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(B) The force per unit length between two parallel current-carrying wires is given by $F = \frac{\mu_0 I_1 I_2}{2 \pi r}$.For wire $A$ and $B$: Both c

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$F_1=-F_2$

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(B) The force per unit length between two parallel current-carrying wires is given by $F = \frac{\mu_0 I_1 I_2}{2 \pi r}$.For wire $A$ and $B$: Both carry current $I$ in the same direction. The force $F_1$ exerted by $B$ on $A$ is attractive,with magnitude $F_1 = \frac{\mu_0 I^2}{2 \pi d}$.For wire $A$ and $C$: Wire $A$ carries current $I$ upwards and wire $C$ carries current $2I$ downwards. Since the currents are in opposite directions,the force $F_2$ exerted by $C$ on $A$ is repulsive,with magnitude $F_2 = \frac{\mu_0 I (2I)}{2 \pi (2d)} = \frac{\mu_0 I^2}{2 \pi d}$.Since the forces $F_1$ and $F_2$ have equal magnitudes but act in opposite directions (one is attractive and the other is repulsive),we have $F_1 = -F_2$.

$A, B$ and $C$ are three parallel conductors of equal lengths and carry currents $I, I$ and $2I$ respectively as shown in the figure. The distance between $A$ and $B$ and between $B$ and $C$ is $d$. If $F_1$ is the force exerted by $B$ on $A$ and $F_2$ is the force exerted by $C$ on $A$,then:

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