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 `Bl V` so we have
`BlV - (k)/(a) (2x+l)l=0`
Or, `I = (Bla)/(k) ((V)/(2x+l))`
This current exerts magnetic force in the wire given by F
So F = `I lB = (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)`
`(F_(0))/(m) -(B^(2)l^(2)a)/(km) ((1)/(2x+l))= (dV)/(dx)`
`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){"In"(2x+l)/(l)}+v_(0)=V`