Question

Two metal strips, each of length l, are elamped parallel to each other on a horizontal floor with a separation b between them. A wire of mass m lies on them perpendicularly as shown in figure (34 - E8). A vertically upward magnetic field of strength B exists in the space. The metal strips are smooth but the coefficient of friction 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

Let 'F' be the force applied due to magnetic field on the wire and 'x' be the covered distance.

So, $F\times l=\mu mg\times x$

$\Rightarrow ibBl=\mu mgx$

$\Rightarrow x=\frac{ilBl}{\mu mg}$