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

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towards $A$

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(A) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $d$ is given by $f = \frac{\mu_0 I_1 I_2}{2 \pi d}$.Parallel currents attract each other,while anti-parallel currents repel each other.Wire $A$ carries $1 	ext{ A}$ downwards,wire $B$ carries $2 	ext{ A}$ downwards,and wire $C$ carries $3 	ext{ A}$ upwards.Force on $B$ due to $A$ $(F_{BA})$: Since currents in $A$ and $B$ are in the same direction,they attract. Thus,$F_{BA}$ is directed towards $A$.Magnitude $F_{BA} \propto (1 	ext{ A} 	imes 2 	ext{ A}) = 2$.Force on $B$ due to $C$ $(F_{BC})$: Since currents in $B$ and $C$ are in opposite directions,they repel. Thus,$F_{BC}$ is directed away from $C$,which is also towards $A$.Magnitude $F_{BC} \propto (2 	ext{ A} 	imes 3 	ext{ A}) = 6$.Since both forces $F_{BA}$ and $F_{BC}$ are directed towards $A$,the resultant force is directed towards $A$.

Three infinite straight wires $A, B$ and $C$ carry currents as shown in the figure. The resultant force on wire $B$ is directed:

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