Question

A straight segment OC (of length L) of a circuit carrying a current I is placed along the .v-axis . Two infinitely long straight wires A and B, each extending from z = - $-\infty to+\infty ,$ are fixed at y= - a and y = + a respectively, as shown in the figure. If the wires A and B each carry a current 1 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 1

(a) Let us assume a segment of wire OC at a point P, a
distancex from the centre of length dx as shown in figure.
Magnetic field at P due to current in wires A and B will be
in the directions perpendicular to AP and BP respectively
as shown.
$|B|=\frac{{\mu }_0}{2\pi }\frac{I}{AP}$
Therefore, net magnetic force at P will be along negative
v-axis as shown
$B_{net}=2|B|cos\theta$
$=2\left(\frac{{\mu }_0}{2\pi }\right)\frac{I}{AP}\left(\frac{x}{AP}\right)$
$B_{net}=\left(\frac{{\mu }_0}{\pi }\right)\frac{1.x}{(AP{)}^2}$
$B_{net}=\frac{{\mu }_0}{\pi }.\frac{1x}{(a^2+x^2)}$
Therefore, force on this element will be
$dF=1\left\{\frac{{\mu }_0}{\pi }\frac{1x}{a^2+x^2}\right\}dx$(in negative z-direction)
$\therefore$ Total force on the wire will be
$F={\int }_{x=0}^{x=L}$ dF=\frac{\mu_0 I^2}{\pi}${\int }_0^L\frac{xdx}{x^2+a^2}$
$=\frac{{\mu }_0{I}^2}{2\pi }In\left(\frac{L^2+a^2}{a^2}\right)$(in negative z-axis)
Hence, $F=-\frac{{\mu }_0{I}^2}{2\pi }In\left(\frac{L^2+a^2}{a^2}\right)\hat{k}$
(b) When direction of current in B is reversed net magnetic
field is along the current. Hence, force is zero.