Question 222 of 276

Two wires $A$ & $B$ are carrying currents ${I}_{1}$ and ${I}_{2}$ as shown in the figure. The separation between them is $d$ . A third wire $C$ carrying a current $I$ is to be kept parallel to them at a distance $x$ from $A$ such that the net force acting on it is zero. The possible values of $x$ are:

A$x=\pm \dfrac{{I}_{1}d}{\left({I}_{1}-{I}_{2}\right)}$B$x=\left(\dfrac{{I}_{1}}{{I}_{1}+{I}_{2}}\right)d$ and $x=\dfrac{{I}_{2}}{\left({I}_{1}-{I}_{2}\right)}d$C$x=\left(\dfrac{{I}_{2}}{{I}_{1}+{I}_{2}}\right)d$ and $x=\left(\dfrac{{I}_{2}}{{I}_{1}-{I}_{2}}\right)d$D$x=\left(\dfrac{{I}_{1}}{{I}_{1}-{I}_{2}}\right)d$ and $x=\dfrac{{I}_{2}}{\left({I}_{1}+{I}_{2}\right)}d$

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Answer: (A) $x=\pm \dfrac{{I}_{1}d}{\left({I}_{1}-{I}_{2}\right)}$

As current in $A$ and $B$ are opposite in direction, net force on $C$ can never be zero if it is kept between
$A$ and $B$ .

$∴ \dfrac{{\mu }_{0}{I}_{1}I}{2\pi x}=\dfrac{{\mu }_{0}I {I}_{2}}{2\pi \left(x\pm d\right)}$
${I}_{1}\left(x\pm d\right)={I}_{2} x$
$\left({I}_{2}-{I}_{1}\right)x=\pm {I}_{1} d$
$x=\pm \dfrac{{I}_{1} d}{{I}_{2}-{I}_{1}}$