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  • As shown in the figure, a current of 2 A flowing in an equilateral triangle of side <img alt="4 \sqrt{3} \mathrm{~cm}" src="https://entrancecorner.oncodecogs.com/gif.latex?4%20%5Csqrt%7B3%7D%20%5Cm

As shown in the figure, a current of 2 A flowing in an equilateral triangle of side 4 \sqrt{3} \mathrm{~cm}. The magnetic field at the centroid 0 of the triangle is

(Neglect the effect of earth's magnetic field)

Option: 1

1.4 \sqrt{3} \times 10^{-5} \mathrm{~T}


Option: 2

4 \sqrt{3} \times 10^{-4} \mathrm{~T}


Option: 3

3 \sqrt{3} \times 10^{-5} \mathrm{~T}


Option: 4

\sqrt{3} \times 10^{-4} \mathrm{~T}


Answers (1)

best_answer

Let d= perpendicular distance from O to any one side of triangle

L=side of triangle

So 

\tan 60^{\circ}=\frac{\frac{L}{2}}{d}=\frac{2 \sqrt{3}}{d}= \sqrt{3}
d=2 \mathrm{~cm}

B=3\left(\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{d}}\right) \sin 60^{\circ}
B=\frac{3 \times 2 \times 10^{-7} \times 2}{2 \times 10^{-2}} \times \frac{\sqrt{3}}{2}
B=3 \sqrt{3} \times 10^{-5} \mathrm{~T}

Posted by

Rakesh

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