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

Let `I_(1) " and " I_(2)` be the currents passing through the two long and parallel straight wires A and B, respectively If the wires are separated by a distance and length of each wire is l , then the magnitude of force acting on the wires due to each other is given by:
` F = mu_(0)/(4pi) (2I_(1)I_(2)l)/r`
Here, `I_(1) = 8.0 A, I_(2) = 5.0 A`
` r = 4.0 cm = 0.04 m`
Here,
`I_(1) = 8.0 A, I_(2) = 5.0 A`
` r = 4.0 cm = 0.04 m`
` l = 10 cm = 0.04 m`
` l = 10 cm = 0.1 m`
` :. " " F = mu_(0)/(4pi) (2 xx 8 xx 5 xx 0.1)/(0.04) `
Thus, the force on a 10 cm section of wire A will be
` F = 2 xx 10^(-5) N`
To find the direction of magnetic force on the wire, we will apply Fleming.s left-hand rule. So we find that the current in the two parallel wires is flowing in the same direction, so the force will be normal to the wire A towards B.