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(B) The force per unit length between two parallel current-carrying wires is given by $f = \frac{\mu_0 i_1 i_2}{2\pi r}$.Since the currents in all thr

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Towards $C$

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(B) The force per unit length between two parallel current-carrying wires is given by $f = \frac{\mu_0 i_1 i_2}{2\pi r}$.Since the currents in all three wires are in the same direction,the force between any two wires is attractive.Let $i_A = 1 	ext{ A}$,$i_B = 2 	ext{ A}$,and $i_C = 3 	ext{ A}$. The distance between adjacent wires is $d$.Force on $B$ due to $A$ $(F_{BA})$ is attractive,so it acts towards $A$: $F_{BA} = \frac{\mu_0 i_A i_B}{2\pi d} = \frac{\mu_0 (1)(2)}{2\pi d} = \frac{\mu_0}{\pi d}$.Force on $B$ due to $C$ $(F_{BC})$ is attractive,so it acts towards $C$: $F_{BC} = \frac{\mu_0 i_B i_C}{2\pi d} = \frac{\mu_0 (2)(3)}{2\pi d} = \frac{3\mu_0}{\pi d}$.Since $F_{BC} > F_{BA}$,the net force on $B$ is $F_{net} = F_{BC} - F_{BA} = \frac{2\mu_0}{\pi d}$ directed towards $C$.

Three long straight wires $A$,$B$,and $C$ are carrying currents as shown in the figure. Then the resultant force on $B$ is directed .....

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