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?

Explanation

Magnetic field at distance `x` from `O`
`B_(1) = B_(2) = (mu_(0)i)/(2pi(a^(2) + x^(2))^(1//2)) = B`
`B_(P) = 2B cos theta = 2 xx (mu_(0)i)/(2pi(a^(2) + x^(2))^(1//2)) xx (x)/((a^(2) + x^(2))^(1//2))`
`= (mu_(0)ix)/(pi(a^(2) + x^(2)))`, downward

Consider a length `dx` of wire at distance `x`,
`dF = B_(P) i dx`
`= (mu_(0)i x i dx)/(pi(a^(2) + x^(2)))`
`F = (mu_(0)i^(2))/(pi) int_(0)^(L) (x dx)/((a^(2) + x^(2)))`
`I = int_(0)^(L) (x dx)/(a^(2) + x^(2))`
Let `a^(2) + x^(2) = t^(2)`
`x dx = t dt`
`= int (t dt)/(t^(2)) = int (dt)/(t) = log_(e) t = |log_(e)sqrt((a^(2) + x^(2)))|_(0)^(L)`
`= log_(e) (a^(2) + L^(2))^(1//2) - log_(e)(a) = log_(e)(sqrt(a^(2) + L^(2))/(a))`
`= (1)/(2) log_(e) ((a^(2) + L^(2))/(a^(2)))`
`F = (mu_(0)i^(2))/(2pi) log_(e) ((a^(2) + L^(2))/(a))` along `-z` direction
If current in wire `B` is reversed

`B_(P) = 2B sin theta` along x- axis.

Since current and magentic field are in same direction, hence force `= 0`.
Note: For medical aspirants, this type of integral can be ignored.