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(C) The magnetic field due to a straight wire of length $L$ at a perpendicular distance $d$ is given by $B = \frac{\mu_0 I}{4\pi d} (\sin 	heta_1 + \

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$3\sqrt{3}$

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(C) The magnetic field due to a straight wire of length $L$ 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 of side $a = 4\sqrt{3} 	ext{ cm}$, the distance $d$ from the centroid to any side is $d = \frac{a}{2\sqrt{3}} = \frac{4\sqrt{3}}{2\sqrt{3}} = 2 	ext{ cm} = 2 	imes 10^{-2} 	ext{ m}$.The angles subtended by the corners at the centroid are $60^{\circ}$ each, so $	heta_1 = 	heta_2 = 60^{\circ}$.The magnetic field due to one side is $B_1 = \frac{\mu_0 I}{4\pi d} (\sin 60^{\circ} + \sin 60^{\circ}) = \frac{10^{-7} 	imes 2}{2 	imes 10^{-2}} (\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}) = 10^{-5} 	imes \sqrt{3} 	ext{ T}$.Since there are $3$ sides, the total magnetic field is $B = 3 	imes B_1 = 3\sqrt{3} 	imes 10^{-5} 	ext{ T}$.Comparing this with $\alpha 	imes 10^{-5} 	ext{ T}$, we get $\alpha = 3\sqrt{3}$.

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