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.

Options

A. `(mu_(0)(II_(0))a)/(pi)log_(e)(2)`
B. `(mu_(0)(II_(0))a)/(pi)log_(e)(3)`
C. `(mu_(0)(II_(0))a)/(pi)log_(e)(5)`
D. `(mu_(0)(II_(0))a)/(pi)log_(e)(4)`

Answer

A

Explanation

`W=U_(f)-U_(i)" "dU_(i)=(dM)B cos (0^(@))`
`implies dU_(i)=I(adx)(mu_(0)I_(0))/(2pix)" "impliesU_(i)=(mu_(0)II_(0)a)/(2pi)int_(a)^(2a)(dx)/(x)`
`impliesU_(i)=(mu_(0)II_(0)a)/(2pi)log_(e)(2)`
similarly
`U_(f)=intdU_(f)=int(dM)Bcos(180^(@))=-(dM)B`
`=-(mu_(0)(II_(0))a)/(2pi)lo_(e)(2)`
`impliesDB=|U_(f)-U_(i)|=(mu_(0)(II_(0))a)/(pi)log_(e)(2)`.