A regular polygon of 6 sides is formed by ben | vedclass

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(A) The perimeter of the regular hexagon is $L = 4 \pi 	ext{ m}$. The length of each side $a$ is given by $a = \frac{L}{6} = \frac{4 \pi}{6} = \frac{

Vedclass · Accepted Answer

$72$

Vedclass · Answer

(A) The perimeter of the regular hexagon is $L = 4 \pi 	ext{ m}$. The length of each side $a$ is given by $a = \frac{L}{6} = \frac{4 \pi}{6} = \frac{2 \pi}{3} 	ext{ m}$.The distance $r$ from the centre to the midpoint of a side is $r = \frac{a}{2 	an(30^{\circ})} = \frac{a}{2 	imes (1/\sqrt{3})} = \frac{a \sqrt{3}}{2}$.Substituting $a = \frac{2 \pi}{3}$,we get $r = \frac{(2 \pi / 3) \sqrt{3}}{2} = \frac{\pi}{\sqrt{3}} 	ext{ m}$.The magnetic field $B$ at the centre due to one side is $B_1 = \frac{\mu_0 I}{4 \pi r} (\sin 30^{\circ} + \sin 30^{\circ}) = \frac{\mu_0 I}{4 \pi r} (1)$.For $6$ sides,the total magnetic field is $B = 6 	imes B_1 = 6 	imes \frac{\mu_0 I}{4 \pi r}$.Given $\mu_0 = 4 \pi 	imes 10^{-7} 	ext{ T m/A}$,$I = 4 \pi \sqrt{3} 	ext{ A}$,and $r = \frac{\pi}{\sqrt{3}} 	ext{ m}$.$B = 6 	imes \frac{4 \pi 	imes 10^{-7} 	imes 4 \pi \sqrt{3}}{4 \pi 	imes (\pi / \sqrt{3})} = 6 	imes \frac{4 \pi 	imes 10^{-7} 	imes 4 \pi \sqrt{3} 	imes \sqrt{3}}{4 \pi 	imes \pi} = 6 	imes 4 	imes 3 	imes 10^{-7} = 72 	imes 10^{-7} 	ext{ T}$.Thus,$x = 72$.

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