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= ρAvwhere is the charge per unit volume. If the cross-section is that of a circle, then i= πR2v. (b) Recalling that a Coulomb per second is an Ampere, we obtain i=×=×−−11 102 017 103325...C/mmm/sA.chbg bgπ 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 PF v=⋅GGmakes it clear that P= 0 if GGF v⊥. This suggests that a radial potential difference and an axial flow of charge will not together produce the
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