question_answer

A current of 1 A is flowing on the sides of an equilateral triangle of sides\[4.5\times {{10}^{-2}}\text{ }m\]. 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)  \[1.2\times {{10}^{-4}}T\]

Correct Answer: B

Solution :

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