Explanation

(a) Force on side $BC$ and $AD$ are equal but opposite so their net will be zero
But $F_{AB} = 10^{- 7} \times \frac{2 \times 2 \times 1}{2 \times 10^{- 2}} \times 15 \times 10^{- 2} = 3 \times 10^{- 6}N$
and $F_{CD} = 10^{- 7} \times \frac{2 \times 2 \times 1}{(12 \times 10^{- 2})} \times 15 \times 10^{- 2} = 0.5 \times 10^{- 6}N$
$\Rightarrow F_{net} = F_{AB} - F_{CD} = 2.5 \times 10^{- 6}N$
$= 25 \times 10^{- 7}N$, towards the wire