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Current-carrying circular loop in a magnetic field

So, in answering this question the book resolves the components of the Force vector in this manner;

  1. $\mathrm dF_x= \mathrm dF \sin \theta$
  2. $\mathrm dF_y= \mathrm dF \cos \theta$.

And then, it integrates from $\theta_0$ to $2\pi-\theta_0$.

I don't understand why are the x and y components reversed and not depending on cos and sin respectively.
I also don't understand why are the limits of integration in this manner, since I thought this way the part of the circle that we are going to be leaving out is the right corner of the circle instead of the top.

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  • $\begingroup$ do you accept the law about voltage beeing the change of the flux, or is it before this where the question begins? $\endgroup$ – Ilja May 1 '16 at 10:07
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Both questions have the same answer: The coordinate system is chosen such that $\theta = 0$ points to the top.

So if $\theta_0 = 0$, then the circle is fully immersed in the magnetic field. The limits of integration will be just $0$ to $2\pi$.

In the other extreme case, where the circle barely touches the magnetic field, one has $\theta_0 = \pi - \epsilon$ where $\epsilon > 0$ is a small number. The limits of integration are then $(\pi-\epsilon, \pi + \epsilon)$. The only force that contributes will come from the portion of the $x$ axis. There, $\sin(\pi) = -1$ and $\cos(\pi) = 0$. So you see that only $\mathrm dF_x$ contributes here.

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