24_Ch 19 College Physics ProblemCH19 Magnetism

24_Ch 19 College Physics ProblemCH19 Magnetism - 164...

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164 CHAPTER 19 19.54 For the rail to move at constant velocity, the net force acting on it must be zero. Thus, the magnitude of the magnetic force must equal that of the friction force giving ( ) k BIL mg µ = , or ( ) ( )( ) ( ) () ( ) 2 2 0.100 0.200 kg 9.80 m s 3.92 10 T 10.0 A 0.500 m mg IL == = × k B 19.55 The magnetic force acting on each type particle supplies the centripetal acceleration for that particle. Thus, 2 qvB mv r = or rm v q B = . After completing one half of the circular paths, the two types of particle are separated by the difference in the diameters of the two paths. Therefore, 21 2 1 5 27 19 2 2 2 .00 10 ms 23.4 20.0 10 kg 1.60 10 C 0.200 T 2.13 10 m 2.13 cm v dr r mm qB ∆= = × =−
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