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?

Thanks for your help.

 

... The second case yields the same equation for dB, however, with the only difference being r. ...

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.

 

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?

 

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?

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.