Current flowing in wire $\mathrm{B}, \mathrm{I}_{\mathrm{B}}=5.0 \mathrm{A}$

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

Length of a section of wire $A, I=10 \mathrm{cm}=0.1 \mathrm{m}$

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

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

Where,

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

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

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

The magnitude of force is $2 \times 10^{-5} \mathrm{N}$. This is an attractive force normal to A towards

B because the direction of the currents in the wires is the same.