A square frame carrying a current `I = 0.9 A` is located in the same plane as a long straght wire carrying a current, `I_(0) = 5.0 A`. The frame side has a length `a = 8.0 cm`. The axis of the frame passing thorugh the midpoints of opposite sides is parallel to the wire and is separated from it by the distance which is `eta = 1.5` times greater than the side of the frame. FInd:
(a) Ampere force acting on the frame,
(b) the mechnical work to be performed in order to turn the frame throguh `180^(@)` about its axis, with the currents maintained constant.

Explanation

(a) As is clear from, the condition, Ampere's forces on the sides (2) and (4) are equal in magnitude but opposite in diraction. Hence the effective froce on the frame is the Now, the Ampere force on (1),
`F_(1) = (mu_(0))/(2pi) (ii_(0))/((eta + (1)/(2)))`
So, the resultant force on the frame
`= F_() - F_(3)`. (as they in nature.)
`= (2 mu_(0) ii_(0))/(pi(4 eta^(2) - 1)) = 0.40 mu N`.
(b) Work done in turning the frame throguh some angle, `A = int i d Phi = i (Phi_(f) - Phi_(i))`, where `Phi_(f)` is the flux through the frame is final position, and `Phi_(p)` that in the initial position. Phi Here, `|Phi_(f)| = |Phi_(i)| = Phi` and `Phi_(i) = Phi_(f)` ,brgt so, `Delta Phi = 2Phi` and `A = i 2 Phi`
Hence, `A = 2i int vec(B).d.vec(S)`
`2i int_(a (eta - (1)/(2)))^(a (eta + (1)/(2))) (mu_(0))/(2pi) (i_(0) a)/(r) dr = (mu_(0) ii_(0) a)/(pi) ln ((2eta + 1)/(2eta - 1))`