NEET Solved Paper 2016 Question 2
Question: 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 :-
Options:
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 } $
Show Answer
Answer:
Correct Answer: A
Solution:
$ F _{AB}=i\ell B(Attractive) $ $ F _{AB}=i(L).\frac{{\mu_0}I}{2\pi ( \frac{L}{2} )}(arrow )=\frac{{\mu_0}il}{\pi }(arrow ) $ $ {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 ( \frac{3L}{2} )}(\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 } $
BETA
