P27_044.PDF

44. (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 15, 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 = ρAv where ρ

is the charge per unit volume. If the cross-section is that of a circle, then i = ρπR 2 v .

(b) Recalling that a Coulomb per second is an Ampere, we obtain

i = 1 . 1

10 − 3 C / m 3 π (0 . 050 m) 2 (2 . 0m / s) = 1 . 7

10 − 5 A .

v . This suggests that a radial potential diﬀerence

and an axial ﬂow of charge will not together produce the needed transfer of energy (into the form

of a spark).

(d) With the assumption that there is (at least) a voltage equal to that computed in problem 57 of

Chapter 25, in the proper direction to enable the transference of energy (into a spark), then we use

our result from that problem in Eq. 27-21:

P = iV = 1 . 7

v makes it clear that P =0if F

⊥

10 − 5 A 7 . 8

10 4 V =1 . 3W .

(e) Recalling that a Joule per second is a Watt, we obtain (1 . 3 W)(0 . 20 s) = 0 . 27 J for the energy that

can be transferred at the exit of the pipe.

(f) This result is greater than the 0 . 15 J needed for a spark, so we conclude that the spark was likely

to have occurred at the exit of the pipe, going into the silo.

of Chapter 25. It might be useful to think of (by analogy) Eq. 7-48; there, the scalar (dot) product

in P = F