A) \[\frac{3{{\mu }_{0}}l}{2\pi d}\]
B) \[\frac{{{\mu }_{0}}l}{2\pi d}\]
C) \[\frac{{{\mu }_{0}}l}{\sqrt{3}\pi d}\]
D) \[\frac{\sqrt{3}{{\mu }_{0}}l}{2\pi d}\]
Correct Answer: D
Solution :
We know that \[{{B}_{net}}=2\left[ \frac{{{\mu }_{0}}i}{4\pi r}(sin{{\theta }_{1}}+sin{{\theta }_{2}}) \right]\] \[=2\left[ \frac{{{\mu }_{0}}}{4\pi }\times \frac{i}{\frac{d\sqrt{3}}{2}}\times (sin{{90}^{o}}+sin{{30}^{o}}) \right]\] \[=2\left[ \frac{{{\mu }_{0}}}{4\pi }\times \frac{2i}{d\sqrt{3}}\times \left( 1+\frac{1}{2} \right) \right]\] \[=2\left[ \frac{{{\mu }_{0}}}{4\pi }\times \frac{2i}{d\sqrt{3}}\times \frac{3}{2} \right]=\frac{\sqrt{3}{{\mu }_{0}}i}{2\pi d}\]You need to login to perform this action.
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