A straight segment `OC`(of length L meter) of a circuit carrying a current `I amp` is placed along the ` x- axis` ( fig.). Two infinetely long straight wires ` A and B` , each extending from ` z = -oo to +oo`, are fixed at ` y = -ameter and y = +a meter` respectively, as shown in the figure.
If the wires ` A and B` each carry a current `I amp` into the plane of the paper, obtain the expression for the force acting on the segment ` OC`. What will be the force on `OC` if the current in the wire `B` is reversed?

Answer

`F =((mu_(0)I^(2))/(2pi))ln" ((L^(2) + a^(2))/(a^(2))) (-hatk), zero`

Explanation

`B =B_(A) cos 0+B_(B) cos 0 =2B_(A) cos0`
`B =2(mu_(0))/(4pi) (2I)/sqrt(a^(2)+x^(2)) xx (X)/(sqrt(a^(2)+X^(2))`
`F =I (dx) B`
`F =(mu_(0)I^(2))/(4pi) 2 underset(0)overset(L)int(2xdx)/((a^(2) +X^(2)))`
`= (mu_(0I^(2)))/(4pi) 2[log_(e) (a^(20) + X^(2))]_(0)^(L)`
`= (mu_(0)I^(2))/(4pi) 2[log_(e) "(a^(2) +L^(2))/(a^(2))]`
`oversetrarrF = (mu_(0)I^(2))/(2pi) log_(e) "(L^(2)+a^(2))/(a^(2)) (-hatk)`
when direaction of current in condutor `B` is reversed the `B` will be along x-axis so force on segment `BC` segment becomes zero .