Current flowing in wire $B , I_{B}=5.0 \,A$

Distance between the two wires, $r=4.0 \,cm =0.04 \,m$

Length of a section of wire A, $l=10\, cm =0.1 \,m$

Force exerted on length $l$ due to the magnetic field is given as:

$B=\frac{\mu_{0} 2 I_{A} I_{B} l}{4 \pi r}$

Where,

$\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-7} \,T\,m\,A ^{-1}$

$B=\frac{4 \pi \times 10^{-7} \times 2 \times 8 \times 5 \times 0.1}{4 \pi \times 0.04}$

$=2 \times 10^{-5} \;N$

The magnitude of force is $2 \times 10^{-5}\,N$. This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.