A regular polygon of $$6$$ sides is formed by bending a wire of length $$4\pi$$ meter. If an electric current of $$4\pi\sqrt{3}$$ A is flowing through the sides of the polygon, the magnetic field at the centre of the polygon would be $$x \times 10^{-7}$$ T. The value of $$x$$ is ______.

A regular hexagon is formed by bending a wire of length $$4\pi$$ m. A current $$I = 4\pi\sqrt{3}$$ A flows through it, and we wish to find the magnetic field at the centre.

A regular hexagon has 6 sides, so its side length is $$a = \frac{4\pi}{6} = \frac{2\pi}{3}$$ m. The perpendicular distance from the centre to each side (the apothem) is $$d = \frac{a\sqrt{3}}{2} = \frac{2\pi}{3} \cdot \frac{\sqrt{3}}{2} = \frac{\pi\sqrt{3}}{3}$$ m.

Each side subtends an angle of $$60°$$ at the centre, giving a half-angle of $$30°$$. The magnetic field due to a finite straight conductor at perpendicular distance $$d$$ is given by $$B_1 = \frac{\mu_0 I}{4\pi d}(\sin\alpha_1 + \sin\alpha_2)$$, and since $$\alpha_1 = \alpha_2 = 30°$$, this becomes $$B_1 = \frac{\mu_0 I}{4\pi d}(2\sin 30°) = \frac{\mu_0 I}{4\pi d}$$.

The total magnetic field from all six sides is $$B = 6 \times \frac{\mu_0 I}{4\pi d} = \frac{6\mu_0 I}{4\pi d} = \frac{6 \times 4\pi \times 10^{-7} \times 4\pi\sqrt{3}}{4\pi \times \frac{\pi\sqrt{3}}{3}} = \frac{6 \times 4\pi\sqrt{3} \times 10^{-7}}{\frac{\pi\sqrt{3}}{3}} = \frac{6 \times 4\pi\sqrt{3} \times 3}{\pi\sqrt{3}} \times 10^{-7} = 72 \times 10^{-7}$$ T.

The value of $$x$$ is 72.