A square wire frame with side `a` and a straight conductor carrying constant current `I` are located in the same plane (Fig). The inductance and the resitance of the frame are equal to `L` and `R` respectively. The frame was turned throgh `180^(@)` about the axis `OO'`separated fromt the current carrying conductor by a distance `b`. FInd the electric charge having flown through the frame.

Explanation

According to `Ohm's` law and Faraday's law of induction, the current `i_(0)` appearing in the frame, during its rotation is determind by the formula,
`i_(0) = (d Phi)/(dt) = (I. d i_(0))/(dt)`
Hence, the required amount of electricity (charge) is,
`q = int i_(0) dt = (1)/(R) int (d Phi + L di_(0)) = - (1)/(R) (Delta Phi + L Delta i_(0))`
Since the frame has been stopped after rotations, the current in it vanishes, and hence `Delta i_(0) = 0`. It remains for us to find the increment of the flux `Delta Phi` through the frame `(Delta Phi - Phi_(2) - Phi_(1))`.
Let us choose the normal `vec(n)` to the plane of the frame, for instane, so that is the final position, `vec(n)` is directed behind the plane of the figure (along `vec(B)`).
Then it can be easily seen that in the final position, `Phi_(2) gt 0`, while in the initial position, `Phi_(1) lt 0` (the normal is opposite to `vec(B)`) and `Delta Phi` turns out to be simply equal to the fulx throgh the surface bounded by the final and initial positions of the frame :
`Delta Phi = Phi_(2) + |Phi_(1)| = int_(b - a)^(b + a) B a dr`,
where `B` is a function of `r`, whose from can be easily found with the help of the theorem of circulation. Finally omitting the minus sign, we obtain,
`q = (Delta Phi)/(R) = (mu_(0) a i)/(2 pi R)` In `(b + a)/(b - a)`