166CHAPTER 19 19.59The magnetic force is very small in comparison to the weight of the ball, so we treat the motion as that of a freely falling body. Then, as the ball approaches the ground, it has velocity components with magnitudes of 020.0 m sxxvv==(), and ()()2202029.80 m s20.0 m19.8 m syyyvvay=+∆=+−−=The velocity of the ball is perpendicular to the magnetic field and, just before it reaches the ground, has magnitude 2228.1 m sxyv=+=()vv. Thus, the magnitude of the magnetic force is ()()66sin5.0010C90.01.4110NmFqvB28.1 m s0.010 0 Tsinθ−−==×° =×19.60We are given that the field at points on the axis of the disk varies as 3Bk h=, where kis a constant and his the distance from the midplane of the disk. At the surface of the disk,
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