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48. (a) Current is the transport of charge; here it is being transported “in bulk” due to the volume rate of flow of the powder. From Chapter 14, we recall that the volume rate of flow is the product of the cross-sectional area (of the stream) and the (average) stream velocity. Thus, i = ρ Av where is the charge per unit volume. If the cross-section is that of a circle, then i = π R 2 v . (b) Recalling that a Coulomb per second is an Ampere, we obtain i =× = × −− 11 10 2 0 17 10 33 2 5 .. . C/m m m/s A. c h b g b g π 0.050 (c) The motion of charge is not in the same direction as the potential difference computed in problem 57 of Chapter 25. It might be useful to think of (by analogy) Eq. 7-48; there, the scalar (dot) product in P F v =⋅ G G makes it clear that P = 0 if G G F v ⊥ . This suggests that a radial potential difference and an axial flow of charge will not together produce the
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