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26_Ch 19 College Physics ProblemCH19 Magnetism

# 26_Ch 19 College Physics ProblemCH19 Magnetism - 166...

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166 CHAPTER 19 19.59 The 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 0 20.0 m s x x v v = = ( ) , and ( ) ( ) 2 2 0 2 0 2 9.80 m s 20.0 m 19.8 m s y y y v v a y = + = + = The velocity of the ball is perpendicular to the magnetic field and, just before it reaches the ground, has magnitude 2 2 28.1 m s x y v = + = ( ) v v . Thus, the magnitude of the magnetic force is ( )( ) 6 6 sin 5.00 10 C 90.0 1.41 10 N m F qvB 28.1 m s 0.010 0 T sin θ = = × ° = × 19.60 We are given that the field at points on the axis of the disk varies as 3 B k h = , where k is a constant and h is the distance from the midplane of the disk. At the surface of the disk,
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