### 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)} $

## Show Answer

#### 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)} $