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[parent] loop example of Biot-Savart law (Example)

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

Figure: Figure 1: Current Loop
\includegraphics[scale=.8]{CurrentLoop.eps}

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

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

$\displaystyle dl = r d\theta$

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

$\displaystyle {\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

$\displaystyle {\bf B} = \frac{\mu_0 I \hat{{\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

Figure: Figure 2: Current Loop
\includegraphics[scale=.8]{AxisCurrentLoop.eps}

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

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

once again the differential is given as

$\displaystyle dl = R d\theta$
, so the integral to get the magnetic field is

$\displaystyle {\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

$\displaystyle r^2 = R^2 + z^2 $

$\displaystyle cos(\phi) = \frac{R}{r} $

this leads to

$\displaystyle r = \sqrt{R^2 + z^2} $

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

substituting these relations into the integral

$\displaystyle {\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

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



"loop example of Biot-Savart law" is owned by bloftin.

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Cross-references: relations, vector, cross product, magnetic field, Biot-Savart law
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This is version 4 of loop example of Biot-Savart law, born on 2006-01-14, modified 2011-01-18.
Object id is 105, canonical name is LoopExampleOfBiotSavartLaw.
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Physics Classification: 41.20.Gz (Magnetostatics; magnetic shielding, magnetic induction, boundary-value problems)
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Discussion
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Question about B-field at center of loop by evanm3 on 2011-01-17 17:15:44
I have a quick question about the answer posted for the magnetic field at the center of a current loop. Everything makes sense up to the final answer, where I believe the factor of pi on the denominator should vanish. When you do the integral of dtheta, from 0 -> 2*pi, you get 2*pi, and thus the two pis cancel out giving (mu-naught*I / 2*r). Please let me know if I am thinking about this properly. Thanks in advance.
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