An infinitely long wire is placed near a square loop as shown in figure. Choose the correct options.

Explanation

`E = B_(1)lv - b_(2)lv = (mu_(0)i)/(2pix)lv - (mu_(0)i)/(2pi(x + b))lv = (mu_(0)ilbv)/(2pix(x + b))`

Alternatively:
Consider a samll segment of width `dy` at a distance `y` from tne wire. Let flux through the segment be `dphi`.

`:.` `dphi = (mu_(0)i)/(2piy)ldy`
`:.` `phi = (mu_(0)il)/(2pi) int_(x)^(x +b)(dy)/(y) = (mu_(0)il)/(2pi)(1n (x + b) - 1n x)`
Now, `(dphi)/(dt) = (mu_(0)il)/(2pi)[(1)/(x + b)(dx)/(dt) - (1)/(x)(dx)/(dt)]`
`= (mu_(0)il)/(2pi)[(-b)/(x(x - b))]v = (-mu_(0)iblv)/(2pix(x + b))`
`:.` Induced emf `= (mu_(0)iblv)/(2pix(x +b))`