Two metal strips, each of length, each of length l, are clamped parallel to each other on a horizontal floor with a separation b between them. A wire of mass m line on them perpendicularly as shown in figure. A vertically upward magnetic field of strenght B exists in the space. The matal strips are smooth but the cofficient of freiction between the wire and the floor is `mu`. A current i is established when the switch S is closed at the instant `t=0`. Discuss the motion of the wire after the switch is closed. How far away from the strips will the wire reach?

Explanation

When current starts flowing in the wire, it experiences a force, F =Bib of constant magnitude due to which it accelerates on the metal strips. There after when wire falls on the floor, it retards due to friction and finally stops. Thus
`a = (F)/(m) = (Bib)/(m)`
The velocity gained by the wire on the strips
`v^2 = 0 + 2al = 2 (Bib l)/(m)`
Let x be the distance moved by the wire on the floor.
Its final velocity becomes zero, and so
`= v^2 -2a.x or x = (v^2)/(2a.)`
Here ` a. = (mu mg)/(m) = mu g`
`therefore x = ([(2Bib l)/(m)])/(2mug) = (Bib l)/(mu mg)`
Note : Try usiing work - energy theorem