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

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

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

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

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

$$ {\mathbf B} = \frac{\mu_0 I \hat}{2 r} $$

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 $$d{\mathbf B}$$. 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

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

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

$$ {\mathbf B} = \hat \frac{\mu_0 I R}{4 \pi r^2} \int_0^{2\pi} \, cos(\phi) \, d \theta $$

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

$$ {\mathbf B} = \hat \frac{\mu_0 I R^2}{4 \pi (R^2 + z^2)^{\frac{3}{2}}} \int_0^{2\pi} \, d \theta $$

Finally, taking the integral gives us the magnetic field

$$ {\mathbf B} = \frac{\mu_0 I R^2 \, \hat}{2 (R^2 + z^2)^{\frac{3}{2}}} $$