# How does $\sin\theta=\theta$ give a right answer even when it is an approximation?

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

Instead of seeing your magnetic field by assuming $$\sin\theta = \theta$$ for small angles, you can rewrite your magnetic field as $$$$B=\frac{\mu In\sin{\frac{\pi}{n}}}{2 \pi R} = \frac{\mu I}{2R} \frac{n\sin{\frac{\pi}{n}}}{\pi}=\frac{\mu I}{2R} \frac{\sin{\frac{\pi}{n}}}{\frac{\pi}{n}}$$$$ In the limit $$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$$ is a valid one and gives an exact answer.
• Ok, but isn't $\sin\theta \rightarrow \theta$ exactly the same as $\frac{\sin\theta}{\theta}\rightarrow1$? – garyp Dec 30 '18 at 18:09
• $\sin \theta \rightarrow \theta$ as $\theta \rightarrow 0$ comes from Taylor's approximation whereas $\frac{\sin \theta}{\theta} \rightarrow 1$ is from limiting geometric argument. I think both are equivalent. However, I guess the OP thought of the limiting case as an approximation rather than an exact solution, and I just wanted to point out another way of writing the limit, which I think is more apparent. – Alpha7200 Dec 30 '18 at 18:20
The source of your confusion is that $$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.