Exercícios resolvidos Hayt
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Exercícios resolvidos Hayt


Disciplina<strong>ele</strong>5 materiais
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CHAPTER 5
5.1. Given the current density J = \u2212104[sin(2x)e\u22122yax + cos(2x)e\u22122yay] kA/m2:
a) Find the total current crossing the plane y = 1 in the ay direction in the region 0 < x < 1,
0 < z < 2: This is found through
I =
Z Z
S
J · n
ØØØ
S
da =
Z 2
0
Z 1
0
J · ay
ØØØ
y=1
dx dz =
Z 2
0
Z 1
0
\u2212104 cos(2x)e\u22122 dx dz
= \u2212104(2)1
2
sin(2x)
ØØØ1
0
e\u22122 = \u22121.23MA
b) Find the total current leaving the region 0 < x, x < 1, 2 < z < 3 by integrating J ·dS over
the surface of the cube: Note first that current through the top and bottom surfaces will
not exist, since J has no z component. Also note that there will be no current through the
x = 0 plane, since Jx = 0 there. Current will pass through the three remaining surfaces,
and will be found through
I =
Z 3
2
Z 1
0
J · (\u2212ay)
ØØØ
y=0
dx dz +
Z 3
2
Z 1
0
J · (ay)
ØØØ
y=1
dx dz +
Z 3
2
Z 1
0
J · (ax)
ØØØ
x=1
dy dz
= 104
Z 3
2
Z 1
0
£
cos(2x)e\u22120 \u2212 cos(2x)e\u22122§ dx dz \u2212 104 Z 3
2
Z 1
0
sin(2)e\u22122y dy dz
= 104
µ
1
2
\u2202
sin(2x)
ØØØ1
0
(3\u2212 2) £1\u2212 e\u22122§+ 104µ1
2
\u2202
sin(2)e\u22122y
ØØØ1
0
(3\u2212 2) = 0
c) Repeat part b, but use the divergence theorem: We find the net outward current through
the surface of the cube by integrating the divergence of J over the cube volume. We have
\u2207 · J = \u2202Jx
\u2202x
+
\u2202Jy
\u2202y
= \u221210\u22124 £2 cos(2x)e\u22122y \u2212 2 cos(2x)e\u22122y§ = 0 as expected
5.2. Given J = \u221210\u22124(yax + xay)A/m2, find the current crossing the y = 0 plane in the \u2212ay
direction between z = 0 and 1, and x = 0 and 2.
At y = 0, J(x, 0) = \u2212104xay, so that the current through the plane becomes
I =
Z
J · dS =
Z 1
0
Z 2
0
\u2212104xay · (\u2212ay) dx dz = 2× 10\u22124 A
58
5.3. Let
J =
400 sin \u3b8
r2 + 4
ar A/m2
a) Find the total current flowing through that portion of the spherical surface r = 0.8,
bounded by 0.1\u3c0 < \u3b8 < 0.3\u3c0, 0 < \u3c6 < 2\u3c0: This will be
I =
Z Z
J · n
ØØØ
S
da =
Z 2\u3c0
0
Z .3\u3c0
.1\u3c0
400 sin \u3b8
(.8)2 + 4
(.8)2 sin \u3b8 d\u3b8 d\u3c6 =
400(.8)22\u3c0
4.64
Z .3\u3c0
.1\u3c0
sin2 d\u3b8
= 346.5
Z .3\u3c0
.1\u3c0
1
2
[1\u2212 cos(2\u3b8)] d\u3b8 = 77.4A
b) Find the average value of J over the defined area. The area is
Area =
Z 2\u3c0
0
Z .3\u3c0
.1\u3c0
(.8)2 sin \u3b8 d\u3b8 d\u3c6 = 1.46m2
The average current density is thus Javg = (77.4/1.46)ar = 53.0ar A/m2.
5.4. If volume charge density is given as \u3c1v = (cos\u3c9t)/r2 C/m3 in spherical coordinates, find J. It
is reasonable to assume that J is not a function of \u3b8 or \u3c6.
We use the continuity equation (5), along with the assumption of no angular variation to
write
\u2207 · J = 1
r2
\u2202
\u2202r
°
r2Jr
¢
= \u2212\u2202\u3c1v
\u2202t
= \u2212 \u2202
\u2202t
µ
cos\u3c9t
r2
\u2202
=
\u3c9 sin\u3c9t
r2
So we may now solve
\u2202
\u2202r
°
r2Jr
¢
= \u3c9 sin\u3c9t
by direct integration to obtain:
J = Jr ar =
\u3c9 sin\u3c9t
r
ar A/m2
where the integration constant is set to zero because a steady current will not be created
by a time-varying charge density.
59
5.5. Let
J =
25
\u3c1
a\u3c1 \u2212 20
\u3c12 + 0.01
az A/m2
a) Find the total current crossing the plane z = 0.2 in the az direction for \u3c1 < 0.4: Use
I =
Z Z
S
J · n
ØØØ
z=.2
da =
Z 2\u3c0
0
Z .4
0
\u221220
\u3c12 + .01
\u3c1 d\u3c1 d\u3c6
= \u2212
µ
1
2
\u2202
20 ln(.01 + \u3c12)
ØØØ.4
0
(2\u3c0) = \u221220\u3c0 ln(17) = \u2212178.0A
b) Calculate \u2202\u3c1v/\u2202t: This is found using the equation of continuity:
\u2202\u3c1v
\u2202t
= \u2212\u2207 · J = 1
\u3c1
\u2202
\u2202\u3c1
(\u3c1J\u3c1) +
\u2202Jz
\u2202z
=
1
\u3c1
\u2202
\u2202\u3c1
(25) +
\u2202
\u2202z
µ \u221220
\u3c12 + .01
\u2202
= 0
c) Find the outward current crossing the closed surface defined by \u3c1 = 0.01, \u3c1 = 0.4, z = 0,
and z = 0.2: This will be
I =
Z .2
0
Z 2\u3c0
0
25
.01
a\u3c1 · (\u2212a\u3c1)(.01) d\u3c6 dz +
Z .2
0
Z 2\u3c0
0
25
.4
a\u3c1 · (a\u3c1)(.4) d\u3c6 dz
+
Z 2\u3c0
0
Z .4
0
\u221220
\u3c12 + .01
az · (\u2212az) \u3c1 d\u3c1 d\u3c6+
Z 2\u3c0
0
Z .4
0
\u221220
\u3c12 + .01
az · (az) \u3c1 d\u3c1 d\u3c6 = 0
since the integrals will cancel each other.
d) Show that the divergence theorem is satisfied for J and the surface specified in part b.
In part c, the net outward flux was found to be zero, and in part b, the divergence of J
was found to be zero (as will be its volume integral). Therefore, the divergence theorem
is satisfied.
5.6. In spherical coordinates, a current density J = \u2212k/(r sin \u3b8)a\u3b8 A/m2 exists in a conducting
medium, where k is a constant. Determine the total current in the az direction that crosses a
circular disk of radius R, centered on the z axis and located at a) z = 0; b) z = h.
Integration over a disk means that we use cylindrical coordinates. The general flux integral
assumes the form:
I =
Z
s
J · dS =
Z 2\u3c0
0
Z R
0
\u2212k
r sin \u3b8
a\u3b8 · az| {z }
\u2212 sin \u3b8
\u3c1 d\u3c1 d\u3c6
Then, using r =
p
\u3c12 + z2, this becomes
I =
Z 2\u3c0
0
Z R
0
k\u3c1p
\u3c12 + z2
= 2\u3c0k
p
\u3c12 + z2
ØØØR
0
= 2\u3c0k
hp
R2 + z2 \u2212 z
i
At z = 0 (part a), we have I(0) = 2\u3c0kR , and at z = h (part b): I(h) = 2\u3c0k
£\u221a
R2 + h2 \u2212 h§.
60
5.7. Assuming that there is no transformation of mass to energy or vice-versa, it is possible to
write a continuity equation for mass.
a) If we use the continuity equation for charge as our model, what quantities correspond to J
and \u3c1v? These would be, respectively, mass flux density in (kg/m2 \u2212 s) and mass density
in (kg/m3).
b) Given a cube 1 cm on a side, experimental data show that the rates at which mass is
leaving each of the six faces are 10.25, -9.85, 1.75, -2.00, -4.05, and 4.45 mg/s. If we
assume that the cube is an incremental volume element, determine an approximate value
for the time rate of change of density at its center. We may write the continuity equation
for mass as follows, also invoking the divergence theorem:Z
v
\u2202\u3c1m
\u2202t
dv = \u2212
Z
v
\u2207 · Jm dv = \u2212
I
s
Jm · dS
where I
s
Jm · dS = 10.25\u2212 9.85 + 1.75\u2212 2.00\u2212 4.05 + 4.45 = 0.550 mg/s
Treating our 1 cm3 volume as di\ufb00erential, we find
\u2202\u3c1m
\u2202t
.= \u22120.550× 10
\u22123 g/s
10\u22126 m3
= \u2212550 g/m3 \u2212 s
5.8. A truncated cone has a height of 16 cm. The circular faces on the top and bottom have radii
of 2mm and 0.1mm, respectively. If the material from which this solid cone is constructed
has a conductivity of 2× 106 S/m, use some good approximations to determine the resistance
between the two circular faces.
Consider the cone upside down and centered on the positive z axis. The 1-mm radius end
is at distance z = ` from the x-y plane; the wide end (2-mm radius) lies at z = `+16 cm.
` is chosen such that if the cone were not truncated, its vertex would occur at the origin.
The cone surface subtends angle \u3b8c from the z axis (in spherical coordinates). Therefore,
we may write
` =
0.1mm
tan \u3b8c
and tan \u3b8c =
2mm
160 + `
Solving these, we find ` = 8.4 mm, tan \u3b8c = 1.19× 10\u22122, and so \u3b8c = 0.68\u25e6, which gives
us a very thin cone! With this understanding, we can assume that the current density is
uniform with \u3b8 and \u3c6 and will vary only with spherical radius, r. So the current density
will be constant over a spherical cap (of constant r) anywhere within the cone. As the
cone is thin, we can also assume constant current density over any flat surface within the
cone at a specifed z. That is, any spherical cap looks flat if the cap radius, r, is large
compared to its radius as measured from the z axis (\u3c1). This is our primary assumption.
Now, assuming constant current density at constant r, and net current, I, we may write
I =
Z 2\u3c0
0
Z \u3b8c
0
J(r)ar · ar r2 sin \u3b8 d\u3b8 d\u3c6 = 2\u3c0r2J(r)(1\u2212 cos \u3b8c)
or
J(r) =
I
2\u3c0r2(1\u2212 cos \u3b8c) ar =
(1.42× 104)I
2\u3c0r2
ar
61
5.8 (continued) The electric field is now
E(r) =
J(r)
\u3c3
=
(1.42× 104)I
2\u3c0r2(2× 106) ar = (7.1× 10
\u22123)
I
2\u3c0r2
ar V/m
The voltage between the ends is now
V0 = \u2212
Z rin
rout
E · ar dr
where rin = `/ cos \u3b8c
.= ` and where rout = (160 + `)/ cos \u3b8c
.= 160 + `. The voltage is
V0 = \u2212
Z `×10\u22123
(160+`)×10\u22123
(7.1× 10\u22123) I
2\u3c0r2
ar · ar dr = (7.1× 10\u22123) I2\u3c0
\u2211
1
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