A) \[\frac{2{{\mu }_{0}}Ii}{3\pi }\]
B) \[\frac{{{\mu }_{0}}Ii}{2\pi }\]
C) \[\frac{2{{\mu }_{0}}IiL}{3\pi }\]
D) \[\frac{{{\mu }_{0}}IiL}{2\pi }\]
Correct Answer: A
Solution :
[a] The net force on the loop \[{{F}_{net}}={{F}_{AB}}+{{F}_{BC}}+{{F}_{CD}}+{{F}_{AD}}\] \[\Rightarrow \,\,\,{{F}_{net}}=\frac{{{\mu }_{0}}iI}{\pi }-\frac{{{\mu }_{0}}iI}{3\pi }=\frac{2{{\mu }_{0}}iI}{3\pi }\]
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