A current of 1,A is passed through a hexagona | vedclass

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(B) The magnetic field $B$ due to a straight wire of length $L$ at a perpendicular distance $r$ is given by $B = \frac{{\mu _0 I}}{{4\pi r}}(\sin 	he

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$\frac{{\sqrt 3 {\mu _0}}}{\pi }$

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(B) The magnetic field $B$ due to a straight wire of length $L$ at a perpendicular distance $r$ is given by $B = \frac{{\mu _0 I}}{{4\pi r}}(\sin 	heta_1 + \sin 	heta_2)$.For a regular hexagon with side $a = 1\,m$,the distance from the centre to any side is $r = a \frac{{\sqrt 3 }}{2} = \frac{{\sqrt 3 }}{2}\,m$.The angles subtended by the ends of each side at the centre are $	heta_1 = 	heta_2 = 30^\circ$.The magnetic field due to one side is $B_1 = \frac{{\mu _0 I}}{{4\pi r}}(\sin 30^\circ + \sin 30^\circ) = \frac{{\mu _0 I}}{{4\pi (\sqrt 3 / 2)}}(1/2 + 1/2) = \frac{{\mu _0 I}}{{2\pi \sqrt 3 }}$.Since there are $6$ sides,the total magnetic field at the centre is $B = 6 	imes B_1 = 6 	imes \frac{{\mu _0 I}}{{2\pi \sqrt 3 }} = \frac{{3\mu _0 I}}{{\pi \sqrt 3 }} = \frac{{\sqrt 3 \mu _0 I}}{\pi }$.Given $I = 1\,A$,we get $B = \frac{{\sqrt 3 \mu _0}}{\pi }\,Wb/m^2$.

$A$ current of $1\,A$ is passed through a hexagonal conducting wire of side $1\,m$. The magnetic induction at its centre $O$ in $Wb/m^2$ will be

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