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  • A current of is flowing through a triangle, of side <img alt="9 \mathrm{~cm}" src="https://entrancecorner.codecogs.com/gif.latex?9%20%5Cmathrm%7B%7

A current of 1.5 \mathrm{~A} is flowing through a triangle, of side 9 \mathrm{~cm} each. The magnetic field at the centroid of the triangle is:
(Assume that the current is flowing in the clockwise direction.)
 
Option: 1 2 \sqrt{3} \times 10^{-5} \mathrm{~T}, inside the plane of triangle
 
Option: 2 2 \sqrt{3} \times 10^{-7} \mathrm{~T}, outside the plane of triangle
 
Option: 3 3 \times 10^{-7} \mathrm{~T}, outside the plane of triangle
 
Option: 4 3 \times 10^{-5} \mathrm{~T}, inside the plane of triangle

Answers (1)

best_answer

l=9cm\\

h=l\sin60^{\circ}=\frac{\sqrt{3}}{2}\times9\times10^{-2}m\\

h=\frac{9\sqrt{3}}{2}\times10^{-2}m\\

a=\frac{1}{3}\times\left ( \frac{9\sqrt{3}}{2}\times10^{-2} \right )\\

a=\frac{3\sqrt{3}}{2}\times10^{-2}m\\

Magnetic field due to each side

B=\frac{\mu_{0}I}{4\pi a}\left ( \sin60^{\circ}+\sin60^{\circ} \right )\\

     =\frac{1.5}{\frac{3\sqrt{3}}{2}\times10^{-2}}\times10^{-7}\left (\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} \right )\\

B=\frac{\sqrt{3}}{\sqrt{3}\times10^{-2}}\times10^{-7} \otimes  =1\times10^{-5}\otimes       

B_{total}=3B\\

            =3\times10^{-5}T\otimes

Posted by

vishal kumar

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