Three infinitely long thin wires each carrying current `I` in the same direction are in `x-y` plane of a gravity free space. The central wire is along the `y`-axis while the other two are along `x=+-d`. If the central wire is slightly displaced along `z`-axis and released, show that it will execute `S.H.M`. The linear mass density of the wire is `lambda`. If the time period of this small oscillation is `(kpid)/(I)sqrt((pilambda)/(mu_(0)))`, then find the value of `k`.

Explanation

Let the central wire is displaced along `z`-axis by a small distance `z` and released.
Net restoring force on the central wire
`F=-2F_(1) costheta` (`F_(1)=` is the magnitude of force on central wire due to either of the other two wires)
`=-2(mu_(0)I^(2))/(2pir)(z)/(r )l`
`=-(mu_(0)I^(2)z)/(pi(d^(2)+z^(2)))l`
`=-(mu_(0)I^(2)z)/(pid^(2))l` (Since `z lt lt drArrz^(2)+d^(2)~~d^(2))`
Acceleration of the central wire
`a=(F)/(lambdal)=-(mu_(0)I^(2))/(pilambdad^(2))z`
comparing this equation with equation of SHM
`a=- omega^(2)x`
`rArr omega=sqrt((mu_(0))/(pilambda))(I)/(d)`
`rArr T=(2pi)/(omega)=(2pid)/(I)sqrt((pilambda)/(mu_(0)))`