PlanetPhysics/Loop Example of Biot Savart Law

Here we will examine two examples of the [[../BiotSavartLaw/|Biot-Savart law]], one simple and the other more challenging. To begin we will find the [[../NeutrinoRestMass/|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

$d𝐥\times \widehat{𝐫}=dlsin(\pi /2)$ $dl=rd\theta$

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

$𝐁=\widehat{𝐤}\frac{{\mu }_{0}I}{4\pi r}\underset{0}{\overset{2\pi }{\int }}d\theta$

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

$𝐁=\frac{{\mu }_{0}I\widehat{𝐤}}{2r}$

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 $d𝐁$. The [[../VectorProduct/|cross product]] of $\widehat{𝐫}$ with $d𝐥$ leads to a [[../Vectors/|vector]] perpendicular to both of them and as you go around the loop, $d𝐁$ will always be off the z axis by an angle $\varphi$. This makes all the horizontal components of $d𝐁$ cancel leaving just the vertical so

$d𝐥\times \widehat{𝐫}=dlcos(\varphi )\widehat{𝐤}$

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

$𝐁=\widehat{𝐤}\frac{{\mu }_{0}IR}{4\pi r^2}\underset{0}{\overset{2\pi }{\int }}cos(\varphi )d\theta$

From the geometry of the problem we see that

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

this leads to

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

substituting these [[../Bijective/|relations]] into the integral

$𝐁=\widehat{𝐤}\frac{{\mu }_{0}I{R}^2}{4\pi (R^2+z^2{)}^{\frac{3}{2}}}\underset{0}{\overset{2\pi }{\int }}d\theta$

Finally, taking the integral gives us the magnetic field

$𝐁=\frac{{\mu }_{0}I{R}^2\widehat{𝐤}}{2(R^2+z^2{)}^{\frac{3}{2}}}$