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(C) The force per unit length between two parallel conductors carrying currents $I_1$ and $I_2$ separated by distance $d$ is given by $F = \frac{\mu_0

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$-\frac{2F}{3}$

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(C) The force per unit length between two parallel conductors carrying currents $I_1$ and $I_2$ separated by distance $d$ is given by $F = \frac{\mu_0 I_1 I_2}{2 \pi d}$.Since the currents are in the same direction,the force is attractive. Let the initial force be $F = k \frac{I_1 I_2}{d}$,where $k = \frac{\mu_0}{2 \pi}$.In the new scenario,the current in one conductor becomes $2I_1$ (or $2I_2$) and the direction is reversed,making the force repulsive (negative sign).The new distance is $d' = 3d$.The new force $F'$ is given by $F' = -k \frac{(2I_1) I_2}{3d} = -\frac{2}{3} \left( k \frac{I_1 I_2}{d} \right) = -\frac{2}{3} F$.

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