A straight segment `OC` (of lengh `L`) of a circuit carrying a current I is placed along the x-axis. Two infinitely long straight wire `A` and `B` each extending from `z=-ooto+oo` are fixed at `y=-a` and `y=+a` respectively as show in the figure. If the wires `A` and `B` each carry a current I 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

Let us take an element of thickness dx at a distance x from
origin on the wire OC. Magnetic field `B_A` produced at P(x,0,0)
due to wires placed at A and B are
`B_A=mu_0I//2piR, B_B=mu_0I//2piR`
Component of `B_A and B_B` along x-axis cancel, while those along
y-axis add up to give total field,
`B=2((mu_0I)/(2piR))cos theta=(2mu_0I)/(2piR) x/R=(mu_0I)/pi x/((a^2+x^2))`
The force dF acting on the current element is `dvecF=I(dveclxxvecB)`
`dF=(mu_0I^2)/pi (xdx)/(a^2+x^2) [:' sin 90^@=1]`
Hence, net force on wire OC, `F=intdF`
i.e., `F=(mu_0I^2)/pi int_0^L(xdx)/(a^2+x^2)=(mu_0I^2)/pi In(a^2+L^2)/(a^2)` along z-direction.

If the current in B is reserved, the magnetic field due to the two
wires would be only x-direction and the force on hte current
carrying wire along x-direction will be zero.