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A current of \[1A\] is flowing on the sides of an equilateral triangle of side\[4.5\times {{10}^{-2}}m\]. The magnetic field at the centre of the triangle will be:                               [JEE Online 15-04-2018 (II)]

A) \[4\times {{10}^{-5}}Wb/{{m}^{2}}\]      

B) Zero                      

C) \[2\times {{10}^{-5}}Wb/{{m}^{2}}\]      

D) \[8\times {{10}^{-5}}Wb/{{m}^{2}}\]

Correct Answer: A

Solution :

\[l=4.5\times {{10}^{-2}}m\] \[\tan 60{}^\circ =\sqrt{3}=\frac{l}{2d}\] \[\Rightarrow d=\frac{l}{2\sqrt{3}}=\left( \frac{4.5\times {{10}^{-2}}}{2\sqrt{3}} \right)m\] \[B=\frac{{{\mu }_{o}}i}{4\pi d}(\cos {{\theta }_{1}}+\cos {{\theta }_{2}})\] \[=\frac{2{{\mu }_{0}}i}{4\pi d}\left( \frac{\sqrt{3}}{2} \right)\] \[=\frac{{{\mu }_{o}}i}{2\pi }\left( \frac{\sqrt{3}}{2} \right)\frac{2\sqrt{3}}{(4.5\times {{10}^{-2}})}\] On solving we will get option A as answer