Two infinite parallel wire, having the cross sectional area `'a'` and resistivity `'k'` are connected at a junction point `'p'` (as shown in the figure). A slide wire of negligible resistance and having mass `'m'` and length `'l'` can slide between the parallel wires, without any frictional resistance. If the system of wires is introduced to a magnetic field of intensity `'B'` (into the plane of paper) and the slide wire is pulled with a force which varies with the velocity of the slide wire as `F=F_(0)v`, then find the velocity of the slide wire as a function of the distance travelled.
(The slide wire is initially at origin and has a velocity `v_(0)`)

Explanation

At any given instance of time the slide wire is at distance `x` from origin, then the resistance of the circuit is `R=(k(2x+l))/(a)`
If the velocity of slide wire is `v`, then the emf generated is `Blv` so we have
`BlV-(k)/(a)(2x+l)I=0`
Or, `I=(Bla)/(k)((v)/(2x+l))`
This current exerts magnetic force in the wire given by `F'`
So, `F'=IlB=(Bla)/(k)((v)/(2x+l))lB=(B^(2)l^(2)a)/(k)((v)/(2x+l))`
Since, `F-F'=(mdv)/(dt)`
So `F_(0)v-(B^(2)l^(2)a)/(k)((v)/(2x+l))=mv(dv)/(dx)`
`rArr (F_(0))/(m)-(B^(2)l^(2)a)/(km)((1)/(2x+l))=(dv)/(dx)`
`rArr int_(0)^(x)[(F_(0))/(m)-(B^(2)l^(2)a)/(km)((1)/(2x+l))]dx=int_(V_(0))^(V)dv`
`(F_(0))/(m)x-(B^(2)l^(2)a)/(2km){ln((2x+l)/(l))}+v_(0)=v`