack of a minus sign since the current is going in the opposite direction so the magnetic field will be out of the web browser

$$ {\bf B} = \hatTemplate:\bf k \frac{\mu_0 I}{4 \pi r} \int_0^{2\pi} \, d \theta $$

taking the integral gives the magnetic field at the center of the loop

$$ {\bf B} = \frac{\mu_0 I \hatTemplate:\bf k}{2 r} $$

The second more challenging example is the magnetic field at a point z above the loop as shown in figure 2

\begin{figure} \includegraphics[scale=.8]{AxisCurrentLoop.eps} \vspace{10 pt} \caption{Figure 2: Current Loop} \end{figure}

The not so obvious hint is the direction of $d{\bf B}$. The \htmladdnormallink{cross product}{http://planetphysics.us/encyclopedia/VectorProduct.html} of ${\bf \hat{r}}$ with $d{\bf l}$ leads to a \htmladdnormallink{vector}{http://planetphysics.us/encyclopedia/Vectors.html} perpendicular to both of them and as you go around the loop, $d{\bf B}$ will always be off the z axis by an angle $\phi$. This makes all the horizontal components of $d{\bf B}$ cancel leaving just the vertical so

$$ d{\bf l} \times {\bf \hat{r}} = dl \, cos(\phi) \hatTemplate:\bf k $$

once again the differential is given as $$dl = R d\theta$$, so the integral to get the magnetic field is

$$ {\bf B} = \hatTemplate:\bf k \frac{\mu_0 I R}{4 \pi r^2} \int_0^{2\pi} \, cos(\phi) \, d \theta $$

From the geometry of the problem we see that

$$ r^2 = R^2 + z^2 $$ $$ cos(\phi) = \frac{R}{r} $$

this leads to

$$ r = \sqrt{R^2 + z^2} $$ $$ cos(\phi) = \frac{R}{\sqrt{R^2 + z^2}} $$

substituting these \htmladdnormallink{relations}{http://planetphysics.us/encyclopedia/Bijective.html} into the integral

$$ {\bf B} = \hatTemplate:\bf k \frac{\mu_0 I R^2}{4 \pi (R^2 + z^2)^{\frac{3}{2}}} \int_0^{2\pi} \, d \theta $$

Finally, taking the integral gives us the magnetic field

$$ {\bf B} = \frac{\mu_0 I R^2 \, \hatTemplate:\bf k}{2 (R^2 + z^2)^{\frac{3}{2}}} $$

\end{document}