P27_047.PDF

47. (a) It is useful to read the whole problem before considering the sketch here in part (a) (which we

do not show, but brieﬂy describe). We ﬁnd in part (d) and part (f), below, that J A >J B which

suggests that the streamlines should be closer together in region A than in B (at least for portions

of those regions which lie close to the pipe). Associated with this (see part (g)) the sketch of the

streamlines should reﬂect that fact that some of the conduction charge-carriers are entering the

pipe walls during the transition from region A to region B .

(b) Eq. 27-16 yields

L =(6 . 0Ω) 0 . 010m 2

=6 . 0

ρ pipe = R A

1 . 0

10 6 m

10 − 8 Ω

m .

Thus in region A , Ohm’s law leads to

i pipe = V ab

R = 8 . 0mV

6 . 0mΩ =1 . 3A .

(d) Using Eq. 27-11 and Eq. 25-42 (in absolute value), we ﬁnd the magnitude of the current density

vector in region A :

J ground

V ab

ρ ground L =

0 . 0080V

(500Ω

m)(1000m) =1 . 6

10 − 8 A / m 2 .

(e) Similarly, in region B we obtain

i pipe = V cd

R = 9 . 5mV

6 . 0mΩ =1 . 6A ,

(f) and

J ground

V cd

ρ ground L =

0 . 0095V

(1000Ω

m)(1000m) =9 . 5

10 − 9 A / m 2

A is the reverse of that discussed in part (g). Here, some current

leaves the pipe walls and joins in the ground-supported telluric ﬂows.

(i) There is no current here, because there is no potential diﬀerence along this section of pipe. The

reason V gh = 0 is best seen using Eq. 27-11 and Eq. 25-18 (and remembering that the scalar dot

product gives zero for perpendicular vectors). The arrows shown in the ﬁgure for current actually

refer, in the technical sense, to the direction of J . We refer to this as the x direction. The pipe

section gh is oriented in what we will refer to as the y direction. Eq. 27-11 implies that J and

E must be in the same direction ( x ). But a nonzero voltage diﬀerence here would require (by

Eq. 25-18)

→

ds vanishes identically.

(j) Our discussion in part (j) serves also to motivat e the fact that the current in section fg is less than

= 0. But since ds = dy for this section of pipe, then E

that in section ef by a factor of cos45 ◦ =1 / √ 2. To see this, one may consider the component of

the electric ﬁeld which would “drive” the current (in the sense of Eq. 27-11) along section fg ;it

is less than the ﬁeld responsible for the current in section ef by exactly the factor just mentioned.

Thus,

i fg = i ef cos45 ◦ = 1 . 0A

√ 2 =0 . 71 A .

(k) The answers to the previous parts indicate that current leaves the pipe at point f and

(l) at point g .

(g) Theseresultssuggestthatthepipewalls, inleavingregion A andenteringregion B , have“absorbed”

some of the current, leaving the current density in the nearby ground somewhat “depleted” of the

telluric ﬂows.

(h) We assume the transition B

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