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 perpendicular 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 wire and the floor is$\mu$. A current $i$ is established when the switch $S$ is closed at the instant $t = 0$. How far away from the strips will the wire reach?

 

When current starts flowing in the wire, it experiences a force,$F = Bib$ of constant magnitude. Due to which it accelerates on the metal strips. Thereafter when wire falls on to the floor, it retards due to friction and finally stops. Thus

                            $a=\frac{F}{m}=\frac{Bib}{m}$

The velocity gained by the wire on the strips

                            $v^2=0+2al=2\frac{Bibl}{m}$

Let $x$ be the distance moved by the wire on the floor, its final velocity becomes zero, and so

                         $0=2v^2 – a\text{'}x \Rightarrow x=\frac{v^2}{2a\text{'}}$

Here $a^{\text{'}}=\frac{mg\mu }{m}=\mu g \Rightarrow x=\frac{Bibl}{\mu mg}$