164CHAPTER 19 19.54For 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 ()kBILmgµ=, or ()()()()()()220.1000.200 kg9.80 m s3.9210T10.0 A0.500 mmgIL−===×kBµ19.55The magnetic force acting on each type particle supplies the centripetal acceleration for that particle. Thus, 2qvBmvr=or rmv qB=. 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, ()()()()()()2121527192222 1.0010 m s23.420.010kg1.6010C0.200 T2.1310m2.13 cmvdrrmmqB−−−∆=−=−×=−×
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