A) \[\frac{2{{\mu }_{0}}li}{3\pi }\]
B) \[\frac{{{\mu }_{0}}li}{2\pi }\]
C) \[\frac{2{{\mu }_{0}}liL}{3\pi }\]
D) \[\frac{{{\mu }_{0}}liL}{2\pi }\]
Correct Answer: A
Solution :
\[{{F}_{AB}}=i\ell B(Attractive)\] \[{{F}_{AB}}=i(L).\frac{{{\mu }_{0}}I}{2\pi \left( \frac{L}{2} \right)}(\leftarrow )=\frac{{{\mu }_{0}}il}{\pi }(\leftarrow )\] \[{{F}_{(BC)}}(\uparrow )\]and \[{{F}_{(AD)}}(\uparrow )\Rightarrow \]cancel each other \[{{F}_{CD}}=i\ell B(Repulsive)\] \[{{F}_{CD}}=i(L)\frac{{{\mu }_{0}}I}{2\pi \left( \frac{3L}{2} \right)}(\to )=\frac{{{\mu }_{0}}iI}{3\pi }(\to )\] \[\Rightarrow \] \[{{F}_{net}}=\frac{{{\mu }_{0}}iI}{\pi }-\frac{{{\mu }_{0}}iI}{3\pi }=\frac{2{{\mu }_{0}}iI}{3\pi }\]You need to login to perform this action.
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