# Two wires A & B are carrying currents $I_{1}$ & $I_{2}$ as shown in the figure.The separation between them is d.A third wire C carrying a currentI 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: Option 1) $x=(\frac{I_{1}}{I_{1}-I_{2}})d$   and     $x=(\frac{I_{2}}{I_{1}+I_{2}})d$ Option 2) $x=(\frac{I_{2}}{I_{1}+I_{2}})d$ and  $x=(\frac{I_{2}}{I_{1}-I_{2}})d$ Option 3) $x=(\frac{I_{1}}{I_{1}+I_{2}})d$ and   $x=(\frac{I_{2}}{I_{1}-I_{2}})d$ Option 4) $x=\pm (\frac{I_{1}}{I_{1}-I_{2}})d$

Force between two parallel current carrying conductors -

$F=\frac{\mu }{4\pi } \frac{2I_{1 I_{2}}}{a} l$

$\frac{F}{l}=\frac{\mu o}{4\pi } \frac{2I_{1 I_{2}}}{a}$

- wherein

I1 and I2 current carrying two parallel wires

a-seperation between two wires

when $I_2>I_1$ ( that is wire 3 is on left side of both  1&2)

For equilibrium,

$F_{31}=F_{32}$

$\frac{\mu _oI_1Il}{2\pi x}=\frac{\mu _oI_2Il}{2\pi (d+x)}$

$\frac{d+x}{x}=\frac{I_2}{I_1}$

=> $x=(\frac{I_1}{I_2-I_1})d$........................(1)

when $I_1>I_2$ ( that is wire 3 is on right side of both  1&2)

$F_{31}=F_{32}$

$\frac{\mu _oI_1Il}{2\pi x}=\frac{\mu _oI_2Il}{2\pi (x-d)}$

$\frac{x-d}{x}=\frac{I_2}{I_1}$

=>$x=(\frac{-I_1}{I_2-I_1})d$   .....................(2)

from (1) & (2)

$x=\pm (\frac{I_1d}{I_1-I_2})$

Option 1)

$x=(\frac{I_{1}}{I_{1}-I_{2}})d$   and     $x=(\frac{I_{2}}{I_{1}+I_{2}})d$

Option 2)

$x=(\frac{I_{2}}{I_{1}+I_{2}})d$ and  $x=(\frac{I_{2}}{I_{1}-I_{2}})d$

Option 3)

$x=(\frac{I_{1}}{I_{1}+I_{2}})d$ and   $x=(\frac{I_{2}}{I_{1}-I_{2}})d$

Option 4)

$x=\pm (\frac{I_{1}}{I_{1}-I_{2}})d$

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