Question

A current $i$ flows in a thin wire in the shape of a regular polygon with $n$ sides. what will be the magnetic field at the centre of polygon? [$R$ is the circum-radius of the polygon]

• A. $\frac{{\mu }_{0}ni}{2\pi R}\tan \frac{\pi }{6}$
• B. $\frac{{\mu }_{0}ni}{2\pi R}\tan \left(\frac{\pi }{n}\right)$
• C. $\frac{{\mu }_{0}i}{2nR}$
• D. Zero

Answer 1

The correct option is B $\frac{{\mu }_{0}ni}{2\pi R}\tan \left(\frac{\pi }{n}\right)$


For $n-$ sided polygon $\angle AOB=\frac{2\pi }{n}$

Perpendicular distance $\text{OC}=Rcos\frac{\pi }{n}$

Now, Magnetic field due to wire $\text{AB}$ is given by,

$B=\frac{{\mu }_{0}i}{4\pi d}(sin{\theta }_1+sin{\theta }_2)$

$B=\frac{{\mu }_{0}i}{4\pi Rcos\frac{\pi }{n}}2sin\frac{\pi }{n}(\because d=Rcos\frac{\pi }{n},{\theta }_1={\theta }_2=\frac{\pi }{n})$

$B$ due to all the wires at the centre $\text{O}$ points perpendicular to the plane and outwards.

So, $B_{net}=nB$

$B_{net}=\frac{{\mu }_{0}ni}{4\pi Rcos\frac{\pi }{n}}2sin\frac{\pi }{n}$

$B_{net}=\frac{{\mu }_{0}ni}{2\pi R}tan\frac{\pi }{n}$

Hence, option $\left(b\right)$ is the correct answer.