Question

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 $

Answer 1

Answer: (B)

$ {{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 $