A square frame carrying a current `I=0.90A` is located in the same plane as a long straight wire carrying a current `I_0=5A`, The frame side has a length `a=8cm`. The axis of the frame passing through the mid-point of the opposite sides is parrallel to the wire and is separated form it by a distance `h=15` times greater tahn the side of the frame.
Find:
(a) ampere force acting on the frame.
(b) The machanical work to be performed in order to turn the frmae through `180^@` about its axis, with the current maintained constant.

Explanation

(a) Force of attraction between parallel currents
`F_1[(mu_0)/(4pi) (2II_0)/((2eta-1)a/2)]a`
Similarly force repulsion between antiparallel currents
`F_2[(mu_0)/(4pi) (2II_0)/((2eta+1)a/2)]a`
Net force of attraction between the square frame and the
long straight wire
`F=F_1-F_2=(2mu_0)/pi (II_0)/(4eta^2-1)`
Putting values, `F=4xx10^-9N=40nN`
(b) Initial energy, `U_1=-intdvec(M).vecB`
`U_1=-int_((2eta-1)a//2)^((2eta+1)a//2) (Iadr) (mu_0I_0)/(2pir)cos0^@`
`=-(mu_0II_0a)/(2pi) int_((2eta-1)a//2)^((2eta+1)a//2) (dr)/r=-(mu_0II_0a)/(2pi)In[(2eta+1)/(2eta-1)]`
Final energy: `U_2=-intdMBcos 180^@`
`=intdM B=(mu_0II_0a)/(2pi)In((2eta+1)/(2eta-1))`

Work done: `W=U_2-U_1`
`=(mu_0II_0a)/(pi) In[(2eta+1)/(2eta-1)]=9.6xx10^-9J=9.6nJ`