Three very long straight current carrying conductors are placed parallel to each other as shown in. The conductors A and C are fixed whereas conductors B is free to move. If the conductor B is pulled towards right through a very small distance r, then what is the net force acting on it and the angular frequency of the resulting oscillation.

Options

A. `(mu_02II_0)/(pid^2)`, `sqrt((mu_0II_0)/(pimd^2))`
B. `(mu_02II_0)/(4pir^2)`, `sqrt((mu_0II_0)/(2pimd^2))`
C. `(mu_0II_0)/(2pir^2)`, `sqrt((mu_0II_0)/(4pimd))`
D. `(mu_0II_0)/(2pi)[((d+r)-(d-r))/(d^2-r^2)]`, `sqrt((mu_0II_0)/(pimd^2))`

Explanation

The net force acting on the conductor after it gets displaced through a very small distance r is given by

`F=|F_1-F_2|=|(mu_0II_0)/(2pi(d-r))-(mu_0II_0)/(2pi(d+r))|`
`=(mu_0II_0)/(2pi)[((d+r)-(d-r))/(d^2-r^2)]`
As it is given that `r lt lt d`, so `F=(mu_0II_0r)/(pid^2)`
Now `mromega^2=(mu_0II_0r)/(pid^2)impliesomega^2=(mu_0II_0)/(pimd^2)`
`omega=sqrt((mu_0II_0)/(pimd^2))`