A wire shaped in a regular hexagon of side 2 | vedclass

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(B) The distance $r$ from the centre of the hexagon to the midpoint of any side of length $a = 2 	ext{ cm}$ is given by:$r = \frac{a/2}{	an 30^{\cir

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$8 \sqrt{3} 	imes 10^{-5} 	ext{ T}$

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(B) The distance $r$ from the centre of the hexagon to the midpoint of any side of length $a = 2 	ext{ cm}$ is given by:$r = \frac{a/2}{	an 30^{\circ}} = \frac{a}{2 	imes (1/\sqrt{3})} = \frac{\sqrt{3} a}{2} = \sqrt{3} 	ext{ cm} = \sqrt{3} 	imes 10^{-2} 	ext{ m}$.The magnetic field $B$ at the centre due to one side of the hexagon is given by the formula for a finite wire:$B_1 = \frac{\mu_0 I}{4 \pi r} (\sin 	heta_1 + \sin 	heta_2)$Here,$	heta_1 = 	heta_2 = 30^{\circ}$,so:$B_1 = \frac{\mu_0 I}{4 \pi r} (2 \sin 30^{\circ}) = \frac{\mu_0 I}{4 \pi r} (2 	imes 0.5) = \frac{\mu_0 I}{4 \pi r}$.Since there are $6$ identical sides,the total magnetic field $B$ at the centre is:$B = 6 	imes B_1 = 6 	imes \frac{\mu_0 I}{4 \pi r} = 6 	imes 10^{-7} 	imes \frac{4}{\sqrt{3} 	imes 10^{-2}} = \frac{24 	imes 10^{-5}}{\sqrt{3}} = 8 \sqrt{3} 	imes 10^{-5} 	ext{ T}$.

$A$ wire shaped in a regular hexagon of side $2 \text{ cm}$ carries a current of $4 \text{ A}$. The magnetic field at the centre of the hexagon is

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