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Consider three infinitely long current carrying wires 1, 2, and 3 kept parallel to each other as shown
Given $i_{1} = i_{2} = i_{3}$
If $F_{1}$ = Net magnetic force per length acting on wire $i_{1}$ due to $i_{2}$ and $i_{3}$
$F_{2}$ = Net magnetic force per length acting on wire $i_{2}$ due to $i_{1}$ and $i_{3}$
$F_{3}$ = Net magnetic force per length acting on wire $i_{3}$ due to $i_{1}$ and $i_{2}$
Then

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$F_{3} = F_{2} > F_{1}$

$F_{2} > F_{3} > F_{1}$

$F_{3} > F_{2} > F_{1}$

$F_{2} > F_{1} > F_{3}$

answer is A.

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Detailed Solution

$\frac{F}{l} = \frac{μ_{0} i_{1} i_{2}}{2 π d}$

Using the concept, parallel currents attract and opposite currents repel,
Force acting on wire $i_{2}$ will be highest followed by $i_{3}$ and then $i_{1}$

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