Question

A current $I$ flows along a thin wire shaped as a regular polygon with $n$ sides which can be inscribed into a circle of radius $R$. Analyse the obtained expression at $n\to \infty$.

• A. for $n\to \infty ,B=\frac{{\mu }_{0}I}{R}$
• B. for $n\to \infty ,B=\frac{2{\mu }_{0}I}{R}$
• C. for $n\to \infty ,B=\frac{3{\mu }_{0}I}{2R}$
• D. for $n\to \infty ,B=\frac{{\mu }_{0}I}{2R}$

Answer 1

The correct option is D for $n\to \infty ,B=\frac{{\mu }_{0}I}{2R}$

For polygon we can write

$P=\left(2R\right)\sin \left(\frac{\alpha }{2}\right)$

$P=\left(2R\right)\sin \left(\frac{2\pi }{2n}\right)$

$P=\left(2R\right)\sin \left(\frac{\pi }{n}\right)$

$P\left(n\right)=n\left(2R\right)\sin \left(\frac{\pi }{n}\right)$

$\frac{P\left(n\right)}{n}=2R\sin \left(\frac{\pi }{n}\right)$

Let, $a=\frac{P\left(n\right)}{n}=2R\sin \left(\frac{\pi }{n}\right)$

$b=R\cos \left(\frac{\pi }{n}\right)$

Therefore,

$B_1=\frac{{\mu }_{0}I}{4\pi b}(\cos {\varphi }_1+\cos {\varphi }_2)$

$B_1=\frac{{\mu }_{0}I}{4\pi b}\left[\frac{a}{\sqrt{b^2+\frac{a^2}{2}}}\right]$

$B_1=\frac{{\mu }_{0}Ia}{4\pi b}\left[\frac{1}{\sqrt{b^2+\frac{a^2}{2}}}\right]$

Magnetic induction due to $n-$sides

$B=n{B}_1$

Using above values, $\frac{1}{\sqrt{b^2+\frac{a^2}{2}}}=1$

$B=n\left[\frac{{\mu }_{0}I}{4\pi }\right]\left[\frac{2R\sin \left(\frac{\pi }{n}\right)}{R\cos \left(\frac{\pi }{n}\right)}\right]$

$B=\left[\frac{n{\mu }_{0}I}{2\pi R}\right]\left[\tan \left(\frac{\pi }{n}\right)\right]$

For $n\to \infty$

$B=\frac{{\mu }_{0}I}{2R}$