Question

Two metal strips, 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 lies on them perpendicularly as shown in fig. 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 from the strips will the wire reach?

Answer 1

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: $a=\frac{F_b}{m}=\frac{IBb}{m}$
Velocity gained by wire till it reaches the end of rails:
$v^2=2al=\frac{2IBbl}{m}$
After this, let wire slips a distance s on the floor:

Work done by friction force= Change in kinetic energy

$\Rightarrow \mu mgs=\frac{1}{2}m{v}^2$

$\Rightarrow \mu mgs=IBbl$

$\Rightarrow s=\frac{IBbl}{\mu mg}$