Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.
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.