A wire X of length 50cm carrying a current of 2A is placed parallel...

An attractive force will act normal to $X$ towards $Y$ because the direction of the currents in the wires is the same.

Force exerted on length $l$ due to the magnetic field is given as $F = {F}_{XY} = {F}_{YX} = {I}_{1}{l}_{1}{B}_{12}$

$⇒ F = \dfrac{{\mu}_{0}{I}_{1}{I}_{2}}{2\pir}{l}_{1}$

Given here, ${I}_{1} = 2A, {I}_{2} = 3A, {l}_{1} = 5m, r = 5cm$.

Putting the values, we have

$F = \dfrac{4\pi \times {10}^{- 7} \times 2 \times 3}{2\pi \times 5 \times {10}^{- 2}} \times 50 \times {10}^{- 2}$

$= 1 . 2 \times {10}^{- 5}N$ directed towards wire $X$.