A metal wire PQ having a mass of 10 g rests on two horizontal metal rails separated by 4.90 cm . A vertically downward magnetic field with a magnitude of 0.800 T is present in the space. As the resistance of the circuit is slowly decreased, it is found that the wire PQ begins sliding on the rails once the resistance drops below 20.0 . Determine the coefficient of friction.

Step-by-step solution

When the resistance drops to 20.0 , the current in the circuit is: I = V R = 6 20.0 = 0.3 A The magnetic force acting on the wire PQ is given by: F_ m = I L B Substituting the given values ( L = 4.90 cm = 4.90 10⁻² m , B = 0.800 T ): F_ m = 0.3 (4.90 10⁻²) 0.800 = 0.01176 N The normal force on the wire is equal to its weight (taking g = 9.8 m/s ^2 ): N = mg = (10 10⁻³ kg ) 9.8 m/s ^2 = 0.098 N The wire begins to slide when the magnetic force equals the maximum static friction force: F_ m = N = F_ m N = 0.01176 0.09