question_answer

The magnitude of the magnetic field at the center of an equilateral triangular loop of side 1m which is carrying a current of 10 A is : \[[\text{Take}\,{{\mu }_{0}}=4\pi \times {{10}^{-7}}N{{A}^{-2}}]\] [JEE Main 10-4-2019 Afternoon]

A) \[18\mu T\]

B) \[3\mu T\]

C) \[1\mu T\]                               

D) \[9\mu T\]

Correct Answer: A

Solution :

            \[B=3\left[ \frac{{{\mu }_{0}}i}{4\pi r}(sin{{60}^{o}}+sin{{60}^{o}}) \right]\] Here, \[r=\frac{a}{2\sqrt{3}}=\frac{1}{2\sqrt{3}}\] \[B=3\left[ \frac{4\pi \times {{10}^{-7}}\times 10\times 2\sqrt{3}}{4\pi \times 1}\left[ \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} \right] \right]\] \[B=18\times {{10}^{-6}}=18\mu T\]