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I'm trying to understand the solution to the magnetic field of a circular loop along its central axis. The solution I'm looking at is page 9 of this document: http://www.physics.umd.edu/courses/Phys270/Jenkins/LectureChapter33pt2.PDF

What I'm confused about is when they break the dB vector into two components, a y component and an x component. I agree that this is possible given that the current vector ds is entirely along the z axis. The problem is that as we sum around the loop, ds is not always along the z axis, so that I don't believe that dB will only have a y and x component, but will also have a z component so that one couldn't say that $dB_x$ = $dB\frac{R}{r}$ as they do in the solution. Am I wrong that dB is not always breakable into an x and y component? If I am correct, then why does the solution still work?

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You are right that those notes could have been written better.

The division of $d\textbf{B}$ into $y$ and $x$ components is only valid for the $d\textbf{s}$ as shown at the top of the loop. At other locations, there will be a non-zero $z$ component to $d\textbf{B}$.

However, when the integration is done, there will be no net $z$ component, just like there will be no net $y$ component: by symmetry, both those integrals will be zero. The only non-zero component is in the $x$ direction.

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  • $\begingroup$ The problem is that you can't say that the angle from the x axis to dB + the angle from dB to the y axis is 90 degrees and use the geometry of the big right triangle do compute the x component if there is a z component, so everything falls apart. $\endgroup$ – user7348 Jul 17 '15 at 22:01
  • $\begingroup$ Just to see if we are on the same page, would you agree that it works for the $ds$ shown where the $z$ component of $dB$ is zero? $\endgroup$ – John1024 Jul 17 '15 at 22:07
  • $\begingroup$ Yes I do agree! All I am saying is that for any ds where dB has 3 components, you can't use that simple trig to argue that the x-component is dB$\frac{R}{r}$. But, it has to be true if we are to sum over the circle. I'd like to see a picture showing how it still works when there are 3 components. $\endgroup$ – user7348 Jul 17 '15 at 23:56
  • $\begingroup$ Very good. The next step is to look at the symmetry. Pick any angle. The y and z components will be different for that angle. By symmetry, however, the x-component will be the same regardless of angle. $\endgroup$ – John1024 Jul 18 '15 at 0:46
  • $\begingroup$ Alternatively, pick any angle. Then, create custom $y$ and $z$ axes aligned with that angle. Call them, if you like, $y'$ and $z'$. Compute the $x$-component of $B$ for these new axes using the same method as in the notes. The result will be that the x-component does not depend on angle. $\endgroup$ – John1024 Jul 18 '15 at 0:48

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