831_Physics ProblemsTechnical Physics

831_Physics ProblemsTechnical Physics - Chapter 29 P29.33...

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Chapter 29 171 P29.33 qV m v a f = 1 2 2 or v m = 2 af . Also, qvB mv r = 2 so r mv qB m qB m mV qB == = 22 2 ∆∆ . Therefore, r eB p p 2 2 2 = r qB eB eB r d d d p p p 2 2 2 2 2 2 2 = F H G I K J = afej a f and r eB eB r p p p α 2 222 2 2 24 2 2 2 2 = F H G I K J = a f . The conclusion is: rr r dp 2. P29.34 (a) We begin with qvB mv R = 2 or qRB mv = . But Lm v Rq RB 2 . Therefore, R L qB ×⋅ ×× −− 400 10 100 10 0050 500 25 19 3 . . .. Js 1.60 10 C T 0 m cm ej e j . (b) Thus, v L mR × 10 0 050 0 878 10 25 31 6 . . . 9.11 kg m ms bg . P29.35 ω × × qB m 160 10 520 167 10 498 10 19 27 8 . . C T kg rad s P29.36 1 2 2 mv q V =∆ so v m = 2 r mv qB = so r mqVm qB = 2 r m q V B 2 2 2 =⋅ and ′= r m q V B a f 2 2 2
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