Question

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?

Answer 1

Correct Answer - `(IBbl)/(mu mg)`
When the switch is closed, current flows in the wire. Due to this current, magnetic force acts on wire towards right. The acceleration of wire : `a=(F_(b))/(m)=(IBb)/(m)`
Velocity gained by wire till it reaches the end of rails:
`v^2=2al=(2IBbl)/m`
After this, let wire slips a distance s on the floor
`mumgs=1/2mv^2 implies mumgs=IBbl implies s=(IBbl)/(mumg)`