A square frame carrying a curren `I=0.90A` is located in the same plane as a long straight 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 through the midpoints of opposite sides is parallel to the wire and is separated from it by the distance which
`eta=1.5` times greater than the side of the frame.
(a)Ampere force (magnetic force) acting on the frame is `9/P muN`,Then find value of `P`.
(b)The magnidute of mechanical work is `(144lnP)xx10^(-9) J` to be performed in order to turn the frame slowly through `180^(@)` about its axis with the currents maintained constant.then find value of `P`

Explanation

(a) Force of attraction between parallel currents
`F_(1) = (mu_0)/(4pi) ( 2I I_0)/((2eta -1 )(a)/(2)`
Similar force of repulsion between antiparallel currents
`F_(2) = (mu_0)/(4pi) (4I I_0)/((2 eta + 1)a)`
Net force of repulsion between the square frame and the long straight wire
`F = F_1 - F_2 = (2 mu_0)/(pi a) (I I_0)/(2 eta^2 -1)`
Putting values, `F = 0.05 mu N.`
b) Work perfomed in turning the frame through `180^@`
`W = int_0^(pi) tau d theta,` But `d tau = bar(dM) xx bar (B)(r)`
`dM = IdA - = 2a dr n B(r) = (mu_0)/(4 pi) (2I_0)/(r) n`.
n = unit normal to the frame into the paper
`therefore d tau = dM B sin theta`
`theta` = angle between B(r) and dA during the rotation process.
`therefore tau = int d tau (mu_0)/(4 pi). 2I I_(0) sin theta int_( eta a - a//2)^(eta a+a//2) (dr//r)`
`= (mu_0)/(4pi) 2 I I_0a sin theta log_e (2eta + 1)/(2 eta-1) therefore` Workdone
` W = int_(0)^(pi) tau d theta = (mu_0)/(4pi)2 I I_0 a log ""(2eta +1)/(2 eta - 1) sin theta d theta`
` = (mu_0)/(pi) I I_0 a log ""(2 eta + 1)/(2 eta -1) = 0.1 44 mu J`