P27_046.PDF

(66 KB) Pobierz
46. (a) Using Eq. 27-11 and Eq. 25-42, we obtain
E
A
J
A
=
ρ
=
|∆V
A
|
40
×
10
−6
V
=
= 2.0
×
10
−8
A/m
2
.
ρL
(100 Ω·m)(20 m)
(b) Similarly, in region
B
we find
J
B
=
60
×
10
−6
V
|∆V
B
|
=
= 3.0
×
10
−8
A/m
2
.
ρL
(100 Ω·m)(20 m)
(c) With
w
= 1.0 m and
d
A
= 3.8 m (so that the cross-section area is
d
A
w)
we have (using Eq. 27-5)
i
A
=
J
A
d
A
w
= 2.0
×
10
−8
A/m
2
(1.0 m)(3.8 m) = 7.6
×
10
−8
A
.
(d) Assuming
i
A
=
i
B
we obtain
d
B
=
i
B
J
B
7.6
×
10
−8
A
= 2.5 m
.
=
(3.0
×
10
−8
A/m
2
) (1.0 m)
w
(e) We do not show the graph-and-figure here, but describe it briefly. To be meaningful (as a function
of
x)
we would plot
V
(x) measured relative to
V
(0) (the voltage at, say, the left edge of the figure,
which we are effectively setting equal to 0). From the problem statement, we note that
V
(x) would
grow linearly in region
A,
increasing by 40
µV
for each 20 m distance. Once we reach the transition
region (between
A
and
B)
we might assume a parabolic shape for
V
(x) as it changes from the
40
µV-per-20
m slope to a 60
µV-per-20
m slope (which becomes its constant slope once we are into
region
B,
where the function is again linear). The figure goes further than region
B,
so as we leave
region
B,
we might assume again a parabolic shape for the function as it tends back down toward
some lower slope value.

Plik z chomika:

Inne pliki z tego folderu:

Inne foldery tego chomika:

Zgłoś jeśli naruszono regulamin