Two long straight parallel current conductors are kept at a distance d from each other in air. The direction of current in both the conductors is the same. Find the magnitude and direction of the force between them. Hence, define one ampere.

Explanation

Consider two straight parallel long current carrying conductors AB and CD carrying currents, `I_1 and I_2` respectively in same direction and let these be separated by a distance d.
Now magnitude field `B_1` developed at a point Q on 2nd conductor due to current `I_1` flowing in 1st conductur is
`B_1 = (mu_0 I_1)/(2 pi d)`
As per right hand rule `B_2` is acting normal to the plane of the paper pointing inward. Thus, conductor CD carrying current `I_2` is in a magnetic field which is perpendicular to its length. Therefore, force experienced by 2nd conductor CD due to `B_1`
`F_(21) = B_1 I_2, l,`
Where l = length of the 2nd condcutor
or `F_(21) = (mu_0 I_1)/(2 pi d) I_2 l = (mu_0 I_1 I_2 l)/(2 pi d)`
and force per unit length
`(F_(21))/(l) = (mu_0 I_1 I_2)/(2 pi d) = (mu_0)/(4pi) cdot (2 I_1 I_2)/(d)`
The force `F_(21)` in accordance with Fleming.s left hand rule is directed towards the conductor AB. In the same way, it is found that force experienced per unit length of wire AB is
`(F_(21))/(l) = (mu_0)/(4pi) cdot (2 I_1 I_2)/(d)`
and is directed towards CD.