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 the figure. 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 µ. 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

Given:
Length of the two metal strips = l
Separation between the strips = b
Mass of the wire = m
Strength magnetic field = B
Coefficient of friction between the wire and the floor = µ
Let the distance covered by the wire be x.
Due to the presence of the magnetic field, a net magnetic force will act on the wire towards the right.

Let the wire has moved through a distance x after travelling the length of the strips.
As the contact between the wire and strip is smooth so coefficient of friction between them is zero. Under the influence of magnetic force,firstly the wire will travel a distance equal to the length of the strips. After this, it travels a distance x and also now,a frictional force will act on the wire in a direction opposite to its direction of motion.
So we can equate the work done by the magnetic force and the frictional force.
Thus, F × l = µmg × x,
where g is the acceleration due to gravity
⇒ ibBl = µmgx
$\Rightarrow x=\frac{ibl\mathrm{B}}{\mathrm{\mu }m\mathrm{g}}$