A bar of mass `m` slides on a smooth horizontal conducting rail. A constant current `I` is maintained in the rod using a constant current generator as shown. There is a long straight conductor carrying current `I_(0)` parallel to the rod. Find the speed of the bar after it has travelled a distance `D`

Answer

`upsilon=sqrt((mu_(0)II_(0)Llog_(e)(1+(D)/(d)))/(pi m))`

Explanation

The magnetic force acting on the sliding bar when tis is at a distance x from the straight long conductor is given as,

`F=(mu_(0)II_(0)L)/(2pix)implies(m upsilond upsilon)/(dx)=(mu_(0)II_(0)L)/(2pix)`
`impliesupsilond upsilon =(mu_(0)II_(0)L)/(2m)(dx)/(x)`
`implies int_(0)^(upsilon)upsilon d upsilon=(mu_(0)II_(0)L)/(2pim)int_(0)^(d+D)(dx)/(x)`
`=(upsilon^(2))/(2)=(mu_(0)II_(0)L)/(2pim)log_(e)(1+(D)/(d))`
`implies upsilon=sqrt((mu_(0)II_(0)Llog_(e)(1+(D)/(d)))/(pim))`