$A, B$ and $C$ are parallel conductors of equal length carrying currents $I, I$ and $2I$ respectively. Distance between $A$ and $B$ is $x$. Distance between $B$ and $C$ is also $x$. ${F_1}$ is the force exerted by $B$ on $A$ and $F_2$ is the force exerted by $B$ on $A$ choose the correct answer
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(d)$F = \frac{{{\mu _0}}}{{4\pi }}\frac{{2{i_1}{i_2}}}{a}$
${F_1} = \frac{{{\mu _0}}}{{4\pi }} = \frac{{2{i^2}}}{x}$ (Attraction)
${F_2} = \frac{{{\mu _0}}}{{4\pi }} = \frac{{2i \times 2i}}{{2x}} = \frac{{{\mu _0}}}{{4\pi }}\frac{{2{i^2}}}{x}$(Repulsion)
Thus ${F_1} = - {F_2}$
${F_1} = \frac{{{\mu _0}}}{{4\pi }} = \frac{{2{i^2}}}{x}$ (Attraction)
${F_2} = \frac{{{\mu _0}}}{{4\pi }} = \frac{{2i \times 2i}}{{2x}} = \frac{{{\mu _0}}}{{4\pi }}\frac{{2{i^2}}}{x}$(Repulsion)
Thus ${F_1} = - {F_2}$
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