question_answer

A square loop ABCD carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be :-                                                                      [NEET - 2016]

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 :

[a]

\[{{F}_{AB}}=i\ell B\,\,(Attractive)\]

\[{{F}_{AB}}=i(L).\frac{{{\mu }_{0}}I}{2\pi \left( \frac{L}{2} \right)}(\leftarrow )=\frac{{{\mu }_{0}}il}{\pi }(\leftarrow )\]

\[{{F}_{(BC)}}(\uparrow )\] and \[{{F}_{(AD)}}(\uparrow )\]

\[\Rightarrow \]   cancel each other

\[{{F}_{CD}}=i\ell B\,\,(Repulsive)\]

\[{{F}_{CD}}=i(L)\frac{{{\mu }_{0}}I}{2\pi \left( \frac{3L}{2} \right)}(\to )=\frac{{{\mu }_{0}}iI}{3\pi }(\to )\]

\[\Rightarrow \]   \[{{F}_{net}}=\frac{{{\mu }_{0}}iI}{\pi }-\frac{{{\mu }_{0}}iI}{3\pi }=\frac{2{{\mu }_{0}}iI}{3\pi }\]