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?

Two metal strips, each of length, each of length l, are clamped parall

1.	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» 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)`

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