Two long thin parallel conductors of the shape shown in Fig. carry direct currents `I_(1)` and width of the right-hand conductor is equal to `b`. With both conductors lying in one plane, find the magnetic interaction force between them reduced to a unit of their length.

Explanation

We know that Ampere's force per unit length on a wire element in a magnetic field is given by
`d vec(F_(n)) = I (hat(n) xx hat(B))` where `hat(n)` is the unit vector along the direction of current...(1)
Now, let us take an element of the conductor `i_(2)` as shwon in the fig. The wire element is in the magnetic field, produced by the current `i_(1)` which is directed normally into the sheet of the paper and its magnitude is given by,
`|vec(B)| = (mu_(0) l_(1))/(2pi r)` ....(2)
From Eqs. (1) and (2)
`d vec(F_(n)) = (I_(2))/(b) dr (hat(n) xx vec(B))`. (becuase the current through the element equals `(l_(2))/(b) dr`)
So, `d vec(F_(n)) = (mu_(0))/(2pi) (l_(1) l_(2))/(b) (dr)/(r)`, towards left (as `hat(n) _|_^(r) vec(B)`)
Hence the magnetic force on the conductor:
`vec(F_(n)) = (mu_(0))/(2pi) (I_(1) I_(2))/(b) int_(a)^(a + b) (dr)/(r)` (towards left) `= (mu_(0) I_(1) I_(2))/(2pi b) ln (a + b)/(a)`