PlanetPhysics/Loop Example of Biot Savart Law

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

$d{\mathbf {l} }\times {\hat {\mathbf {r} }}=dl\,sin(\pi /2)$ $dl=rd\theta$

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

${\mathbf {B} }={\hat {\mathbf {k} }}{\frac {\mu _{0}I}{4\pi r}}\int _{0}^{2\pi }\,d\theta$

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

${\mathbf {B} }={\frac {\mu _{0}I{\hat {\mathbf {k} }}}{2r}}$

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 $d{\mathbf {B} }$. The cross product of ${\mathbf {\hat {r}} }$ with $d{\mathbf {l} }$ leads to a vector perpendicular to both of them and as you go around the loop, $d{\mathbf {B} }$ will always be off the z axis by an angle $\phi$. This makes all the horizontal components of $d{\mathbf {B} }$ cancel leaving just the vertical so

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

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

${\mathbf {B} }={\hat {\mathbf {k} }}{\frac {\mu _{0}IR}{4\pi r^{2}}}\int _{0}^{2\pi }\,cos(\phi )\,d\theta$

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

${\mathbf {B} }={\hat {\mathbf {k} }}{\frac {\mu _{0}IR^{2}}{4\pi (R^{2}+z^{2})^{\frac {3}{2}}}}\int _{0}^{2\pi }\,d\theta$

Finally, taking the integral gives us the magnetic field

${\mathbf {B} }={\frac {\mu _{0}IR^{2}\,{\hat {\mathbf {k} }}}{2(R^{2}+z^{2})^{\frac {3}{2}}}}$