A) \[\frac{{{\mu }_{0}}}{2\pi }.\frac{\sqrt{3}i}{\ell }\]
B) \[\frac{{{\mu }_{0}}}{\pi }.\frac{i}{\ell }\sqrt{3}\]
C) zero
D) \[\frac{{{\mu }_{0}}}{4\pi }.\frac{i}{\ell \sqrt{3}}\]
Correct Answer: B
Solution :
[b] B due to \[AB=\frac{{{\mu }_{0}}}{4\pi }\frac{i}{d}\left[ \sqrt{3} \right]\] \[=\frac{{{\mu }_{0}}}{4\pi }\frac{i}{\ell /2}\left[ \sqrt{3} \right]=\frac{{{\mu }_{0}}i}{4\pi \ell }2\sqrt{3}\] Net magnetic field \[=\frac{{{\mu }_{0}}}{2\pi }\frac{i}{\ell }\sqrt{3}\times 2=\frac{{{\mu }_{0}}}{\pi }\frac{i}{\ell }\sqrt{3}\]
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