26_Chapter 19 HomeworkCH19 Magnetism

26_Chapter 19 HomeworkCH19 Magnetism - 166 CHAPTER 19 19.59...

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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 xx vv == () , and ( ) 22 0 2 0 2 9.80 m s 20.0 m 19.8 m s yy y vva y =+∆ = + = The velocity of the ball is perpendicular to the magnetic field and, just before it reaches the ground, has magnitude 28.1 m s xy v =+ = . Thus, the magnitude of the magnetic force is 6 6 sin 5.00 10 C 90.0 1.41 10 N m Fq v B 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 Bkh = , where k is a constant and
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This document was uploaded on 10/20/2011.

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