A current of i ampere is flowing in an equila | vedclass

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(D) The magnetic field due to a finite straight wire at a perpendicular distance $r$ is given by $B = \frac{\mu_0 i}{4 \pi r} (\sin 	heta_1 + \sin

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$\frac{9 \mu_0 i}{2 \pi a}$

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(D) The magnetic field due to a finite straight wire at a perpendicular distance $r$ is given by $B = \frac{\mu_0 i}{4 \pi r} (\sin 	heta_1 + \sin 	heta_2)$.For an equilateral triangle of side $a$,the distance from the centroid $O$ to any side is $r = \frac{a}{2 \sqrt{3}}$.The angles subtended by the corners at the perpendicular point on the side are $	heta_1 = 60^{\circ}$ and $	heta_2 = 60^{\circ}$.The magnetic field due to one side is $B_1 = \frac{\mu_0 i}{4 \pi (a / 2 \sqrt{3})} (\sin 60^{\circ} + \sin 60^{\circ}) = \frac{\mu_0 i \sqrt{3}}{2 \pi a} (\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}) = \frac{3 \mu_0 i}{2 \pi a}$.Since there are three sides and the magnetic field due to each side is in the same direction (inward),the total magnetic field is $B_{	ext{total}} = 3 	imes B_1 = 3 	imes \frac{3 \mu_0 i}{2 \pi a} = \frac{9 \mu_0 i}{2 \pi a}$.

$A$ current of $i$ ampere is flowing in an equilateral triangle of side $a$. The magnetic induction at the centroid will be

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