Two metal strips, each of length l , are clamped parallel to one...

While the wire is on the metal strips, it experiences a magnetic force due to the current flowing through it. The magnitude of this magnetic force is given by:

$F_m = i b B$

Since the metal strips are smooth, there is no frictional force acting on the wire during this part of the motion. The work done by the magnetic force as the wire moves along the entire length $l$ of the strips is converted into its kinetic energy:

$W_m = F_m \times l = i b B l$

Let $v$ be the velocity of the wire just as it leaves the metal strips. Its kinetic energy is:

$K = i b B l$

Once the wire leaves the strips, it moves on the horizontal floor where the magnetic force is no longer present (as the circuit is broken), but it experiences a kinetic frictional force. The frictional force is:

$f_k = \mu m g$

Let $x$ be the distance the wire travels on the floor before coming to a halt. The work done by friction will dissipate the entire kinetic energy of the wire. By the work-energy theorem:

$W_f = \Delta K$

$-f_k \times x = 0 - K$

$\mu m g x = i b B l$

Solving for $x$, we get the distance the wire reaches away from the strips:

$x = \dfrac{i l b B}{\mu m g}$