Question

A metal wire PQ of mass 10 g lies at rest on two horizontal metal rails separated by 4.90 cm (figure). A vertically-downward magnetic field of magnitude 0.800 T exists in the space. The resistance of the circuit is slowly decreased and it is found that when the resistance goes below 20.0 Ω, the wire PQ starts sliding on the rails. Find the coefficient of friction.

Answer 1

Given:
Mass of the metal wire, M = 10 g
Distance between the two horizontal metal rails, l = 4.90 cm
Vertically-downward magnetic field, B = 0.800 T
As per the question, when the resistance of the circuit is slowly decreased below 20.0 Ω, the wire PQ starts sliding on the rails. At that moment,
current in the wire, i =$\frac{V}{R}=\frac{6}{20}\mathrm{A}$
Using Fleming's left-hand rule, the magnetic force will act towards the right. So, due to this magnetic force, the wire will try to slide on the rails.The frictional force will try to oppose this motion of the wire.When the wire just starts sliding on the rails, the frictional force acting on the wire will just balance the magnetic force acting on the wire.This implies
µR = F, where
µ is the coffiecent of friction
R is the normal reaction force and
F is the magnetic force
⇒ µ × M × g = ilB
µ × 10 × 10⁻³ × 9.8 = $\frac{6}{20}$ × 4.9 × 10⁻² × 0.8
$\mu =\frac{6\times 4.9\times {10}^{-2}\times 0.8}{20\times 10\times {10}^{-3}\times 9.8}$
µ = 0.12