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I have the equation for flux containing mod operation of the angle as an independent variable.

$$ \phi\left(\theta_{R}\right)=-\ell_{1} \ell_{2} B\left(\theta_{R} \bmod \pi-\frac{\pi}{2}\right) $$

The textbook then directly derives the derivative of flux for the calculation of emf as follows:

$$ \xi=-\frac{d \phi}{d t}=\left(\ell_{1} \ell_{2} B\right) \frac{d \theta_{R}}{d t} $$

My question is, where did the mod go after differentiation?

Wolgwang
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TeilZeitGott
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1 Answers1

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Since the derivative is a local operator (you only care about doing small changes), you can just imagine taking $\theta_R$ to be in a small interval around the point you're differentiating, smaller than the $\pi$ interval, so you can effectively ignore the mod $\pi$.

Keshav
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  • so nothing happens when \theta_R = \pi ? like discontinuities. – TeilZeitGott Oct 19 '20 at 03:59
  • No there's nothing special about $\pi$, it's just the same as doing it at $0$. It seems like $B$ is only defined on an interval of size $\pi$ somewhere? You can extend $B$ to be periodic on the whole line, and then take the derivative with no problem. – Keshav Oct 19 '20 at 04:07
  • yes, B is defined in 0 to pi and pi to 2pi separately. – TeilZeitGott Oct 19 '20 at 04:12