Two long and parallel straight wires A and B carrying currents

Current flowing in wire A, IA = 8.0 A

Current flowing in wire B, IB = 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:

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

Where,

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

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

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

The magnitude of force is 2 × 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.