As shown in the figure,a current of 2,A flows | vedclass

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

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

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(D) The magnetic field due to a finite straight wire at a perpendicular distance $d$ is given by $B = \frac{\mu_0 i}{4 \pi d} (\sin 	heta_1 + \sin 	heta_2)$.For an equilateral triangle,the distance $d$ from the centroid to any side is $d = \frac{a}{2 \sqrt{3}}$,where $a = 4 \sqrt{3} \,cm = 4 \sqrt{3} 	imes 10^{-2} \,m$.Thus,$d = \frac{4 \sqrt{3} 	imes 10^{-2}}{2 \sqrt{3}} = 2 	imes 10^{-2} \,m$.The angles at the centroid subtended by the ends of each side are $	heta_1 = 	heta_2 = 60^{\circ}$.The magnetic field due to one side is $B_1 = \frac{\mu_0 i}{4 \pi d} (\sin 60^{\circ} + \sin 60^{\circ}) = \frac{\mu_0 i}{4 \pi d} (2 \sin 60^{\circ}) = \frac{\mu_0 i}{2 \pi d} \sin 60^{\circ}$.The total magnetic field at the centroid due to three sides is $B = 3 	imes B_1 = 3 	imes \frac{\mu_0 i}{2 \pi d} \sin 60^{\circ}$.Substituting the values: $B = 3 	imes \frac{2 	imes 10^{-7} 	imes 2}{2 	imes 10^{-2}} 	imes \frac{\sqrt{3}}{2} = 3 	imes 2 	imes 10^{-5} 	imes \frac{\sqrt{3}}{2} = 3 \sqrt{3} 	imes 10^{-5} \,T$.

As shown in the figure,a current of $2\,A$ flows in an equilateral triangle of side $4 \sqrt{3}\,cm$. The magnetic field at the centroid $O$ of the triangle is:
(Neglect the effect of the earth's magnetic field.)

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As shown in the figure,a current of $2\,A$ flows in an equilateral triangle of side $4 \sqrt{3}\,cm$. The magnetic field at the centroid $O$ of the triangle is:(Neglect the effect of the earth's magnetic field.)