Two long and parallel straight wires A and B carrying currents of `8*0A` and `5*0A` in the same direction are separated by a distance of `4*0cm`. Estimate the force on a `10cm` section of wire A.

Explanation

Current flowing in wire `A,I_(A) = 8.0A`
Current flowing in wire `B,I_(B) = 5.0A`
Distance between the two wires, `r = 4.0 cm = 0.04cm`
Length of a section of wire `A,l =10 cm = 0.1cm`
Force exerted on length l due to the magnetic field is given as:
`B = (mu_(0)2l_(A)l_(B)l)/(4pir)`
Where,
`mu_(0) =` Permeability of free space `= 4n xx 10^(-7) TmA^(-1)`
`B =(4pi xx 10^(-7) xx 2xx 8 xx 5 xx 0.1)/(4pi xx 0.04)`
`=2 xx 10^(-5)N`
The magnitude of force is `2 xx 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.