A current of 1.5 , A is flowing through an eq | vedclass

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(D) The magnetic field 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 +

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$3 	imes 10^{-5} \, T$,inward into the plane of the triangle

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(D) The magnetic field 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 an equilateral triangle,the distance $r$ from the centroid to any side is $r = \frac{L}{2 \sqrt{3}}$,where $L = 9 \, cm = 0.09 \, m$.$r = \frac{0.09}{2 \sqrt{3}} = \frac{0.045}{\sqrt{3}} \, m$.The angles subtended at the centroid by the ends of each side are $	heta_1 = 	heta_2 = 60^{\circ}$.The magnetic field due to one side is $B_1 = \frac{\mu_0 i}{4 \pi r} (\sin 60^{\circ} + \sin 60^{\circ}) = \frac{\mu_0 i}{4 \pi r} (2 \sin 60^{\circ}) = \frac{\mu_0 i}{4 \pi r} (\sqrt{3})$.Since there are three identical sides,the total magnetic field is $B = 3 B_1 = 3 	imes \frac{\mu_0 i}{4 \pi r} \sqrt{3}$.Substituting the values: $\mu_0 = 4 \pi 	imes 10^{-7} \, T \cdot m/A$,$i = 1.5 \, A$,$r = \frac{0.09}{2 \sqrt{3}} \, m$.$B = 3 	imes \frac{4 \pi 	imes 10^{-7} 	imes 1.5}{4 \pi 	imes (0.09 / 2 \sqrt{3})} 	imes \sqrt{3} = 3 	imes \frac{10^{-7} 	imes 1.5 	imes 2 \sqrt{3}}{0.09} 	imes \sqrt{3} = 3 	imes \frac{10^{-7} 	imes 1.5 	imes 2 	imes 3}{0.09} = 3 	imes \frac{9 	imes 10^{-7}}{0.09} = 3 	imes 10^{-5} \, T$.Since the current flows clockwise,by the right-hand rule,the magnetic field is directed inward into the plane of the triangle.

$A$ current of $1.5 \, A$ is flowing through an equilateral triangle of side $9 \, cm$. The magnetic field at the centroid of the triangle is (Assume that the current is flowing in the clockwise direction.)

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