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Magnetic field at the center of circular current carrying loop is given by

$$ B=\frac{\mu I}{2 R} $$

Where $\mu$ is the permeability of free space and $R$ is the radius of loop.

In a question by calculation I got a field at the center of the regular $n$-sided polygon as

$$ B=\frac{\mu I n \sin \frac \pi n} {2 \pi R} $$

Now the question also asks about what when $n$ goes to infinity the polygon is a circle.

With some assumptions

  1. $n$ goes to infinity, hence $\theta$ goes to zero
  2. hence $\sin \theta=\theta$

then we get the magnetic field same as a circle.

My question is when we derive $B$ directly assuming a circle we get an answer. when we derive $B$ of polygon and assume $n=\infty$ and do approximation, why do we get the same correct answer since there is an approximation involved.

the $\theta$ for a 4 sided polygon is shown enter image description here

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Instead of seeing your magnetic field by assuming $\sin\theta = \theta$ for small angles, you can rewrite your magnetic field as \begin{equation} B=\frac{\mu In\sin{\frac{\pi}{n}}}{2 \pi R} = \frac{\mu I}{2R} \frac{n\sin{\frac{\pi}{n}}}{\pi}=\frac{\mu I}{2R} \frac{\sin{\frac{\pi}{n}}}{\frac{\pi}{n}} \end{equation} In the limit $n \rightarrow \infty$ or $\frac{\pi}{n} = \theta \rightarrow 0$, $\lim_{\theta \rightarrow 0}$ $\frac{\sin{\theta}}{\theta}=1$ and hence $B=\frac{\mu I}{2R}$.

This approximation of taking $\theta \rightarrow 0$ as $n \rightarrow \infty$ is a valid one and gives an exact answer.

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  • $\begingroup$ Ok, but isn't $\sin\theta \rightarrow \theta$ exactly the same as $\frac{\sin\theta}{\theta}\rightarrow1$? $\endgroup$ – garyp Dec 30 '18 at 18:09
  • $\begingroup$ Yes this makes sense now. $\endgroup$ – shinigami Dec 30 '18 at 18:14
  • $\begingroup$ $\sin \theta \rightarrow \theta$ as $\theta \rightarrow 0$ comes from Taylor's approximation whereas $\frac{\sin \theta}{\theta} \rightarrow 1$ is from limiting geometric argument. I think both are equivalent. However, I guess the OP thought of the limiting case as an approximation rather than an exact solution, and I just wanted to point out another way of writing the limit, which I think is more apparent. $\endgroup$ – Alpha7200 Dec 30 '18 at 18:20
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polygons

The source of your confusion is that $n=4$ is not very far along the path to your limit $n\to\infty$. For a twenty-sided polygon (shown) the difference between $\theta$ and $\sin\theta$ starts in the third significant figure.

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