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. `(1)/(x^(2))`
B. `(1)/((2x-a)6(2))`
C. `(1)/((2x+a)^(2))`
D. `(1)/((2x-a)(2x+a))`

Answer

D

Explanation

The emf induced in side `1` of frame `e_(1)=B_(1)Vl`
`B_(1)=(mu_(0)^(I))/(2pi(x-a//2))`
the emf induced in side `2` of frame `e_(2)=B_(2)Vl`
`B_(1)=(mu_(0)I)/(2pi(x+a//2))`
Net emf induced in square frame `e=B_(1)Vl-B_(2)Vl`
`=(mu0I)/(2pi(x-a//2))lv-(mu0I)/(2pi(x+a//2)lv`
or, `eprop(1)/((2x-a)(2x+a)`