A square frame with side a and a long straight wire carrying a current `i` are located in the same plane as m shown in figure. The fram translates to the right with a constant velocity `v`. Find the emf induced in the frame as a function of distance `x`.

Explanation

Field, due to the current carrying wire, at a perpendicular distance `x` from it is given by,
`B(x) = (mu_(0))/(2pi) (i)/(x)`
Motional e.m.f is given by `|int - (vec(v) xx vec(B)).d vec(l)|`
There will be no induced e.m.f in the segments (2) and (4)
as, `vec(v) uarr uarr d vec(l)` and magnitude of e.m.f induced in 1 and 3, will be
`xi_(1) = v((mu_(0))/(2pi) (i)/(x))a` and `xi_(2) = v((mu_(0))/(2pi) (i)/((a + x)))a`,
respectively, and thier sense will be in the direction of `(vec(v) xx vec(B))`.
So, e.m.f., induced in the network `= xi_(1) - xi_(2)`[as `xi_(1) gt xi_(2)`]
`= (a v mu_(0) i)/(2pi) [(1)/(x) - (1)/(a + x)] = (v a^(2) mu_(0) i)/(2pi x (a + x))`