Ans: We are given the following:
Current flowing in wire $A_{1} I_{A}=8.0 \mathrm{~A}$
Current flowing in wire $B, I_{B}=5.0 \mathrm{~A}$
Distance between the two wires, $r=4.0 \mathrm{~cm}=0.04 \mathrm{~m}$
Length of a section of wire $\mathrm{A}_{4} \mathrm{I}=10 \mathrm{~cm}=0.1 \mathrm{~m}$
Force exerted on length I due to the magnetic field could be given as:
$$
\begin{array}{l}
\mathrm{B}=\frac{\mu_{0} 2 I_{\mathrm{A}} \mathrm{I}_{\mathrm{A}} \mathrm{I}}{4 \pi r} \\
\text { Where, } \mu_{0}=\text { Permeability of free space }=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}^{-1} \\
\Rightarrow \mathrm{B}=\frac{4 \pi \times 10^{-7} \times 2 \times 8 \times 5 \times 0.1}{4 \pi \times 0.04} \\
\Rightarrow \mathrm{B}=2 \times 10^{-5} \mathrm{~N}
\end{array}
$$
The magnitude of force is found to be $2 \times 10^{-5} \mathrm{~N}$. This is an attractive force that is normal to A towards B because the direction of the currents in the wires are the same.