question_answer

A current of 1A is flowing on the sides of an equilateral triangle of sides \[4.5\times {{10}^{-2}}m.\] The magnetic field at the centroid of the triangle is

A)  \[2\times {{10}^{-5}}T\]                              

B)  \[4\times {{10}^{-5}}T\]

C)  \[8\times {{10}^{-5}}T\]                              

D)  \[1.2\times {{10}^{-5}}T\]

Correct Answer: B

Solution :

                \[{{B}_{1}}=\frac{{{\mu }_{0}}I}{4\pi a}(\sin {{\phi }_{1}}+\sin {{\phi }_{2}})\] \[\tan 30{}^\circ =\frac{a}{\left( \frac{I}{2} \right)}\] \[a=\frac{I}{2\sqrt{3}}\] \[B=\frac{{{\mu }_{0}}I}{4\pi \left( \frac{I}{2\sqrt{3}} \right)}(\sin 60+\sin 60)\] \[{{B}_{1}}=\frac{6{{\mu }_{0}}I}{4\pi I}\] \[{{B}_{total}}=3\,{{B}_{1}}\] \[{{B}_{total}}=4\times {{10}^{-5}}T\]