Question

Magnetic field due to regular polygon of side $n$ made up of total length $2\pi R$ which carries current $I$. Find the magnetic field at the centre of polygon.

• A. $\frac{{\mu }_{0}I}{2\pi R}\frac{\left(\frac{\sin \left(\frac{\pi }{n}\right)}{\frac{\pi }{n}}\right)}{\cot \left(\frac{\pi }{n}\right)}$
• B. $\frac{{\mu }_{0}I}{2R}\frac{{\left(\frac{\sin \left(\frac{\pi }{n}\right)}{\frac{\pi }{n}}\right)}^2}{\cot \left(\frac{\pi }{n}\right)}$
• C. $\frac{{\mu }_{0}I}{4R}\frac{\left(\frac{\cos \left(\frac{\pi }{n}\right)}{\frac{\pi }{n}}\right)}{\cot \left(\frac{\pi }{n}\right)}$
• D. $\frac{{\mu }_0}{2\pi }\frac{I}{\frac{\pi R}{n}}\frac{\sin \left(\frac{\pi }{n}\right)}{\cot \left(\frac{\pi }{n}\right)}$

Answer 1

The correct option is D $\frac{{\mu }_0}{2\pi }\frac{I}{\frac{\pi R}{n}}\frac{\sin \left(\frac{\pi }{n}\right)}{\cot \left(\frac{\pi }{n}\right)}$

As the polygon is of $n$ sides, then length of each side is

$l=\frac{2\pi R}{n}$

Each slide will be making an angle $\theta$ at the centre then

$\theta =\frac{2\pi }{n}$

Therefore, magnetic field at centre due to only one side is

$B=\frac{{\mu }_0}{4\pi }\frac{I}{a}[\sin \frac{\theta }{2}+\sin \frac{\theta }{2}]$


From $△APO$

$\tan \left(\frac{\theta }{2}\right)=\frac{AP}{OP}=\frac{\frac{\pi R}{n}}{OP}$

$\Rightarrow OP=\frac{\pi R}{n}\times \frac{1}{\tan \frac{\theta }{2}}=\frac{\pi R}{n}\left(\cot \frac{\theta }{2}\right)$

$\therefore OP=\frac{\pi R}{n}\left(\cot \frac{\pi }{n}\right)$

$B=\frac{{\mu }_0}{4\pi }\times \frac{I}{\frac{\pi R}{n}\left(\cot \frac{\pi }{n}\right)}\times \left[2\sin \frac{\theta }{2}\right]$

$=\frac{{\mu }_{0}nI}{2{\pi }^{2}R}\times \frac{2\sin \left(\frac{\pi }{n}\right)}{\cot \left(\frac{\pi }{n}\right)}$

Hence, (d) is the correct answer.