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.
Figure

Answer 1

0.12 is the coefficient of friction.

Explanation:

Given:

       Metal wire’s mass, M = 10 g

       Between the two horizontal metal rails, the distance is l = 4.90 cm

       Magnetic field that is vertically-downward, B = 0.800 T

• According to the information provided, when the circuit’s resistance is decreased slowly below 20.0 Ω, the wire PQ begins descending on the railings. At that instant, the wire’s  current, i = VR = 620A
• Towards the right, the magnetic force will act using Fleming’s left-hand rule. Thus, the wire will attempt to slide on the rails due to the magnetic force.
• This motion of the wire will be opposed by the frictional force. When the wire just begins to slide on the railings, there will just be the balance of frictional force and the magnetic force acting on the wire. This suggests

µR = F, where

µ is the friction’s coefficient

R is the standard reaction force and

F is the magnetic force

⇒ µ $\times$ M $\times$ g = ilB

µ $\times$ 10 $\times$ 9.8 $\times$ 10−3  

μ = 0.12