Neet Solved Paper 2015 Question 33
Question: A conducting square frame of side a and a long straight wire carrying current $ I $ are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity V. The emf induced in the frame will be proportional to
Options:
A) $ \frac{1}{x^{2}} $
B) $ \frac{1}{{{(2x-a)}^{2}}} $
C) $ \frac{1}{{{(2x+a)}^{2}}} $
D) $ \frac{1}{(2x+a)(2x+a)} $
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Answer:
Correct Answer: D
Solution:
Potential difference across PQ is $ V _{P}-V _{Q}=B _1(a)v=\frac{{\mu _0}I}{2\pi ( x-\frac{a}{2} )}av $
Potential difference across side RS of frame is
$ V _{S}-V _{R}=B _2(a)v=\frac{{\mu _0}I}{2\pi ( x+\frac{a}{2} )}av $
Hence, the net potential difference in the loop will be
$ V _{net}=(V _{P}-V _{Q})-(V _{S}-V _{R}) $ $ =\frac{{\mu _0}iav}{2\pi }[ \frac{1}{( x-\frac{a}{2} )}-\frac{1}{( x+\frac{a}{2} )} ] $ $ =\frac{{\mu _0}iav}{2\pi }( \frac{a}{( x-\frac{a}{2} )( x+\frac{a}{2} )} ) $
Thus $ V _{net}\propto \frac{1}{(2x-a)(2x+a)} $
BETA
