14_Chapter 19 HomeworkCH19 Magnetism

14_Chapter 19 HomeworkCH19 Magnetism - 154 CHAPTER 19 19.28...

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154 CHAPTER 19 19.28 Since the path is circular, the particle moves perpendicular to the magnetic field, and the magnetic force supplies the centripetal acceleration. Hence, 2 v mq v r = B , or mv qr = B . But the momentum is given by () 2 p mv m KE == , and the kinetic energy of this proton is 19 61 1.60 10 J 10.0 10 eV 1.60 10 J 1 eV KE  ×   2 . We then have ( ) 27 12 12 19 10 2 1.67 10 kg 1.60 10 J 2 7.88 10 T 1.60 10 C 5.80 10 m mKE B qr −− ×× = × 19.29 For the particle to pass through with no deflection, the net force acting on it must be zero. Thus, the magnetic force and the electric force must be in opposite directions and have equal magnitudes. This gives , or which reduces to m FF = e qvB qE = vE B = 19.30 The speed of the particles emerging from the velocity selector is B = (see Problem 29). In the deflection chamber, the magnetic force supplies the centripetal acceleration,
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This document was uploaded on 10/20/2011.

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