# Magnetic Field due to a Current Loop

Discussion in 'Physics' started by shespuzzling, Dec 22, 2009.

1. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
1
Hi,

I have a few questions regarding some examples in my Pysics textbook about finding the magnetic field at a point due to a current loop using the Biot-Savart Law. In the book, there are two examples. The first one shows the magnetic field at the origin of a current loop and the second calculates the B-field along the axis of a current loop. Diagrams showing the same problem can be found here: http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/magnetic/curloo.html#c2

When you first apply the Biot-Savart law to both cases, you end up with the following equation, where you then have to integrate.

dB=(μ/4∏)*(Idl/r^2)

I see that in the first case, r=R and that the magnetic field will be along the z-axis throughout the integration. Integrating the above equation gives you B=(μI/2R). The second case yields the same equation for dB, however, with the only difference being r. Why won't it work to integrate that equation with the value for r in the second case? Isn't that the whole point of integration? Why must you break it up into x and y components? Is it just a product of working in 3 dimensions?

Last edited: Dec 22, 2009
2. ### steveb Senior Member

Jul 3, 2008
2,431
469
No, it isn't the same equation. Note that there is a cross product involved, and this brings in the $\sin \theta$ dependence. The angle depends on the distance along the z-axis. At the center of the loop (z=0), the sine is 1, but at any other point along the axis, it is not 1. You have to work out the geometry exactly.

That reference skips a few steps and makes it hard to see. If you still don't see it, I recommed that you look at the classic book "Electromagnetics" by John D. Kraus.

3. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
1
Hmm, just to make sure we're on the same page, in the diagram on the attachment called "Field on Axis of Current Loop" I thought that the current vector Idl would always be perpendicular to the vector r all the way around the loop, by virtue of the field point being on the axis. If that is the case, then Idl x r(unit vector) would equal Idl(1)=Idl.

I was thinking about it some more...does it maybe have to do with the fact that the vector dB happens to have 2 components, while the vector dB in the first case is only along the positive z axis?

4. ### steveb Senior Member

Jul 3, 2008
2,431
469
The vector dB does have two components, but once integrated the vector B always points in the z-direction. Symmetry allows you to know that the other component is zero, and only the z-direction needs to be considered.