LoopExampleOfBiotSavartLaw 2006-01-14 21:45:06 2006-01-15 00:09:38 Example Biot-Savart law removed pi from the last eq. for the magnetic field at the center of the loop % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) \usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Here we will examine two examples of the Biot-Savart law, one simple and the other more challenging. To begin we will find the magnetic field 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 $$ d{\bf l} \times \hat{{\bf r}} = dl \, sin(\pi/2) $$ $$dl = r d\theta$$ 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 $$ {\bf B} = \hat{{\bf k}} \frac{\mu_0 I}{4 \pi r} \int_0^{2\pi} \, d \theta $$ taking the integral gives the magnetic field at the center of the loop $$ {\bf B} = \frac{\mu_0 I \hat{{\bf k}}}{2 \pi r} $$ 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 cross product of ${\bf \hat{r}}$ with $d{\bf l}$ leads to a vector 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 $$ d{\bf l} \times {\bf \hat{r}} = dl \, cos(\phi) \hat{{\bf k}} $$ once again the differential is given as $$dl = R d\theta$$, so the integral to get the magnetic field is $$ {\bf B} = \hat{{\bf k}} \frac{\mu_0 I R}{4 \pi r^2} \int_0^{2\pi} \, cos(\phi) \, d \theta $$ 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 relations into the integral $$ {\bf B} = \hat{{\bf k}} \frac{\mu_0 I R^2}{4 \pi (R^2 + z^2)^{\frac{3}{2}}} \int_0^{2\pi} \, d \theta $$ Finally, taking the integral gives us the magnetic field $$ {\bf B} = \frac{\mu_0 I R^2 \, \hat{{\bf k}}}{2 (R^2 + z^2)^{\frac{3}{2}}} $$