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Physics Question on Moving charges and magnetism

Two wires AA & BB are carrying currents I1I_1 & I2I_2 as shown in the figure. The separation between them is dd. A third wire CC carrying a current II is to be kept parallel to them at a distance xx from A such that the net force acting on it is zero. The possible values of xx are :

A

x=(I1I1I2)dx=\left(\frac{I_{1}}{I_{1} -I_{2}}\right)d and x=I2(I1+I2)dx = \frac{I_{2}}{\left(I_{1} + I_{2}\right)}d

B

x=±I1d(I1+I2)x = \pm \frac{I_{1} d}{\left(I_{1} + I_{2}\right)}

C

x=(I1I1+I2)dx=\left(\frac{I_{1}}{I_{1} + I_{2}}\right)d and x=I2(I1I2)dx = \frac{I_{2}}{\left(I_{1} - I_{2}\right)}d

D

x=(I1I1+I2)dx=\left(\frac{I_{1}}{I_{1} + I_{2}}\right)d and x=(I2I1+I2)dx = ( \frac{I_{2}}{I_{1} + I_{2}})d

Answer

x=±I1d(I1+I2)x = \pm \frac{I_{1} d}{\left(I_{1} + I_{2}\right)}

Explanation

Solution

Net force on wire carrying current I per unit length is
μ0I1I2πx+μ0I2I2x(dx)=0\frac{\mu_{0} I_{1} I}{2\pi x} + \frac{\mu_{0} I_{2}I }{2x\left(d-x\right)} = 0
I1x=I2xd\frac{I_{1}}{x} = \frac{I_{2}}{x-d}
x=I1dI1I2\Rightarrow x = \frac{I_{1}d}{I_{1} -I_{2}}