Engineering Electromagnetics
NINTH EDITION
William H. Hayt, Jr.
Late Emeritus Professor
Purdue University
John A. Buck
Georgia Institute of Technology
CHAPTER 1 – 9th Edition
1.1. If A represents a vector two units in length directed due west, B represents a vector three units in length
directed due north, and A + B = C * D and 2B * A = C + D, find the magnitudes and directions of
C and D. Take north as the positive y direction:
With north as positive y, west will be -x. We may therefore set up:
C + D = 2B * A = 6ay + 2ax and
C * D = A + B = *2ax + 3ay
Add the equations to find C = 4.5ay (north), and then D = 2ax + 1.5ay (east of northeast).
1.2. Vector A extends from the origin to (1,2,3) and vector B from the origin to (2,3,-2).
a) Find the unit vector in the direction of (A * B): First
A * B = (ax + 2ay + 3az ) * (2ax + 3ay * 2az ) = (*ax * ay + 5az )
˘
⌅
⇧1_2
whose magnitude is A * B = (*ax * ay + 5az ) (*ax * ay + 5az )
= 1 + 1 + 25 =
˘
3 3 = 5.20. The unit vector is therefore
aAB = (*ax * ay + 5az )_5.20
b) find the unit vector in the direction of the line extending from the origin to the midpoint of the
line joining the ends of A and B:
The midpoint is located at
Pmp = [1 + (2 * 1)_2, 2 + (3 * 2)_2, 3 + (*2 * 3)_2)] = (1.5, 2.5, 0.5)
The unit vector is then
(1.5ax + 2.5ay + 0.5az )
amp = ˘
= (1.5ax + 2.5ay + 0.5az )_2.96
(1.5)2 + (2.5)2 + (0.5)2
1.3. The vector from the origin to the point A is given as (6, *2, *4), and the unit vector directed from the
origin toward point B is (2, *2, 1)_3. If points A and B are ten units apart, find the coordinates of
point B.
With A = (6, *2, *4) and B = 13 B(2, *2, 1), we use the fact that B * A = 10, or
(6 * 23 B)ax * (2 * 23 B)ay * (4 + 13 B)az  = 10
Expanding, obtain
36 * 8B + 49 B 2 + 4 * 83 B + 49 B 2 + 16 + 83 B + 19 B 2 = 100
˘
or B 2 * 8B * 44 = 0. Thus B = 8± 64*176
= 11.75 (taking positive option) and so
2
2
2
1
B = (11.75)ax * (11.75)ay + (11.75)az = 7.83ax * 7.83ay + 3.92az
3
3
3
1
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1.4. A circle, centered at the origin with a radius of 2 units, lies in the xy plane. Determine
the unit vector
˘
in rectangular components that lies in the xy plane, is tangent to the circle at ( 3, *1, 0), and is in the
general direction of increasing values of x:
ㅡ
A unit vector tangent to this circle in the general increasing x direction is t = +a . Its x and
˘
y components are tx = a ax = * sin , and ty = a ay = cos . At the point ( 3, *1),
˘
= 330˝ , and so t = * sin 330˝ ax + cos 330˝ ay = 0.5(ax + 3ay ).
1.5. An equilateral triangle lies in the xy plane with its centroid at the origin. One vertex lies on the positive
y axis.
a) Find unit vectors that are directed from the origin to
the three vertices: Referring to the figure,the easy
one is a1 = ay . Then, a2 will have negative x and y
components, and can be constructed as a2 = G(*ax
* tan 30˝ ay ) where G = (1 + tan 30˝ )1_2 = 0.87.
So finally a2 = *0.87(ax + 0.58ay ). Then, a3 is the
same as a2 , but with the x component reversed:
a3 = 0.87(ax * 0.58ay ).
y
a1
a5
a6
x
30°
a
a3
2
b) Find unit vectors that are directed from the origin
a4
to the three sides, intersecting these at right angles:
These will be a4 , a5 , and a6 in the figure, which are in turn just the part a results, oppositely
directed:
a4 = *a1 = *ay , a5 = *a3 = *0.87(ax * 0.58ay ), and a6 = *a2 = +0.87(ax + 0.58ay ).
1.6. Find the acute angle between the two vectors A = 2ax + ay + 3az and B = ax * 3ay + 2az by using
the definition of:
˘
˘
2 + 1 2 + 32 =
a) the dot product:˘
First, A B = 2 *
3
+
6
=
5
=
AB
cos
✓,
where
A
=
2
14,
˘
˝
2
2
2
and where B = 1 + 3 + 2 = 14. Therefore cos ✓ = 5_14, so that ✓ = 69.1 .
b) the cross product: Begin with
Ûa a a Û
Û x
y
zÛ
Û
Û
A ù B = Û 2 1 3 Û = 11ax * ay * 7az
Û
Û
Û 1 *3 2 Û
Û
Û
˘
˘
˘
2
2
2
and then A ù B
⇠˘= 11 ⇡+ 1 + 7 = 171. So now, with A ù B = AB sin ✓ = 171,
find ✓ = sin*1
171_14 = 69.1˝
1.7. Given the field F = xax + yay . If F G = 2xy and F ù G = (x2 * y2 ) az , find G:
Let G = g1 ax + g2 ay + g3 az Then F G = g1 x + g2 y = 2xy, and
Ûa a a Û
Û x y zÛ
Û
Û
F ù G = Û x y 0 Û = g3 y ax * g3 x ay + (g2 x * g1 y) az = (x2 * y2 ) az
Û
Û
Ûg1 g2 g3 Û
Û
Û
From the last equation, it is clear that g3 = 0, and that g1 = y and g2 = x. This is confirmed in the
F G equation. So finally G = yax + xay .
2
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1.8. Demonstrate the ambiguity that results when the cross product is used to find the angle between two
vectors by finding the angle between A = 3ax *2ay + 4az and B = 2ax + ay *2az . Does this ambiguity
exist when the dot product is used?
We use the relation A ù B = AB sin ✓n. With the given vectors we find
L
M
˘ 2ay + az
˘
˘
A ù B = 14ay + 7az = 7 5
= 9 + 4 + 16 4 + 1 + 4 sin ✓ n
˘
5
´≠≠≠≠Ø≠≠≠≠¨
±n
where n is identified as shown; we see that n can be positive or negative, as sin ✓ can be positive or negative. This apparent sign ambiguity is not the real problem, however, as we really want ˘
the magnitude
˘ ˘ of the angle anyway. Choosing the positive sign, we are left with
sin ✓ = 7 5_( 29 9) = 0.969. Two values of ✓ (75.7˝ and 104.3˝ ) satisfy this equation,
and hence the real ambiguity.
˘
In using the dot
product,
we
find
A
B
=
6
*
2
*
8
=
*4
=
AB
cos
✓
=
3
29 cos ✓, or
˘
cos ✓ = *4_(3 29) = *0.248 Ÿ ✓ = *75.7˝ . Again, the minus sign is not important, as we
care only about the angle magnitude. The main point is that only one ✓ value results when using
the dot product, so no ambiguity.
1.9. A field is given as
Find:
G=
25
(xax + yay )
(x2 + y2 )
a) a unit vector in the direction of G at P (3, 4, *2): Have Gp = 25_(9 + 16) ù (3, 4, 0) = 3ax + 4ay ,
and Gp  = 5. Thus aG = (0.6, 0.8, 0).
b) the angle between G and ax at P : The angle is found through aG ax = cos ✓. So cos ✓ =
(0.6, 0.8, 0) (1, 0, 0) = 0.6. Thus ✓ = 53˝ .
c) the value of the following double integral on the plane y = 7:
4
2
0
4
0
2
0
0
G ay dzdx
4
2
25
25
(xax + yay ) ay dzdx =
ù 7 dzdx =
2 + 49
x2 + y2
x
0
0
⌧
⇠ ⇡
4
1
= 350 ù
tan*1
* 0 = 26
7
7
4
0
350
dx
x2 + 49
1.10. By expressing diagonals as vectors and using the definition of the dot product, find the smaller angle
between any two diagonals of a cube, where each diagonal connects diametrically opposite corners,
and passes through the center of the cube:
Assuming a side length, b, two diagonal vectors would be A = b(ax + ay + az ) and B =
˘
˘
b(ax * ay + az ). Now use A B = AB cos ✓, or b2 (1 * 1 + 1) = ( 3b)( 3b) cos ✓ Ÿ
cos ✓ = 1_3 Ÿ ✓ = 70.53˝ . This result (in magnitude) is the same for any two diagonal vectors.
3
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1.11. Given the points M(0.1, *0.2, *0.1), N(*0.2, 0.1, 0.3), and P (0.4, 0, 0.1), find:
a) the vector RMN : RMN = (*0.2, 0.1, 0.3) * (0.1, *0.2, *0.1) = (*0.3, 0.3, 0.4).
b) the dot product RMN RMP : RMP = (0.4, 0, 0.1) * (0.1, *0.2, *0.1) = (0.3, 0.2, 0.2). RMN
RMP = (*0.3, 0.3, 0.4) (0.3, 0.2, 0.2) = *0.09 + 0.06 + 0.08 = 0.05.
c) the scalar projection of RMN on RMP :
RMN aRMP = (*0.3, 0.3, 0.4) ˘
(0.3, 0.2, 0.2)
0.05
= 0.12
=˘
0.09 + 0.04 + 0.04
0.17
d) the angle between RMN and RMP :
*1
✓M = cos
0
RMN RMP
RMN RMP 
H
1
*1
= cos
0.05
˘
˘
0.34 0.17
I
= 78˝
1.12. Write an expression in rectangular components for the vector that extends from (x1 , y1 , z1 ) to (x2 , y2 , z2 )
and determine the magnitude of this vector.
The two points can be written as vectors from the origin:
A1 = x1 ax + y1 ay + z1 az
and
A2 = x2 ax + y2 ay + z2 az
The desired vector will now be the difference:
A12 = A2 * A1 = (x2 * x1 )ax + (y2 * y1 )ay + (z2 * z1 )az
whose magnitude is
A12  =
1.13.
˘
⌅
⇧1_2
A12 A12 = (x2 * x1 )2 + (y2 * y1 )2 + (z2 * z1 )2
a) Find the vector component of F = (10, *6, 5) that is parallel to G = (0.1, 0.2, 0.3):
FG =
(10, *6, 5) (0.1, 0.2, 0.3)
F G
G
=
(0.1, 0.2, 0.3) = (0.93, 1.86, 2.79)
0.01 + 0.04 + 0.09
G2
b) Find the vector component of F that is perpendicular to G:
FpG = F * FG = (10, *6, 5) * (0.93, 1.86, 2.79) = (9.07, *7.86, 2.21)
c) Find the vector component of G that is perpendicular to F:
GpF = G * GF = G *
G F
1.3
F = (0.1, 0.2, 0.3) *
(10, *6, 5) = (0.02, 0.25, 0.26)
2
100 + 36 + 25
F
4
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1.14. Given that A + B + C = 0, where the three vectors represent line segments and extend from a common
origin,
a) must the three vectors be coplanar?
In terms of the components, the vector sum will be
A + B + C = (Ax + Bx + Cx )ax + (Ay + By + Cy )ay + (Az + Bz + Cz )az
which we require to be zero. Suppose the coordinate system is configured so that vectors A and
B lie in the x-y plane; in this case Az = Bz = 0. Then Cz has to be zero in order for the three
vectors to sum to zero. Therefore, the three vectors must be coplanar.
b) If A + B + C + D = 0, are the four vectors coplanar?
The vector sum is now
A + B + C + D = (Ax + Bx + Cx + Dx )ax + (Ay + By + Cy + Dy )ay + (Az + Bz + Cz + Dz )az
Now, for example, if A and B lie in the x-y plane, C and D need not, as long as Cz + Dz = 0. So
the four vectors need not be coplanar to have a zero sum.
1.15. Three vectors extending from the origin are given as r1 = (7, 3, *2), r2 = (*2, 7, *3), and
r3 = (0, 2, 3). Find:
a) a unit vector perpendicular to both r1 and r2 :
ap12 =
r1 ù r2
(5, 25, 55)
=
= (0.08, 0.41, 0.91)
r1 ù r2 
60.6
b) a unit vector perpendicular to the vectors r1 * r2 and r2 * r3 : r1 * r2 = (9, *4, 1) and r2 * r3 =
(*2, 5, *6). So r1 * r2 ù r2 * r3 = (19, 52, 37). Then
ap =
(19, 52, 37)
(19, 52, 37)
=
= (0.29, 0.78, 0.56)
(19, 52, 37)
66.6
c) the area of the triangle defined by r1 and r2 :
1
Area = r1 ù r2  = 30.3
2
d) the area of the triangle defined by the heads of r1 , r2 , and r3 :
1
1
Area = (r2 * r1 ) ù (r2 * r3 ) = (*9, 4, *1) ù (*2, 5, *6) = 33.3
2
2
5
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1.16. In geometrical optics, the path of a light ray is treated as a vector having the usual three components in
a rectangular coordinate system. When light reflects from a plane surface, the effect is to reverse the
vector component of the ray that is normal to the surface. This yields a reflection angle that is equal
to the incidence angle. Explain what happens when light ray reflects from a corner cube reflector,
consisting of three mutually orthogonal surfaces that occupy, for example, the xy, xz, and yz planes.
The ray is incident in an arbitrary direction within the first octant of the coordinate system, and has
negative x, y, and z directions of travel: We can model the incident ray as the vector, Ri = a(*ax ) +
b(*ay ) + c(*az ), where a, b, and c are arbitrary positive values. With the three orthogonal planes
positioned as given, their effect is to reverse all three components of the incident vector, giving Rr =
a(ax ) + b(ay ) + c(az ) = *Ri . So the light propagates in precisely the reverse direction after reflection,
no matter what direction it arrived from.
1.17. Point A(*4, 2, 5) and the two vectors, RAM = (20, 18, *10) and RAN = (*10, 8, 15), define a triangle.
a) Find a unit vector perpendicular to the triangle: Use
ap =
RAM ù RAN
(350, *200, 340)
=
= (0.664, *0.379, 0.645)
RAM ù RAN 
527.35
The vector in the opposite direction to this one is also a valid answer.
b) Find a unit vector in the plane of the triangle and perpendicular to RAN :
aAN =
(*10, 8, 15)
= (*0.507, 0.406, 0.761)
˘
389
Then
apAN = ap ù aAN = (0.664, *0.379, 0.645) ù (*0.507, 0.406, 0.761) = (*0.550, *0.832, 0.077)
The vector in the opposite direction to this one is also a valid answer.
c) Find a unit vector in the plane of the triangle that bisects the interior angle at A: A non-unit
vector in the required direction is (1_2)(aAM + aAN ), where
aAM =
(20, 18, *10)
= (0.697, 0.627, *0.348)
(20, 18, *10)
Now
1
1
(a
+ aAN ) = [(0.697, 0.627, *0.348) + (*0.507, 0.406, 0.761)] = (0.095, 0.516, 0.207)
2 AM
2
Finally,
abis =
(0.095, 0.516, 0.207)
= (0.168, 0.915, 0.367)
(0.095, 0.516, 0.207)
6
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1.18. Given two vector fields, E = (A_r) sin ✓ a✓ and H = (B_r) sin ✓ a , where A and B are constants,
a) evaluate S = E ù H and express the result in rectangular coordinates: Begin with
AB 2
sin ✓ ar
r2
Now, find the three rectangular components by using:
S=EùH=
Sx = S ax =
x(x2 + y2 )
AB 2
sin
✓
a
a
=
AB
r
x
r2
(x2 + y2 + z2 )5_2
where we have used ar ax = sin ✓ cos with r = (x2 + y2 + z2 )1_2 ,
sin ✓ = (x2 + y2 )1_2 _(x2 + y2 + z2 )1_2 and cos = x_(x2 + y2 )1_2 .
Then
Sy = S ay =
y(x2 + y2 )
AB 2
sin
✓
a
a
=
AB
r
y
r2
(x2 + y2 + z2 )5_2
where we have used ar ay = sin ✓ sin
Finally
Sz = S a z =
with sin
= y_(x2 + y2 )1_2 .
z(x2 + y2 )
AB 2
sin
✓
a
a
=
AB
r
z
r2
(x2 + y2 + z2 )5_2
where ar az = cos ✓ = z_(x2 + y2 + z2 )1_2 .
b) Determine S along the x, y, and z axes: Using the part a results, we would have
S(x, 0, 0) =
AB
ax ,
x2
S(0, y, 0) =
AB
ay ,
y2
S(0, 0, z) = 0
1.19. Consider the important inverse-square dependent radial field in spherical coordinates: F = A_r2 ar
where A is a constant.
a) Transform the given field into cylindrical coordinates: Using r2 = ⇢2 + z2 , the given field may
be written:
⇧
A ⌅
F= 2
f
a
+
f
a
+
f
a
⇢
⇢
z
z
⇢ + z2
Now
⇢
⇢
f⇢ = ar a⇢ = sin ✓ = = 2
r
(⇢ + z2 )1_2
f = ar a = 0
z
z
fz = ar az = cos ✓ = = 2
r
(⇢ + z2 )1_2
Substituting all, we find
⇧
A ⌅
F= 2
⇢a
+
za
⇢
z
⇢ + z2
b) Transform the given field
˘ into rectangular coordinates: The quickest way is to use˘the results of
part a by using ⇢ = x2 + y2 and a⇢ = cos ax + sin ay , where cos = x_ x2 + y2 and
˘
sin = y_ x2 + y2 . Incorporating the above leads to:
⌅
⇧
A
F= 2
xax + yay + zaz
2
2
3_2
(x + y + z )
7
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1.20. If the three sides of a triangle are represented by the vectors A, B, and C, all directed counter-clockwise,
show that C2 = (A + B) (A + B) and expand the product to obtain the law of cosines.
With the vectors drawn as described above, we find that C = *(A + B) and so C2 = C 2 = C C =
(A + B) (A + B) So far so good. Now if we expand the product, obtain
(A + B) (A + B) = A2 + B 2 + 2A B
where A B = AB cos(180˝ * ↵) = *AB cos ↵ where ↵ is the interior angle at the junction of A and
B. Using this, we have C 2 = A2 + B 2 * 2AB cos ↵, which is the law of cosines.
1.21. Express in cylindrical components:
a) the vector from C(3, 2, *7) to D(*1, *4, 2):
C(3, 2, *7) ô C(⇢ = 3.61, = 33.7˝ , z = *7) and
D(*1, *4, 2) ô D(⇢ = 4.12, = *104.0˝ , z = 2).
Now RCD = (*4, *6, 9) and R⇢ = RCD a⇢ = *4 cos(33.7) * 6 sin(33.7) = *6.66. Then
R = RCD a = 4 sin(33.7) * 6 cos(33.7) = *2.77. So RCD = *6.66a⇢ * 2.77a + 9az
b) a unit vector at D directed toward C:
RCD = (4, 6, *9) and R⇢ = RDC a⇢ = 4 cos(*104.0) + 6 sin(*104.0) = *6.79. Then R =
RDC a = 4[* sin(*104.0)] + 6 cos(*104.0) = 2.43. So RDC = *6.79a⇢ + 2.43a * 9az
Thus aDC = *0.59a⇢ + 0.21a * 0.78az
c) a unit vector at D directed toward the origin: Start with rD = (*1, *4, 2), and so the vector toward
the origin will be *rD = (1, 4, *2). Thus in cartesian the unit vector is a = (0.22, 0.87, *0.44).
Convert to cylindrical:
a⇢ = (0.22, 0.87, *0.44) a⇢ = 0.22 cos(*104.0) + 0.87 sin(*104.0) = *0.90, and
a = (0.22, 0.87, *0.44) a = 0.22[* sin(*104.0)] + 0.87 cos(*104.0) = 0, so that finally,
a = *0.90a⇢ * 0.44az .
1.22. A sphere of radius a, centered at the origin, rotates about the z axis at angular velocity ⌦ rad/s. The
rotation direction is clockwise when one is looking in the positive z direction.
a) Using spherical components, write an expression for the velocity field, v, which gives the tangential velocity at any point on the sphere surface:
The tangential velocity is the product of the angular velocity and the perpendicular distance from
the rotation axis. With clockwise rotation, and with sphere radius a, we obtain
v(✓) = ⌦a sin ✓ a m_s
b) Derive an expression for the difference in velocities between two points on the surface having
different latitudes, where the latitude difference is ✓ in radians. Assume ✓ is small:
Method 1: The velocity difference between the two points can be approximated as
.
dv Û
v= ✓
Û
d✓ Û✓0
where ✓0 is the mean angle between the two points, and where ✓ is small enough so that v(✓)
can be approximated as a linear function. Using this along with the part a result, we find:
.
v = a⌦ ✓ cos ✓0 m_s
8
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1.22
b) (continued) Method 2 (harder): Write
v = a⌦(sin ✓2 * sin ✓1 )
where ✓1 and ✓2 lie on either side of ✓0 , and where ✓2 * ✓1 = ✓. Using a trig identity, the above
expression becomes
⇠
⇡
⇠
⇡ .
1
1
v = 2a⌦ cos (✓2 + ✓1 ) sin (✓2 * ✓1 ) = a⌦ ✓ cos ✓0 m_s
2
2
´≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠¨ ´≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠¨
.
cos ✓0
= ✓_2
c) Find the difference in velocities at locations ±1.0˝ on either side of 45˝ north latitude on Earth.
Take the Earth’s radius as 6370 km at 45˝ :
We have a = 6370 km, ✓0 = 45˝ (same in spherical and geographic coordinates), ✓ = 2.0˝ =
0.035 rad, and ⌦ = 2⇡_24 hr = 7.3 ù 10*5 rad/s. Using part b, the velocity difference becomes
˘
.
v = (6.37 ù 106 m)(7.3 ù 10*5 rad_s)(0.035 rad)( 2_2) = 11.5 m_s
with speed increasing southward.
1.23. The surfaces ⇢ = 3, ⇢ = 5,
= 100˝ ,
= 130˝ , z = 3, and z = 4.5 define a closed surface.
a) Find the enclosed volume:
Vol =
NOTE: The limits on the
4.5
130˝
3
100˝
5
3
⇢ d⇢ d dz = 6.28
integration must be converted to radians (as was done here, but not shown).
b) Find the total area of the enclosing surface:
Area = 2
+
4.5
3
130˝
100˝
130˝
100˝
5
3
⇢ d⇢ d
5 d dz + 2
+
3
4.5
3
4.5
130˝
5
3
100˝
3 d dz
d⇢ dz = 20.7
c) Find the total length of the twelve edges of the surfaces:
⌧ ˝
30
30˝
Length = 4 ù 1.5 + 4 ù 2 + 2 ù
ù
2⇡
ù
3
+
ù 2⇡ ù 5 = 22.4
360˝
360˝
d) Find the length of the longest straight line that lies entirely within the volume: This will be
between the points A(⇢ = 3, = 100˝ , z = 3) and B(⇢ = 5, = 130˝ , z = 4.5). Performing
point transformations to cartesian coordinates, these become A(x = *0.52, y = 2.95, z = 3)
and B(x = *3.21, y = 3.83, z = 4.5). Taking A and B as vectors directed from the origin, the
requested length is
Length = B * A = (*2.69, 0.88, 1.5) = 3.21
9
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1.24. Two unit vectors, a1 and a2 lie in the xy plane and pass through the origin. They make angles
2 with the x axis respectively.
1 and
a) Express each vector in rectangular components; Have a1 = Ax1 ax +Ay1 ay , so that Ax1 = a1 ax =
cos 1 . Then, Ay1 = a1 ay = cos(90 * 1 ) = sin 1 . Therefore,
a1 = cos
1 ax + sin
1 ay
and similarly, a2 = cos
2 ax + sin
b) take the dot product and verify the trigonometric identity, cos( 1 *
sin 1 sin 2 : From the definition of the dot product,
a1 a2 = (1)(1) cos( 1 *
= (cos
1 ax + sin
2)
1 ay )
(cos
2 ax + sin
2 ay ) = cos
2)
1 cos
2 ay
= cos
1 cos
2 + sin
1 sin
2 +
2
c) take the cross product and verify the trigonometric identity sin( 2 * 1 ) = sin 2 cos 1 *
cos 2 sin 1 : From the definition of the cross product, and since a1 and a2 both lie in the xy plane,
Û a
ay
az ÛÛ
Û x
Û
Û
a1 ù a2 = (1)(1) sin( 1 * 2 ) az = Ûcos 1 sin 1 0 Û
Û
Û
Ûcos 2 sin 2 0 Û
Û
Û
⌅
⇧
= sin 2 cos 1 * cos 2 sin 1 az
thus verified.
1.25. Convert the vector field H = A(x2 + y2 )*1 [x ay * y ax ] into cylindrical coordinates. A is a constant:
The z component is obviously zero, so we are left only with the ⇢ and components to find. First,
⇧
A ⌅
x
a
a
*
y
a
a
y
⇢
x
⇢
x2 + y2
˘
˘
= y_ x2 + y2 and ax a⇢ = cos = x_ x2 + y2 . Thus
H⇢ = H a⇢ =
where ay a⇢ = sin
H⇢ =
Then
where ay a = cos
A
[xy * yx] = 0
(x2 + y2 )3_2
⇧
A ⌅
x
a
a
*
y
a
a
y
x
x2 + y2
˘
˘
= x_ x2 + y2 and ax a = * sin = *y_ x2 + y2 . Thus
H =H a =
H =
⌅ 2
⇧
A
A
A
2
x
+
y
= 2
=
⇢
(x2 + y2 )3_2
(x + y2 )1_2
Finally
H=
A
a
⇢
10
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1.26. Express the uniform vector field, F = 10 ay in
a) cylindrical components:
F⇢ = 10 ay a⇢ = 10 sin , F = 10 ay a = 10 cos , and Fz = 0.
Combining, we obtain F(⇢, ) = 10(sin a⇢ + cos a ).
b) spherical components:
Fr = 10 ay ar = 10 sin ✓ sin ; F✓ = 10 ay a✓ = 10 cos ✓ sin ; F = 10 ay a = 10 cos .
⌅
⇧
Combining, we obtain F(r, ✓, ) = 10 sin ✓ sin ar + cos ✓ sin a✓ + cos a .
1.27. The dipole field is given in spherical coordinates as:
E=
A
(2 cos ✓ ar + sin ✓ a✓ )
r3
where A is a constant and where r > 0.
a) Identify the surface on which the field is entirely perpendicular to the xy plane, and express the
field on that surface in cylindrical coordinates: As there is no a component, the cylindrical
coordinate field components will be a⇢ and az only. We look for the surface on which the a⇢
component is zero. So we set up:
E⇢ = E a⇢ =
A
(2 cos ✓ ar a⇢ + sin ✓ a✓ a⇢ ) = 0
r3
´Ø¨
´Ø¨
cos ✓
sin ✓
From which the condition for zero E⇢ is identified:
3 cos ✓ sin ✓ = 0 Ÿ ✓ = 0, 90˝
The ✓ = 0 option is only a line (the z axis – and the answer to part b), so the surface on which the
field is perpendicular to the xy plane is the xy plane itself, on which ✓ = 90˝ . On that surface:
E(✓ = 90) =
A
A
a✓ = * 3 az
r3
⇢
b) Identify the coordinate axis on which the field is entirely perpendicular to the xy plane and express
the field there in cylindrical coordinates: This will be the ✓ = 0 solution in part a, which is the z
axis. The field on that axis is
2A
E(✓ = 0) = 3 az
z
c) Specify the surface on which the field is entirely parallel to the xy plane: In this case, we require
a zero z component, so we construct:
Ez = E az =
A
(2 cos ✓ ar az + sin ✓ a✓ az ) = 0
r3
´Ø¨
´Ø¨
cos ✓
* sin ✓
This tells us that for a zero z component, we must have
2 cos2 ✓ * sin2 ✓ = 0 Ÿ 3 cos2 ✓ * 1 = 0 Ÿ ✓ = 54.74˝
The surface is a cone of angle 54.74˝ to the z axis, and extending over all values of r within the
limits of validity of the given field.
11
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1.28. State whether or not A = B and, if not, what conditions are imposed on A and B when
a) A ax = B ax : For this to be true, both A and B must be oriented at the same angle, ✓, from
the x axis. But this would allow either vector to lie anywhere along a conical surface of angle ✓
about the x axis. Therefore, A can be equal to B, but not necessarily.
b) A ù ax = B ù ax : This is a more restrictive condition because the cross product gives a vector.
For both cross products to lie in the same direction, A, B, and ax must be coplanar. But if A lies
at angle ✓ to the x axis, B could lie at ✓ or at 180˝ * ✓ to give the same cross product. So again,
A can be equal to B, but not necessarily.
c) A ax = B ax and A ù ax = B ù ax : In this case, we need to satisfy both requirements in parts
a and b – that is, A, B, and ax must be coplanar, and A and B must lie at the same angle, ✓, to
ax . With coplanar vectors, this latter condition might imply that both +✓ and *✓ would therefore
work. But the negative angle reverses the direction of the cross product direction. Therefore both
vectors must lie in the same plane and lie at the same angle to x; i.e., A must be equal to B.
…
d) A C = B C and AùC = BùC where C is any vector except C = 0: This is just the general case
of part c. Since we can orient our coordinate system in any manner we choose, we can arrange it
so that the x axis coincides with the direction of vector C. Thus all the arguments of part c apply,
and again we conclude that A must be equal to B.
1.29. A vector field is expressed as F = 10zaÜ z . Evaluate the components of this field that are a) normal, and
b) tangent to a spherical surface of radius a:
a) The normal component is the radial component, found by projecting F into the ar direction. In
general,
Fr = F ar = 10zaz ar = 10z cos ✓
where, on the sphere surface, z = a cos ✓. Therefore Fr (on surface) = 10a cos2 ✓.
b) The tangential component is found by projecting F into the a✓ or a directions. Note that
az a = 0, so we are left with the tangential component lying in only the direction of a✓ :
F✓ = 10zaz a✓ r=a = *10a cos ✓ sin ✓
12
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1.30. Consider a problem analogous to the varying wind velocities encountered by transcontinental aircraft.
We assume a constant altitude, a plane earth, a flight along the x axis from 0 to 10 units, no vertical
velocity component, and no change in wind velocity with time. Assume ax to be directed to the east
and ay to the north. The wind velocity at the operating altitude is assumed to be:
v(x, y) =
(0.01x2 * 0.08x + 0.66)ax * (0.05x * 0.4)ay
1 + 0.5y2
a) Determine the location and magnitude of the maximum tailwind encountered: Tailwind would
be x-directed, and so we look at the x component only. Over the flight range, this function
maximizes at a value of 0.86_(1 + 0.5y2 ) at x = 10 (at the end of the trip). It reaches a local
minimum of 0.50_(1 + 0.5y2 ) at x = 4, and has another local maximum of 0.66_(1 + 0.5y2 ) at
the trip start, x = 0.
b) Repeat for headwind: The x component is always positive, and so therefore no headwind exists
over the travel range.
c) Repeat for crosswind: Crosswind will be found from the y component, which is seen to maximize
over the flight range at a value of 0.4_(1 + 0.5y2 ) at the trip start (x = 0).
d) Would more favorable tailwinds be available at some other latitude? If so, where? Minimizing
the denominator accomplishes this; in particular, the lattitude associated with y = 0 gives the
strongest tailwind.
13
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CHAPTER 2 – 9th Edition
˘
2.1. Three positive point charges of equal magnitude q are located at x = *2, y = 2, and y = * 2. Find
the coordinates of a fourth positive charge, also of magnitude q, that will yield a zero net electric field
at the origin: The field at the origin that arises from the three charges can be expressed as
L
H
I M
⇧
q
q ⌅
1
1
1
Eo =
ax + ˘
* 2 ay =
ax + ay
2
4⇡✏0 2
16⇡✏0
( 2)2 2
The magnitude of this field is
Eo  = Eo Eo
1_2
˘
=
2q
16⇡✏0
˝ line in the first quadrant. Its
To counter this field, the fourth charge must be positioned
t ˘along a 45
˘
distance from the origin along this line will be d = 4_ 2 = 21_4 2 = 1.68. This translates into
equal x and y coordinates of 21_4 = 1.19. Therefore the fourth charge of positive magnitude q is
located at (1.19, 1.19)
2.2. Point charges of 1nC and -2nC are located at (0,0,0) and (1,1,1), respectively, in free space. Determine
the vector force acting on each charge.
First, the electric field intensity associated with the 1nC charge, evalutated at the -2nC charge
location is:
H
I
1
1
E12 =
ax + ay + az
nC_m
˘
4⇡✏0 (3)
3
˘
in which the distance between charges is 3 m. The force on the -2nC charge is then
F12 = q2 E12 =
*2
*1
ax + ay + az =
a + ay + az
˘
10.4
⇡✏0 x
12 3 ⇡✏0
nN
The force on the 1nC charge at the origin is just the opposite of this result, or
F21 =
+1
a + ay + az
10.4 ⇡✏0 x
nN
2.3. Point charges of 50nC each are located at A(1, 0, 0), B(*1, 0, 0), C(0, 1, 0), and D(0, *1, 0) in free
space. Find the total force on the charge at A.
The force will be:
4
5
RDA
RBA
(50 ù 10*9 )2 RCA
F=
+
+
4⇡✏0
RCA 3 RDA 3 RBA 3
where RCA = ax * ay , RDA = ax + ay , and RBA = 2ax . The magnitudes are RCA  = RDA  =
˘
2, and RBA  = 2. Substituting these leads to
L
M
(50 ù 10*9 )2
1
1
2
F=
ax = 21.5ax N
˘ + ˘ +
4⇡✏0
2 2 2 2 8
where distances are in meters.
14
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2.4. Eight identical point charges of Q C each are located at the corners of a cube of side length a, with
one charge at the origin, and with the three nearest charges at (a, 0, 0), (0, a, 0), and (0, 0, a). Find an
expression for the total vector force on the charge at P (a, a, a), assuming free space:
」
The total electric field at P (a, a, a) that produces a force on the charge there will be the sum of the
fields from the other seven charges. This is written below, where the charge locations associated
with each term are indicated:
Enet (a, a, a)
:
범
b
c
f
g
g
q ff ax + ay + az ay + az ax + az ax + ay
g
=
+
+
+
+
a
+
a
+
a
˘
˘
˘
˘
x
y
z
4⇡✏0 a2 f
´Ø¨ ´Ø¨ ´Ø¨g
3
3
2
2
2
2
2
2
f´≠≠≠≠≠Ø≠≠≠≠≠¨ ´Ø¨ ´Ø¨ ´Ø¨ (0,a,a)
g
(a,a,0)
(a,0,a)
f
g
(0,0,0)
(a,0,0)
(0,a,0)
(0,0,a)
d
e
,
The force is now the product of this field and the charge at (a, a, a). Simplifying, we obtain
F(a, a, a) = qEnet (a, a, a)
L
M
q2
1.90 q 2
1
1
=
+
+
1
a
+
a
+
a
=
ax + ay + az
˘
˘
x
y
z
4⇡✏0 a2 3 3
4⇡✏0 a2
2
in which the magnitude is F = 3.29 q 2 _(4⇡✏0 a2 ).
2.5. A point charge of 3nC is located at (1,1,1) in free space. What charge must be located at (1,3,2) to
cause the y component of E to be zero at the origin?
For two point charges, we may write:
E=
q1 (r * r1® )
q2 (r * r2® )
+
4⇡✏0 r * r1® 3 4⇡✏0 r * r2® 3
where q1 = 3nC, and where q2 is to be found. With q1 located at (1,1,1), r1® = ax + ay + az . The
position vector for q2 is then r2® = ax + 3ay + 2az . Because the observation point is at the origin, we
have r = 0. The field now becomes:
4
5
*3(ax + ay + az ) *q2 (ax + 3ay + 2az )
1
E=
+
4⇡✏0 (12 + 12 + 12 )3_2
(12 + 32 + 22 )3_2
For a zero y component, we thus find q2 = *(14)3_2 _33_2 = *10.1 nC.
15
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2.6. Two point charges of equal magnitude q are positioned at z = ±d_2.
a) find the electric field everywhere on the z axis: For a point charge at any location, we have
E=
q(r * r ® )
4⇡✏0 r * r ® 3
In the case of two charges, we would therefore have
ET =
q1 (r * r1® )
4⇡✏0 r * r1® 3
+
q2 (r * r2® )
(1)
4⇡✏0 r * r2® 3
In the present case, we assign q1 = q2 = q, the observation point position vector as r = zaz , and
the charge position vectors as r1® = (d_2)az , and r2® = *(d_2)az Therefore
r * r1® = [z * (d_2)]az ,
then
r * r1 3 = [z * (d_2)]3
r * r2® = [z + (d_2)]az ,
and
r * r2 3 = [z + (d_2)]3
Substitute these results into (1) to obtain:
4
5
q
1
1
ET (z) =
+
a
4⇡✏0 [z * (d_2)]2 [z + (d_2)]2 z
V_m
(2)
b) find the electric field everywhere on the xy plane: We proceed as in part a, except that now r lies
in the xy plane. For simplicity, we can choose the x axis on which to evaluate the field, so that
r = xax . Eq. (1) becomes
4
5
xax * (d_2)az
xax + (d_2)az
q
ET (x) =
+
(3)
4⇡✏0 xax * (d_2)az 3 xax + (d_2)az 3
where
⌅
⇧1_2
xax * (d_2)az  = xax + (d_2)az  = x2 + (d_2)2
Therefore (3) becomes
ET (x) =
2qx ax
⌅
⇧3_2
4⇡✏0 x2 + (d_2)2
This result can be generalized to apply anywhere in the xy plane by noting that the problem
exhibits cylindrical symmetry – any rotation of the x axis about the z axis will produce no change.
Therefore, we may use cylindrical coordinates, and replace the x variable by the radial variable
⇢, and use a⇢ instead of ax . The field then becomes:
ET (⇢) =
2q⇢ a⇢
⌅
⇧3_2
4⇡✏0 ⇢2 + (d_2)2
16
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2.7. Two point charges of equal magnitude but of opposite sign are positioned with charge +q at z = d_2,
and charge *q at z = *d_2. The pair form an electric dipole.
a) Find the electric field intensity everywhere on the z axis: For the two charges we would write in
general:
q(r * r+® )
q(r * r*® )
E=
*
4⇡✏0 r * r+® 3 4⇡✏0 r * r*® 3
where r = zaz , r+® = +d_2 az , and r*® = *d_2 az . Using these substitutions, we find:
q
E=
4⇡✏0
L
(z * d2 )az
z * d2 3
*
(z + d2 )az
M
z + d2 3
b) Evaluate your part a result at the origin: We set z = 0 in the above result to obtain
ㅡ
E(z = 0) =
*2q az ⌧ 2 2 *2q az
=
4⇡✏0 d
⇡✏0 d 2
as expected
c) Find the electric field intensity everywhere on the xy plane, expressing your result as a function
of radius ⇢ in cylindrical coordinates: This will begin with the same initial setup as in part a,
except now r = ⇢a⇢ describes the observation point in the xy plane. With this change, we have
4
5
⇢a⇢ * (d_2)az
⇢a⇢ + (d_2)az
*qd az
q
E=
* 2
=
2
2
3_2
2
3_2
2
4⇡✏0 [⇢ + (d_2) ]
[⇢ + (d_2) ]
4⇡✏0 [⇢ + (d_2)2 ]3_2
d) Evaluate your part c result at the origin: Setting ⇢ = 0 in the part c expression, we find:
E(⇢ = 0) =
*2q az
⇡✏0 d 2
:
as in part b, * and as expected
e) Simplify your part c result for the case in which ⇢ >> d: With this requirement, we find
. *qd az
E(⇢ >> d) =
4⇡✏0 ⇢3
2.8. A crude device for measuring charge consists of two small insulating spheres of radius a, one of which
is fixed in position. The other is movable along the x axis, and is subject to a restraining force kx,
where k is a spring constant. The uncharged spheres are centered at x = 0 and x = d, the latter fixed.
If the spheres are given equal and opposite charges of Q coulombs:
a) Obtain the expression by which Q may be found as a function of x: The spheres will attract, and
so the movable sphere at x = 0 will move toward the other until the spring and Coulomb forces
balance. This will occur at location x for the movable sphere. With equal and opposite forces,
we have
Q2
= kx
4⇡✏0 (d * x)2
˘
from which Q = 2(d * x) ⇡✏0 kx.
17
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2.8
b) Determine the maximum charge that can be measured in terms of ✏0 , k, and d, and state the
separation of the spheres then:
With increasing charge, the spheres move toward each other until they just touch at xmax = d * 2a.
˘
Using the part a result, we find the maximum measurable charge: Qmax = 4a ⇡✏0 k(d * 2a).
Presumably some form of stop mechanism is placed at x = x*
max to prevent the spheres from
actually touching.
c) What happens if a larger charge is applied? No further motion is possible, so nothing happens.
2.9. A 100 nC point charge is located at A(*1, 1, 3) in free space.
a) Find the locus of all points P (x, y, z) at which Ex = 500 V/m: The total field at P will be:
EP =
100 ù 10*9 RAP
4⇡✏0
RAP 3
where RAP = (x+1)ax +(y*1)ay +(z*3)az , and where RAP  = [(x+1)2 +(y*1)2 +(z*3)2 ]1_2 .
The x component of the field will be
4
5
(x + 1)
100 ù 10*9
Ex =
= 500 V_m
4⇡✏0
[(x + 1)2 + (y * 1)2 + (z * 3)2 ]1.5
And so our condition becomes:
(x + 1) = 0.56 [(x + 1)2 + (y * 1)2 + (z * 3)2 ]1.5
b) Find y1 if P (0, y1 , 3) lies on that locus: At point P , the condition of part a becomes
⌅
⇧3
3.19 = 1 + (y1 * 1)2
from which (y1 * 1)2 = 0.47, or y1 = 1.69 or 0.31
2.10. A configuration of point charges consists of a single charge of value *2q at the origin, and two charges
of value +q at locations z = *d and +d. The charges as positioned form an electric quadrupole,
equivalent to two dipoles of opposite orientation that are separated by distance d along the z axis.
a) Find the electric field intensity E everywhere in the xy plane, expressing your result as a function
of cylindrical radius ⇢: We begin by applying the general formula for the point charge field, where
the three terms apply to the three charges:
E=
®
q(r * rlower
)
®
4⇡✏0 r * rlower

*
3
®
2q(r * rmiddle
)
®
4⇡✏0 r * rmiddle

+
3
®
q(r * rupper
)
®
4⇡✏0 r * rupper
3
®
®
®
The position vectors will be r = ⇢a⇢ , rlower
= *daz , rupper
= +daz , and rmiddle
= 0. With these
substitutions, the field expression becomes:
4
5
4
5
⇢a⇢ + daz
2⇢a⇢
⇢a⇢ * daz
*qa⇢
q
1
E=
* 3 + 2
=
1*
4⇡✏0 (⇢2 + d 2 )3_2
⇢
(⇢ + d 2 )3_2
2⇡✏0 ⇢2
(1 + d 2 _⇢2 )3_2
18
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2.10
b) Specialize your part a result for large distances, ⇢ >> d: Under this condition, we may use the
expansion:
1
(1 + d 2 _⇢2 )3_2
= 1 + d 2 _⇢2
*1
1 + d 2 _⇢2
*1_2
.
= 1 * d 2 _⇢2
1 * d 2 _2⇢2
Carrying out the product and neglecting the term involving d 4 _⇢4 , we find:
1 * d 2 _⇢2
from which
.
3 d2
1 * d 2 _2⇢2 = 1 *
2 ⇢2
4
5
*3qd 2 a⇢
. *qa⇢
3 d2
E(⇢ >> d) =
1
*
1
+
=
2 ⇢2
2⇡✏0 ⇢2
4⇡✏0 ⇢4
2.11. A charge Q0 located at the origin in free space produces a field for which Ez = 1 kV/m at point
P (*2, 1, *1).
a) Find Q0 : The field at P will be
4
5
Q0 *2ax + ay * az
EP =
4⇡✏0
61.5
Since the z component is of value 1 kV/m, we find Q0 = *4⇡✏0 61.5 ù 103 = *1.63 C.
b) Find E at M(1, 6, 5) in cartesian coordinates: This field will be:
4
5
*1.63 ù 10*6 ax + 6ay + 5az
EM =
4⇡✏0
[1 + 36 + 25]1.5
or EM = *30.11ax * 180.63ay * 150.53az .
c) Find E at M(1, 6, 5) in cylindrical coordinates: At M, ⇢ =
80.54˝ , and z = 5. Now
E⇢ = EM a⇢ = *30.11 cos
˘
1 + 36 = 6.08,
* 180.63 sin
E = EM a = *30.11(* sin ) * 180.63 cos
= tan*1 (6_1) =
= *183.12
= 0 (as expected)
so that EM = *183.12a⇢ * 150.53az .
˘
d) Find E at M(1, 6, 5) in spherical coordinates: At M, r = 1 + 36 + 25 = 7.87, = 80.54˝ (as
before), and ✓ = cos*1 (5_7.87) = 50.58˝ . Now, since the charge is at the origin, we expect to
obtain only a radial component of EM . This will be:
Er = EM ar = *30.11 sin ✓ cos
* 180.63 sin ✓ sin
* 150.53 cos ✓ = *237.1
19
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2.12. Electrons are in random motion in a fixed region in space. During any 1 s interval, the probability of
finding an electron in a subregion of volume 10*15 m2 is 0.27. What volume charge density, appropriate
for such time durations, should be assigned to that subregion?
The finite probabilty effectively reduces the net charge quantity by the probability fraction. With
e = *1.602 ù 10*19 C, the density becomes
⇢v = *
0.27 ù 1.602 ù 10*19
= *43.3 C_m3
10*15
2.13. A uniform volume charge density of 0.2 C_m3 is present throughout the spherical shell extending
from r = 3 cm to r = 5 cm. If ⇢v = 0 elsewhere:
a) find the total charge present throughout the shell: This will be
Q=
2⇡
0
⇡
0
.05
.03
4
5.05
r3
0.2 r sin ✓ dr d✓ d = 4⇡(0.2)
= 8.21 ù 10*5 C = 82.1 pC
3 .03
2
b) find r1 if half the total charge is located in the region 3 cm < r < r1 : If the integral over r in part
a is taken to r1 , we would obtain
4
5r
r3 1
4⇡(0.2)
= 4.105 ù 10*5
3 .03
Thus
4
3 ù 4.105 ù 10*5
r1 =
+ (.03)3
0.2 ù 4⇡
51_3
= 4.24 cm
2.14. The electron beam in a certain cathode ray tube possesses cylindrical symmetry, and the charge density is represented by ⇢v = *0.1_(⇢2 + 10*8 ) pC_m3 for 0 < ⇢ < 3 ù 10*4 m, and ⇢v = 0 for
⇢ > 3 ù 10*4 m.
a) Find the total charge per meter along the length of the beam: We integrate the charge density
over the cylindrical volume having radius 3 ù 10*4 m, and length 1m.
q=
1
0
3ù10*4
2⇡
0
*0.1
(⇢2 + 10*8 )
0
⇢ d⇢ d dz
From integral tables, this evaluates as
q = *0.2⇡
⇠ ⇡
*4
1
Û3ù10
ln ⇢2 + 10*8 Û
= 0.1⇡ ln(10) = *0.23⇡ pC_m
Û0
2
b) if the electron velocity is 5ù107 m/s, and with one ampere defined as 1C/s, find the beam current:
Current = charge_mùv = *0.23⇡ [pC_m]ù5ù107 [m_s] = *11.5⇡ù106 [pC_s] = *11.5⇡ A
20
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2.15. A spherical volume having a 2 m radius contains a uniform volume charge density of 105 C_m3 .
a) What total charge is enclosed in the spherical volume?
This will be Q = (4_3)⇡(2 ù 10*6 )3 ù 105 = 3.35 ù 10*12 C.
b) Now assume that a large region contains one of these little spheres at every corner of a cubical
grid 3mm on a side, and that there is no charge between spheres. What is the average volume
charge density throughout this large region? Each cube will contain the equivalent of one little
sphere. Neglecting the little sphere volume, the average density becomes
⇢v,avg =
3.35 ù 10*12
= 1.24 ù 10*4 C_m3
(0.003)3
2.16. Within a region of free space, charge density is given as ⇢v = ⇢0 r_a cos ✓ C_m3 , where ⇢0 and a are
constants. Find the total charge lying within:
a) the sphere, r f a: This will be
Qa =
2⇡
0
⇡
0
a
0
⇢0 r
cos ✓ r2 sin ✓ dr d✓ d = 0
a
It is the integral over ✓, performed first, that gives the zero result.
b) the cone, r f a, 0 f ✓ f 0.1⇡:
Qb =
2⇡
0
0.1⇡
0
a
0
⇧
⇢0 r
⇢ a3 ⌅
cos ✓ r2 sin ✓ dr d✓ d = ⇡ 0
1 * cos2 (0.1⇡) = 0.024⇡⇢0 a3
a
4
c) the region, r f a, 0 f ✓ f 0.1⇡, 0 f
Qc =
0.2⇡
0
0.1⇡
0
f 0.2⇡.
⇠
⇡
⇢0 r
0.2⇡
cos ✓ r2 sin ✓ dr d✓ d = 0.024⇡⇢0 a3
= 0.0024⇡⇢0 a3
a
2⇡
a
0
2.17. A length d of line charge lies on the z axis in free space. The charge density on the line is
⇢L = +⇢0 C/m (0 < z < d_2) and ⇢L = *⇢0 C/m (*d_2 < z < 0), where ⇢0 is a positive constant.
a) Find the electric field intensity E everywhere in the xy plane, expressing your result as a function
of cylindrical radius ⇢: Begin by constructing the differential field at radius ⇢ in the xy plane that
arises from a point charge dq = ⇢L dz on the z axis. To do this, use the general expression:
dE =
dq(r * r ® )
4⇡✏0 r * r ® 3
where in this case, r = ⇢a⇢ and r ® = zaz . With these substitutions, we find
dE =
±⇢0 dz(⇢a⇢ * zaz )
4⇡✏0 (⇢2 * z2 )3_2
where the positive sign applies to the region z > 0; the negative sign to z < 0. The total field at
radius ⇢ is then found by integrating dE over the total charge length:
E=
0
*⇢0 dz(⇢a⇢ * zaz )
+
2
2 3_2
*d_2 4⇡✏0 (⇢ * z )
+d_2 +⇢ dz(⇢a * za )
0
⇢
z
0
4⇡✏0 (⇢2 * z2 )3_2
21
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2.17.
a) (continued) Note that the radial component will integrate to zero through odd parity, leaving only
the z component, as would be expected. The integral simplifies to:
L
M
d_2
*⇢0 zaz dz
⇢0 az
1
Ûd_2 *⇢0 az
E=2
=
=
1* ˘
Û
2⇡✏0 ⇢
4⇡✏0 (⇢2 + z2 )3_2
2⇡✏0 (⇢2 + z2 )1_2 Û0
0
1 + (d_2⇢)2
b) Simplify your part a result for the case in which radius ⇢ >> d, and express this result in terms
of charge q = ⇢0 d_2: At large radii, we can use the binomial expansion to the first two terms:
1
.
1
= 1*
˘
2
1 + (d_2⇢)2
with which
.
E =
*⇢0 d 2
16⇡✏0 ⇢3
az =
0
d
2⇢
12
*qd
az
4⇡✏0 ⇢3
where q = ⇢0 d_2.
2.18.
a) Find E in the plane z = 0 that is produced by a uniform line charge, ⇢L , extending along the z
axis over the range *L < z < L in a cylindrical coordinate system: We find E through
E=
⇢L dz(r * r ® )
® 3
*L 4⇡✏0 r * r 
L
where the observation point position vector is r = ⇢a⇢ (anywhere in the x-y plane), and where
the position vector that locates any differential charge element on the z axis is r ® = zaz . So
r * r ® = ⇢a⇢ * zaz , and r * r ®  = (⇢2 + z2 )1_2 . These relations are substituted into the integral
to yield:
L ⇢ dz(⇢a * za )
⇢L ⇢ a⇢ L
L
⇢
z
dz
E=
=
= E⇢ a⇢
2 + z2 )3_2
2 + z2 )3_2
4⇡✏
4⇡✏
(⇢
(⇢
*L
*L
0
0
Note that the second term in the left-hand integral (involving zaz ) has effectively vanished because it produces equal and opposite sign contributions when the integral is taken over symmetric
limits (odd parity). Evaluating the integral results in
E⇢ =
⇢L ⇢
⇢L
⇢L
z
L
1
ÛL
=
Û =
˘
˘
˘
Û
*L
4⇡✏0 ⇢2 ⇢2 + z2
2⇡✏0 ⇢ ⇢2 + L2
2⇡✏0 ⇢ 1 + (⇢_L)2
Note that as L ô ÿ, the expression reduces to the expected field of the infinite line charge in
free space, ⇢L _(2⇡✏0 ⇢).
b) if the finite line charge is approximated by an infinite line charge (L ô ÿ), by what percentage
is E⇢ in error if ⇢ = 0.5L? The percent error in this situation will be
L
% error = 1 * ˘
1
1 + (⇢_L)2
M
ù 100
For ⇢ = 0.5L, this becomes % error = 10.6 %
c) repeat b with ⇢ = 0.1L. For this value, obtain % error = 0.496 %.
22
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2.19. A line having charge density ⇢0 z C/m and of length l is oriented along the z axis at *l_2 < z < l_2.
a) Find the electric field intensity E everywhere in the xy plane, expressing your result in cylindrical
coordinates: As the problem exhibits cylindrical symmetry, we may write the position vector for
the observation point in the xy plane as r = ⇢a⇢ . Then, with r ® = zaz , we may write
E=
l_2 ⇢ zdz(⇢a * za )
0
⇢
z
⇢L (z)dz(r * r ® )
=
® 3
*l_2 4⇡✏0 r * r 
l_2
*l_2
4⇡✏0 (⇢2 + z2 )3_2
mm
Note that the second term in the integrand (the z component) is zero, because of odd parity.
We are left with
E=
⇢0 ⇢ a⇢
4⇡✏0
2⇢0 ⇢ a⇢
zdz
=
2
2 3_2
4⇡✏0
*l_2 (⇢ + z )
l_2
l_2
0
*⇢0 ⇢ a⇢
zdz
1
Ûl_2
=
Û
2⇡✏0 (⇢2 + z2 )1_2 Û0
(⇢2 + z2 )3_2
Evaluating the limits the final result can be written as
L
M
⇢ 0 a⇢
1
E=
1* ˘
V_m
2⇡✏0
1 * (l_2⇢)2
b) Evaluate your result of part a in the limit as l (not z) approaches infinity: In this limit, the second
term in the bracket tends to zero, and we have
E(l ô ÿ) =
⇢ 0 a⇢
2⇡✏0
V_m
thus exhibiting no radial variation!
2.20. A line charge of uniform charge density ⇢0 C_m and of length l, is oriented along the z axis at
*l_2 < z < l_2.
a) Find the electric field strength, E, in magnitude and direction at any position along the x axis:
This follows the method in Problem 2.18. We find E through
E=
l_2
⇢0 dz(r * r ® )
® 3
*l_2 4⇡✏0 r * r 
where the observation point position vector is r = xax (anywhere on the x axis), and where
the position vector that locates any differential charge element on the z axis is r ® = zaz . So
r * r ® = xax * zaz , and r * r ®  = (x2 + z2 )1_2 . These relations are substituted into the integral
to yield:
l_2
⇢0 dz(xax * zaz ) ⇢0 x ax l_2
dz
E=
=
= Ex ax
2
2
3_2
2
4⇡✏0 *l_2 (x + z2 )3_2
*l_2 4⇡✏0 (x + z )
Note that the second term in the left-hand integral (involving zaz ) has effectively vanished because it produces equal and opposite sign contributions when the integral is taken over symmetric
limits (odd parity). Evaluating the integral results in
Ex =
⇢0 x
⇢0
⇢0
l_2
z
1
Ûl_2
=
=
Û
˘
˘
˘
4⇡✏0 x2 x2 + z2 Û*l_2 2⇡✏0 x x2 + (l_2)2
2⇡✏0 x 1 + (2x_l)2
23
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2.20.
b) with the given line charge in position, find the force acting on an identical line charge that is
oriented along the x axis at l_2 < x < 3l_2: The differential force on an element of the xdirected line charge will be dF = dqE = (⇢0 dx)E, where E is the field as determined in part
a. The net force is then the integral of the differential force over the length of the horizontal line
charge, or
3l_2
⇢20
1
F=
dx ax
˘
2⇡✏
x
l_2
0
1 + (2x_l)2
This can be re-written and then evaluated using integral tables as
L
M3l_2
˘
a
2 + (l_2)2
⇢20 l ax 3l_2
*⇢20 l ax ` 1
l_2
+
x
dx
r
s
F=
=
ln
˘
s
4⇡✏0 l_2 x x2 + (l_2)2
4⇡✏0 r (l_2)
x
l_2 q
p
⇠
˘ ⇡c
L
b
˘ M
*⇢20 ax f (l_2) 1 + 10 g ⇢20 ax
0.55⇢20
3(1 + 2)
=
ln f
ln
=
a N
⇠
˘
˘ ⇡g =
2⇡✏0
2⇡✏0 x
1 + 10
f 3(l_2) 1 + 2 g 2⇡✏0
d
e
2.21. A charged filament forms a circle of radius a in the xy plane with center at the origin. The filament
carries uniform line charge density +⇢0 C/m for *⇡_2 < < ⇡_2, and *⇢0 C/m for ⇡_2 < < 3⇡_2.
Find the electric field intensity at the origin:
The field at the origin arising from a differential length ad of charge on the ring will be
dEo =
±⇢0 ad
4⇡✏0 a2
a⇢
where the positive sign applies to the negative charge contribution, the negative sign to the positive
charge contribution. Now the total field will be the piecewise integral of the differential field over the
two semi-circles. Using a⇢ = cos ax + sin ay (as is required to include all dependence), we have
Eo =
*⇢0 d
(cos ax + sin ay ) +
*⇡_2 4⇡✏0 a
⇡_2
3⇡_2
⇡_2
+⇢0 d
(cos ax + sin ay )
4⇡✏0 a
Noting that the two terms in the above expression are equal to each other, we may evaluate the first
integral and introduce a factor of 2:
b
c
f
g
*⇢0 ax
*2⇢0 f
Û⇡_2
Û⇡_2
Eo =
sin Û
ax * cos Û
ay g =
V_m
Û*⇡_2
Û*⇡_2 g
4⇡✏0 a f
⇡✏0 a
´≠≠Ø≠≠¨ g
f´≠≠Ø≠≠¨
d
e
2
0
2.22. Two identical uniform sheet charges with ⇢s = 100 nC_m2 are located in free space at z = ±2.0 cm.
What force per unit area does each sheet exert on the other?
The field from the top sheet is E = *⇢s _(2✏0 ) az V/m. The differential force produced by this
field on the bottom sheet is the charge density on the bottom sheet times the differential area
there, multiplied by the electric field from the top sheet: dF = ⇢s daE. The force per unit area is
then just F = ⇢s E = (100 ù 10*9 )(*100 ù 10*9 )_(2✏0 ) az = *5.6 ù 10*4 az N_m2 .
24
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2.23. A disk of radius a in the xy plane carries surface charge of density ⇢s = ⇢0 _⇢, where ⇢0 is a constant.
Find the electric field strength, E, everywhere on the z axis.
We find the field through
⇢s da(r * r ® )
E=
4⇡✏0 r * r ® 3
where the integral is taken over the surface of the disk, and where r = zaz and r ® = ⇢a⇢ . The
integral then becomes
E=
a (⇢ _⇢) ⇢ d⇢ d
0
2⇡
0
zaz * ⇢a⇢
4⇡✏0 (z2 + ⇢2 )3_2
0
In evaluating this integral, we need to introduce the dependence in a⇢ by writing it as a⇢ =
cos ax + sin ay , where ax and ay are invariant in their orientation as varies. So the integral
now simplifies to
L
Ma
2⇡⇢0 z az ÿ
2⇡⇢0 z az
2⇡a⇢0 az
d⇢
⇢
E=
=
=
˘
4⇡✏0
4⇡✏0
(z2 + ⇢2 )3_2
4⇡✏0 z2 [1 + (a_z)2 ]1_2
0
z2 z2 + ⇢2
⇢=0
2.24.
a) Find the electric field on the z axis produced by an annular ring of uniform surface charge density
⇢s in free space. The ring occupies the region z = 0, a f ⇢ f b, 0 f f 2⇡ in cylindrical
coordinates: We find the field through
⇢s da(r * r ® )
E=
4⇡✏0 r * r ® 3
where the integral is taken over the surface of the annular ring, and where r = zaz and r ® = ⇢a⇢ .
The integral then becomes
E=
2⇡
0
b ⇢
a
s ⇢ d⇢ d
zaz * ⇢a⇢
4⇡✏0 (z2 + ⇢2 )3_2
In evaluating this integral, we first note that the term involving ⇢a⇢ integrates to zero over the
integration range of 0 to 2⇡. This is because we need to introduce the dependence in a⇢
by writing it as a⇢ = cos ax + sin ay , where ax and ay are invariant in their orientation as
varies. So the integral now simplifies to
L
Mb
2⇡⇢s z az b
⇢s z az
⇢ d⇢
*1
E=
=
˘
2
2 3_2
4⇡✏0
2✏0
a (z + ⇢ )
z2 + ⇢ 2 a
L
M
⇢
1
1
= s ˘
*˘
az
2✏0
1 + (a_z)2
1 + (b_z)2
b) from your part a result, obtain the field of an infinite uniform sheet charge by taking appropriate
limits. The infinite sheet is obtained by letting a ô 0 and b ô ÿ, in which case E ô ⇢s _(2✏0 ) az
as expected.
25
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2.25. A disk of radius a in the xy plane carries surface charge of density ⇢s1 = +⇢s0 _⇢ C_m2 for 0 <
and ⇢s2 = *⇢s0 _⇢ C_m2 for ⇡ < < 2⇡, where ⇢s0 is a constant.
a) Find the electric field intensity E everywhere on the
z axis:
z
z
With the setup shown at right, the field can be found
through the general relation
E=
⇢s da(r * r ® )
E=
+
a +(⇢ _⇢)[za * ⇢ cos
s0
z
⇡
0
⇡
2⇡
0
r-r
r
® 3
disk area 4⇡✏0 r * r 
where r = zaz and r ® = ⇢a⇢ . Because the charge is
-dependent, we need to include the dependence
in all terms. This means that we must use a⇢ =
ax cos +ay sin . Then, with r*r ® 3 = (z2 +⇢2 )3_2
and with the positive and negative charges accounted
for, the integral is performed piecewise over the two
halves of the disk:
< ⇡,
�s2
�s1
r
da
y
x
ax * ⇢ sin ay ] ⇢d⇢d
4⇡✏0 (z2 + ⇢2 )3_2
a *(⇢ _⇢)[za * ⇢ cos a * ⇢ sin a ] ⇢d⇢d
s0
z
x
y
4⇡✏0 (z2 + ⇢2 )3_2
0
Noting that the z components will cancel, and performing the integration first, we find:
⌅
⇧
a * sin ⇡ a + cos ⇡ a + sin 2⇡ a * cos 2⇡ a ⇢d⇢
⇢s0
⇡ x
⇡ y
0 x
0 y
E=
4⇡✏0 0
(z2 + ⇢2 )3_2
L
M
4
5a
*4⇢s0 ay a
*⇢s0 ay
*⇢
a
⇢d⇢
s0
y
*1
1
=
=
=
1* ˘
2 + ⇢2 )3_2
2 + ⇢2 )3_2
4⇡✏0
⇡✏
⇡✏
z
(z
(z
0
0
0
0
1 + a2 _z2
.
b) Specialize your part a result for distances z >> a: With this condition we have (1 + a2 _z2 )*1_2 =
1 * a2 _2z2 and thus
4
5
*⇢s0 a2 ay
. *⇢s0 ay
a2
E(z >> a) =
1*1+ 2 =
⇡✏0 z
2z
2⇡✏0 z3
Note the inverse distance cubed dependence that is characteristic of a dipole field.
26
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2.26
a) Find the electric field intensity on the z axis produced by a cone surface that carries charge density
⇢s (r) = ⇢0 _r C_m2 in free space. The cone has its vertex at the origin and occupies the region
✓ = ↵, 0 < r < a, and 0 <
< 2⇡ in spherical coordinates. Differential area in spherical
coordinates is given as da = r sin ↵drd : E is found using the general surface integral
⇢s da(r * r ® )
E=
® 3
cone 4⇡✏0 r * r 
where in this case r = zaz and r ® = rar .
⌅
⇧1_2 ⌅
⇧1_2 ˘
Then r * r ®  = (r * r ® ) (r * r ® )
= (zaz * rar ) (zaz * rar )
= z2 * 2rz cos ↵ + r2 ,
where az ar = cos ↵ has been used. The integral now becomes:
2⇡
E=
0
a
0
(⇢0 _r)r sin ↵drd (zaz * rar )
4⇡✏0 (z2 * 2rz cos ↵ + r2 )3_2
It is necessary to include the dependence in ar .
We substitute ar = sin ↵ cos ax + sin ↵ sin ay + cos ↵az . The first two terms of this, involving
sin and cos , will integrate to zero when taking from 0 to 2⇡. Only the z component survives,
and the integral becomes, after performing the integration:
a
E = 2⇡
⇢0 sin ↵(z * r cos ↵)dr az
4⇡✏0 (z2 * 2rz cos ↵ + r2 )3_2
4
5
2⇡⇢0 sin ↵az a
zdr
r cos ↵dr
=
*
4⇡✏0
(z2 * 2rz cos ↵ + r2 )3_2 (z2 * 2rz cos ↵ + r2 )3_2
0
0
The two integrals evaluate as
a
0
a
0
zdr
=
(z2 * 2rz cos ↵ + r2 )3_2
(r * z cos ↵)z
Ûa
Û
z2 sin ↵(z2 * 2rz cos ↵ + r2 )1_2 Û0
2
*(z * r cos ↵)z cos ↵
r cos ↵dr
Ûa
=
Û
(z2 * 2rz cos ↵ + r2 )3_2
z2 sin2 ↵(z2 * 2rz cos ↵ + r2 )1_2 Û0
Combining these, cancelling terms, and rearranging, finally leads to
E=
2⇡a⇢0 sin ↵ az
4⇡✏0 z(z2 * 2az cos ↵ + a2 )1_2
b) Find the total charge on the cone: This will be
Qc =
cone
⇢s da =
2⇡
0
a
0
⇢0
r sin ↵ drd = 2⇡a⇢0 sin ↵
r
c) Specialize your result of part a to the case in which ↵ = 90˝ , at which the cone flattens to a disk
in the xy plane. Compare this result to the answer to Problem 2.23.
When ↵ = 90˝ , cos ↵ = 0, and the field expression in part a can be expressed as:
E(↵ = 0) =
2⇡a⇢0 az
2
4⇡✏0 z [1 + (a_z)2 ]1_2
which is the same result as found in Problem 2.23.
27
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2.26
d) Show that your part a result becomes a point charge field when z >> a.
Using the expression for the total charge as found in part b, and rearranging terms, the field of
part a takes the form:
E=
Qc az
. Qc az
=
4⇡✏0 z2 [1 * (2a_z) cos ↵ + (a_z)2 ]1_2
4⇡✏0 z2
(z >> a)
where we recognize the second equality as the point charge field.
e) Show that your part a result becomes an inverse-z-dependent E field when z << a.
.
If z << a, we have [1 * (2a_z) cos ↵ + (a_z)2 ]1_2 = a_z and thus
E=
Qc az
2
4⇡✏0 z [1 * (2a_z) cos ↵ + (a_z)2 ]1_2
. Q aa
= c z
4⇡✏0 z
2.27. Given the electric field E = (4x * 2y)ax * (2x + 4y)ay , find:
a) the equation of the streamline that passes through the point P (2, 3, *4): We write
Ey
dy
*(2x + 4y)
=
=
dx Ex
(4x * 2y)
Thus
2(x dy + y dx) = y dy * x dx
or
2 d(xy) =
So
1
1
d(y2 ) * d(x2 )
2
2
1
1
C1 + 2xy = y2 * x2
2
2
or
y2 * x2 = 4xy + C2
Evaluating at P (2, 3, *4), obtain:
9 * 4 = 24 + C2 , or C2 = *19
Finally, at P , the requested equation is
y2 * x2 = 4xy * 19
b) a unit vector specifying the direction of E at Q(3, *2, 5): Have EQ = [4(3) + 2(2)]ax * [2(3) *
˘
4(2)]ay = 16ax + 2ay . Then E = 162 + 4 = 16.12 So
aQ =
16ax + 2ay
16.12
= 0.99ax + 0.12ay
28
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2.28 An electric dipole (discussed in detail in Sec. 4.7) consists of two point charges of equal and opposite
magnitude ±Q spaced by distance d. With the charges along the z axis at positions z = ±d_2 (with
the positive
⌅ charge at the⇧ ⌅positive z location),⇧ the electric field in spherical coordinates is given by
E(r, ✓) = Qd_(4⇡✏0 r3 ) 2 cos ✓ar + sin ✓a✓ , where r >> d. Using rectangular coordinates, determine expressions for the vector force on a point charge of magnitude q:
a) at (0, 0, z): Here, ✓ = 0, ar = az , and r = z. Therefore
F(0, 0, z) =
qQd az
4⇡✏0 z3
N
b) at (0, y, 0): Here, ✓ = 90˝ , a✓ = *az , and r = y. The force is
F(0, y, 0) =
*qQd az
4⇡✏0 y3
N
2.29. If E = 20e*5y cos 5xax * sin 5xay , find:
a) E at P (⇡_6, 0.1, 2): Substituting this point, we obtain EP = *10.6ax * 6.1ay , and so EP  =
12.2.
b) a unit vector in the direction of EP : The unit vector associated with E is cos 5xax * sin 5xay ,
which evaluated at P becomes aE = *0.87ax * 0.50ay .
c) the equation of the direction line passing through P : Use
dy * sin 5x
=
= * tan 5x Ÿ dy = * tan 5x dx
dx
cos 5x
Thus y = 15 ln cos 5x + C. Evaluating at P , we find C = 0.13, and so
y=
1
ln cos 5x + 0.13
5
2.30. For fields that do not vary with z in cylindrical coordinates, the equations of the streamlines are obtained by solving the differential equation E⇢ _E = d⇢(⇢d ). Find the equation of the line passing
through the point (2, 30˝ , 0) for the field E = ⇢ cos 2 a⇢ * ⇢ sin 2 a :
E⇢
E
=
d⇢
*⇢ cos 2
=
⇢d
⇢ sin 2
= * cot 2
Ÿ
4
5
Integrate to obtain
C
2 ln ⇢ = ln sin 2 + ln C = ln
sin 2
d⇢
= * cot 2 d
⇢
Ÿ ⇢2 =
C
sin 2
˘
˝ ) Ÿ C = 4 sin 60˝ = 2 3. Finally, the equation for
At the given point, we have
4
=
C_
sin(60
˘
the streamline is ⇢2 = 2 3_ sin 2 .
29
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CHAPTER 3 – 9th Edition
3.1. Suppose that the Faraday concentric sphere experiment is performed in free space using a central
charge at the origin, Q1 , and with hemispheres of radius a. A second charge Q2 (this time a point
charge) is located at distance R from Q1 , where R >> a.
a) What is the force on the point charge before the hemispheres are assembled around Q1 ? This
will be simply the force beween two point charges, or
F=
Q1 Q2
ar
4⇡✏0 R2
b) What is the force on the point charge after the hemispheres are assembled but before they are
discharged? The answer will be the same as in part a because induced charge Q1 now resides
as a surface charge layer on the sphere exterior. This produces the same electric field at the Q2
location as before, and so the force acting on Q2 is the same.
c) What is the force on the point charge after the hemispheres are assembled and after they are
discharged? Discharging the hemispheres (connecting them to ground) neutralizes the positive
outside surface charge layer, thus zeroing the net field outside the sphere. The force on Q2 is now
zero.
d) Qualitatively, describe what happens as Q2 is moved toward the sphere assembly to the extent
that the condition R >> a is no longer valid. Q2 itself begins to induce negative surface charge
on the sphere. An attractive force thus begins to strengthen as the charge moves closer. The point
charge field approximation used in parts a through c is no longer valid.
3.2. An electric field in free space is E = (5z2 _✏0 ) aÇ z V/m. Find the total charge contained within a
cube, centered at the origin, of 4-m side length, in which all sides are parallel to coordinate axes (and
therefore each side intersects an axis at ±2.
The flux density is D = ✏0 E = 5z2 az . As D is z-directed only, it will intersect only the top and
bottom surfaces (both parallel to the x-y plane). From Gauss’ law, the charge in the cube is equal
to the net outward flux of D, which in this case is
Qencl =
Õ
D n da =
2
2
*2
*2
5(2)2 az az dx dy +
2
2
*2
*2
5(*2)2 az (*az ) dx dy = 0
where the first and second integrals on the far right are over the top and bottom surfaces
respectively.
30
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3.3. Consider an electric dipole in free space, consisting of a point charge q at location z = +d_2, and a
point charge *q at location z = *d_2. The electric field intensity in the xy plane is (see Problem 2.7):
E=
*qd az
2
4⇡✏0 [⇢ + (d_2)2 ]3_2
where ⇢ is the radius from the origin in cylindrical coordinates.
a) Determine the net electric flux associated with this field that penetrates the xy plane:
The flux will be
e =
xy
D dS where D = ✏0 E and dS = *az ⇢d⇢d
Over the entire xy plane, the flus is then:
2⇡
ÿ
ÿ
*qdaz (*az )
2⇡qd
⇢d⇢d =
2 + (d_2)2 ]3_2
4⇡
4⇡[⇢
0
0
ÿ
qd
*1
Û
=
Û =q
2
2
1_2
2 [⇢ + (d_2) ] Û0
e =
0
⇢d⇢
[⇢2 + (d_2)2 ]3_2
b) Interpret your result as it relates to Gauss’ Law: In the dipole configuration, all field lines terminate on the two charges. Therefore, a closed surface that is drawn around the upper charge, and
which includes the entire xy plane will show outward flux that passes through the xy plane only.
This flux will equal the enclosed charge of q.
3.4. An electric field in free space is E = (5z3 _✏0 ) aÇ z V/m. Find the total charge contained within a sphere
of 3-m radius, centered at the origin. Using Gauss’ law, we set up the integral in free space over the
sphere surface, whose outward unit normal is ar :
Q=
Õ
✏0 E n da =
2⇡
0
⇡
0
5z3 az ar (3)2 sin ✓ d✓ d
where in this case z = 3 cos ✓ and (in all cases) az ar = cos ✓. These are substituted to yield
Q = 2⇡
⇡
0
5(3)5 cos4 ✓ sin ✓d✓ = *2⇡(5)(3)5
⇠ ⇡
1
Û⇡
cos5 ✓ Û = 972⇡
Û0
5
31
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3.5. A volume charge distribution in free space is characterized by the density:
q
exp(*z_d)
2Ad
⇢v =
where d is a distance along z, A is the area of a surface parallel to the xy plane, and q is a fixed charge
quantity. The charge distribution exists everywhere.
a) Find the electric field intensity E everywhere: Choose a Gaussian surface of thickness 2z0 , with
surfaces of area A located at z = ±z0 . From symmetry, the presumed D field will be directed
along +az for z > 0, and along *az for z < 0. It will therefore penetrate only the aforementioned
surfaces. Gauss’ Law takes the form:
Õ
D dS = 2A1 Dz  = Qencl = 2A1
z0
0
q
exp(*z_d) dz
2Ad
Therefore
Dz  =
⇧
⇧
q ⌅
q ⌅
1 * exp(*z0 _d) Ÿ E(z) = ±
1 * exp(*z_d) az V_m
2A
2✏0 A
where the positive sign applies for z > 0, the negative sign for z < 0.
b) What is the interpretation of q? q would be the charge contained within a volume of infinite
length in z, and of cross-sectional area A.
3.6. In free space, volume charge of constant density ⇢v = ⇢0 exists within the region *ÿ < x < ÿ,
*ÿ < y < ÿ, and *d_2 < z < d_2. Find D and E everywhere.
From the symmetry of the configuration, we surmise that the field will be everywhere z-directed,
and will be uniform with x and y at fixed z. For finding the field inside the charge, an appropriate
Gaussian surface will be that which encloses a rectangular region defined by *1 < x < 1,
*1 < y < 1, and z < d_2. The outward flux from this surface will be limited to that through
the two parallel surfaces at ±z:
in =
Õ
D dS = 2
1
1
*1
*1
Dz dxdy = Qencl =
z
1
1
*z
*1
*1
⇢0 dxdydz®
where the factor of 2 in the second integral account for the equal fluxes through the two surfaces.
The above readily simplifies, as both Dz and ⇢0 are constants, leading to
Din = ⇢0 z az C_m2 (z < d_2), and therefore Ein = (⇢0 z_✏0 ) az V_m (z < d_2).
Outside the charge, the Gaussian surface is the same, except that the parallel boundaries at ±z
occur at z > d_2. As a result, the calculation is nearly the same as before, with the only change
being the limits on the total charge integral:
out =
Õ
D dS = 2
1
1
*1
*1
Dz dxdy = Qencl =
d_2
1
1
*d_2
*1
*1
⇢0 dxdydz®
Solve for Dz to find the constant values:
<
<
(⇢0 d_2) az (z > d_2)
(⇢0 d_2✏0 ) az (z > d_2)
2
Dout =
C_m and Eout =
V_m
*(⇢0 d_2) az (z < d_2)
*(⇢0 d_2✏0 ) az (z < d_2)
32
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3.7. A spherically symmetric charge distribution in free space is characterized by the charge density
⇢v =
qb
exp(*br) C_m3
2
r
(0 < r < ÿ)
a) Find the electric field intensity, E(r), everywhere: By symmetry, the field will be radially directed, so Gauss’ Law can be applied to a spherical surface of radius r:
Õr
D dS = 4⇡r2 Dr =
= 4⇡qb
Therefore
r
0
encl. vol
2⇡
⇢v dv =
0
⇡
0
r
0
qb
exp[*b(r® )2 ] (r® )2 sin ✓drd✓d
(r® )2
exp[*b(r® )2 ] = 4⇡q[1 * exp(*br)]
4
5
4
5
1 * exp(*br)
q 1 * exp(*br)
2
D(r) = q
ar C_m Ÿ E(r) =
ar V_m
✏0
r2
r2
b) Find the total charge present: From the right hand side of Gauss’ Law, we found the charge
enclosed within a sphere of radius r, which is Q(r) = 4⇡q[1 * exp(*br)]. The total charge
present will be this result evaluated as r ô ÿ, or Qnet = 4⇡q.
3.8. Use Gauss’s law in integral form to show that an inverse distance field in spherical coordinates, D =
Aar _r, where A is a constant, requires every spherical shell of 1 m thickness to contain 4⇡A coulombs
of charge. Does this indicate a continuous charge distribution? If so, find the charge density variation
with r.
The net outward flux of this field through a spherical surface of radius r is
=
Õ
2⇡
D dS =
⇡
0
0
A
a a r2 sin ✓ d✓ d = 4⇡Ar = Qencl
r r r
We see from this that with every increase in r by one m, the enclosed charge increases by 4⇡A
(done). It is evident that the charge density is continuous, and we can find the density indirectly
by constructing the integral for the enclosed charge, in which we already found the latter from
Gauss’s law:
Qencl = 4⇡Ar =
2⇡
0
⇡
0
r
0
⇢(r® ) (r® )2 sin ✓ dr® d✓ d = 4⇡
r
0
⇢(r® ) (r® )2 dr®
To obtain the correct enclosed charge, the integrand must be ⇢(r) = A_r2 .
33
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3.9. A sphere of radius a in free space contains charge of density ⇢v = ⇢0 r_a, where ⇢0 is a constant.
a) Find the electric field intensity, EI , inside the sphere: Noting from symmetry that D will be
radially directed and will vary only with radius, we apply Gauss’ Law to a spherical surface of
radius r < a, within the charge. On that surface we have
Õs
DI dS = 4⇡r2 DI = Qencl =
So
DI =
v
r
⇢v dv = 4⇡
0
⇢ 0 r® ® 2
r4
(r ) dr = 4⇡⇢0
a
4a
⇢0 r2
⇢ r2
C_m2 Ÿ EI = 0 ar V_m (r < a)
4a
4✏0 a
b) Find the electric field intensity, EII , outside the sphere: The Gaussian surface of radius r is now
positioned outside the sphere. Gauss’ Law for that surface becomes:
Õs
DII dS = 4⇡r2 DII = Qtot =
which gives
DII =
v
⇢0 a3
4r
a
⇢v dv = 4⇡
C_m2 Ÿ EII =
2
⇢0 a3
4✏0 r2
0
⇢0 r® ® 2
a3
(r ) dr = 4⇡⇢0
a
4
ar V_m (r > a)
Note that EI = EII at r = a, which will be true as long as both regions have the same permittivity
(more about this in Chapter 5).
c) A spherical shell of radius b (where b > a) is positioned concentrically around the sphere. What
surface charge density, ⇢s , must exist on the shell so that the electric field at locations r > b
is zero? For zero field in the region r > b, a concentric Gaussian surface drawn in that region
must enclose zero net charge. This means that the volume and surface charges must be equal and
opposite to each other. We express this through the condition:
4⇡b2 ⇢s = *Qtot = *⇡⇢0 a3 Ÿ ⇢s =
*⇢0 a3
4b2
C_m2
Note that this result is equivalent to *DII at r = b.
d) What electrostatic force per unit area is exerted by the solid sphere on the spherical shell? This
will be the charge per unit area on the shell times the field from the interior charge evaluated at
the shell radius:
*⇢20 a6
*⇢0 a3 ⇢0 a3
F
Û
= ⇢s EII Û =
ar =
ar N_m2
Ûr=b
A
4b2 4✏0 b2
16✏0 b4
34
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3.10. An infinitely long cylindrical dielectric of radius b contains charge within its volume of density ⇢v =
a⇢2 , where a is a constant. Find the electric field strength, E, both inside and outside the cylinder.
Inside, we note from symmetry that D will be radially-directed, in the manner of a line charge
field. So we apply Gauss’ law to a cylindrical surface of radius ⇢, concentric with the charge
distribution, having unit length in z, and where ⇢ < b. The outward normal to the surface is a⇢ .
Õ
D n da =
1
2⇡
0
0
D⇢ a⇢ a⇢ ⇢ d dz = Qencl =
1
2⇡
0
⇢
0
0
a(⇢® )2 ⇢® d⇢® d dz
in which the dummy variable ⇢® must be used in the far-right integral because the upper radial
limit is ⇢. D⇢ is constant over the surface and can be factored outside the integral. Evaluating
both integrals leads to
2⇡(1)⇢D⇢ = 2⇡a
⇠ ⇡
a⇢3
a⇢3
1 4
⇢ Ÿ D⇢ =
or Ein =
a (⇢ < b)
4
4
4✏0 ⇢
To find the field outside the cylinder, we apply Gauss’ law to a cylinder of radius ⇢ > b. The
setup now changes only by the upper radius limit for the charge integral, which is now the charge
radius, b:
Õ
1
D n da =
0
2⇡
0
D⇢ a⇢ a⇢ ⇢ d dz = Qencl =
1
0
2⇡
0
b
0
a⇢2 ⇢ d⇢ d dz
where the dummy variable is no longer needed. Evaluating as before, the result is
D⇢ =
ab4
ab4
or Eout =
a (⇢ > b)
4⇢
4✏0 ⇢ ⇢
3.11. Consider a cylindrical charge distribution having an infinite length in z, but which has a radial dependence in charge density given by the gaussian, ⇢v (⇢) = ⇢0 exp[*(⇢_b)2 ] C_m3 , where b is a constant.
a) Find the electric field intensity E at large radii, ⇢ >> b. This enables the enclosed charge integral
in Gauss’ Law to be approximated using an infinite upper limit in radius: Choose a cylindrical
Gaussian surface of radius ⇢ >> b and of unit length in z, that is concentric with the charge.
Also assume that D = D⇢ a⇢ is uniform over the surface. Gauss’ Law is thus written:
Õ
D dS =
which becomes
Therefore
1
0
2⇡
0
.
D⇢ ⇢d dz = 2⇡(1)⇢D⇢ = Qencl =
.
2⇡⇢D⇢ = 2⇡⇢0
ÿ
0
2
2
1
0
2⇡
0
⇢e*⇢ _b d⇢ = 2⇡⇢0
ÿ
0
2
2
⇢0 e*⇢ _b ⇢d⇢d dz
b2
2
. ⇢ b2
. ⇢ b2
D⇢ = 0 and thus E = 0 a⇢ V_m (⇢ >> b)
2⇢
2✏0 ⇢
35
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3.11
b) Compare your result to the field outside a charged cylinder of radius b containing uniform charge
density ⇢0 : The setup is the same as in part a, except the charge integral is taken only to radius
b, over the constant charge density:
2⇡⇢D⇢ = Qencl =
1
0
and thus E is the same as in part a!
2⇡
0
b
0
⇢0 ⇢d⇢d dz = ⇡⇢0 b2 Ÿ D⇢ =
⇢0 b2
2⇢
3.12. The sun radiates a total power of about 3.86 ù 1026 watts (W). If we imagine the sun’s surface to be
marked off in latitude and longitude and assume uniform radiation,
a) What power is radiated by the region lying between latitude 50˝ N and 60˝ N and longitude 12˝
W and 27˝ W?
50˝ N lattitude and 60˝ N lattitude correspond respectively to ✓ = 40˝ and ✓ = 30˝ . 12˝ and 27˝
correspond directly to the limits on . Since the sun for our purposes is spherically-symmetric,
the flux density emitted by it is I = 3.86 ù 1026 _(4⇡r2 ) ar W_m2 . The required power is now
found through
27˝
40˝
3.86 ù 1026
ar ar r2 sin ✓ d✓ d
2
˝
˝
4⇡r
12
30
⇠
⇡
⇧
3.86 ù 1026 ⌅
2⇡
=
cos(30˝ ) * cos(40˝ ) (27˝ * 12˝ )
= 8.1 ù 1023 W
4⇡
360
P1 =
b) What is the power density on a spherical surface 93,000,000 miles from the sun in W_m2 ?
First, 93,000,000 miles = 155,000,000 km = 1.55 ù 1011 m. Use this distance in the flux density
expression above to obtain
I=
3.86 ù 1026
ar = 1200 ar W_m2
4⇡(1.55 ù 1011 )2
3.13. Spherical surfaces at r = 2, 4, and 6 m carry uniform surface charge densities of 20 nC_m2 , *4 nC_m2 ,
and ⇢s0 , respectively.
a) Find D at r = 1, 3 and 5 m: Noting that the charges are spherically-symmetric, we ascertain that
D will be radially-directed and will vary only with radius. Thus, we apply Gauss’ law to spherical
shells in the following regions: r < 2: Here, no charge is enclosed, and so Dr = 0.
2 < r < 4 : 4⇡r2 Dr = 4⇡(2)2 (20 ù 10*9 ) Ÿ Dr =
So Dr (r = 3) = 8.9 ù 10*9 C_m2 .
80 ù 10*9
C_m2
2
r
4 < r < 6 : 4⇡r2 Dr = 4⇡(2)2 (20 ù 10*9 ) + 4⇡(4)2 (*4 ù 10*9 ) Ÿ Dr =
So Dr (r = 5) = 6.4 ù 10*10 C_m2 .
16 ù 10*9
r2
b) Determine ⇢s0 such that D = 0 at r = 7 m. Since fields will decrease as 1_r2 , the question could
be re-phrased to ask for ⇢s0 such that D = 0 at all points where r > 6 m. In this region, the total
field will be
2
16 ù 10*9 ⇢s0 (6)
Dr (r > 6) =
+
r2
r2
*9
Requiring this to be zero, we find ⇢s0 = *(4_9) ù 10 C_m2 .
36
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3.14. A certain light-emitting diode (LED) is centered at the origin with its surface in the xy plane. At far
distances, the LED appears as a point, but the glowing surface geometry produces a far-field radiation
pattern that follows a raised cosine law: That is, the optical power (flux) density in Watts/m2 is given
in spherical coordinates by
cos2 ✓
Pd = P0
ar
Watts_m2
2⇡r2
where ✓ is the angle measured with respect to the normal to the LED surface (in this case, the z axis),
and r is the radial distance from the origin at which the power is detected.
a) Find, in terms of P0 , the total power in Watts emitted in the upper half-space by the LED: We
evaluate the surface integral of the power density over a hemispherical surface of radius r:
Pt =
2⇡
0
⇡_2
0
P0
P
cos2 ✓
Û⇡_2 P0
ar ar r2 sin ✓ d✓ d = * 0 cos3 ✓ Û
=
2
Û0
3
3
2⇡r
b) Find the cone angle, ✓1 , within which half the total power is radiated; i.e., within the range
0 < ✓ < ✓1 : We perform the same integral as in part a except the upper limit for ✓ is now ✓1 .
The result must be one-half that of part a, so we write:
⇠
⇡
Pt
P
P
1
Û✓1 P
= 0 = * 0 cos3 ✓ Û = 0 1 * cos3 ✓1 Ÿ ✓1 = cos*1 1_3 = 37.5˝
Û0
2
6
3
3
2
c) An optical detector, having a 1 mm2 cross-sectional area, is positioned at r = 1 m and at ✓ = 45˝ ,
such that it faces the LED. If one nanowatt (stated in error as 1mW) is measured by the detector,
what (to a very good estimate) is the value of P0 ? Start with
Pd (45˝ ) = P0
P0
cos2 (45˝ )
a
=
ar
r
2⇡r2
4⇡r2
Then the detected power in a 1-mm2 area at r = 1 m approximates as
. P
.
P [W ] = 0 ù 10*6 = 10*9 Ÿ P0 = 4⇡ ù 10*3 W
4⇡
If the originally stated 1mW value is used for the detected power, the answer would have been
4⇡ kW (!).
3.15. Volume charge density is located as follows: ⇢v = 0 for ⇢ < 1 mm and for ⇢ > 2 mm, ⇢v = 4⇢ C_m3
for 1 < ⇢ < 2 mm.
a) Calculate the total charge in the region 0 < ⇢ < ⇢1 , 0 < z < L, where 1 < ⇢1 < 2 mm: We find,
Q=
2⇡
L
0
0
⇢1
.001
4⇢ ⇢ d⇢ d dz =
8⇡L 3
[⇢1 * 10*9 ] C
3
b) Use Gauss’ law to determine D⇢ at ⇢ = ⇢1 : Gauss’ law states that 2⇡⇢1 LD⇢ = Q, where Q is the
result of part a. So, with ⇢1 in meters,
D⇢ (⇢1 ) =
4(⇢31 * 10*9 )
3⇢1
C_m2
37
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3.15
c) Evaluate D⇢ at ⇢ = 0.8 mm, 1.6 mm, and 2.4 mm: At ⇢ = 0.8 mm, no charge is enclosed by a
cylindrical gaussian surface of that radius, so D⇢ (0.8mm) = 0. At ⇢ = 1.6 mm, we evaluate the
part b result at ⇢1 = 1.6 to obtain:
D⇢ (1.6mm) =
4[(.0016)3 * (.0010)3 ]
= 3.6 ù 10*6 C_m2
3(.0016)
At ⇢ = 2.4, we evaluate the charge integral of part a from .001 to .002, and Gauss’ law is written
as
8⇡L
2⇡⇢LD⇢ =
[(.002)2 * (.001)2 ] C
3
from which D⇢ (2.4mm) = 3.9 ù 10*6 C_m2 .
3.16. An electric flux density is given by D = D0 a⇢ , where D0 is a given constant.
a) What charge density generates this field? Charge density is found by taking the divergence: With
radial D only, we have
D
1 d
⇢v = ( D =
(⇢D0 ) = 0 C_m3
⇢ d⇢
⇢
b) For the specified field, what total charge is contained within a cylinder of radius a and height b,
where the cylinder axis is the z axis? We can either integrate the charge density over the specified
volume, or integrate D over the surface that contains the specified volume:
Q=
2⇡
b
0
0
a
0
D0
⇢ d⇢ d dz =
⇢
2⇡
b
0
0
D0 a⇢ a⇢ a d dz = 2⇡abD0 C
3.17. In a region having spherical symmetry, volume charge is distributed according to:
⇢v (r) = ⇢0
sin(⇡r_a)
C_m2
r2
Find the surfaces on which E = 0: By Gauss’ Law, these will be spherical surfaces, of radii r0 say,
within which the total charge is zero. We have
Qencl =
v
⇢v dv = 4⇡
r0
0
⇢0
⌅
⇧
sin(⇡r_a) 2
r
dr
=
4a⇢
1
*
cos(⇡r
_a)
C
0
0
r2
This will be zero when cos(⇡r0 _a) = +1 or when r0 = 2ma where m = 1, 2, 3....
38
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3.18. State whether the divergence of the following vector fields is positive, negative, or zero:
a) the thermal energy flow in J_(m2 * s) at any point in a freezing ice cube: One way to visualize
this is to consider that heat is escaping through the surface of the ice cube as it freezes. Therefore
the net outward flux of thermal energy through the surface is positive. Calling the thermal flux
density F, the divergence theorem says
Õs
F dS =
v
( F dv
and so if we identify the left integral as positive, the right integral (and its integrand) must also
be positive. Answer: positive.
b) the current density in A_m2 in a bus bar carrying direct current: In this case, we have no accumulation or dissipation of charge within any small volume, since current is dc; this also means
that the net outward current flux through the surface that surrounds any small volume is zero.
Therefore the divergence must be zero.
c) the mass flow rate in kg_(m2 * s) below the surface of water in a basin, in which the water is
circulating clockwise as viewed from above: Here again, taking any small volume in the water,
the net outward flow through the surface that surrounds the small volume is zero; i.e., there is
no accumulation or dissipation of mass that would result in a change in density at any point.
Divergence is therefore zero.
3.19. A spherical surface of radius 3 mm is centered at P (4, 1, 5) in free space. Let D = xax C_m2 . Use the
.
results of Sec. 3.4 to estimate the net electric flux leaving the spherical surface: We use = ( D v,
where in this case ( D = ()_)x)x = 1 C_m3 . Thus
. 4
= ⇡(.003)3 (1) = 1.13 ù 10*7 C = 113 nC
3
3.20. A radial electric field distribution in free space is given in spherical coordinates as:
r⇢0
E1 =
a
(r f a)
3✏0 r
(2a3 * r3 )⇢0
E2 =
ar
(a f r f b)
3✏0 r2
(2a3 * b3 )⇢0
E3 =
ar
(r g b)
3✏0 r2
where ⇢0 , a, and b are constants.
a) Determine the volume charge density in the entire region (0 f r f ÿ) by appropriate use of
( D = ⇢v . We find ⇢v by taking the divergence of D in all three regions, where D = ✏0 E. As D
has only a radial component, the divergences become:
0 3 1
1 d 2
1 d r ⇢0
⇢v1 = ( D1 = 2
r D1 = 2
= ⇢0
(r f a)
3
r dr
r dr
⇠
⇡
1 d 2
1 d 1 3
⇢v2 = 2
r D2 = 2
(2a * r3 )⇢0 = *⇢0
(a f r f b)
r dr
r dr 3
⇠
⇡
1 d 2
1 d 1 3
⇢v3 = 2
r D3 = 2
(2a * b3 )⇢0 = 0
(r g b)
r dr
r dr 3
39
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3.20
b) Find, in terms of given parameters, the total charge, Q, within a sphere of radius r where r > b.
We integrate the charge densities (piecewise) over the spherical volume of radius b:
Q=
2⇡
0
⇡
0
a
0
⇢0 r2 sin ✓ dr d✓ d *
2⇡
0
⇡
0
b
a
4
⇢0 r2 sin ✓ dr d✓ d = ⇡ 2a3 * b3 ⇢0
3
3.21. In a region exhibiting spherical symmetry, electric flux density is found to be
D1 = ⇢0 r_3 ar (0 < r < a), D2 = 0 (a < r < b), and D3 = (a3 ⇢0 )_(3r2 ) ar (r > b).
a) Find the charge configuration that would produce the given field: We use Gauss’ Law in point
form to find:
⇠ r⇢ ⇡
1 d
⇢v1 = ( D1 = 2
r2 0 = ⇢0 C_m3
3
r dr
⇢v2 = ( D2 = 0
⇢v3 = ( D3 = 0
To produce zero field in region 2, the enclosed charge within a Gaussian sphere of radius a <
r < b must be zero. This is accomplished by a surface charge layer at r = a whose net charge is
equal and opposite to the total volume charge in the region r < a. The required surface charge
density is
(4_3)⇡a3 ⇢0
a⇢
⇢s1 (r = a) = *
= * 0 C_m2
2
3
4⇡a
Then an additional spherical surface charge layer at r = b is needed to produce D3 . Applying
Gauss’ Law to a spherical surface at r > b, we find
4⇡r2 D3 = Qencl = 4⇡b2 ⇢s2 Ÿ ⇢s2 =
a3 ⇢0
r2
D
=
3
b2
3b2
or just D3 evaluated at r = b. Note that the enclosed charge is just that of the outer layer because
the volume charge at r < a and the surface charge at r = a combine to give zero.
b) What total charge is present? Again, because the volume and first layer surface charges add to
zero, we are left only with the second layer charge, which is
Qnet = 4⇡b2 ⇢s2 =
4⇡a3 ⇢0
C
3
40
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3.22.
a) A flux density field is given as F1 = 5az . Evaluate the outward flux of F1 through the hemispherical surface, r = a, 0 < ✓ < ⇡_2, 0 < < 2⇡.
The flux integral is
1 =
hem.
F1 dS =
2⇡
0
⇡_2
0
5 az ar a2 sin ✓ d✓ d = *2⇡(5)a2
´Ø¨
cos2 ✓ Û⇡_2
= 5⇡a2
Û
2 Û0
cos ✓
b) What simple observation would have saved a lot of work in part a? The field is constant, and so
the inward flux through the base of the hemisphere (of area ⇡a2 ) would be equal in magnitude to
the outward flux through the upper surface (the flux through the base is a much easier calculation).
c) Now suppose the field is given by F2 = 5zaz . Using the appropriate surface integrals, evaluate
the net outward flux of F2 through the closed surface consisting of the hemisphere of part a and
its circular base in the xy plane:
Note that the integral over the base is zero, since F2 = 0 there. The remaining flux integral is
that over the hemisphere:
2 =
2⇡
0
= 10⇡a3
⇡_2
0
5z az ar a2 sin ✓ d✓ d =
⇡_2
0
2⇡
0
⇡_2
0
5(a cos ✓) cos ✓ a2 sin ✓ d✓ d
10
Û⇡_2 10 3
cos2 ✓ sin ✓ d✓ d = * ⇡a3 cos3 ✓ Û
= ⇡a
Û0
3
3
d) Repeat part c by using the divergence theorem and an appropriate volume integral:
The divergence of F2 is just dF2 _dz = 5. We then integrate this over the hemisphere volume,
which in this case involves just multiplying 5 by (2_3)⇡a3 , giving the same answer as in part c.
41
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3.23.
a) A point charge Q lies at the origin. Show that div D is zero everywhere except at the origin. For
a point charge at the origin we know that D = Q_(4⇡r2 ) ar . Using the formula for divergence in
spherical coordinates (see problem 3.21 solution), we find in this case that
0
1
1 d
2 Q
( D= 2
r
=0
r dr
4⇡r2
The above is true provided r > 0. When r = 0, we have a singularity in D, so its divergence is
not defined.
b) Replace the point charge with a uniform volume charge density ⇢v0 for 0 < r < a. Relate ⇢v0 to
Q and a so that the total charge is the same. Find div D everywhere: To achieve the same net
charge, we require that (4_3)⇡a3 ⇢v0 = Q, so ⇢v0 = 3Q_(4⇡a3 ) C_m3 . Gauss’ law tells us that
inside the charged sphere
Qr3
4
4⇡r2 Dr = ⇡r3 ⇢v0 = 3
3
a
Thus
0
1
Qr
Qr3
3Q
1 d
2
Dr =
C_m
and
(
D
=
=
3
2
3
dr
4⇡a
r
4⇡a
4⇡a3
as expected. Outside the charged sphere, D = Q_(4⇡r2 ) ar as before, and the divergence is zero.
3.24. In a region in free space, electric flux density is found to be:
<
⇢0 (z + 2d) az C_m2
(*2d f z f 0)
D=
2
*⇢0 (z * 2d) az C_m
(0 f z f 2d)
Everywhere else, D = 0.
a) Using ( D = ⇢v , find the volume charge density as a function of position everywhere: Use
<
dDz
⇢0 (*2d f z f 0)
⇢v = ( D =
=
dz
*⇢0 (0 f z f 2d)
b) determine the electric flux that passes through the surface defined by z = 0, *a f x f a,
*b f y f b: In the x-y plane, D evaluates as the constant D(0) = 2d⇢0 az . Therefore the flux
passing through the given area will be
=
a
b
*a
*b
2d⇢0 dxdy = 8abd ⇢0 C
c) determine the total charge contained within the region *a f x f a, *b f y f b, *d f z f d:
From part a, we have equal and opposite charge densities above and below the x-y plane. This
means that within a region having equal volumes above and below the plane, the net charge is
zero.
d) determine the total charge contained within the region *a f x f a, *b f y f b, 0 f z f 2d.
In this case,
Q = *⇢0 (2a) (2b) (2d) = *8abd ⇢0 C
This is equivalent to the net inward flux of D into the volume, as was found in part b.
42
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3.25. Within the spherical shell, 3 < r < 4 m, the electric flux density is given as
D = 5(r * 3)3 ar C_m2 .
a) What is the volume charge density at r = 4? In this case we have
1 d
5
⇢v = ( D = 2 (r2 Dr ) = (r * 3)2 (5r * 6) C_m3
r
r dr
which we evaluate at r = 4 to find ⇢v (r = 4) = 17.50 C_m3 .
b) What is the electric flux density at r = 4? Substitute r = 4 into the given expression to
find D(4) = 5 ar C_m2
c) How much electric flux leaves the sphere r = 4? Using the result of part b, this will be
4⇡(4)2 (5) = 320⇡ C
=
d) How much charge is contained within the sphere, r = 4? From Gauss’ law, this will be the same
as the outward flux, or again, Q = 320⇡ C.
3.26. If we have a perfect gas of mass density ⇢m kg_m3 , and assign a velocity U m/s to each differential
element, then the mass flow rate is ⇢m U kg_(m2 * s). Physical reasoning then leads to the continuity
equation, ( (⇢m U) = *)⇢m _)t.
a) Explain in words the physical interpretation of this equation: The quantity ⇢m U is the flow (or
flux) density of mass. Then the divergence of ⇢m U is the outward mass flux per unit volume at a
point. This must be equivalent to the rate of depletion of mass per unit volume at the same point,
as the continuity equation states.
b) Show that ós ⇢m U dS = *dM_dt, where M is the total mass of the gas within the constant
closed surface, S, and explain the physical significance of the equation.
Applying the divergence theorem, we have
Õs
⇢m U dS =
v
( (⇢m U) dv =
v
*
)⇢m
d
dv = *
)t
dt
v
⇢m dv = *
dM
dt
This states in large-scale form what was already stated in part a. That is – the net outward mass
flow (in kg/s) through a closed surface is equal to the negative time rate of change in total mass
within the enclosed volume.
3.27. Consider a slab of material containing a volume charge distribution throughout. The slab is of length
d in the z direction, and its dimensions in x and y represent a cross-sectional area of A. Free space
permittivity exists throughout. The electric field in the slab is given by
⇢
E = 0 exp(*↵z) az V_m
✏0 ↵
where ⇢0 is a positive constant (as is ↵). Note that the implication here is that the charge density
extends essentially to infinity in the x and y directions, which would then lead to a field having only a
z component, We are considering a sub-volume of this charge of cross-section A.
a) Find the volume charge density ⇢v in the slab:
⇠
⇡
d ⇢0 *↵z
⇢v = ( D =
e
= *⇢0 e*↵z
dz ↵
so we have negative charge of decreasing magnitude with z.
43
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3.27
b) Find the total charge in the slab: This is found by integrating the charge density over the slab
volume, given as
d
*A⇢0
Q=A
*⇢0 e*↵z dz =
(1 * e*↵d )
↵
0
c) Verify your result of part b by evaluating the net outward flux of D through the slab surfaces: We
need to evaluate
D dS
Õs
where in this case only two surfaces contribute to the integral: those at z = 0 and z = d, each
of which is of area A, and which have outward normal vectors *az and +az , respectively. We
write:
*A⇢0
D dS =
D(0) (*az )da +
D(d) (+az )da =
(1 * e*↵d )
Õs
↵
A
A
as before.
3.28. Repeat Problem 3.8, but use ( D = ⇢v and take an appropriate volume integral.
We begin by finding the charge density directly through
⇠
⇡
1 d
A
A
⇢v = ( D = 2
r2
= 2
r
r dr
r
Then, within each spherical shell of unit thickness, the contained charge is
Q(1) = 4⇡
r+1
r
A ®2 ®
(r ) dr = 4⇡A(r + 1 * r) = 4⇡A
(r® )2
3.29. In the region of free space that includes the volume 2 < x, y, z < 3,
2
(yz ax + xz ay * 2xy az ) C_m2
2
z
a) Evaluate the volume integral side of the divergence theorem for the volume defined above: In
cartesian, we find ( D = 8xy_z3 . The volume integral side is now
3
3
3
⇡
⇠
8xy
1 1
( D dv =
dxdydz
=
(9
*
4)(9
*
4)
*
= 3.47 C
4 9
z3
vol
2
2
2
D=
b) Evaluate the surface integral side for the corresponding closed surface: We call the surfaces at
x = 3 and x = 2 the front and back surfaces respectively, those at y = 3 and y = 2 the right and
left surfaces, and those at z = 3 and z = 2 the top and bottom surfaces. To evaluate the surface
integral side, we integrate D n over all six surfaces and sum the results. Note that since the x
component of D does not vary with x, the outward fluxes from the front and back surfaces will
cancel each other. The same is true for the left and right surfaces, since Dy does not vary with y.
This leaves only the top and bottom surfaces, where the fluxes are:
3
3
3
3
⇠
⇡
*4xy
*4xy
1 1
D dS =
dxdy
*
dxdy
=
(9
*
4)(9
*
4)
*
= 3.47 C
Õ
4 9
32
22
2
2
2
2
´≠≠≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠≠≠¨ ´≠≠≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠≠≠¨
top
bottom
44
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3.30.
a) Use Maxwell’s first equation, ( D = ⇢v , to describe the variation of the electric field intensity
with x in a region in which no charge density exists and in which a non-homogeneous dielectric
has a permittivity that increases exponentially with x. The field has an x component only: The
permittivity can be written as ✏(x) = ✏1 exp(↵1 x), where ✏1 and ↵1 are constants. Then
4
5
⇧
dEx
d ⌅ ↵1 x
( D = ( [✏(x)E(x)] =
✏1 e Ex (x) = ✏1 ↵1 e↵1 x Ex + e↵1 x
=0
dx
dx
This reduces to
dEx
+ ↵1 Ex = 0 Ÿ Ex (x) = E0 e*↵1 x
dx
where E0 is a constant.
b) Repeat part a, but with a radially-directed electric field (spherical coordinates), in which again
⇢v = 0, but in which the permittivity decreases exponentially with r. In this case, the permittivity
can be written as ✏(r) = ✏2 exp(*↵2 r), where ✏2 and ↵2 are constants. Then
4
5
1 d ⌅ 2 *↵2 r ⇧ ✏2
2
2 dEr
( D = ( [✏(r)E(r)] = 2
r ✏2 e
Er = 2 2rEr * ↵2 r Er + r
e*↵2 r = 0
dr
r dr
r
This reduces to
⇡
dEr ⇠ 2
+
* ↵2 Er = 0
dr
r
whose solution is
4
Er (r) = E0 exp *
⇠
⇡ 5
⌅
⇧ E
2
* ↵2 dr = E0 exp *2 ln r + ↵2 r = 20 e↵2 r
r
r
where E0 is a constant.
3.31. Given the flux density
16
cos(2✓) a✓ C_m2 ,
r
use two different methods to find the total charge within the region 1 < r < 2 m, 1 < ✓ < 2 rad,
1 < < 2 rad: We use the divergence theorem and first evaluate the surface integral side. We are
evaluating the net outward flux through a curvilinear “cube”, whose boundaries are defined by the
specified ranges. The flux contributions will be only through the surfaces of constant ✓, however,
since D has only a ✓ component. On a constant-theta surface, the differential area is da = r sin ✓drd ,
where ✓ is fixed at the surface location. Our flux integral becomes
Õ
D dS = *
D=
2
2
2
2
16
16
cos(2) r sin(1) drd +
cos(4) r sin(2) drd
r
r
1
1
1
1
´≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠¨ ´≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠¨
✓=1
= *16 [cos(2) sin(1) * cos(4) sin(2)] = *3.91 C
✓=2
We next evaluate the volume integral side of the divergence theorem, where in this case,
⌧
⌧
1 d
1 d 16
16 cos 2✓ cos ✓
( D=
(sin ✓ D✓ ) =
cos 2✓ sin ✓ = 2
* 2 sin 2✓
r sin ✓ d✓
r sin ✓ d✓ r
sin ✓
r
45
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3.31 (continued) We now evaluate:
vol
( D dv =
2
1
2
1
2
1
⌧
16 cos 2✓ cos ✓
* 2 sin 2✓ r2 sin ✓ drd✓d
sin ✓
r2
The integral simplifies to
2
1
2
1
2
1
16[cos 2✓ cos ✓ * 2 sin 2✓ sin ✓] drd✓d = 8
2
1
[3 cos 3✓ * cos ✓] d✓ = *3.91 C
46
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CHAPTER 4 – 9th Edition
4.1. Given E = Ex ax + Ey ay + Ez az , where Ex , Ey , and Ez are constants, determine the incremental work
required to move charge q through a distance :
a) along the positive x axis... This will be dWa = *q E ax = *q Ex J.
b) in a direction at 45 degrees
˘ from the x axis in the first quadrant...
˘ This will be dWb = *q E ab ,
where ab = (ax + ay )_ 2. Finally, dWb = *q (Ex + Ey )_ 2 J.
c) along a line in the first octant having equal x, y, and z components, and moving
away from the
˘
origin... This will be dWc = *q E ac , where ac = (ax + ay + az )_ 3, which results in
˘
dWc = *q (Ex + Ey + Ez )_ 3 J.
4.2. A positive point charge of magnitude q1 lies at the origin. Derive an expression for the incremental
work done in moving a second point charge q2 through a distance dx from the starting position (x, y, z),
in the direction of *ax : The incremental work is given by
dW = *q2 E12 dL
where E12 is the electric field arising from q1 evaluated at the location of q2 , and where dL = *dx ax .
Taking the location of q2 at spherical coordinates (r, ✓, ), we write:
dW =
*q2 q1
ar (*dx)ax
4⇡✏0 r2
where r2 = x2 + y2 + z2 , and where ar ax = sin ✓ cos . So
˘
q2 q1
q2 q1 x dx
x2 + y2
x
dW =
dx =
˘
˘
2
2
2
4⇡✏0 (x + y + z ) x2 + y2 + z2 x2 + y2
4⇡✏0 (x2 + y2 + z2 )3_2
´≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠¨ ´≠≠Ø≠≠¨
cos
sin ✓
4.3. Given E = E⇢ a⇢ + E a + Ez az V/m, where E⇢ , E , and Ez are constants:
a) find the incremental work done in moving charge q through distance in a direction˘having
equal ⇢ and components... this will be dWa = *q E a1 , where a1 = (a⇢ + a )_ 2. So
˘
dWa = *q (E⇢ + E )_ 2 J.
b) if the initial charge [location] in part a was at radius ⇢ = b, what change in angle occurred
in moving the charge? We may write dW = *qE dL where dL = d⇢a⇢ + ⇢d a . With
˘
equal distances
˘ along the two directions, we would have d⇢ = ⇢d = _ 2. So, at ⇢ = b,
d = _(b 2) rad
47
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4.4. An electric field in free space is given by E = xax + yay + zaz V_m. Find the work done in moving a
1 C charge through this field
a) from (1,1,1) to (0,0,0): The work will be
L
W = *q
0
*6
E dL = *10
1
xdx +
0
1
0
ydy +
1
M
zdz J = 1.5 J
b) from (⇢ = 2, = 0) to (⇢ = 2, = 90˝ ): The path involves changing
therefore dL = ⇢ d a . We set up the integral for the work as
W = *10*6
⇡_2
0
(xax + yay + zaz )
with ⇢ and z fixed, and
⇢d a
where ⇢ = 2, ax a = * sin , ay a = cos , and az a = 0. Also, x = 2 cos
y = 2 sin . Substitute all of these to get
W = *10*6
⇡_2 ⌅
0
*(2)2 cos sin
+ (2)2 cos sin
⇧
and
d =0
Given that the field is conservative (and so work is path-independent), can you see a much easier
way to obtain this result?
c) from (r = 10, ✓ = ✓0 ) to (r = 10, ✓ = ✓0 + 180˝ ): In this case, we are moving only in the a✓
direction. The work is set up as
W = *10*6
✓0 +⇡
✓0
(xax + yay + zaz )
r d✓ a✓
Now, substitute the following relations: r = 10, x = r sin ✓ cos , y = r sin ✓ sin , z = r cos ✓,
ax a✓ = cos ✓ cos , ay a✓ = cos ✓ sin , and az a✓ = * sin ✓. Obtain
W = *10*6
where we use cos2
✓0 +⇡
✓0
⌅
(10)2 sin ✓ cos ✓ cos2
+ sin2
+ sin ✓ cos ✓ sin2
⇧
* cos ✓ sin ✓ d✓ = 0
= 1.
48
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4.5. Consider the vector field G = (A_⇢) a , where A is a constant.
a) Evaluate the line integral of G over a circular path segment of radius b (with center at the origin),
along which the change in is ↵:
1
G dL =
↵
0
A
a
b
a b d = A↵
b) Same as part a, except the path is at radius c > b and the change in
2
G dL =
0
↵
A
a
c
is *↵:
a c d = *A↵
c) Evaluate ó G dL over a four-segment path that includes those of parts a and b as two of the
segments: We can choose the remaining two segments as straight line paths that involve changes
in radius from b to c and from c to b. The path is counter-clockwise, looking down at the xy
plane. We would therefore have
Õ
G dL =
4
…
i=1
i
G dL = A↵ * A↵ +
c
b
A
a
⇢
a⇢ d⇢ +
b
c
A
a
⇢
a⇢ d⇢ = 0
Note that the last two integrals are each zero because the scalar products within them are.
4.6. An electric field in free space is given as E = x aÇ x + 4z aÇ y + 4y aÇ z . Given V (1, 1, 1) = 10 V. Determine
V (3, 3, 3). The potential difference is expressed as
V (3, 3, 3) * V (1, 1, 1) = *
3,3,3
1,1,1
L
=*
3
1
(x aÇ x + 4z aÇ y + 4y aÇ z )
x dx +
3
1
4z dy +
3
1
(dx ax + dy ay + dz az )
M
4y dz
We choose the following path: 1) move along x from 1 to 3; 2) move along y from 1 to 3, holding x at
3 and z at 1; 3) move along z from 1 to 3, holding x at 3 and y at 3. The integrals become:
L
M
V (3, 3, 3) * V (1, 1, 1) = *
So
3
1
x dx +
3
1
4(1) dy +
3
1
4(3) dz = *36
V (3, 3, 3) = *36 + V (1, 1, 1) = *36 + 10 = *26 V
49
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4.7. A rectangular waveguide (presented in Chapter 13) is a hollow conducting pipe oriented along the z
axis, with an interior cross-section enclosed by conducting planes at x = 0 and a, and y = 0 and b
(see Figure 13.7 for an illustration). The transverse electric mode field, TE11 , defined for (0 < x < a),
(0 < y < b), is given by
⇠ ⇡
⇠ ⇡y ⇡
⇠ ⇡
⇠ ⇡y ⇡
⇡
⇡x
⇡
⇡x
E11 = E0 cos
sin
ax * E0 sin
cos
ay
b
a
b
a
a
b
where E0 is a constant.
a) Find the potential difference between B(0, 0) and the center of the guide, A(a_2, b_2): Choose
a path that has segment 1 going from 0 to a_2 in x, with y = 0; then segment 2 going from 0 to
b_2 in y, with x = a_2. So we have
A
a_2
b_2
Û
Û
Ex Û dx *
Ey Û
dy
Û
Ûx=a_2
y=0
B
0
0
a_2
b_2
⇠ ⇡
⇠ ⇡
⇠ ⇡
⇠ ⇡y ⇡
⇡
⇡x
⇡0
⇡
⇡a
=*
E0 cos
sin
dx +
E0 sin
cos
dy
b
a
b
a
2a
b
0
0
b_2
⇠ ⇡y ⇡
⇡
b
=
E0 cos
dy = E0 V
a
b
a
0
VA * VB = *
E dL = *
b) Using the given field, provide a simple demonstration that the guide walls form an equipotential
surface: Briefly, at all four surfaces locations, the field expression reduces to components that
are completely normal to the surfaces. Therefore no work is done in moving a charge along the
surfaces, and we have an equipotential.
4.8. Given E = *xax + yay , a) find the work involved in moving a unit positive charge on a circular arc,
˘
the circle centered at the origin, from x = a to x = y = a_ 2.
In moving along the arc, we start at
E dL = *
W = *q
= 0 and move to
⇡_4
E ad a = *
0
= ⇡_4. The setup is
⇡_4
0
(*x ax a + y ay a )a d
´Ø¨
´Ø¨
* sin
=*
⇡_4
0
2a2 sin cos d = *
⇡_4
0
a2 sin(2 ) d =
cos
a2
a2
Û⇡_4
cos(2 )Û
=*
Û0
2
2
where q = 1, x = a cos , and y = a sin .
Note that the field is conservative, so we would get the same result by integrating along a twosegment path over x and y as shown:
˘
˘
L
M
W =*
E dL = *
a_ 2
a
(*x) dx +
a_ 2
0
y dy = *a2 _2
b) Verify that the work done in moving the charge around the full circle from x = a is zero: In this
case, the setup is the same, but the integration limits change:
W =*
2⇡
0
a2 sin(2 ) d =
a2
Û2⇡
cos(2 )Û = 0
Û0
2
50
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4.9. An electric field intensity in spherical coordinates is given as
E=
V0 *r_a
e
V_m
d
where V0 and a are constants, and where the field exists everywhere.
a) Find the potential field, V (r), using a zero reference at infinity: We evaluate
V (r) = *
r
ÿ
E dL = *
r
V0 *r® _a
e
dr = V0 e*r_a V
a
ÿ
b) What is the significance of V0 ? From part a we identify V0 as the potential at the origin.
4.10. A sphere of radius a carries a surface charge density of ⇢s0 C_m2 .
a) Find the absolute potential at the sphere surface: The setup for this is
V0 = *
where, from Gauss’ law:
E=
So
V0 = *
a
a2 ⇢s0
ÿ
✏0 r
a
ÿ
a2 ⇢s0
✏0 r2
E dL
ar
a ar dr =
2 r
V_m
a2 ⇢s0 Ûa
a⇢s0
V
Û =
✏0 r Ûÿ
✏0
b) A grounded conducting shell of radius b where b > a is now positioned around the charged
sphere. What is the potential at the inner sphere surface in this case? With the outer sphere
grounded, the field exists only between the surfaces, and is zero for r > b. The potential is then
V0 = *
a
b
a2 ⇢s0
✏0 r
a ar dr =
2 r
a2 ⇢s0 Ûa a2 ⇢s0 ⌧ 1 1
*
V
Û =
✏0 r Ûb
✏0
a b
51
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4.11. At large distances from a dipole antenna (to be discussed in Chapter 14), the electric field amplitude
that it radiates assumes the simplified form:
E=
A
sin ✓ a✓ V_m
r
where A is a constant. A second dipole antenna, receiving radiation from the first, is located at distance
r from the first, has length L, and is oriented along the a✓ direction, thus presenting its full length to
the transmitting antenna at the origin. The angular position of the receiving antenna is ✓ = ✓0 . As
observed from the transmitting antenna, the receiving antenna subtends angle ✓.
a) Find the voltage amplitude across the length of the receiving antenna, and express your result in
terms of A, ✓0 , L, and r. Using the given field, we write:
✓0 * ✓_2
A
sin ✓ a✓ a✓ rd✓
L
✓0 + ✓_2 r
⌧ ⇠
⇡
⇠
⇡
✓
✓
= A cos ✓0 *
* cos ✓0 +
= 2A sin ✓0 sin( ✓_2) V
2
2
V =*
E dL = *
where ✓ = tan*1 (L_r).
.
b) Specialize your part a result for the case in which L << r (the usual case): In this case, ✓ = L_r
.
.
and sin( ✓_2) = ✓_2. Therefore, V = A(L_r) sin ✓0 V, which could be obtained from the
given field just by assuming constant E over small changes in angle.
4.12. In spherical coordinates, E = 2r_(r2 + a2 )2 ar V/m. Find the potential at any point, using the reference
a) V = 0 at infinity: We write in general
V (r) = *
2r dr
(r2 + a2 )2
+C =
1
r2 + a2
+C
With a zero reference at r ô ÿ, C = 0 and therefore V (r) = 1_(r2 + a2 ).
b) V = 0 at r = 0: Using the general expression, we find
V (0) =
1
1
+C =0 Ÿ C =* 2
2
a
a
V (r) =
1
1
*r2
*
=
r2 + a2 a2
a2 (r2 + a2 )
Therefore
c) V = 100V at r = a: Here, we find
V (a) =
Therefore
V (r) =
1
1
+ C = 100 Ÿ C = 100 * 2
2
2a
2a
1
1
a2 * r2
*
+
100
=
+ 100
r2 + a2 2a2
2a2 (r2 + a2 )
52
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4.13. Three identical point charges of 4 pC each are located at the corners of an equilateral triangle 0.5 mm
on a side in free space. How much work must be done to move one charge to a point equidistant from
the other two and on the line joining them? This will be the magnitude of the charge times the potential
difference between the finishing and starting positions, or
W =
(4 ù 10*12 )2 ⌧ 1
1
*
ù 104 = 5.76 ù 10*10 J = 576 pJ
2⇡✏0
2.5 5
4.14. Given the electric field E = (y + 1)ax + (x * 1)ay + 2az , find the potential difference between the points
a) (2,-2,-1) and (0,0,0): We choose a path along which motion occurs in one coordinate direction at
a time. Starting at the origin, first move along x from 0 to 2, where y = 0; then along y from 0
to *2, where x is 2; then along z from 0 to *1. The setup is
Vb * Va = *
2
Û
(y + 1)Û dx *
Ûy=0
0
*2
0
Û
(x * 1)Û dy *
Ûx=2
*1
0
2 dz = 2
b) (3,2,-1) and (-2,-3,4): Following similar reasoning,
Vb * Va = *
3
Û
(y + 1)Û
dx *
Ûy=*3
*2
2
Û
(x * 1)Û dy *
Ûx=3
*3
*1
4
2 dz = 10
4.15. Two uniform line charges, 8 nC/m each, are located at x = 1, z = 2, and at x = *1, y = 2 in free
space. If the potential at the origin is 100 V, find V at P (4, 1, 3): The net potential function for the two
charges would in general be:
V =*
At the origin, R1 = R2 =
˘
⇢l
⇢
ln(R1 ) * l ln(R2 ) + C
2⇡✏0
2⇡✏0
5, and V = 100 V. Thus, with ⇢l = 8 ù 10*9 ,
(8 ù 10*9 ) ˘
ln( 5) + C Ÿ C = 331.6 V
2⇡✏0
˘
˘
At P (4, 1, 3), R1 = (4, 1, 3) * (1, 1, 2) = 10 and R2 = (4, 1, 3) * (*1, 2, 3) = 26. Therefore
100 = *2
VP = *
˘
(8 ù 10*9 ) ⌧ ˘
ln( 10) + ln( 26) + 331.6 = *68.4 V
2⇡✏0
53
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4.16. A spherically-symmetric charge distribution in free space (with a < r < ÿ) is known to have a
potential function V (r) = V0 a2 _r2 , where V0 and a are constants.
a) Find the electric field intensity: This is found through
E = *(V = *
dV
a2
ar = 2V0 3 ar V_m
dr
r
b) Find the volume charge density: Use Maxwell’s first equation:
4 0
15
1 d
a2
a2
2
⇢v = ( D = ( (✏0 E) = 2
r 2✏0 V0 3
= *2✏0 V0 4 C_m3
r dr
r
r
c) Find the charge contained inside radius a: Here, we do not know the charge density inside radius
a, but we do know the flux density at that radius. We use Gauss’ law to integrate D over the
spherical surface at r = a to find the charge enclosed:
0
1
a2
2
2
Qencl =
D dS = 4⇡a Dr=a = 4⇡a 2✏0 V0 3 = 8⇡✏0 aV0 C
Õr=a
a
d) Find the total energy stored in the charge (or equivalently, in its electric field) in the region
(a < r < ÿ). Use Eq. (42):
10
1
2⇡
⇡
ÿ0
1
1
a2
a2
WE =
⇢ V dv =
*2✏0 V0 4
V0 2 r2 sin ✓ dr d✓ d
2 v v
2 0
r
r
0
a
(1)
4
2
= * ⇡✏0 aV0 J
3
If we integrate the energy density in the field over the region, the result is
2⇡
⇡
ÿ
1
a4
D E dv =
2✏0 V02 6 r2 sin ✓ dr d✓ d
r
v 2
0
0
a
ÿ
1
8
Û
= *8⇡ V02 ✏0 a4 3 Û = ⇡✏0 aV02 J
3
3r Ûa
We =
(2)
The apparent discrepancy between the two results can be explained by noting that (1) is interpreted as the energy needed to assemble the negative charge density around the already assembled
positive charge distribution. The latter lies in the region (r < a). The second result, Eq. (2), includes part of the energy associated with the central charge (r < a), and so is not specific to the
negative charge distribution. The answer given in (1) would be the correct one. The energy is
negative because the negative charge, being moved in from infinity, is attracted to the existing
positive charge, and so negative work is done.
This relationship can be demonstrated by the following exercise: Take the central charge as a
surface density, ⇢s , which covers the surface of a conducting sphere of radius a, where ⇢s =
Q_4⇡a2 , with Q as found in part c. The energy in that charge is found through
WE® =
1
2
s
⇢s V0 da = 4⇡✏0 aV02
Add this result to the answer in Eq. (1) of part d to obtain the answer in Eq. (2) (which in this
case would be the total system energy).
54
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4.17. Uniform surface charge densities of 6 and 2 nC_m2 are present at ⇢ = 2 and 6 cm respectively, in free
space. Assume V = 0 at ⇢ = 4 cm, and calculate V at:
a) ⇢ = 5 cm: Since V = 0 at 4 cm, the potential at 5 cm will be the potential difference between
points 5 and 4:
V5 = *
5
4
E dL = *
5
4
a⇢sa
(.02)(6 ù 10*9 ) ⇠ 5 ⇡
d⇢ = *
ln
= *3.026 V
✏0 ⇢
✏0
4
b) ⇢ = 7 cm: Here we integrate piecewise from ⇢ = 4 to ⇢ = 7:
V7 = *
6
4
a⇢sa
d⇢ *
✏0 ⇢
7
6
(a⇢sa + b⇢sb )
d⇢
✏0 ⇢
With the given values, this becomes
4
5 ⇠ ⇡ 4
5 ⇠ ⇡
(.02)(6 ù 10*9 )
(.02)(6 ù 10*9 ) + (.06)(2 ù 10*9 )
6
7
V7 = *
ln
*
ln
✏0
4
✏0
6
= *9.678 V
4.18. Find the potential at the origin produced by a line charge ⇢L = kx_(x2 + a2 ) extending along the x
axis from x = a to +ÿ, where a > 0. Assume a zero reference at infinity.
Think of the line charge as an array of point charges, each of charge dq = ⇢L dx, and each having
potential at the origin of dV = ⇢L dx_(4⇡✏0 x). The total potential at the origin is then the sum
of all these potentials, or
V =
ÿ
a
⇢L dx
=
4⇡✏0 x
ÿ
a
⇠ ⇡ÿ
⌧
k dx
k
k
⇡ ⇡
k
*1 x
=
tan
=
*
=
2
2
a a
4⇡✏0 a 2 4
16✏0 a
4⇡✏0 (x + a ) 4⇡✏0 a
55
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4.19. Volume charge density is given as ⇢v = ⇢0 e*r _r C_m3 , valid everywhere in free space.
a) Find the potential at the origin, using Eq. (17): This equation states:
V (r) =
⇢v (r ® )dv®
®
vol 4⇡✏0 r * r 
Because we are evaluating the potential at the origin, r = 0, r ® = rar , and so
V (0) =
2⇡
0
ÿ
⇡
0
⇢0 e*r
4⇡✏0 r
0
r2 sin ✓ drd✓d
2
=
⇢0
✏0
ÿ
0
e*r dr =
⇢0
✏0
b) (Harder) Find the potential by first finding E from Gauss’ Law, then line-integrating that result
from infinity to the origin: In view of the charge symmetry, D will be entirely radially-directed,
and will vary only with radius. Gauss’ Law, applied to a spherical surface of radius r takes the
form:
Õs
D dS = 4⇡r2 D(r) =
Solving, find
D=
Now,
V (0) = *
v
⇢v dv = 4⇡
r
0
®
⇢0
® Ûr
e*r ® 2 ®
(r ) dr = *4⇡⇢0 (1 + r® )e*r Û
®
Û0
r
⇧
⇢0 ⌅
D
1 * (1 + r)e*r ar and E =
2
✏0
r
0
ÿ
E dL = *
0
⇧
⇢0 ⌅
1 * (1 + r)e*r dr
2
ÿ ✏0 r
To evaluate the integral, we use the form:
e*r
e*r
dr
=
*
*
r
r2
So the potential integral becomes
L
0
0 *r
*⇢0
dr
e
e*r Û0
V (0) =
+
dr
+
Û *
2
✏0
r Ûÿ
ÿ r
ÿ r
e*r
dr
r
M
4
5
*⇢0
e*r
1 Û0
e*r Û0
dr =
* Û +
Û
✏0
r Ûÿ
r Ûÿ
ÿ r
0
.
For the evaluation, use the small argument approximation: e*r = 1 * r. Finally,
⇢
. *⇢0 ⌧ 1 1 * r
V (0) =
* +
= 0 Q.E.D.
✏0
r
r 0 ✏0
56
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4.20. In a certain medium, the electric potential is given by
V (x) =
⇢0
(1 * e*ax )
a✏0
where ⇢0 and a are constants.
a) Find the electric field intensity, E:
E = *(V = *
4
5
⇢
d ⇢0
*ax
*
e
(1
) ax = * 0 e*ax ax V_m
dx a✏0
✏0
b) find the potential difference between the points x = d and x = 0:
Vd0 = V (d) * V (0) =
⇢0
1 * e*ad V
a✏0
c) if the medium permittivity is given by ✏(x) = ✏0 eax , find the electric flux density, D, and the
volume charge density, ⇢v , in the region:
0
1
⇢0 *ax
ax
D = ✏E = ✏0 e
* e ax = *⇢0 ax C_m2
✏0
Then ⇢v = ( D = 0.
d) Find the stored energy in the region (0 < x < d), (0 < y < 1), (0 < z < 1):
1
We =
D E dv =
v 2
1
0
1
0
d
0
⇢20
2✏0
*ax
e
dx dy dz =
*⇢20
2✏0 a
e
*ax Û
d
⇢20
1 * e*ad J
Û =
Û0 2✏0 a
57
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4.21. A capacitor is formed in free space from two concentric spherical shells, arranged concentrically with
their centers at the origin; the shell radii are a and b, where b > a. With the inner shell at potential V0
and the outer shell grounded, the potential field (using methods to be discussed in Chapter 6) is found
to be:
1
* 1b
r
V (r) = V0 1 1 (a < r < b)
*b
a
The charge present is in the form of surface densities ⇢sa and ⇢sb on the inner and outer surfaces
respectively. The net charges on the inner and outer spheres are equal and opposite (omitted from the
problem statement).
a) Find E between spheres: This is found through the negative gradient of the given potential field:
E = *(V = *
V
dV
ar = ⇠ 0 ⇡ ar V_m
dr
r2 1 * 1
a
b
b) Find E for r < a and for r > b: For r < a applying Gauss’ Law would give zero field there, as
there is no enclosed charge, and the charge densities are uniform. For r > b, the two enclosed
charges, 4⇡a2 ⇢sa and 4⇡b2 ⇢sb add to give zero, so that the field in the region r > b must be zero.
c) Find the electric field energy density and the stored energy in the system:
1
1
w E = D E = ✏0 E 2 =
2
2
WE =
v
wE dv = 4⇡
b
a
✏0 V02
3
⇠
⇡2 J_m
1
1
2r4 a * b
✏0 V02
2⇡✏0 V02
2
r
dr
=
⇠
⇡J
⇠
⇡2
1
1
1
1
4
*
2r a * b
a
b
d) Knowing that the energy can be found through 12 îs ⇢s da, use your part c result to find the charge
density ⇢sa on the inner sphere and thereby determine the relationship between ⇢sa and E(r = a):
We have
WE =
Therefore
1
2
2⇡✏0 V02
⇢sa V0 da = 2⇡a2 ⇢sa V0 = ⇠
⇡ (as found in part c)
1
1
inner
*b
a
⇢sa =
✏0 V 0
Û
⇠
⇡ = ✏0 EÛ ar C_m2
Ûa
1
1
a2 a * b
e) What is the potential at the origin? Starting at r = a at which the potential is V0 , one would take
the negative line integral of E in the sphere interior from r = a to r = 0. As there is no field in
the region r < a, the potential at the origin is just that on the inner sphere surface, or V0 , as is the
case everywhere inside the inner sphere.
58
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4.22. A line charge of infinite length lies along the z axis, and carries a uniform linear charge density of ⇢l
C/m. A perfectly-conducting cylindrical shell, whose axis is the z axis, surrounds the line charge. The
cylinder (of radius b), is at ground potential. Under these conditions, the potential function inside the
cylinder (⇢ < b) is given by
⇢
V (⇢) = k * l ln(⇢)
2⇡✏0
where k is a constant.
a) Find k in terms of given or known parameters: At radius b,
⇢
⇢
V (b) = k * l ln(b) = 0 Ÿ k = l ln(b)
2⇡✏0
2⇡✏0
b) find the electric field strength, E, for ⇢ < b:
4
5
⇢l
⇢l
⇢l
d
Ein = *(V = *
ln(b) *
ln(⇢) a⇢ =
a V_m
d⇢ 2⇡✏0
2⇡✏0
2⇡✏0 ⇢ ⇢
c) find the electric field strength, E, for ⇢ > b: Eout = 0 because the cylinder is at ground potential.
d) Find the stored energy in the electric field per unit length in the z direction within the volume
defined by ⇢ > a, where a < b:
1
2⇡
b
⇠ ⇡
⇢2l
⇢2l
1
b
We =
D E dv =
⇢
d⇢
d
dz
=
ln
J
2 ✏ ⇢2
2
4⇡✏
a
8⇡
v
0
0
a
0
0
4.23. In free space, an electric potential is given in cylindrical coordinates by
V (⇢) =
a) Find the electric field intensity, E:
d
E = *(V = *
d⇢
b) Find the volume charge density:
⇢v = ( D = ✏0 ( E =
a2 ⇢0 *⇢_a
e
✏0
0 2
1
a ⇢0 *⇢_a
a⇢
e
a⇢ = 0 e*⇢_a a⇢ V_m
✏0
✏0
a⇢ ⇠
⇢ ⇡ *⇢_a
a d
⇢⇢0 e*⇢_a = 0 1 *
e
C_m3
⇢ d⇢
⇢
a
c) Using Eq. (42), find the stored energy in the region (0 < ⇢ < ÿ), (0 < < 2⇡), (0 < z < 1):
Use
54
5
1
2⇡
ÿ4
a⇢0 ⇠
⇢ ⇡ *⇢_a a2 ⇢0 *⇢_a
1
1
WE =
⇢ V dv =
1*
e
e
⇢d⇢d dz
2 v v
2 0 0
⇢
a
✏0
0
40
1
5ÿ
⇡a3 ⇢20 ÿ ⇠
⇡a4 ⇢20
⇡a4 ⇢20
⇢ ⇡ *2⇢_a
2⇢ *2⇢_a
=
1*
e
d⇢ = *
1*
e
=
J
✏0
a
4✏0
a
4✏0
0
0
d) Repeat part c, but use Eq. (44): Here we have...
1
2⇡
ÿ a2 ⇢2
1
1
1
0 *2⇢_a
e
⇢d⇢d dz
D Edv =
✏0 E 2 dv =
2 v
2 v
2 0 0
✏
0
0
0
1
4 2
⇡a2 ⇢20 ÿ *2⇢_a
⇡a4 ⇢20
2⇢ *2⇢_a Ûÿ ⇡a ⇢0
=
⇢e
d⇢ = *
1+
e
=
J
Û
Û0
✏0
4✏0
a
4✏0
0
WE =
as before.
59
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4.24. A certain spherically-symmetric charge configuration in free space produces an electric field given in
spherical coordinates by:
<
(⇢0 r2 )_(100✏0 ) ar V_m
(r f 10)
E(r) =
2
(100⇢0 )_(✏0 r ) ar V_m
(r g 10)
where ⇢0 is a constant.
a) Find the charge density as a function of position:
1 d
⇢v (r f 10) = ( (✏0 E1 ) = 2
r dr
0
21
⇢0 r
C_m3
100
25
0
1
1 d
2 100⇢0
⇢v (r g 10) = ( (✏0 E2 ) = 2
r
=0
r dr
r2
r
2 ⇢0 r
=
b) find the absolute potential as a function of position in the two regions, r f 10 and r g 10:
V (r f 10) = *
10
⇢0 (r® )2
ar ar dr®
100✏
✏
r
ÿ
10
0
0
⇧
100⇢0 Û10 ⇢0 (r® )3 Ûr
10⇢0 ⌅
=
4 * (10*3 r3 ) V
Û *
Û =
✏0 r Ûÿ
300✏0 Û10
3✏0
V (r g 10) = *
r
100⇢0
ar ar dr *
2
100⇢0
® 2
ÿ ✏0 (r )
ar ar dr® =
r
100⇢0 Ûr
100⇢0
V
Û =
®
✏0 r Ûÿ
✏0 r
c) check your result of part b by using the gradient:
4
5
⇧
10⇢0 2
⇢ r2
d 10⇢0 ⌅
*3 3
E1 = *(V (r f 10) = *
4 * (10 r ) ar =
(3r )(10*3 ) ar = 0 ar
dr 3✏0
3✏0
100✏0
4
5
100⇢0
d 100⇢0
E2 = *(V (r g 10) = *
ar =
ar
dr ✏0 r
✏0 r2
d) find the stored energy in the charge by an integral of the form of Eq. (42):
4
5
2⇡
⇡
10
⇧ 2
1
1 ⇢0 r 10⇢0 ⌅
*3 3
⇢ V dv =
4 * (10 r ) r sin ✓ dr d✓ d
We =
2 v v
2 25
3✏0
0
0
0
5
4
510
10 4
4⇡⇢20
4⇡⇢20
⇢2
r6
r7
3
4
3 0
=
40r *
dr =
10r *
= 7.18 ù 10
150✏0 0
100
150✏0
700 0
✏0
e) Find the stored energy in the field by an integral of the form of Eq. (44).
We =
=
=
1
D1 E1 dv +
(rf10) 2
2⇡
0
2⇡⇢20
✏0
0
L
2⇡
⇡
ÿ 104 ⇢2
0 2
2
r
sin
✓
dr
d✓
d
+
r sin ✓ dr d✓ d
4
(2 ù 104 )✏0
2✏
r
0
0
10
0
M
10
ÿ
2⇡⇢20 ⌧ 1 3
⇢2
dr
3
3 0
r6 dr + 104
=
(10
)
+
10
=
7.18
ù
10
2
✏0 7
✏0
0
10 r
10
⇡
0
10*4
1
D2 E2 dv
(rg10) 2
⇢20 r4
60
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4.25. Consider an electric field intensity in free space that exhibits a gaussian function of radius in spherical
coordinates:
⇢ a 2 2
E = 0 e*r _a ar V_m
✏0
where ⇢0 and a are constants, and where the field exists everywhere.
a) What charge density would produce this field?
1
⇠
⇡ 2⇢ a 0
2 2
1 d
r2
0
2 *r2 _a2
⇢v = ( D = ✏0 ( E = 2
⇢0 ar e
=
1 * 2 e*r _a C_m3
dr
r
r
a
Note that we have positive charge inside r = a, and negative charge outside the region.
b) What total charge is present?
The Easy Way: Use Gauss’ Law applied to a spherical shell of infinite radius:
Qencl =
Õrôÿ
2
2
D dS = lim 4⇡r2 ⇢0 ae*r _a = 0
rôÿ
The Hard Way: Integrate the charge density found in part a over the infinite volume:
0
1
ÿ
2⇢0 a
2 2
r2
Qencl = ⇢v dv = 4⇡
1 * 2 e*r _a r2 dr
r
a
v
0
b
c
f
g
ÿ 3
f ÿ *r2 _a2
r *r2 _a2 g
= 8⇡⇢0 a f
re
dr *
e
drg = 0
a2
0
0
f´≠≠≠≠≠≠≠
Ø≠≠≠≠≠≠≠¨ ´≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠¨gg
f
d
e
a2 _2
a2 _2
c) What is the potential at the origin? Line-integrate the given field from infinity to zero along a
radial line:
˘
0
ÿ
⇢0 a3 ⇡
⇢0 a ÿ *r2 _a2
V0 = *
E dL = +
E dL =
e
dr =
V
✏0 0
2✏0
ÿ
0
d) What is the net stored energy in the field? Integrate the energy density over an infinite spherical
volume:
u
ÿ ⇢2 a2
⇡a5 ⇢20 ⇡
1
1 2
0
*2r2 _a2 2
WE =
D Edv =
E dv = 4⇡
e
r dr =
J
2✏0
4✏0
2
v 2
v 2
0
61
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4.26. Let us assume that we have a very thin, square, imperfectly conducting plate 2m on a side, located in
the plane z = 0 with one corner at the origin such that it lies entirely within the first quadrant. The
potential at any point in the plate is given as V = *e*x sin y.
a) An electron enters the plate at x = 0, y = ⇡_3 with zero initial velocity; in what direction is its
initial movement? We first find the electric field associated with the given potential:
E = *(V = *e*x [sin y ax * cos y ay ]
Since we have an electron,
˘ its motion is opposite that of the field, so the direction on entry is that
of *E at (0, ⇡_3), or 3_2 ax * 1_2 ay .
b) Because of collisions with the particles in the plate, the electron achieves a relatively low velocity
and little acceleration (the work that the field does on it is converted largely into heat). The
electron therefore moves approximately along a streamline. Where does it leave the plate and in
what direction is it moving at the time? Considering the result of part a, we would expect the
exit to occur along the bottom edge of the plate. The equation of the streamline is found through
Ey
Ex
=
dy
cos y
=*
Ÿ x=*
dx
sin y
tan y dy + C = ln(cos y) + C
At the entry point (0, ⇡_3), we have 0 = ln[cos(⇡_3)] + C, from which C = 0.69. Now, along
the bottom edge (y = 0), we find x = 0.69, and so the exit point is (0.69, 0). From the field
expression evaluated at the exit point, we find the direction on exit to be *ay .
4.27. By performing an appropriate line integral, show that Eq. (33) can be found from Eq. (35): To find the
absolute potential, a line integral can be taken from a point at an infinite distance, moving to a general
location (r, ✓). A two-segment path is chosen, starting from point B(r = ÿ, ✓ = ⇡_2), and integrating
along r from ÿ to r, holding ✓ fixed at ⇡_2; then integrating along the second segment from ⇡_2 to
✓, holding radius at r. This ends at point A(r, ✓).
4 r
5
A
✓
2 cos ✓⇡_2
qd
sin ✓ ®
®
®
V (r, ✓) = *
E dL = *
ar ar dr +
a✓ a✓ rd✓
3
4⇡✏0 ÿ
(r® )3
B
⇡_2 r
The first integral term is zero (cos(⇡_2)), so this leaves the second term, which evaluates as
V (r, ✓) =
qd
qd cos ✓
Û✓
cos ✓ ® Û
=
Q.E.D.
2
Û⇡_2
4⇡✏0 r
4⇡✏0 r2
62
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4.28. Use the electric field intensity of the dipole (Sec. 4.7, Eq. (36)) to find the difference in potential
between points at ✓a and ✓b , each point having the same r and coordinates. Under what conditions
does the answer agree with Eq. (34), for the potential at ✓a ?
We perform a line integral of Eq. (36) along an arc of constant r and :
⇧
qd ⌅
2
cos
✓
a
+
sin
✓
a
a✓ r d✓ = *
r
✓
3
✓b 4⇡✏0 r
⇧
qd ⌅
=
cos ✓a * cos ✓b
2
4⇡✏0 r
Vab = *
✓a
✓a
✓b
qd
sin ✓ d✓
4⇡✏0 r2
This result agrees with Eq. (34) if ✓a (the ending point in the path) is 90˝ (the xy plane). Under
this condition, we note that if ✓b > 90˝ , positive work is done when moving (against the field) to
the xy plane; if ✓b < 90˝ , negative work is done since we move with the field.
4.29. A dipole having a moment p = 3ax * 5ay + 10az nC m is located at Q(1, 2, *4) in free space. Find
V at P (2, 3, 4): We use the general expression for the potential in the far field:
V =
p (r * r ® )
4⇡✏0 r * r ® 3
where r * r ® = P * Q = (1, 1, 8). So
VP =
(3ax * 5ay + 10az ) (ax + ay + 8az ) ù 10*9
4⇡✏0 [12 + 12 + 82 ]1.5
= 1.31 V
4.30. A dipole for which p = 10✏0 az C m is located at the origin. What is the equation of the surface on
which Ez = 0 but E ë 0?
First we find the z component:
⇧
⇧
10 ⌅
5 ⌅
2
2
2
cos
✓
(a
a
)
+
sin
✓
(a
a
)
=
2
cos
✓
*
sin
✓
r
z
✓
z
4⇡r3
2⇡r3
⌅
⇧
This will be zero when 2 cos2 ✓ * sin2 ✓ = 0. Using identities, we write
Ez = E az =
1
2 cos2 ✓ * sin2 ✓ = [1 + 3 cos(2✓)]
2
The above becomes zero on the cone surfaces, ✓ = 54.7˝ and ✓ = 125.3˝ .
63
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4.31. A potential field in free space is expressed as V = 20_(xyz) V.
a) Find the total energy stored within the cube 1 < x, y, z < 2. We integrate the energy density over
the cube volume, where wE = (1_2)✏0 E E, and where
4
5
1
1
1
E = *(V = 20 2 ax + 2 ay +
az V_m
x yz
xy z
xyz2
The energy is now
WE = 200✏0
2
1
24
2
1
1
5
1
1
1
+
+
dx dy dz
x4 y2 z2 x2 y4 z2 x2 y2 z4
The integral evaluates as follows:
52
⇠ ⇡
1
1
1
1
WE = 200✏0
*
*
*
dy dz
3 x3 y2 z2 xy4 z2 xy2 z4 1
1
1
5
2
2 4⇠
⇡
⇠ ⇡
⇠ ⇡
7
1
1
1
1
1
= 200✏0
+
+
dy dz
24 y2 z2
2 y4 z 2
2 y2 z 4
1
1
52
24 ⇠
⇡
⇠ ⇡
⇠ ⇡
7
1
1
1
1 1
= 200✏0
*
*
*
dz
24 yz2
6 y3 z2
2 yz4 1
1
2 ⌧⇠
⇡
⇠ ⇡
⇠ ⇡
7 1
7 1
1 1
= 200✏0
+
+
dz
2
2
48 z
48 z
4 z4
1
⌧
7
= 200✏0 (3)
= 387 pJ
96
2
24
b) What value would be obtained by assuming a uniform energy density equal to the value at the
center of the cube? At C(1.5, 1.5, 1.5) the energy density is
4
5
1
wE = 200✏0 (3)
= 2.07 ù 10*10 J_m3
(1.5)4 (1.5)2 (1.5)2
This, multiplied by a cube volume of 1, produces an energy value of 207 pJ.
64
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4.32. Using Eq. (36), a) find the energy stored in the dipole field in the region r > a:
We start with
E(r, ✓) =
⇧
qd ⌅
2
cos
✓
a
+
sin
✓
a
r
✓
4⇡✏0 r3
Then the energy will be
We =
=
=
2⇡
1
✏0 E E dv =
vol 2
*2⇡(qd)2 1 Ûÿ
Û
32⇡ 2 ✏0 3r3 Ûa
(qd)2
12⇡✏0 a3
0
ÿ
⇡
0
a
⇧
(qd)2 ⌅
4 cos2 ✓ + sin2 ✓ r2 sin ✓ dr d✓ d
2
6
32⇡ ✏0 r ´≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠¨
⇡⌅
0
⇧
3 cos2 ✓ + 1 sin ✓ d✓ =
3 cos2 ✓+1
(qd)2 ⌅
⇧⇡
3
*
cos
✓
*
cos
✓
48⇡ 2 ✏0 a3 ´≠≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠≠¨0
4
J
b) Why can we not let a approach zero as a limit? From the above result, a singularity in the energy
occurs as a ô 0. More importantly, a cannot be too small, or the original far-field assumption
used to derive Eq. (36) (a >> d) will not hold, and so the field expression will not be valid.
4.33. A copper sphere of radius 4 cm carries a uniformly-distributed total charge of 5 C in free space.
a) Use Gauss’ law to find D external to the sphere: with a spherical Gaussian surface at radius r, D
will be the total charge divided by the area of this sphere, and will be ar -directed. Thus
D=
Q
5 ù 10*6
a
=
ar C_m2
r
4⇡r2
4⇡r2
b) Calculate the total energy stored in the electrostatic field: Use
2⇡
1
D E dv =
vol 2
0
⇠ ⇡ (5 ù 10*6 )2
1
= (4⇡)
2
16⇡ 2 ✏0
WE =
⇡
0
ÿ
ÿ
1 (5 ù 10*6 )2 2
r sin ✓ dr d✓ d
2
4
.04 2 16⇡ ✏0 r
dr 25 ù 10*12 1
=
= 2.81 J
2
8⇡✏0 .04
.04 r
c) Use WE = Q2 _(2C) to calculate the capacitance of the isolated sphere: We have
C=
(5 ù 10*6 )2
Q2
=
= 4.45 ù 10*12 F = 4.45 pF
2WE
2(2.81)
65
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4.34. A sphere of radius a contains volume charge of uniform density ⇢0 C_m3 . Find the total stored energy
by applying
a) Eq. (43): We first need the potential everywhere inside the sphere. The electric field inside and
outside is readily found from Gauss’s law:
E1 =
⇢0 r
⇢ a3
ar r f a and E2 = 0 2 ar r g a
3✏0
3✏0 r
The potential at position r inside the sphere is now the work done in moving a unit positive point
charge from infinity to position r:
V (r) = *
a
ÿ
E2 ar dr *
r
a
E1 ar dr® = *
a
ÿ
⇢0 a3
3✏0 r
dr *
2
r
a
⇢0 r® ®
⇢
dr = 0 3a2 * r2
3✏0
6✏0
Now, using this result in (43) leads to the energy associated with the charge in the sphere:
We =
1
2
2⇡
⇡
0
0
a
0
⇢20
6✏0
3a2 * r2 r2 sin ✓ dr d✓ d =
⇡⇢0
3✏0
a
0
3a2 r2 * r4 dr =
4⇡a5 ⇢20
15✏0
b) Eq. (45): Using the given fields we find the energy densities
⇢ 2 r2
⇢2 a6
1
1
we1 = ✏0 E1 E1 = 0
r f a and we2 = ✏0 E2 E2 = 0 4 r g a
2
18✏0
2
18✏0 r
We now integrate these over their respective volumes to find the total energy:
We =
2⇡
0
a ⇢ 2 r2
0
2
⇡
0
0
18✏0
r sin ✓ dr d✓ d +
2⇡
0
ÿ
⇡
0
a
⇢20 a6
18✏0 r
r2 sin ✓ dr d✓ d =
4
4⇡a5 ⇢20
15✏0
66
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4.35. Four 0.8 nC point charges are located in free space at the corners of a square 4 cm on a side.
a) Find the total potential energy stored: This will be given by
4
1…
WE =
q V
2 n=1 n n
where Vn in this case is the potential at the location of any one of the point charges that arises
from the other three. This will be (for charge 1)
L
M
q
1
1
1
V1 = V21 + V31 + V41 =
+
+
˘
4⇡✏0 .04 .04 .04 2
Taking the summation produces a factor of 4, since the situation is the same at all four points.
Consequently,
L
M
*9 )2
(.8
ù
10
1
1
WE = (4)q1 V1 =
2+ ˘
= 7.79 ù 10*7 J = 0.779 J
2
2⇡✏0 (.04)
2
b) A fifth 0.8 nC charge is installed at the center of the square. Again find the total stored energy:
This will be the energy found in part a plus the amount of work done in moving the fifth charge
into position from infinity. The latter is just the potential at the square center arising from the
original four charges, times the new charge value, or
WE =
4(.8 ù 10*9 )2
= .813 J
˘
4⇡✏0 (.04 2_2)
The total energy is now
WE net = WE (part a) + WE = .779 + .813 = 1.59 J
4.36 Surface charge of uniform density ⇢s lies on a spherical shell of radius b, centered at the origin in free
space.
a) Find the absolute potential everywhere, with zero reference at infinity: First, the electric field,
found from Gauss’ law, is
b2 ⇢s
E=
ar V_m
✏0 r2
Then
r
r
b2 ⇢s
b2 ⇢s
®
V (r) = *
E dL = *
dr
=
V
® 2
✏0 r
ÿ
ÿ ✏0 (r )
b) find the stored energy in the sphere by considering the charge density and the potential in a twodimensional version of Eq. (42):
We =
1
2
S
⇢s V (b) da =
1
2
2⇡
⇡
0
0
⇢s
2⇡⇢2s b3
b2 ⇢s 2
b sin ✓ d✓ d =
✏0 b
✏0
c) find the stored energy in the electric field and show that the results of parts b and c are identical.
We =
1
D E dv =
v 2
2⇡
0
ÿ
⇡
0
b
4 2
2 3
2⇡⇢s b
1 b ⇢s 2
r
sin
✓
dr
d✓
d
=
2 ✏0 r4
✏0
67
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CHAPTER 5 – 9th Edition
5.1. Given the current density J = *104 [sin(2x)e*2y ax + cos(2x)e*2y ay ] 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
2
1
Û
Û
J nÛ da =
J ay Û dx dz =
Û
Ûy=1
S
S
0
0
1
1
Û
= *104 (2) sin(2x)Û e*2 = *1.23 MA
Û0
2
I=
2
0
1
0
*104 cos(2x)e*2 dx dz
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=
3
2
= 104
1
0
Û
J (*ay )Û dx dz +
Ûy=0
3
1⌅
3
1
2
0
⇧
*2
3
3
2
1
1
0
Û
J (ax )Û dy dz
Ûx=1
dx dz * 104
sin(2)e*2y dy dz
2
0
⇠ ⇡
⇠ ⇡
⌅
⇧
1
1
Û1
Û1
= 104
sin(2x)Û (3 * 2) 1 * e*2 + 104
sin(2)e*2y Û (3 * 2) = 0
Û
Û0
0
2
2
2
0
cos(2x)e*0 * cos(2x)e
Û
J (ay )Û dx dz +
Ûy=1
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
( J=
⌅
⇧
)Jx )Jy
+
= *10*4 2 cos(2x)e*2y * 2 cos(2x)e*2y = 0 as expected
)x
)y
5.2. Given J = *10*4 (yax + xay ) A_m2 , find the current crossing the y = 0 plane in the *ay direction
between z = 0 and 1, and x = 0 and 2.
At y = 0, J(x, 0) = *104 xay , so that the current through the plane becomes
I=
J dS =
1
0
2
0
*104 x ay (*ay ) dx dz = 2 ù 10*4 A
68
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5.3. A solid sphere of radius b carries charge Q uniformly distributed throughout the sphere volume. The
sphere rotates about the z axis at angular velocity ⌦ rad/s. What total current is flowing? Begin with
J = ⇢v v where v is the velocity of any point in the sphere, and is given by v = ⌦r sin ✓ a m/s. Then
the charge density
⇢v =
Q
Q
Q
=
Ÿ J=
⌦r sin ✓ a A_m2
3
vol (4_3)⇡b3
(4_3)⇡b
Now the total current is the integral of J over the half disk, extending from 0 to ⇡ in ✓, and from 0 to
b in r:
⇡
b
3Q⌦r sin ✓
Q⌦
I=
rdrd✓ =
A
3
2⇡
4⇡b
0
0
Note error in Appendix F answer.
5.4. If volume charge density is given as ⇢v = (cos !t)_r2 C_m3 in spherical coordinates, find J. It is
reasonable to assume that J is not a function of ✓ or .
We use the continuity equation (5), along with the assumption of no angular variation to write
( J=
⇠
⇡
)⇢v
1 ) 2
) cos !t
! sin !t
r
J
=
*
=
*
=
r
2
2
)t
)t
r )r
r
r2
So we may now solve
by direct integration to obtain:
) 2
r Jr = ! sin !t
)r
J = Jr a r =
! sin !t
ar A_m2
r
where the integration constant is set to zero because a steady current will not be created by a
time-varying charge density.
69
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5.5. Consider the following time-varying current density, which is in the form of a propagating wave at
frequency f , traveling at velocity v (more about this in Chapter 10):
J(z, t) = J0 cos(!t * z) az A_m2
where J0 is a constant amplitude, ! = 2⇡f is the radian frequency, and is a constant to be found.
Determine as a function of ! and v such that the given wave form satisfies both Eqs. (3) and (5).
Begin with Eq. (5): Applying Eq. (5) (equation of continuity), we have
( J=
So
⇢v = *
dJz
)⇢
= J0 sin(!t * z) = * v
dz
)t
J0 sin(!t * z)dt + C =
J cos(!t * z)
! 0
where the integration constant is set to zero. Now, from Eq. (3):
J = ⇢v v Ÿ v =
!
Ÿ
=
!
v
5.6. In spherical coordinates, a current density J = *k_(r sin ✓) a✓ 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:
2⇡
R
*k
I = J dS =
a✓ az ⇢ d⇢ d
r
sin
✓ ´Ø
s
0
0
¨
* sin ✓
˘
Then, using r = ⇢2 + z2 , this becomes
I=
2⇡
0
R
0
˘
⌧˘
˘
ÛR
= 2⇡k ⇢2 + z2 Û = 2⇡k
R2 + z 2 * z
Û
0
2
2
⇢ +z
k⇢
At z = 0 (part a), we have I(0) = 2⇡kR , and at z = h (part b): I(h) = 2⇡k
⌧˘
R2 + h 2 * h .
70
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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 ⇢v ?
These would be, respectively, mass f lux density in (kg_m2 * 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:
)⇢m
dv = *
( Jm dv = *
J dS
Õs m
v )t
v
where
Õs
Jm dS = 10.25 * 9.85 + 1.75 * 2.00 * 4.05 + 4.45 = 0.550 mg_s
Treating our 1 cm3 volume as differential, we find
)⇢m . 0.550 ù 10*3 g_s
=*
= *550 g_m3 * s
*6
3
)t
10 m
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 ù 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 = l from the x-y plane; the wide end (2-mm radius) lies at z = l + 16 cm. l is chosen
such that if the cone were not truncated, its vertex would occur at the origin. The cone surface
subtends angle ✓c from the z axis (in spherical coordinates). Therefore, we may write
l=
0.1mm
2 mm
and tan ✓c =
tan ✓c
160 + l
Solving these, we find l = 8.4 mm, tan ✓c = 1.19 ù 10*2 , and so ✓c = 0.68˝ , which gives us
a very thin cone! With this understanding, we can assume that the current density is uniform
with ✓ and 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 (⇢). This is our primary assumption.
Now, assuming constant current density at constant r, and net current, I, we may write
I=
2⇡
0
✓c
0
J (r) ar ar r2 sin ✓ d✓ d = 2⇡r2 J (r)(1 * cos ✓c )
or
J(r) =
(1.42 ù 104 )I
I
a
=
ar
r
2⇡r2 (1 * cos ✓c )
2⇡r2
71
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5.8 (continued) The electric field is now
E(r) =
J(r)
=
(1.42 ù 104 )I
I
ar = (7.1 ù 10*3 )
ar V_m
2⇡r2 (2 ù 106 )
2⇡r2
The voltage between the ends is now
V0 = *
rin
rout
E ar dr
.
.
where rin = l_ cos ✓c = l and where rout = (160 + l)_ cos ✓c = 160 + l. The voltage is
V0 = *
=
lù10*3
(160+l)ù10*3
(7.1 ù 10*3 )
0.40
I
⇡
⌧
I
1
1
*3 I
a
a
dr
=
(7.1
ù
10
)
*
r
r
2
2⇡ .0084 .1684
2⇡r
from which we identify the resistance as R = 0.40_⇡ = 0.128 ohms.
A second method uses the idea that we can construct the cone from a stack of thin circular plates of
linearly-increasing radius, ⇢. Assuming each plate is of differential thickness, dz, the differential
resistance of a plate will be
dz
dR =
⇡⇢2
where ⇢ = z tan ✓c = z(1.19 ù 10*2 ). The cone resistance will be the resistance of the stack of
plates (in series), found through
(160+l)ù10*3
(160+l)ù10*3
dz
=
*3
⇡⇢2
lù10
lù10*3
⌧
*3
3.53 ù 10
1
1
=
*
= 0.127 ohms
⇡
.0084 .1684
R=
5.9.
dR =
dz
(2 ù 106 )⇡z2 (1.19 ù 10*2 )2
a) Using data tabulated in Appendix C, calculate the required diameter for a 2-m long nichrome
wire that will dissipate an average power of 450 W when 120 V rms at 60 Hz is applied to it:
The required resistance will be
R=
Thus the diameter will be
u
d = 2a = 2
v
lP
=2
⇡V 2
V2
l
=
P
(⇡a2 )
2(450)
= 2.8 ù 10*4 m = 0.28 mm
(106 )⇡(120)2
b) Calculate the rms current density in the wire: The rms current will be I = 450_120 = 3.75 A.
Thus
3.75
J=
= 6.0 ù 107 A_m2
2
*4
⇡ 2.8 ù 10 _2
72
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5.10. A large brass washer has a 2-cm inside diameter, a 5-cm outside diameter, and is 0.5 cm thick. Its
conductivity is = 1.5 ù 107 S/m. The washer is cut in half along a diameter, and a voltage is applied
between the two rectangular faces of one part. The resultant electric field in the interior of the halfwasher is E = (0.5_⇢) a V/m in cylindrical coordinates, where the z axis is the axis of the washer.
a) What potential difference exists between the two rectangular faces? First, we orient the washer
in the x-y plane with the cut faces aligned with the x axis. To find the voltage, we integrate E
over a circular path of radius ⇢ inside the washer, between the two cut faces:
E dL = *
V0 = *
0
⇡
0.5
a
⇢
a ⇢ d = 0.5⇡ V
b) What total current is flowing? First, the current density is J = E, so
J=
1.5 ù 107 (0.5)
7.5 ù 106
a =
a A_m2
⇢
⇢
Current is then found by integrating J over any transverse plane in the washer (the rectangular
cross-section):
I=
s
J dS =
0.5ù10*2
2.5ù10*2
10*2
0
= 0.5 ù 10*2 (7.5 ù 106 ) ln
⇠
7.5 ù 106
a
⇢
a d⇢ dz
⇡
2.5
= 3.4 ù 104 A
1
c) What is the resistance between the two faces?
R=
V0
0.5⇡
=
= 4.6 ù 10*5 ohms
I
3.4 ù 104
5.11. Two perfectly-conducting cylindrical surfaces of length l are located at ⇢ = 3 and ⇢ = 5 cm. The total
current passing radially outward through the medium between the cylinders is 3 A dc.
a) Find the voltage and resistance between the cylinders, and E in the region between the cylinders,
if a conducting material having = 0.05 S_m is present for 3 < ⇢ < 5 cm: Given the current,
and knowing that it is radially-directed, we find the current density by dividing it by the area of
a cylinder of radius ⇢ and length l:
J=
3
a A_m2
2⇡⇢l ⇢
73
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5.11
a) (continued)
Then the electric field is found by dividing this result by :
E=
3
9.55
a⇢ =
a V_m
2⇡ ⇢l
⇢l ⇢
The voltage between cylinders is now:
3
V =*
5
E dL =
Now, the resistance will be
⇠ ⇡
9.55
9.55
5
4.88
a⇢ a⇢ d⇢ =
ln
=
V
⇢l
l
3
l
5
3
R=
V
4.88 1.63
=
=
⌦
I
3l
l
b) Show that integrating the power dissipated per unit volume over the volume gives the total dissipated power: We calculate
P =
v
E J dv =
2⇡
l
0
0
.05
.03
⇠ ⇡
32
32
5
14.64
⇢
d⇢
d
dz
=
ln
=
W
2
2
2
2⇡(.05)l
3
l
(2⇡) ⇢ (.05)l
We also find the power by taking the product of voltage and current:
P =VI =
4.88
14.64
(3) =
W
l
l
which is in agreement with the power density integration.
5.12. Two identical conducting plates, each having area A, are located at z = 0 and z = d. The region
between plates is filled with a material having z-dependent conductivity, (z) = 0 e*z_d , where 0 is
a constant. Voltage V0 is applied to the plate at z = d; the plate at z = 0 is at zero potential. Find, in
terms of the given parameters:
a) the resistance of the material: We start with the differential resistance of a thin slab of the material
of thickness dz, which is
dR =
dz
ez_d dz
=
so that R =
A
0A
dR =
d
0
ez_d dz
d
1.72d
=
(e * 1) =
⌦
0A
0A
0A
b) the total current flowing between plates: We use
I=
V0
AV
= 0 0
R
1.72 d
c) the electric field intensity E within the material: First the current density is
J=*
* 0 V0
*V0 ez_d
I
J
az =
az so that E =
=
a V_m
A
1.72 d
(z)
1.72 d z
74
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5.13. A hollow cylindrical tube with a rectangular cross-section has external dimensions of 0.5 in by 1 in
and a wall thickness of 0.05 in. Assume that the material is brass, for which = 1.5 ù 107 S/m. A
current of 200 A dc is flowing down the tube.
a) What voltage drop is present across a 1m length of the tube? Converting all measurements to
meters, the tube resistance over a 1 m length will be:
R1 =
(1.5 ù 107 )
1
⌅
⇧
*4
(2.54)(2.54_2) ù 10 * 2.54(1 * .1)(2.54_2)(1 * .2) ù 10*4
= 7.38 ù 10*4 ⌦
The voltage drop is now V = IR1 = 200(7.38 ù 10*4 ) = 0.147 V.
b) Find the voltage drop if the interior of the tube is filled with a conducting material for which
= 1.5 ù 105 S/m: The resistance of the filling will be:
R2 =
1
(1.5 ù 105 )(1_2)(2.54)2 ù 10*4 (.9)(.8)
= 2.87 ù 10*2 ⌦
The total resistance is now the parallel combination of R1 and R2 :
RT = R1 R2 _(R1 + R2 ) = 7.19 ù 10*4 ⌦, and the voltage drop is now V = 200RT = .144 V.
5.14. A rectangular conducting plate lies in the xy plane, occupying the region 0 < x < a, 0 < y < b. An
identical conducting plate is positioned directly above and parallel to the first, at z = d. The region
between plates is filled with material having conductivity (x) = 0 e*x_a , where 0 is a constant.
Voltage V0 is applied to the plate at z = d; the plate at z = 0 is at zero potential. Find, in terms of the
given parameters:
a) the electric field intensity E within the material: We know that E will be z-directed, but the conductivity varies with x. We therefore expect no z variation in E, and also note that the line integral
of E between the bottom and top plates must always give V0 . Therefore E = *V0 _d az V_m.
b) the total current flowing between plates: We have
J = (x)E =
* 0 e*x_a V0
az
d
Using this, we find
I=
J dS =
b
0
a
0
* 0 e*x_a V0
az (*az ) dx dy =
d
0 abV0
d
(1 * e*1 ) =
0.63ab 0 V0
A
d
c) the resistance of the material: We use
R=
V0
d
=
I
0.63 ab
0
⌦
75
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5.15. A conducting medium is in the shape of a hemispherical shell, having inner and outer radii a and
b, respectively. The conductivity varies radially as (r) = 0 a_r S/m, where 0 is a constant. The
surfaces r = a and b are coated with silver (essentially infinite conductivity for this problem). The
inner surface is raised to potential V0 ; the outer surface is grounded. A radial current, I0 amps, flows
between surfaces.
a) Find the current density J A_m2 , in terms of I0 : J = I0 _(2⇡r) ar A_m2
b) Find E between surfaces in terms of I0 :
E=
J
I
= 0
2⇡r
0
1
r
a 0
ar =
I0
a V_m
2⇡a 0 r r
c) Find V0 in terms of I0 :
V0 = *
a
b
E dL = *
d) Find the resistance, R:
a
b
I0
I0
dr =
ln(b_a) V
2⇡a 0 r
2⇡a 0
V0
ln(b_a)
=
ohms
I0
2⇡a 0
R=
5.16. A coaxial transmission line has inner and outer conductor radii a and b. Between conductors (a < ⇢ <
b) lies a conductive medium whose conductivity is (⇢) = 0 _⇢, where 0 is a constant. The inner
conductor is charged to potential V0 , and the outer conductor is grounded.
a) Assuming dc radial current I per unit length in z, determine the radial current density field J in
A_m2 : This will be the current divided by the cross-sectional area that is normal to the current
direction:
I
J=
a A_m2
2⇡⇢(1) ⇢
b) Determine the electric field intensity E in terms of I and other parameters, given or known:
E=
J
=
I⇢
I
a =
a V_m
2⇡ 0 ⇢ ⇢ 2⇡ 0 ⇢
c) by taking an appropriate line integral of E as found in part b, find an expression that relates V0
to I:
a
a
I(b * a)
I
V0 = *
E dL = *
a⇢ a⇢ d⇢ =
V
2⇡ 0
b
b 2⇡ 0
d) find an expression for the conductance of the line per unit length, G:
G=
2⇡ 0
I
=
S_m
V0
(b * a)
76
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5.17. Consider the setup as in Problem 5.15, except find R by considering the medium as made up of a
stack of hemispherical shells, each of thickness dr, and where the overall resistance will be the series
combination of the shell resistances: Assuming uniform conductivity over the differential thickness,
we may use the relation R = L_( A) to construct the differential resistance of each layer:
dR =
dr
dr
dr
=
=
(r)(2⇡r2 ) ( 0 a_r)(2⇡r2 ) 2⇡ar 0
where (r) = 0 a_r as in Problem 5.15. The net resistance will then be the series combination,
expressed through the integral:
R=
dR =
b
a
ln(b_a)
dr
=
ohms
2⇡ar 0
2⇡a 0
as before.
5.18. Two parallel circular plates of radius a are located at z = 0 and z = d. The top plate (z = d) is raised
to potential V0 ; the bottom plate is grounded. Between the plates is a conducting material having
radial-dependent conductivity, (⇢) = 0 ⇢, where 0 is a constant.
a) Find the ⇢-independent electric field strength, E, between plates: The integral of E between plates
must give V0 , independent of position on the plates. Therefore, it must be true that
V
E = * 0 az V_m
d
(0 < ⇢ < a)
b) Find the current density, J between plates:
V ⇢
J = E = * 0 0 az A_m2
d
c) Find the total current, I, in the structure:
I=
2⇡
0
a
0
V ⇢
2⇡a3 0 V0
* 0 0 az (*az ) ⇢ d⇢ d =
A
d
3d
d) Find the resistance between plates:
R=
V0
3d
=
ohms
I
2⇡a3 0
77
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5.19. Consider the setup as in Problem 5.18, except find R by considering medium as made up of concentric
cylindrical shells, each of thickness d⇢. The resistance will be the parallel combination of the shell
resistances: Assuming uniform conductivity throughout the shell volume, we may use R = L_( A)
to write the shell resistance:
d
dR =
⇢(2⇡⇢d⇢)
0
Now
1
=
R
1
=
dR
a
0
2⇡ 0 ⇢2 d⇢ 2⇡ 0 a3
3d
ohms
=
Ÿ R=
d
3d
2⇡a3 0
as before.
5.20. Consider the basic image problem of a point charge q at z = d, suspended over an infinite conducting
plane at z = 0.
a) Apply Eq. (10) in Chapter 2 to find E and D everywhere on the conductor surface as functions
of cylindrical radius, ⇢: Start with the field evaluated on the surface:
E=
q(r * r1® )
q(r * r2® )
*
4⇡✏0 r * r1® 3 4⇡✏0 r * r2® 3
´≠≠≠≠≠≠Ø≠≠≠≠≠≠¨ ´≠≠≠≠≠≠Ø≠≠≠≠≠≠¨
upper charge
image charge
where r = ⇢ a⇢ , r1® = d az , and r2® = *d az . and where r * r1®  = r * r2®  =
the above substitutions, we find
E(⇢) =
˘
⇢2 + d 2 . With
q(⇢a⇢ * daz )
q(⇢a⇢ + daz )
*qdaz
*
=
4⇡✏0 (⇢2 + d 2 )3_2 4⇡✏0 (⇢2 + d 2 )3_2
2⇡✏0 (⇢2 + d 2 )3_2
b) Use your result from part a to find the charge density, ⇢s , and the total induced charge on the
conductor: The charge density is found through:
*qdaz
*qd
Û
Û
⇢s = D nÛ = ✏0 E nÛ =
az =
C_m2
2
2
3_2
2
Ûs
Ûs 2⇡(⇢ + d )
2⇡(⇢ + d 2 )3_2
The total charge is now
Qind =
s
⇢s da =
*1
2⇡
0
ÿ
0
*qd
⇢d⇢d = *qd
2
2⇡(⇢ + d 2 )3_2
Ûÿ
= *qd ˘
Û = *q
⇢2 + d 2 Û0
ÿ
0
⇢d⇢
(⇢2 + d 2 )3_2
No surprise.
78
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5.21. Let the surface y = 0 be a perfect conductor in free space. Two uniform infinite line charges of 30
nC/m each are located at x = 0, y = 1, and x = 0, y = 2.
a) Let V = 0 at the plane y = 0, and find V at P (1, 2, 0): The line charges will image across the
plane, producing image line charges of -30 nC/m each at x = 0, y = *1, and x = 0, y = *2.
We find the potential at P by evaluating the work done in moving a unit positive charge from the
y = 0 plane (we choose the origin) to P : For each line charge, this will be:
4
5
⇢l
f inal distance f rom charge
VP * V0,0,0 = *
ln
2⇡✏0
initial distance f rom charge
where V0,0,0 = 0. Considering the four charges, we thus have
L
H˘ I
H˘ I
H ˘ IM
⇠ ⇡
⇢l
10
17
2
1
VP = *
ln
+ ln
* ln
* ln
2⇡✏0
2
1
1
2
L
H
I
H ˘ IM
L˘ ˘ M
⇠˘ ⇡
⇢l
17
10 17
1
30 ù 10*9
=
ln (2) + ln ˘
+ ln
10 + ln
=
ln
˘
2⇡✏0
2
2⇡✏0
2
2
= 1.20 kV
b) Find E at P : Use
4
⇢l
(1, 2, 0) * (0, 1, 0) (1, 2, 0) * (0, 2, 0)
+
2⇡✏0
(1, 1, 0)2
(1, 0, 0)2
5
(1, 2, 0) * (0, *1, 0) (1, 2, 0) * (0, *2, 0)
*
*
(1, 3, 0)2
(1, 4, 0)2
4
5
⇢l
(1, 1, 0) (1, 0, 0) (1, 3, 0) (1, 4, 0)
=
+
*
*
= 723 ax * 18.9 ay V_m
2⇡✏0
2
1
10
17
EP =
79
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5.22. The line segment x = 0, *1 f y f 1, z = 1, carries a linear charge density ⇢L = ⇡y C/m. Let
z = 0 be a conducting plane and determine the surface charge density at: (a) (0,0,0); (b) (0,1,0).
We consider the line charge to be made up of a string of differential segments of length, dy® , and
of charge dq = ⇢L dy® . A given segment at location (0, y® , 1) will have a corresponding image
charge segment at location (0, y® , *1). The differential flux density on the y axis that is associated
with the segment-image pair will be
dD =
⇢L dy® [(y * y® ) ay * az ]
4⇡[(y * y® )2 + 1]3_2
*
⇢L dy® [(y * y® ) ay + az ]
4⇡[(y * y® )2 + 1]3_2
=
*⇢L dy® az
2⇡[(y * y® )2 + 1]3_2
In other words, each charge segment and its image produce a net field in which the y components
have cancelled. The total flux density from the line charge and its image is now
D(y) =
dD =
14
1
*⇡y®  az dy®
® 2
3_2
*1 2⇡[(y * y ) + 1]
y®
5
az
y®
=*
+
dy®
2 0 [(y * y® )2 + 1]3_2 [(y + y® )2 + 1]3_2
4
51
az
y(y * y® ) + 1
y(y + y® ) + 1
=
+
2 [(y * y® )2 + 1]1_2 [(y + y® )2 + 1]1_2 0
4
5
az
y(y * 1) + 1
y(y + 1) + 1
2
1_2
=
+
* 2(y + 1)
2 [(y * 1)2 + 1]1_2
[(y + 1)2 + 1]1_2
Now, at the origin (part a), we find the charge density through
L
M
az
1
1
Û
2
⇢s (0, 0, 0) = D az Û
=
˘ + ˘ * 2 = *0.29 C_m
Ûy=0
2
2
2
Then, at (0,1,0) (part b), the charge density is
L
M
az
3
Û
⇢s (0, 1, 0) = D az Û
=
1 + ˘ * 2 = *0.24 C_m2
Ûy=1
2
5
5.23. A dipole with p = 0.1az C m is located at A(1, 0, 0) in free space, and the x = 0 plane is perfectlyconducting.
a) Find V at P (2, 0, 1). We use the far-field potential for a z-directed dipole:
V =
p cos ✓
p
z
=
4⇡✏0 [x2 + y2 + z2 ]1.5
4⇡✏0 r2
The dipole at x = 1 will image in the plane to produce a second dipole of the opposite orientation
at x = *1. The potential at any point is now:
4
5
p
z
z
V =
*
4⇡✏0 [(x * 1)2 + y2 + z2 ]1.5 [(x + 1)2 + y2 + z2 ]1.5
Substituting P (2, 0, 1), we find
.1 ù 106
V =
4⇡✏0
L
1
1
˘ * ˘
2 2 10 10
M
= 289.5 V
80
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5.23
b) Find the equation of the 200-V equipotential surface in cartesian coordinates: We just set the
potential exression of part a equal to 200 V to obtain:
4
5
z
z
*
= 0.222
[(x * 1)2 + y2 + z2 ]1.5 [(x + 1)2 + y2 + z2 ]1.5
5.24. At a certain temperature, the electron and hole mobilities in intrinsic germanium are given as 0.43 and
0.21 m2 _V s, respectively. If the electron and hole concentrations are both 2.3 ù 1019 m*3 , find the
conductivity at this temperature.
With the electron and hole charge magnitude of 1.6 ù 10*19 C, the conductivity in this case can
be written:
= ⇢e  e + ⇢h h = (1.6 ù 10*19 )(2.3 ù 1019 )(0.43 + 0.21) = 2.36 S_m
5.25. Electron and hole concentrations increase with temperature. For pure silicon, suitable expressions are
⇢h = *⇢e = 6200T 1.5 e*7000_T C_m3 . The functional dependence of the mobilities on temperature is
given by h = 2.3 ù 105 T *2.7 m2 _V s and e = 2.1 ù 105 T *2.5 m2 _V s, where the temperature, T ,
is in degrees Kelvin. The conductivity will thus be
⌅
⇧
= *⇢e e + ⇢h h = 6200T 1.5 e*7000_T 2.1 ù 105 T *2.5 + 2.3 ù 105 T *2.7
⇧
1.30 ù 109 *7000_T ⌅
=
e
1 + 1.095T *.2 S_m
T
Find
at:
a) 0˝ C: With T = 273˝ K, the expression evaluates as (0) = 4.7 ù 10*5 S_m.
b) 40˝ C: With T = 273 + 40 = 313, we obtain (40) = 1.1 ù 10*3 S_m.
c) 80˝ C: With T = 273 + 80 = 353, we obtain (80) = 1.2 ù 10*2 S_m.
5.26. A semiconductor sample has a rectangular cross-section 1.5 by 2.0 mm, and a length of 11.0 mm. The
material has electron and hole densities of 1.8 ù 1018 and 3.0 ù 1015 m*3 , respectively. If e = 0.082
m2 _V s and h = 0.0021 m2 _V s, find the resistance offered between the end faces of the sample.
Using the given values along with the electron charge, the conductivity is
⌅
⇧
= (1.6 ù 10*19 ) (1.8 ù 1018 )(0.082) + (3.0 ù 1015 )(0.0021) = 0.0236 S_m
The resistance is then
R=
l
0.011
=
= 155 k⌦
A (0.0236)(0.002)(0.0015)
81
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5.27. Atomic hydrogen contains 5.5ù1025 atoms/m3 at a certain temperature and pressure. When an electric
field of 4 kV/m is applied, each dipole formed by the electron and positive nucleus has an effective
length of 7.1 ù 10*19 m.
a) Find P: With all identical dipoles, we have
P = Nqd = (5.5 ù 1025 )(1.602 ù 10*19 )(7.1 ù 10*19 ) = 6.26 ù 10*12 C_m2 = 6.26 pC_m2
b) Find ✏r : We use P = ✏0 e E, and so
e =
Then ✏r = 1 +
P
6.26 ù 10*12
=
= 1.76 ù 10*4
✏0 E
(8.85 ù 10*12 )(4 ù 103 )
e = 1.000176.
5.28. Find the dielectric constant of a material in which the electric flux density is four times the polarization.
First we use D = ✏0 E + P = ✏0 E + (1_4)D. Therefore D = (4_3)✏0 E, so we identify ✏r = 4_3.
5.29. A coaxial conductor has radii a = 0.8 mm and b = 3 mm and a polystyrene dielectric for which
✏r = 2.56. If P = (2_⇢)a⇢ nC_m2 in the dielectric, find:
a) D and E as functions of ⇢: Use
E=
(2_⇢) ù 10*9 a⇢
P
144.9
=
=
a V_m
✏0 (✏r * 1) (8.85 ù 10*12 )(1.56)
⇢ ⇢
Then
D = ✏0 E + P =
b) Find Vab and
2 ù 10*9 a⇢ ⌧ 1
3.28 ù 10*9 a⇢
3.28a⇢
+1 =
C_m2 =
nC_m2
⇢
1.56
⇢
⇢
e : Use
Vab = *
0.8
3
⇠ ⇡
144.9
3
d⇢ = 144.9 ln
= 192 V
⇢
0.8
e = ✏r * 1 = 1.56, as found in part a.
c) If there are 4 ù 1019 molecules per cubic meter in the dielectric, find p(⇢): Use
p=
(2 ù 10*9 _⇢)
P
5.0 ù 10*29
=
a
=
a⇢ C m
⇢
N
⇢
4 ù 1019
82
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5.30. Consider a composite material made up of two species, having number densities N1 and N2
molecules_m3 respectively. The two materials are uniformly mixed, yielding a total number density
of N = N1 + N2 . The presence of an electric field E, induces molecular dipole moments p1 and p2
within the individual species, whether mixed or not. Show that the dielectric constant of the composite
material is given by ✏r = f ✏r1 + (1 * f )✏r2 , where f is the number fraction of species 1 dipoles in the
composite, and where ✏r1 and ✏r2 are the dielectric constants that the unmixed species would have if
each had number density N.
We may write the total polarization vector as
0
Ptot = N1 p1 + N2 p2 = N
N1
N
p1 + 2 p2
N
N
1
⌅
⇧
= N f p1 + (1 * f )p2 = f P1 + (1 * f )P2
⌅
In terms of the susceptibilities, this becomes Ptot = ✏0 f
are evaluated at the composite number density, N. Now
e1 + (1 * f ) e2
⇧
E, where
e1 and
e2
⌅
⇧
D = ✏r ✏0 E = ✏0 E + Ptot = ✏0 1 + f e1 + (1 * f ) e2 E
´≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠¨
✏r
Identifying ✏r as shown, we may rewrite it by adding and subracting f :
⌅
⇧ ⌅
✏r = 1 + f * f + f e1 + (1 * f ) e2 = f (1 +
⌅
⇧
= f ✏r1 + (1 * f )✏r2 Q.E.D.
e1 ) + (1 * f )(1 +
e2 )
⇧
5.31. A circular cylinder of radius a and dielectric constant ✏r is positioned with its axis on the z axis. An
electric field E = E0 ax is incident from free space onto the cylinder surface. E0 is a constant.
E
a
x
a) Find the electric field in the cylinder interior, and express the result in rectangular components:
Referring to the figure, the first step is to transform the given field into cylindrical coordinates.
We have
Eout = E0 ax = E0 cos a⇢ * E0 sin a = E⇢ out a⇢ + E out a
Now, the requirement that the tangential component of E is continuous at the boundary is stated
as E ,in = E ,out . Then, the requirement of normal continuity of D at the boundary is stated as
D⇢,in = D⇢,out Ÿ ✏r ✏0 E⇢,in = ✏0 E⇢,out
Therefore
Ein (⇢, ) =
E0
cos a⇢ * E0 sin a
✏r
83
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5.31
a) (continued). Converting to rectangular components, we find
and
E0
E
cos a⇢ = 0 cos2
✏r
✏r
ax + cos sin ay
*E0 sin a = E0 sin2
ax * cos sin ay
Combining the two components, we may write
Ein =
E0 ⌅
cos2
✏r
+ ✏r sin2
ax + (1 * ✏r ) sin cos ay
⇧
As a check, note that this expression reduces to Ein = E0 ax if ✏r = 1.
b) Consider a field line incident at = 135˝ . Find ✏r so that ˘
the interior field line will be oriented
˝
˝
at 22.5 from the x axis: At 135 , sin = * cos = 1_ 2, and the interior field expression
becomes:
E ⌧1
1
Ein ( = 135˝ ) = 0
✏ + 1 ax +
✏ * 1 ay
✏r 2 r
2 r
The angle of the interior field from the x axis will be
®
So now, we set
*1
= tan
4
Ey,in
Ex,in
5
*1
= tan
4
✏r * 1
✏r + 1
5
˘
✏r * 1
= tan(22.5˝ ) = 2 * 1
✏r + 1
Solve to find ✏r = 2.41. The 22.5˝ angle means that the interior field line will exit the cylinder
at = 90˝ (at the positive y axis).
5.32. Two equal but opposite-sign point charges of 3 C are held x meters apart by a spring that provides
a repulsive force given by Fsp = 12(0.5 * x) N. Without any force of attraction, the spring would be
fully-extended to 0.5m.
a) Determine the charge separation: The Coulomb and spring forces must be equal in magnitude.
We set up
(3 ù 10*6 )2
9 ù 10*12
=
= 12(0.5 * x)
4⇡✏0 x2
4⇡(8.85 ù 10*12 )x2
which leads to the cubic equation:
x3 * 0.5x2 + 6.74 ù 10*3
.
whose solution, found using a calculator, is x = 0.136 m.
b) what is the dipole moment?
Dipole moment magnitude will be p = qd = (3 ù 10*6 )(0.136) = 4.08 ù 10*7 C-m.
84
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5.33. Two perfect dielectrics have relative permittivities ✏r1 = 2 and ✏r2 = 8. The planar interface between
them is the surface x * y + 2z = 5. The origin lies in region 1. If E1 = 100ax + 200ay * 50az V/m,
find E2 : We need to find the components of E1 that are normal and tangent to the boundary, and then
apply the appropriate boundary conditions. The normal component will be EN1 = E1 n. Taking
f = x * y + 2z, the unit vector that is normal to the surface is
n=
⇧
(f
1 ⌅
= ˘ ax * ay + 2az
(f 
6
This normal will point in the direction of increasing f , which will be away from the origin, or into
region 2 (you can visualize a portion of the surface as a triangle
˘ whose vertices are on the three coordinate axes at x = 5, y = *5, and z = 2.5). So EN1 = (1_ 6)[100 * 200 * 100] = *81.7 V_m.
Since the magnitude is negative, the normal component points into region 1 from the surface. Then
H
I
1
EN1 = *81.65 ˘
[ax * ay + 2az ] = *33.33ax + 33.33ay * 66.67az V_m
6
Now, the tangential component will be ET 1 = E1 *EN1 = 133.3ax +166.7ay +16.67az . Our boundary
conditions state that ET 2 = ET 1 and EN2 = (✏r1 _✏r2 )EN1 = (1_4)EN1 . Thus
1
E2 = ET 2 + EN2 = ET 1 + EN1 = 133.3ax + 166.7ay + 16.67az * 8.3ax + 8.3ay * 16.67az
4
= 125ax + 175ay V_m
85
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5.34. A sphere of radius b and dielectric constant ✏r is centered at the origin in free space. An electric field
E = E0 ax is incident from free space onto the sphere surface. E0 is a constant.
a) Find the electric field in the sphere interior and express the result in rectangular components:
First transform the given field into spherical coordinates: We have
Eout (r, ✓, ) = E0 ax = E0 sin ✓ cos ar + cos ✓ cos a✓ * sin a
Because the two tangential components, E✓ and E , are continuous across the sphere boundary,
and because the normal component, Er , is discontinuous by a factor of 1_✏r , the interior field
will be:
0
1
1
Ein (r, ✓, ) = E0
sin ✓ cos ar + cos ✓ cos a✓ * sin a
✏r
This field is now transformed back to rectangular components through:
4
0
1
5
1
2
2
2
2
Ein,x = Ein ax = E0 cos
sin ✓ + cos ✓ + sin
✏r
4
0
1
5
1
2
2
Ein,y = Ein ay = E0 sin cos
sin ✓ + cos ✓ * sin cos
✏r
0
1
1
Ein,z = Ein az = E0 cos ✓ sin ✓ cos
*1
✏r
Combining these and simplifying leads to the final result:
⇧
sin2 ✓ + ✏r cos2 ✓ + ✏r sin2 ax
⌅
⇧
* sin cos sin2 ✓(✏r * 1) ay * cos ✓ sin ✓ cos (✏r * 1)az
Ein =
E0 ⌅ 2
cos
✏r
b) Specialize your result from part a for the case in which
Ein ( = ⇡) =
= ⇡: With this substitution, find
⇧
E0 ⌅
(sin2 ✓ + ✏r cos2 ✓)ax + cos ✓ sin ✓(✏r * 1)az
✏r
c) Specialize your result from part a for the case in which ✓ = ⇡_2 (sphere equator). Compare both
results (parts b and c) to the answer to Problem 5.31. With this substitution, find
Ein (✓ = ⇡_2) =
E0 ⌅
(cos2
✏r
+ ✏r sin2 )ax + (1 * ✏r ) sin cos ay
⇧
Note that the part c result is identical to the answer to Problem 5.31, as the geometry is the same.
The part b answer is seen also to be identical to the 5.31 result when one realizes that the role of
is in that case played by ✓ = * 90˝ .
86
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5.35. Let the cylindrical surfaces ⇢ = 4 cm and ⇢ = 9 cm enclose two wedges of perfect dielectrics, ✏r1 = 2
for 0 < < ⇡_2, and ✏r2 = 5 for ⇡_2 < < 2⇡. If E1 = (2000_⇢)a⇢ V/m, find:
a) E2 : The interfaces between the two media will lie on planes of constant , to which E1 is parallel.
Thus the field is the same on either side of the boundaries, and so E2 = E1 .
b) the total electrostatic energy stored in a 1m length of each region: In general we have
wE = (1_2)✏r ✏0 E 2 . So in region 1:
WE1 =
1
0
9
⇡_2
0
4
⇠ ⇡
(2000)2
1
⇡
9
(2)✏0
⇢ d⇢ d dz = ✏0 (2000)2 ln
= 45.1 J
2
⇢2
2
4
In region 2, we have
WE2 =
1
0
2⇡
⇡_2
9
4
⇠ ⇡
(2000)2
1
15⇡
9
(5)✏0
⇢ d⇢ d dz =
✏0 (2000)2 ln
= 338 J
2
⇢2
4
4
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CHAPTER 6 – 9th Edition
6.1. Consider a coaxial capacitor having inner radius a, outer radius b, unit length, and filled with a material
with dielectric constant, ✏r . Compare this to a parallel-plate capacitor having plate width, w, plate
separation d, filled with the same dielectric, and having unit length. Express the ratio b_a in terms of
the ratio d_w, such that the two structures will store the same energy for a given applied voltage.
Storing the same energy for a given applied voltage means that the capacitances will be equal.
With both structures having unit length and containing the same dielectric (permittivity ✏), we
equate the two capacitances:
⇠
⇡
2⇡✏
✏w
b
d
=
Ÿ
= exp 2⇡
ln(b_a)
d
a
w
6.2. Let S = 100 mm2 , d = 3 mm, and ✏r = 12 for a parallel-plate capacitor.
a) Calculate the capacitance:
C=
✏r ✏0 A 12✏0 (100 ù 10*6 )
=
= 0.4✏0 = 3.54 pf
d
3 ù 10*3
b) After connecting a 6 V battery across the capacitor, calculate E, D, Q, and the total stored
electrostatic energy: First,
E = V0 _d = 6_(3 ù 10*3 ) = 2000 V_m, then D = ✏r ✏0 E = 2.4 ù 104 ✏0 = 0.21 C_m2
The charge in this case is
Q = D ns = DA = 0.21 ù (100 ù 10*6 ) = 0.21 ù 10*4 C = 21 pC
Finally, We = (1_2)QV0 = 0.5(21)(6) = 63 pJ.
c) With the source still connected, the dielectric is carefully withdrawn from between the plates.
With the dielectric gone, re-calculate E, D, Q, and the energy stored in the capacitor.
E = V0 _d = 6_(3 ù 10*3 ) = 2000 V_m, as before. D = ✏0 E = 2000✏0 = 17.7 nC_m2
The charge is now Q = DA = 17.7 ù (100 ù 10*6 ) nC = 1.8 pC.
Finally, We = (1_2)QV0 = 0.5(1.8)(6) = 5.4 pJ.
d) If the charge and energy found in (c) are less than that found in (b) (which you should have discovered), what became of the missing charge and energy? In the absence of friction in removing
the dielectric, the charge and energy have returned to the battery that gave it.
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6.3. Capacitors tend to be more expensive as their capacitance and maximum voltage, Vmax , increase. The
voltage Vmax is limited by the field strength at which the dielectric breaks down, EBD . Which of these
dielectrics will give the largest CVmax product for equal plate areas: (a) air: ✏r = 1, EBD = 3 MV/m;
(b) barium titanate: ✏r = 1200, EBD = 3 MV/m; (c) silicon dioxide: ✏r = 3.78, EBD = 16 MV/m; (d)
polyethylene: ✏r = 2.26, EBD = 4.7 MV/m? Note that Vmax = EBDd, where d is the plate separation.
Also, C = ✏r ✏0 A_d, and so Vmax C = ✏r ✏0 AEBD , where A is the plate area. The maximum CVmax
product is found through the maximum ✏r EBD product. Trying this with the given materials yields the
winner, which is barium titanate.
6.4. An air-filled parallel-plate capacitor with plate separation d and plate area A is connected to a battery
which applies a voltage V0 between plates. With the battery left connected, the plates are moved apart
to a distance of 10d. Determine by what factor each of the following quantities changes:
a) V0 : Remains the same, since the battery is left connected.
b) C: As C = ✏0 A_d, increasing d by a factor of ten decreases C by a factor of 0.1.
c) E: We require E ù d = V0 , where V0 has not changed. Therefore, E has decreased by a
factor of 0.1.
d) D: As D = ✏0 E, and since E has decreased by 0.1, D decreases by 0.1.
e) Q: Since Q = CV0 , and as C is down by 0.1, Q also decreases by 0.1.
f) ⇢S : As Q is reduced by 0.1, ⇢S reduces by 0.1. This is also consistent with D having been
reduced by 0.1.
g) We : Use We = 1_2 CV02 , to observe its reduction by 0.1, since C is reduced by that factor.
Repeat the exercise for the case in which the battery is disconnected before the plate separation is
increased: The ordering of parameters is changed over that in the original problem, as the progression
of thought on the matter is different.
a) Q: Remains the same, since with the battery disconnected, the charge has nowhere to go.
b) ⇢S : As Q is unchanged, ⇢S is also unchanged, since the plate area is the same.
c) D: As D = ⇢S , it will remain the same also.
d) E: Since E = D_✏0 , and as D is not changed, E will also remain the same.
e) V0 : We require E ù d = V0 , where E has not changed. Therefore, V0 has increased by a
factor of 10.
f) C: As C = ✏0 A_d, increasing d by a factor of ten decreases C by a factor of 0.1. The same result
occurs because C = Q_V0 , where V0 is increased by 10, whereas Q has not changed.
g) We : Use We = 1_2 CV02 = 1_2 QV0 , to observe its increase by a factor of 10.
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6.5. A parallel-plate capacitor having plate area A and plate spacing d is partially filled with dielectric of
relative permittivity ✏r . The material fills the capacitor volume from the lower plate (at z = 0) up to
location z = b, where b < d. The remaining volume is free space. The dielectric has the effect of
increasing the capacitance by a certain factor over the value found when the capacitor is completely
air-filled. Find this factor in terms of b, ✏r , and other known parameters:
We use the fact that the total capacitance is the series combination of the air- and dielectric-filled
capacitances:
✏ A
✏✏ A
1
1
1
=
+
where Cair = 0
and Cdiel = r 0
CT
Cair Cdiel
d*b
b
So
✏ (d * b) + b
1
d*b
b
=
+
= r
CT
✏0 A
✏r ✏0 A
✏r ✏0 A
Then, knowing that the completely air-filled capacitance will be C0 = ✏0 A_d, the increase factor
becomes:
C
d
f= T =
C0
(d * b) + b_✏r
6.6. A parallel-plate capacitor is made using two circular plates of radius a, with the bottom plate on the
xy plane centered at the origin. The top plate is located at z = d,, with its center on the z axis. Charge
Q is on the top plate; *Q is on the bottom plate. Dielectric having z-dependent permittivity fills the
region between plates. The permittivity is given by ✏(z) = ✏0 (1 + z_ d 2 ). Find:
a) D: With Q on the top plate of area A = ⇡a2 , we would have a uniform surface charge density,
⇢s = Q_A C_m2 , and therefore D (uniform throughout the volume between plates) is:
D = *⇢s az = *
Q
az C_m2
2
⇡a
D is constant with z because the electric flux (the surface integral of D over any cross-sectional
area within the capacitor) must be constant.
b) E:
E=
*Q az
D
= 2
V_m
✏
⇡a ✏0 (1 + z2 _d 2 )
c) V0 : We will assume the top plate carries voltage V0 , and the bottom plate is grounded. Therefore
V0 = *
d
0
E dL = *
d
0
*Qaz az dz
⇡a2 ✏0 (1 + z2 _d 2 )
=
⇠ ⇡ d
Qd
Qd
*1 z Û
tan
Û = 2 V
2
d Û0 4a ✏0
⇡a ✏0
d) C: Using the result of part c, we construct:
C=
4a2 ✏0
Q
=
F
V0
d
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6.7. For the capacitor of Problem 6.6, consider the dielectric as made up of a stack of layers, each having
differential thickness dz, and where each layer (at location z) has dielectric constant ✏r (z) = (1+z_ d 2 ).
Evaluate the capacitance by considering the structure as series combination of the layer capacitances
and evaluating the appropriate integral: The differential capacitance of a single layer can be written as
✏r ✏0 A ✏0 (1 + z2 _d 2 )⇡a2
=
dz
dz
As the layers are in series, their reciprocal capacitances will sum, leading to
dC =
1
=
C
d
0
1
=
dC
d
0
dz
1 ⇡d
=
⇡a2 ✏0 (1 + z2 _d 2 ) ⇡a2 ✏0 4
Therefore C = 4a2 ✏0 _d F as before.
6.8. A parallel-plate capacitor is made using two circular plates of radius a, with the bottom plate on the xy
plane, centered at the origin. The top plate is located at z = d, with its center on the z axis. Potential
V0 is on the top plate; the bottom plate is grounded. Dielectric having radially-dependent permittivity
fills the region between plates. The permittivity is given by ✏(⇢) = ✏0 (1 + ⇢2 _a2 ). Find:
a) E: Since ✏ does not vary in the z direction, and since we must always obtain V0 when integrating
E between plates, it must follow that E = *V0 _d az V_m.
b) D: D = ✏E = *[✏0 (1 + ⇢2 _a2 )V0 _d] az C_m2 .
c) Q: Here we find the integral of the surface charge density over the top plate:
Q=
S
2⇡✏0 V0
=
d
2⇡
*✏0 (1 + ⇢2 _a2 )V0
az (*az ) ⇢ d⇢ d
d
0
0
a
3⇡✏0 a2
(⇢ + ⇢3 _a2 ) d⇢ =
V0
2d
0
D dS =
a
d) C: We use C = Q_V0 and our previous result to find
C=
3⇡✏0 a2
F
2d
6.9. For the capacitor of Problem 6.8, consider the dielectric as made up of concentric cylindrical shells,
each of thickness d⇢, and where the dielectric constant of each shell (at radius ⇢) is ✏r = (1 + ⇢2 _a2 ).
Evaluate the capacitance by considering the structure as a parallel combination of the shell capacitances, and by evaluating an appropriate integral: The capacitance of a shell of differential thickness
d⇢, radius ⇢, and length d will be
dC =
✏r ✏0 (2⇡⇢)d⇢
⇢2
where ✏r = 1 + 2
d
a
The parallel combination of shell capacitors that make up the whole structure will yield a net capacitance given by the sum of all the shell capacitances, found as the integral:
0
1
0
1
a
2⇡✏0
2⇡✏0 a2
3⇡✏0 a2
⇢2
a4
C=
dC =
1 + 2 ⇢d⇢ =
+ 2 =
F
d
d
2
2d
a
4a
0
as before.
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6.10. A coaxial cable has conductor dimensions of a = 1.0 mm and b = 2.7 mm. The inner conductor is
supported by dielectric spacers (✏r = 5) in the form of washers with a hole radius of 1 mm and an
outer radius of 2.7 mm, and with a thickness of 3.0 mm. The spacers are located every 2 cm down the
cable.
a) By what factor do the spacers increase the capacitance per unit length? The net capacitance can
be constructed as a composite quantity, composed of weighted contributions from the air-filled
and dielectric-filled regions:
Cnet =
2⇡✏0
2⇡✏r ✏0
f1 +
f
ln(b_a)
ln(b_a) 2
where f1 = (2 * 0.3)_2 and f2 = 0.3_2 are the filling factors for air and dielectric. Substituting
these gives
2⇡✏0 ⌧ 1.7
Cnet =
+ 0.15✏r
ln(b_a) 2
´≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠¨
f
where the bracketed term, f , is the capacitance increase factor that we seek. Substituting ✏r = 5
gives f = 1.6.
b) If 100V is maintained across the cable, find E at all points:
Method 1: We recall the expression for electric field in a coaxial line from Gauss’ Law:
E=
a⇢s
a
✏⇢ ⇢
where ⇢s is the surface charge density on the inner conductor. We also note that electric field
will be the same both inside and outside the dielectric rings because the integral of E between
conductors must always give 100 V. Another way to justify this is through the dielectric boundary
condition that requires tangential electric field to be continuous across an interface between two
dielectrics. This again means that E in our case will be the same inside and outside the rings. We
can therefore set up the integral for the voltage:
V0 = *
Then
a
b
E dL = *
a
b
a⇢s
a⇢
d⇢ = s ln(b_a) = 100
✏⇢
✏
a⇢s
100
101
=
= 101 Ÿ E =
a V_m
✏
ln(2.7)
⇢ ⇢
Method 2: Solve Laplace’s equation. This was done for the cylindrical geometry in Example 6.3,
which gave the potential distribution between conductors, Eq. (35):
V (⇢) = V0
ln(b_⇢)
ln(b_a)
from which
E = *(V = *
V0
dV
100
101
a =
a =
a =
a V_m
d⇢ ⇢ ⇢ ln(b_a) ⇢ ⇢ ln(2.7) ⇢
⇢ ⇢
as before.
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6.11. A coaxial transmission line has inner and outer conductors of radii a and b. The volume between conductors is filled with two dielectrics having relative permittivities ✏r1 (a < ⇢ < c) and
✏r2 (c < ⇢ < b).
a) If c = (a + b)_2, find the ratio of ✏r1 to ✏r2 such that the maximum electric field intensities in the
two regions are equal: Assuming a surface charge density ⇢s on the inner conductor and applying
Gauss’ Law in the usual manner, we would have:
D=
a⇢s
a C_m2 for a < ⇢ < b
⇢ ⇢
The maximum electric field intensity in region 1 is
Emax1 =
⇢
1
Û
DÛ
= s a⇢ V_m
Û
✏r1 ✏0 ⇢=a ✏r1 ✏0
The maximum electric field intensity in region 2 is
Emax2 =
⇢s a
1
Û
DÛ
=
a V_m
Û
⇢=c
✏r2 ✏0
c ✏r2 ✏0 ⇢
But c = (a + b)_2, so that
Emax2 =
2⇢s a
a
✏r2 ✏0 (a + b) ⇢
So now with the requirement
Emax1 = Emax2 Ÿ
⇢s
2⇢s a
✏
a+b
=
Ÿ r1 =
✏r1 ✏0
✏r2 ✏0 (a + b)
✏r2
2a
b) Find the capacitance per unit length under the conditions of part a: Begin by finding the voltage
between conductors:
a
c
a
⇠ ⇡
⇠ ⇡
a⇢s
a⇢s
a⇢s
a⇢s
b
c
V0 = *
E dL = *
d⇢ *
d⇢ =
ln
+
ln
V
✏r2 ✏0
c
✏r1 ✏0
a
b
b ✏r2 ✏0 ⇢
c ✏r1 ✏0 ⇢
Then with c = (a + b)_2, this becomes:
4
0
1
0 15
⇠
⇡
⇠
⇡5 a⇢ 4
a⇢s 1
✏
2b
1
a+b
1
b ✏r2
1
s
V0 =
ln
+
ln
=
ln
+
ln r1
✏0 ✏r2
a+b
✏
2a
✏
✏r2
a ✏r1
✏r1
✏r2
4
0 r1
1 0 15 0
⇠
⇡
a⇢
✏
1
b
1
1
= s
ln
+
*
ln r1
✏0 ✏r2
a
✏r1 ✏r2
✏r2
Finally, with the charge per unit length, Q = 2⇡a(1)⇢s , the capacitance is:
C=
2⇡a⇢s
2⇡✏ ✏
Q
=
= ⇠ ⇡ ⇠ r2 0 ⇡ ⇠ ⇡ F_m
✏
✏
V0
V0
ln b * 1 * r2 ln r1
a
✏r1
✏r2
Note that ✏r2 should appear in the numerator as shown here, whereas the answer in Appendix F
reads ✏r1 . Note also that C = 2⇡✏r1 ✏0 _ ln(b_a) as expected when ✏r2 = ✏r1 .
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6.12.
a) Determine the capacitance of an isolated conducting sphere of radius a in free space (consider
an outer conductor existing at r ô ÿ). If we assume charge Q on the sphere, the electric field
will be
Q
E=
ar V_m (a < r < ÿ)
4⇡✏0 r2
The potential on the sphere surface will then be
V0 = *
a
ÿ
E dL = *
Capacitance is then
C=
a
Q dr
Q
=
2
4⇡✏0 a
ÿ 4⇡✏0 r
Q
= 4⇡✏0 a F
V0
b) The sphere is to be covered with a dielectric layer of thickness d and dielectric contant ✏r . If
✏r = 3, find d in terms of a such that the capacitance is twice that of part a: Let the dielectric
radius be b, where b = d + a. The potential at the conductor surface is then
4
5
b
a
Q
Q
Q 1 1 ⇠1 1⇡
V0 = *
dr *
dr =
+
*
2
2
4⇡✏0 b ✏r a b
ÿ 4⇡✏0 r
b 4⇡✏r ✏0 r
Then, substituting capacitance, C = Q_V0 , and solving for b, we find, after a little algebra:
b=
a(✏r * 1)
a✏r 4⇡✏0 _C * 1
Then, substitute ✏r = 3 and C = 8⇡✏0 a (twice the part a result) to obtain:
b = 4a Ÿ d = b * a = 3a
6.13. With reference to Fig. 6.5, let b = 6 m, h = 15 m, and the conductor potential be 250 V. Take ✏ = ✏0 .
Find values for K1 , ⇢L , a, and C: We have
L
K1 =
h+
˘
h2 + b2
b
We then have
⇢L =
Next, a =
˘
h2 * b2 =
˘
L
M2
=
15 +
˘
(15)2 + (6)2
6
M2
= 23.0
4⇡✏0 V0
4⇡✏0 (250)
=
= 8.87 nC_m
ln K1
ln(23)
(15)2 * (6)2 = 13.8 m. Finally,
C=
2⇡✏0
2⇡✏
=
= 35.5 pF
*1
cosh (h_b) cosh*1 (15_6)
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6.14. Two #16 copper conductors (1.29-mm diameter) are parallel with a separation d between axes. Determine d so that the capacitance between wires in air is 30 pF/m.
We use
2⇡✏0
C
= 60 pF_m =
L
cosh*1 (h_b)
The above expression evaluates the capacitance of one of the wires suspended over a plane at
mid-span, h = d_2. Therefore the capacitance of that structure is doubled over that required
(from 30 to 60 pF/m). Using this,
0
1
⇠
⇡
2⇡✏0
h
2⇡ ù 8.854
= cosh
= cosh
= 1.46
b
C_L
60
Therefore, d = 2h = 2b(1.46) = 2(1.29_2)(1.46) = 1.88 mm.
6.15. A 2 cm diameter conductor is suspended in air with its axis 5 cm from a conducting plane. Let the
potential of the cylinder be 100 V and that of the plane be 0 V. Find the surface charge density on the:
a) cylinder at a point nearest the plane: The cylinder will image across the plane, producing an
equivalent two-cylinder problem, with the second one at location 5 cm below the plane. We will
take the plane as the zy plane, with
˘ the cylinder positions at x = ±5. Now b = 1 cm, h = 5
cm, and V0 = 100 V. Thus a = h2 * b2 = 4.90 cm. Then K1 = [(h + a)_b]2 = 98.0, and
⇢L = (4⇡✏0 V0 )_ ln K1 = 2.43 nC_m. Now
4
5
⇢L (x + a)ax + yay (x * a)ax + yay
D = ✏0 E = *
*
2⇡ (x + a)2 + y2
(x * a)2 + y2
and
4
5
⇢L h * b + a
h*b*a
Û
⇢s, max = D (*ax )Û
=
*
= 473 nC_m2
Ûx=h*b,y=0 2⇡ (h * b + a)2 (h * b * a)2
b) plane at a point nearest the cylinder: At x = y = 0,
⇢ ⌧ aax *aax
⇢ 2
D(0, 0) = * L
* 2
= * L ax
2
2⇡ a
2⇡ a
a
from which
⇢
⇢s = D(0, 0) ax = * L = *15.8 nC_m2
⇡a
c) Find the capacitance per unit length. This will be C = ⇢L _V0 = 2.43 [nC_m]_100 = 24.3 pF_m.
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6.16. Consider an arrangement of two isolated conducting surfaces of any shape that form a capacitor. Use
the definitions of capacitance (Eq. (2) in this chapter) and resistance (Eq. (14) in Chapter 5) to show
that when the region between the conductors is filled with either conductive material (conductivity )
or with a perfect dielectric (permittivity ✏), the resulting resistance and capacitance of the structures
are related through the simple formula RC = ✏_ . What basic properties must be true about both the
dielectric and the conducting medium for this condition to hold for certain?
Considering the two surfaces, with location a one surface, and location b on the other, the definitions are written as:
a
ó ✏E dS
* îb E dL
C= sa
and R =
ós E dS
* îb E dL
Note that the integration surfaces in the two definitions are (in this case) closed, because each
must completely surround one of the two conductors. The surface integrals would thus yield the
total charge on – or the total current flowing out of – the conductor inside.
The two formulas are muliplied together to give
RC =
ós ✏E dS
ós E dS
=
✏
The far-right result is valid provided that ✏ and are constant-valued over the integration surfaces.
Since in prinicple any surface can be chosen over which to integrate, the safest (and correct)
conclusion is that ✏ and must be constants over the capacitor (or resistor) volume; i.e., the
medium must be homogeneous. A little more subtle points are that ✏ and generally cannot vary
with field orientation (isotropic medium), and cannot vary with field intensity (linear medium),
for the simple relation RC = ✏_ to work.
6.17 Construct a curvilinear square map for a coaxial capacitor of 3-cm inner radius and 8-cm outer radius.
These dimensions are suitable for the drawing.
a) Use your sketch to calculate the capacitance per meter length, assuming ✏R = 1: The sketch is
shown below. Note that only a 9˝ sector was drawn, since this would then be duplicated 40 times
around the circumference to complete the drawing. The capacitance is thus
. NQ
40
C = ✏0
= ✏0
= 59 pF_m
NV
6
b) Calculate an exact value for the capacitance per unit length: This will be
2⇡✏0
C=
= 57 pF_m
ln(8_3)
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6.18 Construct a curvilinear-square map of the potential field about two parallel circular cylinders, each of
2.5 cm radius, separated by a center-to-center distance of 13cm. These dimensions are suitable for the
actual sketch if symmetry is considered. As a check, compute the capacitance per meter both from
your sketch and from the exact formula. Assume ✏R = 1.
Symmetry allows us to plot the field lines and equipotentials over just the first quadrant, as is done in the
sketch below (shown to one-half scale). The capacitance is found from the formula C = (NQ _NV )✏0 ,
where NQ is twice the number of squares around the perimeter of the half-circle and NV is twice the
number of squares between the half-circle and the left vertical plane. The result is
C=
NQ
NV
✏0 =
32
✏ = 2✏0 = 17.7 pF_m
16 0
We check this result with that using the exact formula:
C=
⇡✏0
*1
cosh (d_2a)
=
⇡✏0
*1
cosh (13_5)
= 1.95✏0 = 17.3 pF_m
97
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6.19. Construct a curvilinear square map of the potential field between two parallel circular cylinders, one
of 4-cm radius inside one of 8-cm radius. The two axes are displaced by 2.5 cm. These dimensions
are suitable for the drawing. As a check on the accuracy, compute the capacitance per meter from the
sketch and from the exact expression:
C=
*1
cosh
2⇡✏
⌅
⇧
2
(a + b2 * D2 )_(2ab)
where a and b are the conductor radii and D is the axis separation.
The drawing is shown below. Use of the exact expression above yields a capacitance value of C =
11.5✏0 F_m. Use of the drawing produces:
. 22 ù 2
C=
✏0 = 11✏0 F_m
4
98
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6.20. A solid conducting cylinder of 4-cm radius is centered within a rectangular conducting cylinder with
a 12-cm by 20-cm cross-section.
a) Make a full-size sketch of one quadrant of this configuration and construct a curvilinear-square
map for its interior: The result below could still be improved a little, but is nevertheless sufficient
for a reasonable capacitance estimate. Note that the five-sided region in the upper right corner has
been partially subdivided (dashed line) in anticipation of how it would look when the next-level
subdivision is done (doubling the number of field lines and equipotentials).
b) Assume ✏ = ✏0 and estimate C per meter length: In this case NQ is the number of squares
around the full perimeter of the circular conductor, or four times the number of squares shown
in the drawing. NV is the number of squares between the circle and the rectangle, or 5. The
capacitance is estimated to be
C=
NQ
NV
✏0 =
.
4 ù 13
✏0 = 10.4✏0 = 90 pF_m
5
99
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6.21. The inner conductor of the transmission line shown in Fig. 6.14 has a square cross-section 2a ù
2a, while the outer square is 5a ù 5a. The axes are displaced as shown. (a) Construct a good-sized
drawing of the transmission line, say with a = 2.5 cm, and then prepare a curvilinear-square plot of
the electrostatic field between the conductors. (b) Use the map to calculate the capacitance per meter
length if ✏ = 1.6✏0 . (c) How would your result to part b change if a = 0.6 cm?
a) The plot is shown below. Some improvement is possible, depending on how much time one
wishes to spend.
b) From the plot, the capacitance is found to be
. 16 ù 2
.
C=
(1.6)✏0 = 12.8✏0 = 110 pF_m
4
c) If a is changed, the result of part b would not change, since all dimensions retain the same relative
scale.
100
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6.22. A parallel-plate capacitor is air-filled and has plate area A and plate separation d. Sufficient dielectric
material of relative permittivity ✏r is available to fill half the capacitor volume. How should the material
be used to maximize the capacitance, and by what factor will the capacitance increase over the airfilled case? The two choices would be either positioning the two regions side by side, or stacking them
vertically. For the vertical positioning, the individual capacitances would be Cdiel = ✏r ✏0 A_(d_2) and
Cair = ✏0 A_(d_2). As the two capacitors are in series, reciprocals add to give
2✏r ✏0 A
d(1 + ✏r )
1
1
1
=
+
=
Ÿ Cvert =
Cvert
Cair Cdiel
2✏r ✏0 A
d(1 + ✏r )
For the regions arranged side by side, the two resulting capacitances add, giving
Choriz =
✏r ✏0 (A_2) ✏0 (A_2) ✏0 A
+
=
(1 + ✏r )
d
d
2d
Now take the ratio
Cvert
4✏r
=
Choriz
(1 + ✏r )2
This result is less than unity if ✏r > 1. Therefore, the side by side configuration yields the larger
capacitance. The increase factor over the air-filled case will be
finc =
Choriz
✏ A(1 + ✏r )_(2d) 1 + ✏r
= 0
=
C0
✏0 A_d
2
As an aside, note that the increase factor achieved when the vertical arrangement is used will be
finc,vert =
Cvert
2✏r
=
C0
(1 + ✏r )
This provides an increase as well, but maximizes at a value of 2.
6.23. A two-wire transmission line consists of two parallel perfectly-conducting cylinders, each having a
radius of 0.2 mm, separated by center-to-center distance of 2 mm. The medium surrounding the wires
has ✏r = 3 and = 1.5 mS/m. A 100-V battery is connected between the wires. Calculate:
a) the magnitude of the charge per meter length on each wire: Use
C=
⇡✏
⇡ ù 3 ù 8.85 ù 10*12
=
= 3.64 ù 10*9 C_m
*1
*1
cosh (h_b)
cosh (1_0.2)
Then the charge per unit length will be
Q = CV0 = (3.64 ù 10*11 )(100) = 3.64 ù 10*9 C_m = 3.64 nC_m
b) the battery current: Use
RC =
Then
✏
Ÿ R=
I=
3 ù 8.85 ù 10*12
= 486 ⌦
(1.5 ù 10*3 )(3.64 ù 10*11 )
V0
100
=
= 0.206 A = 206 mA
R
486
101
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6.24. A potential field in free space is given in spherical coordinates as:
<
[⇢0 _(6✏0 )] [3a2 * r2 ] (r f a)
V (r) =
(a3 ⇢0 )_(3✏0 r) (r g a)
where ⇢0 and a are constants.
a) Use Poisson’s equation to find the volume charge density everywhere: Inside r = a, we apply
Poisson’s equation to the potential there:
0
1
⇢v
⇢ 1 d 2
⇢
1 d
2
2 dV1
( V1 = * = 2
r
= 0 2
r (*2r) = * 0
✏0
dr
6✏0 r dr
✏0
r dr
from which we identify ⇢v = ⇢0
(r f a).
Outside r = a, we use
⇢
1 d
( 2 V2 = * v = 2
✏0
r dr
from which ⇢v = 0 (r g a).
0
r2
dV2
dr
1
=
1 d
r2 dr
0 0 3 11
*a ⇢0
r2
=0
3✏0 r2
b) find the total charge present: We have a constant charge density confined within a spherical
volume of radius a. The total charge is therefore Q = (4_3)⇡a3 ⇢0 C.
6.25. A capacitor is formed from concentric spherical conductors having radii a and b, where b > a. The
inner conductor is raised to potential V0 ; the outer conductor is grounded. Under these conditions,
derive Eq. (39) using Laplace’s equation: Begin with
⇠
⇡
1 d
dV
(2 V = 2
r2
= 0 (assuming radial variation only)
dr
r dr
Multiply through by r2 , and integrate over r to obtain
C
C
dV
= 21 then a second integration gives V = 2 + C3
dr
r
r
where C2 = *C1 . Next apply the boundary conditions, which are: 1) V = 0 at r = b,
and 2) V = V0 at r = a.
Applying 1:
0=
Applying 2:
V0 =
C2
*C2
+ C3 Ÿ C3 =
b
b
V
C2
C
C
+ C3 = 2 * 2 Ÿ C2 = ⇠ 0 ⇡
1
a
a
b
*1
a
b
Substituting C2 and C3 as found, into the general solution gives
⇠
⇡
1
1
*
r
b
V (r) = V0 ⇠
⇡
1
1
*
a
b
which is Eq. (39).
102
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6.26. Given the spherically-symmetric potential field in free space, V = V0 e*r_a , find:
a) ⇢v at r = a; Use Poisson’s equation, (2 V = *⇢v _✏, which in this case becomes
⇠
⇡ *V
⇡
⇢
*V0 ⇠
1 d
dV
d 2 *r_a
r *r_a
* v = 2
r2
= 20
r e
=
2*
e
✏0
dr
ar
a
r dr
ar dr
from which
⇢v (r) =
⇡
✏0 V0 ⇠
✏ V
r *r_a
2*
e
Ÿ ⇢v (a) = 0 2 0 e*1 C_m3
ar
a
a
b) the electric field at r = a; this we find through the negative gradient:
E(r) = *(V = *
V
V
dV
ar = 0 e*r_a ar Ÿ E(a) = 0 e*1 ar V_m
dr
a
a
c) the total charge: The easiest way is to first find the electric flux density, which from part b is
D = ✏0 E = (✏0 V0 _a)e*r_a ar . Then the net outward flux of D through a sphere of radius r would
be
(r) = Qencl (r) = 4⇡r2 D = 4⇡✏0 V0 r2 e*r_a C
As r ô ÿ, this result approaches zero, so the total charge is therefore Qnet = 0.
6.27. Let V (x, y) = 4e2x + f (x) * 3y2 in a region of free space where ⇢v = 0. It is known that both Ex and
V are zero at the origin. Find f (x) and V (x, y): Since ⇢v = 0, we know that (2 V = 0, and so
(2 V =
Therefore
Now
d 2f
)2V
)2V
2x
+
=
16e
+
*6=0
)x2
)y2
dx2
d 2f
df
2x
=
*16e
+
6
Ÿ
= *8e2x + 6x + C1
dx
dx2
and at the origin, this becomes
Ex =
df
)V
= 8e2x +
)x
dx
Ex (0) = 8 +
df Û
= 0(as given)
Û
dx Ûx=0
Thus df _dx x=0 = *8, and so it follows that C1 = 0. Integrating again, we find
f (x, y) = *4e2x + 3x2 + C2
which at the origin becomes f (0, 0) = *4 + C2 . However, V (0, 0) = 0 = 4 + f (0, 0). So f (0, 0) = *4
and C2 = 0. Finally, f (x, y) = *4e2x + 3x2 , and V (x, y) = 4e2x * 4e2x + 3x2 * 3y2 = 3(x2 * y2 ).
103
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6.28. Show that in a homogeneous medium of conductivity , the potential field V satisfies Laplace’s equation if any volume charge density present does not vary with time: We begin with the continuity
equation, Eq. (5), Chapter 5:
)⇢
( J=* v
)t
where J = E, and where, in our homogeneous medium, is constant with position. Now write
( J = ( ( E) = ( (*(V ) = * (2 V = *
This becomes
)⇢v
)t
(2 V = 0 (Laplace® s equation)
when ⇢v is time-independent. Q.E.D.
6.29. What total charge must be located within a unit sphere centered at the origin in free space in order to
produce the potential field V (r) = *6r5 _✏0 for r f 1? Use Poisson’s equation with radial variation
only:
0 0
11
⇠
⇡
⇢
1 d
dV
1 d
*30r4
180r3
(2 V = 2
r2
= 2
r2
=*
= * v Ÿ ⇢v = 180r3 C_m3
dr
✏0
✏0
✏0
r dr
r dr
The total charge in the sphere is then:
Q=
v
⇢v dv = 4⇡
1
0
180r3 rd r = 120⇡ C
104
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6.30. A parallel-plate capacitor has plates located at z = 0 and z = d. The region between plates is filled
with a material containing volume charge of uniform density ⇢0 C_m3 , and which has permittivity ✏.
Both plates are held at ground potential.
a) Determine the potential field between plates: We solve Poisson’s equation, under the assumption
that V varies only with z:
(2 V =
⇢0
*⇢0 z2
d 2V
=
*
Ÿ
V
=
+ C1 z + C2
✏
2✏
dz2
At z = 0, V = 0, and so C2 = 0. Then, at z = d, V = 0 as well, so we find C1 = ⇢0 d_2✏.
Finally, V (z) = (⇢0 z_2✏)[d * z].
b) Determine the electric field intensity, E between plates: Taking the answer to part a, we find E
through
⌧
⇢
dV
d ⇢0 z
E = *(V = *
az = *
(d * z) = 0 (2z * d) az V_m
dz
dz 2✏
2✏
c) Repeat a and b for the case of the plate at z = d raised to potential V0 , with the z = 0 plate
grounded: Begin with
*⇢0 z2
V (z) =
+ C1 z + C2
2✏
with C2 = 0 as before, since V (z = 0) = 0. Then
V (z = d) = V0 =
*⇢0 d 2
V
⇢ d
+ C1 d Ÿ C1 = 0 + 0
2✏
d
2✏
So that
V0
⇢ z
z + 0 (d * z)
d
2✏
We recognize this as the simple superposition of the voltage as found in part a and the voltage
of a capacitor carrying voltage V0 , but without the charged dielectric. The electric field is now
V (z) =
E=*
*V0
⇢
dV
az =
az + 0 (2z * d) az V_m
dz
d
2✏
105
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6.31. For the parallel-plate capacitor shown in Fig. 6.3, find the potential field in the interior if the upper
plate (at z = d) is raised to potential V0 , while the lower plate (at z = 0) is grounded. Do this by
solving Laplace’s equation separately in the two dielectrics. These solutions, as well as the electric
flux density, must be continuous across the dielectric interface. Take the interface to lie at z = b:
Referring to the figure, we have ✏r = ✏r1 for 0 < z < b (region 1), and ✏r = ✏r2 for b < z < d (region
2). Assuming the potential varies only with z, Laplace’s equation solves to give the general solutions:
V1 (z) = Az + B (0 < z < b) and V2 (z) = Cz + D (b < z < d)
The four boundary conditions (needed to solve for the four unknown coefficients) are:
1. V1 = 0 at z = 0
2. V2 = V0 at z = d
dV Û
dV Û
3. D1 = D2 at z = b Ÿ ✏r1 dz1 Û = ✏r2 dz2 Û
Ûb
Ûb
4. V1 = V2 at z = b
The boundary conditions are now applied in order. Beginning with the first condition:
0 = A(0) + B Ÿ B = 0
Second condition:
V0 = Cd + D Ÿ D = V0 * Cd
Third condtion:
✏r1 A = ✏r2 C Ÿ C =
Fourth condition:
Ab + B = Cb + D Ÿ Ab + 0 =
Solving the above for A gives:
A=
✏r1
A
✏r2
✏r1
✏
Ab + V0 * r1 Ad
✏r2
✏r2
V0
✏r1
b + ✏ (d * b)
r2
Making all substitutions back into the general solutions, we finally get:
V1 =
and
V0 z
✏r1
b + ✏ (d * b)
r2
(0 < z < b)
b
c
(d * z) g
f
V2 (z) = V0 1 * ✏r2
(b < z < d)
f
b + (d * b) g
✏r1
d
e
106
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6.32. A uniform volume charge has constant density ⇢v = ⇢0 C_m3 , and fills the region r < a, in which
permittivity ✏ as assumed. A conducting spherical shell is located at r = a, and is held at ground
potential. Find:
a) the potential everywhere: Inside the sphere, we solve Poisson’s equation, assuming radial variation only:
⇠
⇡ *⇢
*⇢0 r2 C1
1 d
dV
0
(2 V = 2
r2
=
Ÿ V (r) =
+
+ C2
dr
✏
6✏0
r
r dr
We require that V is finite at the orgin (or as r ô 0), and so therefore C1 = 0. Next, V = 0 at
r = a, which gives C2 = ⇢0 a2 _6✏. Outside, r > a, we know the potential must be zero, since the
sphere is grounded. To show this, solve Laplace’s equation:
(2 V =
⇠
⇡
C
1 d
2 dV
r
= 0 Ÿ V (r) = 1 + C2
2
dr
r
r dr
Requiring V = 0 at both r = a and at infinity leads to C1 = C2 = 0. To summarize
T⇢
0
(a2 * r2 ) r < a
6✏
V (r) =
0 r>a
b) the electric field intensity, E, everywhere: Use
E = *(V =
⇢ r
*dV
ar = 0 ar r < a
dr
3✏
Outside (r > a), the potential is zero, and so E = 0 there as well.
6.33. The functions V1 (⇢, , z) and V2 (⇢, , z) both satisfy Laplace’s equation in the region a < ⇢ < b,
0 f < 2⇡, *L < z < L; each is zero on the surfaces ⇢ = b for *L < z < L; z = *L for a < ⇢ < b;
and z = L for a < ⇢ < b; and each is 100 V on the surface ⇢ = a for *L < z < L.
a) In the region specified above, is Laplace’s equation satisfied by the functions V1 + V2 , V1 * V2 ,
V1 + 3, and V1 V2 ? Yes for the first three, since Laplace’s equation is linear. No for V1 V2 .
b) On the boundary surfaces specified, are the potential values given above obtained from the functions V1 + V2 , V1 * V2 , V1 + 3, and V1 V2 ? At the 100 V surface (⇢ = a), No for all. At the 0 V
surfaces, yes, except for V1 + 3.
c) Are the functions V1 + V2 , V1 * V2 , V1 + 3, and V1 V2 identical with V1 ? Only V2 is, since it
is given as satisfying all the boundary conditions that V1 does. Therefore, by the uniqueness
theorem, V2 = V1 . The others, not satisfying the boundary conditions, are not the same as V1 .
107
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6.34. Consider the parallel-plate capacitor of Problem 6.30, but this time the charged dielectric exists only
between z = 0 and z = b, where b < d. Free space fills the region b < z < d. Both plates are at
ground potential. No surface charge exists at z = b, so that both V and D are continuous there. By
solving Laplace’s and Poisson’s equations, find:
a) V (z) for 0 < z < d: In Region 1 (z < b), we solve Poisson’s equation, assuming z variation
only:
*⇢0
*⇢0 z
d 2 V1
dV1
=
Ÿ
=
+ C1 (z < b)
2
✏
dz
✏
dz
In Region 2 (z > b), we solve Laplace’s equation, assuming z variation only:
d 2 V2
dV2
=0 Ÿ
= C1® (z > b)
2
dz
dz
At this stage we apply the first boundary condition, which is continuity of D across the interface
at z = b. Knowing that the electric field magnitude is given by dV _dz, we write
✏
*⇢0 b
dV1 Û
dV Û
✏
= ✏0 2 Û
Ÿ *⇢0 b + ✏C1 = ✏0 C1® Ÿ C1® =
+ C1
Û
dz Ûz=b
dz Ûz=b
✏0
✏0
Substituting the above expression for C1® , and performing a second integration on the Poisson
and Laplace equations, we find
⇢ z2
V1 (z) = * 0 + C1 z + C2 (z < b)
2✏
and
⇢ bz
✏
V2 (z) = * 0 + C1 z + C2® (z > b)
2✏0
✏0
Next, requiring V1 = 0 at z = 0 leads to C2 = 0. Then, the requirement that V2 = 0 at z = d
leads to
⇢ bd
⇢ bd
✏
✏
0 = * 0 + C1 d + C2® Ÿ C2® = 0 * C1 d
✏0
✏0
✏0
✏0
With C2 and C2® known, the voltages now become
⇢ z2
⇢ b
✏
V1 (z) = * 0 + C1 z and V2 (z) = 0 (d * z) * C1 (d * z)
2✏
✏0
✏0
Finally, to evaluate C1 , we equate the two voltage expressions at z = b:
4
5
⇢0 b b + 2✏r (d * b)
V1 z=b = V2 z=b Ÿ C1 =
2✏ b + ✏r (d * b)
where ✏r = ✏_✏0 . Substituting C1 as found above into V1 and V2 leads to the final expressions for
the voltages:
40
1
5
⇢0 bz
b + 2✏r (d * b)
z
V1 (z) =
*
(z < b)
2✏
b + ✏r (d * b)
b
4
5
⇢0 b2
d*z
V2 (z) =
(z > b)
2✏0 b + ✏r (d * b)
108
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6.34
b) the electric field intensity for 0 < z < d: This involves taking the negative gradient of the final
voltage expressions of part a. We find
4
0
15
⇢0
dV1
b b + 2✏r (d * b)
E1 = *
a =
z*
az V_m (z < b)
dz z
✏
2 b + ✏r (d * b)
4
5
⇢0 b2
dV2
1
E2 = *
a =
a V_m (z > b)
dz z
2✏0 b + ✏r (d * b) z
6.35. In spherical coordinates, a potential is known to be a function of ✓ only.
a) Find the function V (✓) if v = 10 V at ✓ = 90˝ and E = *500 a✓ V/m at ✓ = 30˝ , r = 04 m:
Begin with the general solution of Laplace’s equation for theta variation (Eq. (41)):
⇠ ⇡
✓
V (✓) = A ln tan
+B
2
where the constants, A and B, are to be found. Using the first boundary condition, V = 10 at
✓ = 90˝ , we have 10 = A ln tan(45˝ ) + B. The first term is zero, so B = 10. The electric field is
found through
1 dV
A
E = *(V = *
a =*
a
r d✓ ✓
r sin ✓ ✓
Next apply the second boundary condition:
E(r = 0.4, ✓ = 30˝ ) = *
A
a = *5Aa✓ = *500a✓ Ÿ A = 100 V
0.4 sin 30˝ ✓
With the constants found, we construct the final answer:
⇠ ⇡
✓
V (✓) = 100 ln tan
+ 10 V
2
b) Find the electric field intensity in rectangular coordinates at ✓ = 90˝ , r = 1 m: Use the previous
result
A
E=*
a so that E(r = 1, ✓ = 90˝ ) = *100a✓
r sin ✓ ✓
As ✓ = 90˝ , the field is evaluated in the xy plane, on which a✓ = *ay .
Therefore E(r = 1, ✓ = 90˝ ) = +100ay V_m.
6.36. The derivation of Laplace’s and Poisson’s equations assumed constant permittivity, but there are cases
of spatially-varying permittivity in which the equations will still apply. Consider the vector identity,
( ( G) = G ( + ( G, where and G are scalar and vector functions, respectively. Determine
a general rule on the allowed directions in which ✏ may vary with respect to the electric field.
In the original derivation of Poisson’s equation, we started with ( D = ⇢v , where D = ✏E.
Therefore
( D = ( (✏E) = *( (✏(V ) = *(V (✏ * ✏(2 V = ⇢v
We see from this that Poisson’s equation, (2 V = *⇢v _✏, results when (V (✏ = 0. In words, ✏
is allowed to vary, provided it does so in directions that are normal to the local electric field.
109
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6.37. Coaxial conducting cylinders are located at ⇢ = 0.5 cm and ⇢ = 1.2 cm. The region between the
cylinders is filled with a homogeneous perfect dielectric. If the inner cylinder is at 100V and the outer
at 0V, find:
a) the location of the 20V equipotential surface: From Eq. (35) we have
V (⇢) = 100
ln(.012_⇢)
V
ln(.012_.005)
We seek ⇢ at which V = 20 V, and thus we need to solve:
20 = 100
ln(.012_⇢)
.012
Ÿ ⇢=
= 1.01 cm
ln(2.4)
(2.4)0.2
b) E⇢ max : We have
E⇢ = *
)V
dV
100
=*
=
)⇢
d⇢
⇢ ln(2.4)
whose maximum value will occur at the inner cylinder, or at ⇢ = .5 cm:
E⇢ max =
100
= 2.28 ù 104 V_m = 22.8 kV_m
.005 ln(2.4)
c) ✏r if the charge per meter length on the inner cylinder is 20 nC/m: The capacitance per meter
length is
2⇡✏0 ✏r
Q
C=
=
ln(2.4) V0
We solve for ✏r :
✏r =
(20 ù 10*9 ) ln(2.4)
= 3.15
2⇡✏0 (100)
6.38. Repeat Problem 6.37, but with the dielectric only partially filling the volume, within 0 <
with free space in the remaining volume.
< ⇡, and
We note that the dielectric changes with , and not with ⇢. Also, since E is radially-directed and
varies only with radius, Laplace’s equation for this case is valid (see Problem 6.36) and is the
same as that which led to the potential and field in Problem 6.37. Therefore, the solutions to parts
a and b are unchanged from Problem 6.37. Part c, however, is different. We write the charge per
unit length as the sum of the charges along each half of the center conductor (of radius a)
Q = ✏r ✏0 E⇢,max (⇡a) + ✏0 E⇢,max (⇡a) = ✏0 E⇢,max (⇡a)(1 + ✏r ) C_m
Using the numbers given or found in Problem 6.37, we obtain
1 + ✏r =
20 ù 10*9 C_m
= 6.31 Ÿ ✏r = 5.31
(8.852 ù 10*12 )(22.8 ù 103 V_m)(0.5 ù 10*2 m)⇡
We may also note that the average dielectric constant in this problem, (✏r + 1)_2, is the same as
that of the uniform dielectric constant found in Problem 6.37.
110
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6.39. The two conducting planes illustrated in Fig. 6.14 are defined by 0.001 < ⇢ < 0.120 m, 0 < z < 0.1 m,
= 0.179 and 0.188 rad. The medium surrounding the planes is air. For region 1, 0.179 < < 0.188,
neglect fringing and find:
a) V ( ): The general solution to Laplace’s equation will be V = C1 + C2 , and so
20 = C1 (.188) + C2 and 200 = C1 (.179) + C2
Subtracting one equation from the other, we find
*180 = C1 (.188 * .179) Ÿ C1 = *2.00 ù 104
Then
20 = *2.00 ù 104 (.188) + C2 Ÿ C2 = 3.78 ù 103
Finally, V ( ) = (*2.00 ù 104 ) + 3.78 ù 103 V.
b) E(⇢): Use
E(⇢) = *(V = *
1 dV
2.00 ù 104
=
a V_m
⇢d
⇢
c) D(⇢) = ✏0 E(⇢) = (2.00 ù 104 ✏0 _⇢) a C_m2 .
d) ⇢s on the upper surface of the lower plane: We use
2.00 ù 104
Û
⇢s = D nÛ
=
a
Ûsurf ace
⇢
a =
2.00 ù 104
C_m2
⇢
e) Q on the upper surface of the lower plane: This will be
Qt =
.1
0
.120
.001
2.00 ù 104 ✏0
d⇢ dz = 2.00 ù 104 ✏0 (.1) ln(120) = 8.47 ù 10*8 C = 84.7 nC
⇢
f) Repeat a) to c) for region 2 by letting the location of the upper plane be = .188 * 2⇡, and then
find ⇢s and Q on the lower surface of the lower plane. Back to the beginning, we use
20 = C1® (.188 * 2⇡) + C2® and 200 = C1® (.179) + C2®
Subtracting one from the other, we find
*180 = C1® (.009 * 2⇡) Ÿ C1® = 28.7
Then 200 = 28.7(.179) + C2® Ÿ C2® = 194.9. Thus V ( ) = 28.7 + 194.9 in region 2. Then
E=*
28.7✏0
28.7
a V_m and D = *
a C_m2
⇢
⇢
⇢s on the lower surface of the lower plane will now be
⇢s = *
28.7✏0
a
⇢
(*a ) =
28.7✏0
C_m2
⇢
The charge on that surface will then be Qb = 28.7✏0 (.1) ln(120) = 122 pC.
111
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6.39
g) Find the total charge on the lower plane and the capacitance between the planes: Total charge
will be Qnet = Qt + Qb = 84.7 nC + 0.122 nC = 84.8 nC. The capacitance will be
C=
Qnet
84.8
=
= 0.471 nF = 471 pF
V
200 * 20
6.40. A parallel-plate capacitor is made using two circular plates of radius a, with the bottom plate on the xy
plane, centered at the origin. The top plate is located at z = d, with its center on the z axis. Potential
V0 is on the top plate; the bottom plate is grounded. Dielectric having radially-dependent permittivity
fills the region between plates. The permittivity is given by ✏(⇢) = ✏0 (1 + ⇢2 _a2 ). Find:
a) V (z): Since ✏ varies in the direction normal to E, Laplace’s equation applies, and we write
(2 V =
d 2V
= 0 Ÿ V (z) = C1 z + C2
dz2
With the given boundary conditions, C2 = 0, and C1 = V0 _d. Therefore V (z) = V0 z_d V.
b) E: This will be E = *(V = *dV _dz az = *(V0 _d) az V_m.
c) Q: First we find the electric flux density: D = ✏E = *✏0 (1 + ⇢2 _a2 )(V0 _d) az C_m2 . The charge
density on the top plate is then ⇢s = D *az = ✏0 (1 + ⇢2 _a2 )(V0 _d) C_m2 . From this we find
the charge on the top plate:
Q=
2⇡
0
a
0
✏0 (1 + ⇢2 _a2 )(V0 _d) ⇢ d⇢ d =
3⇡a2 ✏0 V0
C
2d
d) C. The capacitance is C = Q_V0 = 3⇡a2 ✏0 _(2d) F.
6.41. Concentric conducting spheres are located at r = 5 mm and r = 20 mm. The region between the
spheres is filled with a perfect dielectric. If the inner sphere is at 100 V and the outer sphere at 0 V:
a) Find the location of the 20 V equipotential surface: Solving Laplace’s equation gives us
1
V (r) = V0 1r
a
* 1b
* 1b
where V0 = 100, a = 5 and b = 20. Setting V (r) = 20, and solving for r produces r = 12.5 mm.
b) Find Er,max : Use
E = *(V = *
V a
dV
ar = ⇠ 0 r ⇡
dr
r2 1 * 1
a
Er,max = E(r = a) =
b
V0
100
=
= 26.7 V_mm = 26.7 kV_m
a(1 * (a_b)) 5(1 * (5_20))
c) Find ✏r if the surface charge density on the inner sphere is 1.0 C_m2 : ⇢s will be equal in magnitude to the electric flux density at r = a. So ⇢s = (2.67 ù 104 V_m)✏r ✏0 = 10*6 C_m2 . Thus
✏r = 4.23. Note, in the first printing, the given charge density was 100 C_m2 , leading to a
ridiculous answer of ✏r = 423.
112
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6.42. The hemisphere 0 < r < a, 0 < ✓ < ⇡_2, is composed of homogeneous conducting material of
conductivity . The flat side of the hemisphere rests on a perfectly-conducting plane. Now, the material
within the conical region 0 < ✓ < ↵, 0 < r < a, is drilled out, and replaced with material that is
perfectly-conducting. An air gap is maintained between the r = 0 tip of this new material and the
plane. What resistance is measured between the two perfect conductors? Neglect fringing fields.
With no fringing fields, we have ✓-variation only in the potential. Laplace’s equation is therefore:
⇠
⇡
1
d
dV
(2 V = 2
sin ✓
=0
d✓
r sin ✓ d✓
This reduces to
C
dV
= 1 Ÿ V (✓) = C1 ln tan (✓_2) + C2
d✓
sin ✓
We assume zero potential on the plane (at ✓ = ⇡_2), which means that C2 = 0. On the cone (at
✓ = ↵), we assume potential V0 , and so V0 = C1 ln tan(↵_2)
Ÿ C1 = V0 _ ln tan(↵_2) The potential function is now
V (✓) = V0
ln tan(✓_2)
↵ < ✓ < ⇡_2
ln tan(↵_2)
The electric field is then
E = *(V = *
V0
1 dV
a✓ = *
a V_m
r d✓
r sin ✓ ln tan(↵_2) ✓
The total current can now be found by integrating the current density, J = E, over any crosssection. Choosing the lower plane at ✓ = ⇡_2, this becomes
I=
2⇡
0
a
0
*
V0
2⇡a V0
a✓ a✓ r dr d = *
A
r sin(⇡_2) ln tan(↵_2)
ln tan(↵_2)
The resistance is finally
V0
ln tan(↵_2)
=*
ohms
I
2⇡a
Note that R is in fact positive (despite the minus sign) since ln tan(↵_2) is negative when ↵ < ⇡_2
(which it must be).
R=
113
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6.43. Two concentric spherical conducting surfaces exist at r = a and r = b, where a < b. Between
conductors are two dielectrics having relative permittivities ✏r1 (a < r < c) and ✏r2 (c < r < b). The
inner conductor is raised to potential V0 ; the outer conductor is grounded. Solve Laplace’s equation
for the potential everywhere between conductors. This is the spherical coordinate version of Problem
6.31, and the procedure is the same. As a test of your final result, the potential should reduce to that
given in Eq. (39) when ✏r1 = ✏r2 : In spherical coordinates, Laplace’s equation with radial variation
only is:
⇠
⇡
1 d
dV
(2 V = 2
r2
=0
dr
r dr
which we work with to find:
⇠
⇡
⇠
⇡
C
C
d
dV
dV
dV
r2
= 0 Ÿ r2
= Ca Ÿ
= 2a Ÿ V (r) = * a + Cb
dr
dr
dr
dr
r
r
where the integration constants, Ca and Cb , are found by applying boundary conditions. As we have
two regions, with differing dielectric constants, we need to solve Laplace’s equation separately in each
region, and connect the solutions across the boundary between regions. We therefore write:
V1 (r) =
C
C1
+ C2 (a < r < c) and V2 (r) = 3 + C4 (c < r < b)
r
r
To evaluate the four integration constants, we need four boundary conditions. These are:
1. V2 = 0 at r = b
2. V1 = V0 at r = a
3. D1 r=c = D2 r=c (normal components of D are continuous at the boundary).
4. V1 (c) = V2 (c) (potential is continuous across the boundary).
The boundary conditions are now applied in order. For the first, we have
0=
C3
C
+ C4 Ÿ C4 = * 3
b
b
The second one gives:
C1
C
+ C2 Ÿ C2 = V0 * 1
a
a
As the electric field is the negative gradient of the potential field, the third condition can be written as:
0
1
0
1
*C3
dV1 Û
dV2 Û
*C1
✏1
= ✏2
Ÿ ✏1 C1 = ✏2 C3
Û = ✏2
Û Ÿ ✏1
dr Ûc
dr Ûc
c2
c2
V0 =
Finally, the fourth condition gives:
C
C
C
C1
C1
C
+ C2 = 3 + C4 Ÿ
+ V0 * 1 = 3 * 3
c
c
c
a
c
b
Using the result of the third condition, the above becomes:
C1
⇠
⇡
⇠
⇡
✏
1 1
1 1
*
+ V0 = 1 C1
*
c a
✏2
c b
114
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6.43 (continued) The constants can now be tabulated:
C1 =
✏1
✏2
⇠
1
* 1b
c
V0
⇡ ⇠
⇡
+ a1 * 1c
b
c
f
g
1_a
C2 = V0 f1 * ⇠
⇡ ⇠
⇡g
✏1 1
1
1
1
f
*b + a*c g
✏2 c
d
e
C3 = ⇠
C4 = ⇠
1
* 1b
c
1
* 1b
c
⇡
V0
✏
+ ✏2
1
⇠
1
* 1c
a
⇡
*V0 _b
⇡
⇠
⇡
✏
+ ✏2 a1 * 1c
1
Substituting the four constants into the earlier voltage expressions gives the final results:
⇠
⇡
b
c
1
1
*
f
g
a
r
V1 (r) = V0 f1 * ⇠
⇡ ⇠
⇡ g (a f r f c)
✏1 1
f
* 1b + a1 * 1c g
✏2 c
d
e
⇠
⇡
V0 1r * 1b
V2 (r) = ⇠
⇡
⇠
⇡ (c f r f b)
✏2 1
1
1
1
+
*
*
c
b
✏
a
c
1
6.44. A potential field in free space is given as V = 100 ln tan(✓_2) + 50 V.
a) Find the maximum value of E✓  on the surface ✓ = 40˝ for 0.1 < r < 0.8 m, 60˝ <
First
E=*
< 90˝ .
1 dV
100
100
100
a✓ = *
a✓ = *
a✓ = *
a
2
r d✓
2r sin(✓_2) cos(✓_2)
r sin ✓ ✓
2r tan(✓_2) cos (✓_2)
This will maximize at the smallest value of r, or 0.1:
Emax (✓ = 40˝ ) = E(r = 0.1, ✓ = 40˝ ) = *
100
a = 1.56 a✓ kV_m
0.1 sin(40) ✓
b) Describe the surface V = 80 V: Set 100 ln tan ✓_2+50 = 80 and solve for ✓: Obtain ln tan ✓_2 =
0.3 Ÿ tan ✓_2 = e.3 = 1.35 Ÿ ✓ = 107˝ (the cone surface at ✓ = 107 degrees).
115
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6.45. In free space, let ⇢v = 200✏0 _r2.4 .
a) Use Poisson’s equation to find V (r) if it is assumed that r2 Er ô 0 when r ô 0, and also that
V ô 0 as r ô ÿ: With r variation only, we have
⇠
⇡
⇢
1 d
dV
(2 V = 2
r2
= * v = *200r*2.4
dr
✏
r dr
or
Integrate once:
or
⇠
⇡
d
dV
r2
= *200r*.4
dr
dr
⇠
r2
dV
dr
⇡
=*
200 .6
r + C1 = *333.3r.6 + C1
.6
C
dV
= *333.3r*1.4 + 21 = (V (in this case) = *Er
dr
r
Our first boundary condition states that r2 Er ô 0 when r ô 0 Therefore C1 = 0. Integrate again
to find:
333.3 *.4
V (r) =
r + C2
.4
From our second boundary condition, V ô 0 as r ô ÿ, we see that C2 = 0. Finally,
V (r) = 833.3r*.4 V
b) Now find V (r) by using Gauss’ Law and a line integral: Gauss’ law applied to a spherical surface
of radius r gives:
r
200✏0 ® 2
r.6
4⇡r2 Dr = 4⇡
(r
)
dr
=
800⇡✏
0
® 2.4
.6
0 (r )
Thus
Er =
800⇡✏0 r.6
Dr
=
= 333.3r*1.4 V_m
✏0
.6(4⇡)✏0 r2
Now
V (r) = *
r
ÿ
333.3(r® )*1.4 dr® = 833.3r*.4 V
116
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6.46. By appropriate solution of Laplace’s and Poisson’s equations, determine the absolute potential at the
center of a sphere of radius a, containing uniform volume charge of density ⇢0 . Assume permittivity ✏0
everywhere. HINT: What must be true about the potential and the electric field at r = 0 and at r = a?
With radial dependence only, Poisson’s equation (applicable to r f a) becomes
0
1
⇢
⇢ r2 C
1 d
2
2 dV1
( V1 = 2
r
= * 0 Ÿ V1 (r) = * 0 + 1 + C2 (r f a)
dr
✏0
6✏0
r
r dr
For region 2 (r g a) there is no charge and so Laplace’s equation becomes
0
1
C
dV
1 d
(2 V2 = 2
r2 2 = 0 Ÿ V2 (r) = 3 + C4 (r g a)
dr
r
r dr
Now, as r ô ÿ, V2 ô 0, so therefore C4 = 0. Also, as r ô 0, V1 must be finite, so therefore
C1 = 0. Then, V must be continuous across the boundary, r = a:
⇢ a2
C
C
⇢ a2
Û
Û
V1 Û = V2 Û
Ÿ * 0 + C2 = 3 Ÿ C2 = 3 + 0
Ûr=a
Ûr=a
6✏0
a
a
6✏0
So now
V1 (r) =
⇢0 2
C
C
(a * r2 ) + 3 and V2 (r) = 3
6✏0
a
r
Finally, since the permittivity is ✏0 everywhere, the electric field will be continuous at r = a.
This is equivalent to the continuity of the voltage derivatives:
⇢ a
C
⇢ a3
dV1 Û
dV2 Û
Ÿ * 0 = * 23 Ÿ C3 = 0
Û =
Û
dr Ûr=a
dr Ûr=a
3✏0
3✏0
a
So the potentials in their final forms are
V1 (r) =
⇢0
⇢ a3
(3a2 * r2 ) and V2 (r) = 0
6✏0
3✏0 r
The requested absolute potential at the origin is now V1 (r = 0) = ⇢0 a2 _(2✏0 ) V.
117
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CHAPTER 7 – 9th Edition
7.1 a) Find H in rectangular components at P (2, 3, 4) if there is a current filament on the z axis carrying
8 mA in the az direction:
Applying the Biot-Savart Law, we obtain
Ha =
ÿ
IdL ù aR
=
4⇡R2
*ÿ
ÿ Idz a ù [2a + 3a + (4 * z)a ]
z
x
y
z
*ÿ
4⇡(z2 * 8z + 29)3_2
=
Idz[2ay * 3ax ]
ÿ
2
3_2
*ÿ 4⇡(z * 8z + 29)
Using integral tables, this evaluates as
4
5ÿ
I 2(2z * 8)(2ay * 3ax )
I
Ha =
=
(2a * 3ax )
2
1_2
4⇡ 52(z * 8z + 29)
26⇡ y
*ÿ
Then with I = 8 mA, we finally obtain Ha = *294ax + 196ay A_m
b) Repeat if the filament is located at x = *1, y = 2: In this case the Biot-Savart integral becomes
Hb =
ÿ Idz a
z ù [(2 + 1)ax + (3 * 2)ay + (4 * z)az ]
4⇡(z2 * 8z + 26)3_2
*ÿ
=
ÿ
Idz[3ay * ax ]
*ÿ 4⇡(z
2 * 8z + 26)3_2
Evaluating as before, we obtain with I = 8 mA:
4
5ÿ
I 2(2z * 8)(3ay * ax )
I
Hb =
=
(3a * ax ) = *127ax + 382ay A_m
2
1_2
4⇡ 40(z * 8z + 26)
20⇡ y
*ÿ
c) Find H if both filaments are present: This will be just the sum of the results of parts a and b, or
HT = Ha + Hb = *421ax + 578ay A_m
This problem can also be done (somewhat more simply) by using the known result for H from an
infinitely-long wire in cylindrical components, and transforming to cartesian components. The
Biot-Savart method was used here for the sake of illustration.
7.2. A filamentary conductor is formed into an equilateral triangle with sides of length l carrying current
I. Find the magnetic field intensity at the center of the triangle.
I will work this one from scratch, using the Biot-Savart law. Consider one side of the triangle,
oriented along the z axis, with its end points at z = ±l_2. Then consider a point, x0 , on the x axis,
which would correspond to the center of the triangle, and t
at which we want to find H associated
with the wire segment. We thus have IdL = Idz az , R =
The differential magnetic field at x0 is now
dH =
z2 + x20 , and aR = [x0 ax * z az ]_R.
I dz x0 ay
Idz az ù (x0 ax * z az )
IdL ù aR
=
=
4⇡R2
4⇡(x20 + z2 )3_2
4⇡(x20 + z2 )3_2
where ay would be normal to the plane of the triangle. The magnetic field at x0 is then
H=
l_2
I dz x0 ay
2
*l_2 4⇡(x0 + z2 )3_2
=
I z ay
Il ay
Ûl_2
=
Û
t
t
Û*l_2
4⇡x0 x20 + z2
2⇡x0 l 2 + 4x20
118
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7.2. (continued). Now, x0 lies at the center of the
˘equilateral triangle, and from the geometry of the triangle,
˝
we find that x0 = (l_2) tan(30 ) = l_(2 3). Substituting this result into the just-found expression
for H leads to H = 3I_(2⇡l) ay . The contributions from the other two sides of the triangle effectively
multiply the above result by three. The final answer is therefore Hnet = 9I_(2⇡l) ay A_m. It is
also possible to work this problem (somewhat more easily) by using Eq. (9), applied to the triangle
geometry.
7.3. A circular current filament lies in the xy plane with its center at the origin. The loop carries current I
in the positive a direction, and is of radius a. Find H everywhere on the z axis: Use the Biot-Savart
Law:
IdL ù aR
H=
4⇡R2
loop
˘
˘
where in this case, dL = ad a , R = a2 + z2 , and aR = (*aa⇢ + zaz )_ a2 + z2 . With these
substitutions, we get
H=
2⇡ Iad
a ù (*aa⇢ + zaz )
4⇡(a2 + z2 )3_2
0
=
2⇡ Ia2 a
0
z + Iaza⇢
4⇡(a2 + z2 )3_2
d
Note that when substituting a⇢ = ax cos + ay sin , the second term in the last integral will integrate
to zero, leaving
⇡a2 Iaz
m
H=
=
2
2
3_2
2
2⇡(a + z )
2⇡(a + z2 )3_2
where the magnetic moment is defined by m = ⇡a2 Iaz .
7.4. Two circular current loops are centered on the z axis at z = ±h. Each loop has radius a and carries
current I in the a direction.
a) Find H on the z axis over the range *h < z < h: As a first step, we find the magnetic field
on the z axis arising from a current loop of radius a, centered at the origin in the plane z = 0.
It carries a current I in the a direction. Using the Biot-Savart law, we have IdL = Iad a ,
˘
˘
R = a2 + z2 , and aR = (zaz * aa⇢ )_ a2 + z2 . The field on the z axis is then
H=
2⇡ Iad
a ù (zaz * aa⇢ )
4⇡(a2 + z2 )3_2
0
=
2⇡
0
Ia2 d az
4⇡(a2 + z2 )3_2
=
a2 I
2(a2 + z2 )3_2
az A_m
In obtaining this result, the term involving a ù zaz = za⇢ has integrated to zero, when taken
over the range 0 < < 2⇡. Substitute a⇢ = ax cos + ay sin to show this.
We now have two loops, displaced from the x-y plane to z = ±h. The field is now the superposition of the two loop fields, which we can construct using displaced versions of the H field we
just found:
L
M
H=
a2 I
2
1
1
⇧3_2 + ⌅
⇧3_2
⌅
2
2
2
(z * h) + a
(z + h) + a2
az A_m
119
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7.4 (continued) We can rewrite this in terms of normalized distances, z_a, and h_a:
L0
1*3_2 0⇠
1*3_2 M
⇠
⇡
⇡
I
z h 2
z h 2
aH =
*
+1
+
+
+1
az A
2
a a
a a
Take I = 1 A and plot H as a function of z_a if:
b) h = a_4,
c) h = a_2:
d) h = a.
Which choice for h gives the most uniform field? From the results, h = a_2 is evidently the best. This
is the Helmholtz coil configuration – in which the spacing is equal to the coil radius.
120
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7.5. The parallel filamentary conductors shown in Fig. 8.21 lie in free space. Plot H versus y, *4 < y < 4,
along the line x = 0, z = 2: We need an expression for H in cartesian coordinates. We can start with the
known H in cylindrical for an infinite filament along the z axis: H = I_(2⇡⇢) a , which we transform
to cartesian to obtain:
*Iy
Ix
H=
ax +
ay
2⇡(x2 + y2 )
2⇡(x2 + y2 )
If we now rotate the filament so that it lies along the x axis, with current flowing in positive x, we
obtain the field from the above expression by replacing x with y and y with z:
H=
Iy
*Iz
ay +
az
2
2
2⇡(y + z )
2⇡(y2 + z2 )
Now, with two filaments, displaced from the x axis to lie at y = ±1, and with the current directions as
shown in the figure, we use the previous expression to write
4
5
4
5
I(y * 1)
I(y + 1)
Iz
Iz
H=
*
ay +
*
az
2⇡[(y + 1)2 + z2 ] 2⇡[(y * 1)2 + z2 ]
2⇡[(y * 1)2 + z2 ] 2⇡[(y + 1)2 + z2 ]
˘
We now evaluate this at z = 2, and find the magnitude ( H H), resulting in
I
H =
2⇡
L0
2
2
* 2
2
y + 2y + 5 y * 2y + 5
12
0
+
(y * 1)
(y + 1)
* 2
2
y * 2y + 5 y + 2y + 5
12 M1_2
This function is plotted below
7.6. A disk of radius a lies in the xy plane, with the z axis through its center. Surface charge of uniform
density ⇢s lies on the disk, which rotates about the z axis at angular velocity ⌦ rad/s. Find H at any
point on the z axis.
We use the Biot-Savart
law in the form of Eq. (6), with the following parameters: K = ⇢s v =
˘
2
2
⇢s ⇢⌦ a , R = z + ⇢ , and aR = (z az * ⇢ a⇢ )_R. The differential field at point z is
dH =
⇢s ⇢ ⌦ a ù (z az * ⇢ a⇢ )
⇢s ⇢ ⌦ (z a⇢ + ⇢ az )
Kda ù aR
=
⇢ d⇢ d =
⇢ d⇢ d
2
2
2
3_2
4⇡R
4⇡(z + ⇢ )
4⇡(z2 + ⇢2 )3_2
121
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7.6. (continued). On integrating the above over
symmetry, leaving us with
2⇡
around a complete circle, the a⇢ components cancel from
a
⇢s ⌦ ⇢3 az
⇢
d⇢
d
=
d⇢
2
2 3_2
2
2 3_2
0
0 4⇡(z + ⇢ )
0 2(z + ⇢ )
⇠
⇡
˘
L
Ma
b 2
2 1*
2 _z2 c
a
+
2z
1
+
a
⇢⌦ ˘ 2
⇢ ⌦f
g
z2
= s
z + ⇢2 + ˘
az = s f
˘
g az A_m
2
2z f
1 + a2 _z2
z2 + ⇢2 0
g
d
e
H(z) =
a
⇢s ⇢ ⌦ ⇢ az
7.7. A filamentary conductor carrying current I in the az direction extends along the entire negative z axis.
At z = 0 it connects to a copper sheet that fills the x > 0, y > 0 quadrant of the xy plane.
a) Set up the Biot-Savart law and find H everywhere on the z axis (Hint: express a in terms of ax
and ay and angle in the integral): First, the contribution to the field at z from the current on the
negative z axis will be zero, because the cross product, IdL ù aR = 0 for all current elements on
the z axis. This leaves the contribution of the current sheet in the first quadrant. On exiting the
origin, current fans out over the first quadrant in the a⇢ direction and is uniform at a given radius.
The surface current density can therefore be written as K(⇢) = 2I_(⇡⇢) a⇢ A_m2 over the region
(0 < < ⇡_2). The Biot-Savart law applicable to surface current is written as
K ù aR
dA
2
s 4⇡R
˘
˘
where R = z2 + ⇢2 and aR = (zaz * ⇢a⇢ )_ z2 + ⇢2 Substituting these and integrating over
the first quadrant yields the setup:
H=
H=
ÿ 2I a
⇡_2
0
0
⇢ ù (zaz * ⇢a⇢ )
4⇡ 2 ⇢(z2 + ⇢2 )3_2
⇢ d⇢ d =
Following the hint, we now substitute a = ay cos
ÿ
⇡_2
0
0
*Iz a
2⇡ 2 (z2 + ⇢2 )3_2
* ax sin , and write:
⇡_2
ÿ (a cos * a sin )
y
x
*Iz
Iz
d⇢ d = 2 (ax * ay )
2
2
2
3_2
2⇡ 0
(z + ⇢ )
2⇡
0
⇢
Iz
I
Ûÿ
= 2 (ax * ay ) ˘
Û = 2 (ax * ay ) A_m
2⇡
2⇡ z
z 2 z 2 + ⇢ 2 Û0
H=
d⇢ d
ÿ
0
d⇢
(z2 + ⇢2 )3_2
b) repeat part a, but with the copper sheet occupying the entire xy plane. In this case, the limits
are (0 < < 2⇡). The cos and sin terms would then integrate to zero, so the answer is just
that: H = 0.
122
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7.8. For the finite-length current element on the z axis, as shown in Fig. 8.5, use the Biot-Savart law to
derive Eq. (9) of Sec. 8.1: The Biot-Savart law reads:
H=
z2
z1
IdL ù aR
=
4⇡R2
⇢ tan ↵2 Idza ù (⇢a * za )
z
⇢
z
4⇡(⇢2 + z2 )3_2
⇢ tan ↵1
⇢ tan ↵2
=
⇢ tan ↵1
I⇢a dz
4⇡(⇢2 + z2 )3_2
The integral is evaluated (using tables) and gives the desired result:
L
M
Iza
tan ↵2
tan ↵1
I
Û⇢ tan ↵2
H=
=
*˘
a
Û
˘
˘
1 + tan2 ↵
1 + tan2 ↵
4⇡⇢ ⇢2 + z2 Û⇢ tan ↵1 4⇡⇢
2
I
=
(sin ↵2 * sin ↵1 )a
4⇡⇢
1
7.9. A uniform surface charge density ⇢s = ⇢0 C_m2 is in the shape of a sphere of radius a. The sphere
rotates about the z axis at angular velocity ⌦ rad/s. Find
a) the total current flowing: The surface current density will be K = ⇢s v = ⇢0 ⌦a sin ✓ a A/m. The
total current will be
I=
K ndL =
⇡
0
⇢0 a⌦ sin ✓ a
a ad✓ = 2a2 ⌦⇢0 A
b) H at the origin. Make appropriate use of Eq. (6): The setup is
H=
K ù aR
da
2
s 4⇡R
where R = a and aR = *ar . Substituting these, and the given current density, we have
H=
⇡ ⇢ ⌦a sin ✓ a
0
2⇡
0
0
ù (*ar )
4⇡a2
a2 sin ✓d✓d =
2⇡a3 ⇢0 ⌦
4⇡a2
⇡
0
* sin2 ✓ a✓ d✓
Now, the ✓ dependence in a✓ must be included – done by converting a✓ to cylindrical components:
a✓ = a⇢ cos ✓ * az sin ✓. With this substitution, we have
⇡
⇧
a⇢0 ⌦ ⇡ ⌅ 3
*a⇢0 ⌦ ⌧ 1
sin ✓ az * cos ✓ sin2 ✓ a⇢ d✓ =
cos ✓(sin2 ✓ + 2) az
2
2
3
0
0
2a⇢0 ⌦
=
az A_m
3
H=
Note that in terms of the total current (from part a), this is H = I_(3a) az A/m.
123
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7.10. A uniform volume charge density ⇢v = ⇢0 C_m3 is in the shape of a sphere of radius a. The sphere
rotates about the z axis at angular velocity ⌦ rad/s. Find
a) the total current flowing: This will be found through
J dS where J = ⇢v v = ⇢0 ⌦r sin ✓ a A_m2
I=
from which:
I=
⇡
0
a
0
⇢0 ⌦r sin ✓ a
a rdrd✓ = 2⇢0 ⌦
a
0
r2 dr =
2⇢0 ⌦a3
A
3
b) the magnetic field, H, at the origin: Use the volume current version of the Biot-Savart Law:
2⇡
⇡
a ⇢ ⌦r sin ✓ a ù (*a )
J ù aR dv
0
r 2
=
r sin ✓ drd✓d
2
2
4⇡R
4⇡r
v
0
0
0
⇡
a
*2⇡⇢0 ⌦r sin2 ✓ a✓
=
drd✓d
4⇡
0
0
H=
where we used R = r and aR = *ar as observed from the origin. Next, include the ✓ dependence
in a✓ by converting it to cylindrical components: a✓ = a⇢ cos ✓ * az sin ✓. The field expression
now becomes:
⇧
⇢ ⌦ ⇡ a⌅
⇢ ⌦a2
H=* 0
r sin3 ✓ az * r cos ✓ sin2 ✓ a⇢ drd✓ = * 0
2 0 0
4
⌧
2
2
⇡
a ⇢0 ⌦ 1
a ⇢0 ⌦
=*
cos ✓(sin2 ✓ + 2) az =
az A_m
4
3
3
0
⇡
0
sin3 ✓ az d✓
Note that in terms of the total current (from part a), this is H = I_(2a) az A/m.
124
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7.11. A solenoid of length L (needs inclusion in the problem statement) and radius a is oriented with its axis
along the z axis over the range *d_2 < z < d_2. The coil is tightly wound with wire having density
n turns per meter of length, where each turn carries current I.
a) Find H at the origin. Begin with the current loop field (solution to Problem 7.3) modified by
replacing the current in that result with Is = nIdz. Is is the current in a circular band of width
dz on the coil; the solenoid is made up of these band of current over the entire length. Applying
the Problem 7.3 answer, we write the differential magnetic field contribution at the origin from
a current band at location z:
⇡a2 nIdz az
dH =
2⇡(a2 + z2 )3_2
The total H field at the origin is then found by summing the contributions of all bands of current
that make up the solenoid:
H=
dH =
L_2
⇡a2 nIdz az
=
2
2 3_2
*L_2 2⇡(a + z )
⇡a2 nIaz
nIaz
z
ÛL_2
=˘
A_m
Û
˘
2⇡ a2 a2 + z2 Û*L_2
1 + 4a2 _L2
b) Show that the midpoint field of part a reduces to nI az as the solenoid length approaches infinity:
From the part a result, letting L ô ÿ clearly yields H ô nI az .
7.12. In Fig. 8.22, let the regions 0 < z < 0.3 m and 0.7 < z < 1.0 m be conducting slabs carrying
uniform current densities of 10 A_m2 in opposite directions as shown. The problem asks you to find
H at various positions. Before continuing, we need to know how to find H for this type of current
configuration. The sketch below shows one of the slabs (of thickness D) oriented with the current
coming out of the page. The problem statement implies that both slabs are of infinite length and width.
To find the magnetic field inside a slab, we apply Ampere’s circuital law to the rectangular path of
height d and width w, as shown, since by symmetry, H should be oriented horizontally. For example,
if the sketch below shows the upper slab in Fig. 8.22, current will be in the positive y direction. Thus
H will be in the positive x direction above the slab midpoint, and will be in the negative x direction
below the midpoint.
Hout
Hout
125
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7.12. (continued). In taking the line integral in Ampere’s law, the two vertical path segments will cancel
each other. Ampere’s circuital law for the interior loop becomes
Õ
H dL = 2Hin ù w = Iencl = J ù w ù d Ÿ Hin =
Jd
2
The field outside the slab is found similarly, but with the enclosed current now bounded by the slab
thickness, rather than the integration path height:
2Hout ù w = J ù w ù D Ÿ Hout =
JD
2
where Hout is directed from right to left below the slab and from left to right above the slab (right hand
rule). Reverse the current, and the fields, of course, reverse direction. We are now in a position to
solve the problem. Find H at:
a) z = *0.2m: Here the fields from the top and bottom slabs (carrying opposite currents) will
cancel, and so H = 0.
b) z = 0.2m. This point lies within the lower slab above its midpoint. Thus the field will be oriented
in the negative x direction. Referring to Fig. 8.22 and to the sketch on the previous page, we find
that d = 0.1. The total field will be this field plus the contribution from the upper slab current:
H=
*10(0.1)
10(0.3)
ax *
ax = *2ax A_m
2
2
´≠≠≠≠Ø≠≠≠≠¨ ´≠≠Ø≠≠¨
lower slab
upper slab
c) z = 0.4m: Here the fields from both slabs will add constructively in the negative x direction:
H = *2
10(0.3)
ax = *3ax A_m
2
d) z = 0.75m: This is in the interior of the upper slab, whose midpoint lies at z = 0.85. Therefore
d = 0.2. Since 0.75 lies below the midpoint, magnetic field from the upper slab will lie in the
negative x direction. The field from the lower slab will be negative x-directed as well, leading
to:
*10(0.2)
10(0.3)
H=
ax *
ax = *2.5ax A_m
2
2
´≠≠≠≠Ø≠≠≠≠¨ ´≠≠Ø≠≠¨
upper slab
lower slab
e) z = 1.2m: This point lies above both slabs, where again fields cancel completely: Thus H = 0.
126
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7.13. A hollow cylindrical shell of radius a is centered on the z axis and carries a uniform surface current
density of Ka a .
a) Show that H is not a function of or z: Consider this situation as illustrated in Fig. 8.11. There
(sec. 8.2) it was stated that the field will be entirely z-directed. We can see this by applying
Ampere’s circuital law to a closed loop path whose orientation we choose such that current is
enclosed by the path. The only way to enclose current is to set up the loop (which we choose to be
rectangular) such that it is oriented with two parallel opposing segments lying in the z direction;
one of these lies inside the cylinder, the other outside. The other two parallel segments lie in the
⇢ direction. The loop is now cut by the current sheet, and if we assume a length of the loop in z
of d, then the enclosed current will be given by Kd A. There will be no variation in the field
because where we position the loop around the circumference of the cylinder does not affect the
result of Ampere’s law. If we assume an infinite cylinder length, there will be no z dependence in
the field, since as we lengthen the loop in the z direction, the path length (over which the integral
is taken) increases, but then so does the enclosed current – by the same factor. Thus H would
not change with z. There would also be no change if the loop was simply moved along the z
direction.
b) Show that H and H⇢ are everywhere zero. First, if H were to exist, then we should be able to
find a closed loop path that encloses current, in which all or or portion of the path lies in the
direction. This we cannot do, and so H must be zero. Another argument is that when applying
the Biot-Savart law, there is no current element that would produce a component. Again, using
the Biot-Savart law, we note that radial field components will be produced by individual current
elements, but such components will cancel from two elements that lie at symmetric distances in
z on either side of the observation point.
c) Show that Hz = 0 for ⇢ > a: Suppose the rectangular loop was drawn such that the outside
z-directed segment is moved further and further away from the cylinder. We would expect Hz
outside to decrease (as the Biot-Savart law would imply) but the same amount of current is always
enclosed no matter how far away the outer segment is. We therefore must conclude that the field
outside is zero.
d) Show that Hz = Ka for ⇢ < a: With our rectangular path set up as in part a, we have no
path integral contributions from the two radial segments, and no contribution from the outside
z-directed segment. Therefore, Ampere’s circuital law would state that
Õ
H dL = Hz d = Iencl = Ka d Ÿ Hz = Ka
where d is the length of the loop in the z direction.
e) A second shell, ⇢ = b, carries a current Kb a . Find H everywhere: For ⇢ < a we would have
both cylinders contributing, or Hz (⇢ < a) = Ka + Kb . Between the cylinders, we are outside the
inner one, so its field will not contribute. Thus Hz (a < ⇢ < b) = Kb . Outside (⇢ > b) the field
will be zero.
127
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7.14. A toroid having a cross section of rectangular shape is defined by the following surfaces: the cylinders
⇢ = 2 and ⇢ = 3 cm, and the planes z = 1 and z = 2.5 cm. The toroid carries a surface current density
of *50az A_m on the surface ⇢ = 3 cm. Find H at the point P (⇢, , z): The construction is similar to
that of the toroid of round cross section as done on p.239. Again, magnetic field exists only inside the
toroid cross section, and is given by
H=
Iencl
a
2⇡⇢
(2 < ⇢ < 3) cm, (1 < z < 2.5) cm
where Iencl is found from the given current density: On the outer radius, the current is
Iouter = *50(2⇡ ù 3 ù 10*2 ) = *3⇡ A
This current is directed along negative z, which means that the current on the inner radius (⇢ = 2) is
directed along positive z. Inner and outer currents have the same magnitude. It is the inner current
that is enclosed by the circular integration path in a within the toroid that is used in Ampere’s law.
So Iencl = +3⇡ A. We can now proceed with what is requested:
a) PA (1.5cm, 0, 2cm): The radius, ⇢ = 1.5 cm, lies outside the cross section, and so HA = 0.
b) PB (2.1cm, 0, 2cm): This point does lie inside the cross section, and the
matter. We find
3a
I
HB = encl a =
= 71.4 a A_m
2⇡⇢
2(2.1 ù 10*2 )
c) PC (2.7cm, ⇡_2, 2cm): again,
and z values do not
and z values make no difference, so
HC =
3a
2(2.7 ù 10*2 )
= 55.6 a A_m
d) PD (3.5cm, ⇡_2, 2cm). This point lies outside the cross section, and so HD = 0.
7.15. A hollow spherical conducting shell of radius a has filamentary connections made at the top (r = a,
✓ = 0) and bottom (r = a, ✓ = ⇡). A direct current I flows down the upper filament, down the
spherical surface, and out the lower filament. Find H in spherical coordinates (a) inside and (b) outside
the sphere.
Applying Ampere’s circuital law, we use a circular contour, centered on the z axis, and find
that within the sphere, no current is enclosed, and so H = 0 when r < a. The same contour
drawn outside the sphere at any z position will always enclose I amps, flowing in the negative z
direction, and so
I
I
H=*
a =*
a A_m (r > a)
2⇡⇢
2⇡r sin ✓
128
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7.16. A current filament carrying I in the *az direction lies along the entire positive z axis. At the origin,
it connects to a conducting sheet that forms the xy plane.
a) Find K in the conducting sheet: The current fans outward radially with uniform surface current
density at a fixed radius. The current density at radius ⇢ will be the total current, I, divided by
the circumference at radius ⇢:
I
K=
a A_m
2⇡⇢ ⇢
b) Use Ampere’s circuital law to find H everywhere for z > 0: Circular lines of H are expected,
centered on the z axis – in the *a direction. Ampere’s law is set up by considering a circular
path integral taken around the wire at fixed z. The enclosed current is that which passes through
any surface that is bounded by the line integration path:
Õ
H dL = 2⇡⇢H = Iencl
If the surface is that of the disk whose perimeter is the integration path, then the enclosed current
is just I, and the magnetic field becomes
H =*
I
2⇡⇢
Ÿ
H=*
I
a A_m
2⇡⇢
But the disk surface can be “stretched” so that it forms a balloon shape. Suppose the “balloon”
is a right circular cylinder, with its open top circumference at the path integral location. The
cylinder extends downward, intersecting the surface current in the x-y plane, with the bottom
of the cylinder below the x-y plane. Now, the path integral is unchanged from before, and the
enclosed current is the radial current in the x-y plane that passes through the side of the cylinder.
This current will be I = 2⇡⇢[I_(2⇡⇢)] = I, as before. So the answer given above for H applies
to anywhere in the region z > 0.
c) Find H for z < 0: Consider the same cylinder as in part b, except take the path integral of H
around the bottom circumference (below the x-y plane). The enclosed current now consists of
the filament current that enters through the top, plus the radial current that exits though the side.
The two currents are equal magnitude but opposite in sign. Therefore, the net enclosed current
is zero, and thus H = 0 (z < 0).
129
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7.17. A solid cylindrical wire of radius a carries total current I in the positive z direction. The conductivity
in the wire is radially-dependent, and is given by (⇢) = 0 ⇢ S/m, where 0 is a constant. The electric
field intensity, constant throughout the wire volume, is given by E = E0 az V/m.
a) Find J in terms of E0 ,
0 , and ⇢:
2
0 E0 ⇢ az A_m
This will be J = E =
b) Find E0 in terms of I:
2⇡
J n da =
I=
0
a
0
0 E0 ⇢ az
az ⇢d⇢d = 2⇡ 0 E0
a3
3I
Ÿ E0 =
V_m
3
2⇡ 0 a3
c) Express J in terms of I and other known parameters, excluding E0 : Using the results of part b,
we have again:
3I⇢
J= E=
az A_m2
2⇡a3
d) Using your part c result, find the magnetic field intensity, H, inside and outside the wire.
Inside the wire, we apply Ampere’s law to a circular loop of radius ⇢ < a:
Õ
Hin dL = 2⇡⇢H =
J n da =
So that
Outside (⇢ g a), we have
Hin = H in a =
2⇡
0
⇢
0
3I⇢®
⇢3
®
®
a
a
⇢
d⇢
d
=
I
z
z
2⇡a3
a3
I⇢2
a A_m2 (⇢ f a)
2⇡a3
2⇡⇢H out = I Ÿ Hout =
I
a A_m (⇢ g a)
2⇡⇢
e) Verify your result of part d by applying ( ù H = J: The curl in cylindrical coordinates will in
this case (with H -directed and varying only with ⇢) be the single term:
(ùH=
Inside the wire, we have
1 d
( ù Hin =
⇢ d⇢
0
I⇢3
2⇡a3
1 d(⇢H )
az
⇢ d⇢
1
Outside the wire, we have
1 d
( ù Hout =
⇢ d⇢
az =
3I⇢
az = J (as found in part c)
2⇡a3
0
1
I⇢
2⇡⇢
az = 0 (as expected)
130
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7.18. A wire of 3-mm radius is made up of an inner material (0 < ⇢ < 2 mm) for which = 107 S/m, and
an outer material (2mm < ⇢ < 3mm) for which = 4 ù 107 S/m. If the wire carries a total current of
100 mA dc, determine H everywhere as a function of ⇢.
Since the materials have different conductivities, the current densities within them will differ.
Electric field, however is constant throughout. The current can be expressed as
⌅
⇧
I = ⇡(.002)2 J1 + ⇡[(.003)2 * (.002)2 ]J2 = ⇡ (.002)2 1 + [(.003)2 * (.002)2 ] 2 E
Solve for E to obtain
E=
0.1
= 1.33 ù 10*4 V_m
⇡[(4 ù 10*6 )(107 ) + (9 ù 10*6 * 4 ù 10*6 )(4 ù 107 )]
We next apply Ampere’s circuital law to a circular path of radius ⇢, where ⇢ < 2mm:
2⇡⇢H 1 = ⇡⇢2 J1 = ⇡⇢2 1 E Ÿ H 1 =
1 E⇢
2
= 663 A_m
Next, for the region 2mm < ⇢ < 3mm, Ampere’s law becomes
2⇡⇢H 2 = ⇡[(4 ù 10*6 )(107 ) + (⇢2 * 4 ù 10*6 )(4 ù 107 )]E
Ÿ H 2 = 2.7 ù 103 ⇢ *
8.0 ù 10*3
A_m
⇢
Finally, for ⇢ > 3mm, the field outside is that for a long wire:
H 3=
I
0.1
1.6 ù 10*2
=
=
A_m
2⇡⇢ 2⇡⇢
⇢
7.19. In spherical coordinates, the surface of a solid conducting cone is described by ✓ = ⇡_4 and a conducting plane by ✓ = ⇡_2. Each carries a total current I. The current flows as a surface current
radially inward on the plane to the vertex of the cone, and then flows radially-outward throughout the
cross-section of the conical conductor.
a) Express the surface current density as a function of r: This will be the total current divided by
the circumference of a circle of radius r in the plane, directed toward the origin:
K(r) = *
I
a A_m2
2⇡r r
(✓ = ⇡_2)
b) Express the volume current density inside the cone as a function of r: This will be the total
current divided by the area of the spherical cap subtending angle ✓ = ⇡_4:
L
M*1
2⇡
⇡_4
I ar
2
J(r) = I
r2 sin ✓ ® d✓ ® d
ar =
˘ A_m (0 < ✓ < ⇡_4)
2
0
0
2⇡r (1 * 1_ 2)
c) Determine H as a function of r and ✓ in the region between the cone and the plane: From symmetry, we expect H to be -directed and uniform at constant r and ✓. Ampere’s circuital law can
therefore be stated as:
Õ
H dL = 2⇡r sin ✓ H = I Ÿ H =
I
a A_m (⇡_4 < ✓ < ⇡_2)
2⇡r sin ✓
131
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7.19
d) Determine H as a function of r and ✓ inside the cone: Again, -directed H is anticipated, so we
apply Ampere’s law in the following way:
Õ
H dL = 2⇡r sin ✓ H =
This becomes
or
s
2⇡
J dS =
0
2⇡r sin ✓ H = *2⇡
I ar
✓
2⇡r2 (1 * 1_
0
˘
2)
ar r2 sin ✓ ® d✓ ® d
✓
®Û
˘ cos ✓ ÛÛ0
2⇡(1 * 1_ 2)
I
4
5
(1 * cos ✓)
H=
a A_m
˘
sin ✓
2⇡r(1 * 1_ 2)
I
(0 < ✓ < ⇡_4)
As a test of this, note that the inside and outside fields (results of parts c and d) are equal at the
cone surface (✓ = ⇡_4) as they must be.
7.20. A solid conductor of circular cross-section with a radius of 5 mm has a conductivity that varies with
radius. The conductor is 20 m long and there is a potential difference of 0.1 V dc between its two ends.
Within the conductor, H = 105 ⇢2 a A/m.
a) Find as a function of ⇢: Start by finding J from H by taking the curl. With H -directed, and
varying with radius only, the curl becomes:
J=(ùH=
1 d
⇢H
⇢ d⇢
az =
1 d
105 ⇢3 az = 3 ù 105 ⇢ az A_m2
⇢ d⇢
Then E = 0.1_20 = 0.005 az V/m, which we then use with J = E to find
=
3 ù 105 ⇢
J
=
= 6 ù 107 ⇢ S_m
E
0.005
b) What is the resistance between the two ends? The current in the wire is
I=
s
J dS = 2⇡
a
0
(3 ù 105 ⇢) ⇢ d⇢ = 6⇡ ù 105
⇠
⇡
1 3
a = 2⇡ ù 105 (0.005)3 = 0.079 A
3
Finally, R = V0 _I = 0.1_0.079 = 1.3 ⌦
132
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7.21. A cylindrical wire of radius a is oriented with the z axis down its center line. The wire carries a
non-uniform current down its length of density J = b⇢ az A_m2 , where b is a constant.
a) What total current flows in the wire? We integrate the current density over the wire cross-section:
Itot =
s
2⇡
J dS =
0
a
0
b⇢ az az ⇢ d⇢ d =
2⇡ba3
A
3
b) find Hin (0 < ⇢ < a), as a function of ⇢: From the symmetry, -directed H (= H a ) is expected
in the interior; this will be constant at a fixed radius, ⇢. Apply Ampere’s circuital law to a circular
path of radius ⇢ inside:
Õ
Hin dL = 2⇡⇢H ,in =
So that
Hin =
s
2⇡
J dS =
b⇢2
a
3
⇢
0
0
b⇢® az az ⇢® d⇢® d =
2⇡b⇢3
3
A_m (0 < ⇢ < a)
c) find Hout (⇢ > a), as a function of ⇢ Same as part b, except the path integral is taken at a radius
outside the wire:
Õ
Hout dL = 2⇡⇢H ,out =
So that
s
Hout =
J dS =
ba3
a
3⇢
2⇡
0
a
0
b⇢ az az ⇢ d⇢ d =
A_m (⇢ > a)
d) verify your results of parts b and c by using ( ù H = J: With a
only with ⇢, the curl in cylindrical coordinates reduces to
J=(ùH=
Apply this to the inside field to get
1 d
Jin =
⇢ d⇢
0
For the outside field, we find
1 d
Jout =
⇢ d⇢
component of H only, varying
1 d
(⇢H ) az
⇢ d⇢
b⇢3
3
0
2⇡ba3
3
1
⇢ba3
3⇢
az = b⇢ az
1
az = 0
as expected.
133
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7.22. A solid cylinder of radius a and length L, where L >> a, contains volume charge of uniform density
⇢0 C_m3 . The cylinder rotates about its axis (the z axis) at angular velocity ⌦ rad/s.
a) Determine the current density J, as a function of position within the rotating cylinder: Use J =
⇢0 v = ⇢0 ⇢⌦ a A_m2 .
b) Determine the magnetic field intensity H inside and outside: It helps initially to obtain the field
on-axis. To do this, we use the result of Problem 8.6, but give the rotating charged disk in that
problem a differential thickness, dz. We can then evaluate the on-axis field in the rotating cylinder
as the superposition of fields from a stack of disks which exist between ±L_2. Here, we make
the problem easier by letting L ô ÿ (since L >> a) thereby specializing our evaluation to
positions near the half-length. The on-axis field is therefore:
⇠
⇡
˘
b 2
2 1*
2 _z2 c
a
+
2z
1
+
a
ÿ
⇢0 ⌦ f
g
Hz (⇢ = 0) =
˘
f
g dz
*ÿ 2z f
1 + a2 _z2
g
d
e
L
M
ÿ
⇢0 ⌦
a2
2z2
=2
+˘
* 2z dz
˘
2
2
2
2
2
0
z +a
z +a
4 2
5ÿ
˘
˘
z˘ 2
a2
z2
a
2
2
2
2
2
= 2⇢0 ⌦
ln(z + z + a ) +
z +a *
ln(z + z + a ) *
2
2
2
2 0
⌧ ˘
⌧ ˘
ÿ
= ⇢0 ⌦ z z2 + a2 * z2
= ⇢0 ⌦ z z2 + a2 * z2
0
zôÿ
Using the large z approximation in the radical, we obtain
4 0
1
5
⇢ ⌦a2
a2
2
2
Hz (⇢ = 0) = ⇢0 ⌦ z 1 + 2 * z = 0
2
2z
To find the field as a function of radius, we apply Ampere’s circuital law to a rectangular loop,
drawn in two locations described as follows: First, construct the rectangle with one side along
the z axis, and with the opposite side lying at any radius outside the cylinder. In taking the
line integral of H around the rectangle, we note that the two segments that are perpendicular to
the cylinder axis will have their path integrals exactly cancel, since the two path segments are
oppositely-directed, while from symmetry the field should not be different along each segment.
This leaves only the path segment that coindides with the axis, and that lying parallel to the axis,
but outside. Choosing the length of these segments to be l, Ampere’s circuital law becomes:
Õ
H dL = Hz (⇢ = 0)l + Hz (⇢ > a)l = Iencl =
s
l
J dS =
a
0
0
⇢0 ⇢⌦ a
a d⇢ dz
⇢ ⌦a2
=l 0
2
But we found earlier that Hz (⇢ = 0) = ⇢0 ⌦a2 _2. Therefore, we identify the outside field,
Hz (⇢ > a) = 0. Next, change the rectangular path only by displacing the central path component
off-axis by distance ⇢, but still lying within the cylinder. The enclosed current is now somewhat
less, and Ampere’s law becomes
Õ
H dL = Hz (⇢)l + Hz (⇢ > a)l = Iencl =
s
J dS =
l
0
a
⇢
⇢0 ⇢® ⌦ a
a d⇢ dz
⇢ ⌦
⇢ ⌦
= l 0 (a2 * ⇢2 ) Ÿ H(⇢) = 0 (a2 * ⇢2 ) az A_m
2
2
134
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7.22
c) Check your result of part b by taking the curl of H. With H z-directed, and varying only with ⇢,
the curl in cylindrical coordinates becomes
(ùH=*
dHz
a = ⇢0 ⌦⇢ a A_m2 = J
d⇢
as expected.
7.23. A solid conductor is in the shape of a circular cylinder with its axis along the z axis. A uniform electric
field E = E0 az V/m is applied, resulting in a uniform magnetic field inside the conductor, H = H0 a ,
where E0 and H0 are constants.
a) Using the cylindrical coordinate form of ( ù H = J, determine the required functional form of
the medium conductivity, , in order for the preceding condition to be met: For a constant H that
is -directed, the curl of H in cylindrical coordinates reduces to
(ùH=
We therefore identify:
=
1 d(⇢H )
1
az = H0 az = J = E0 az
⇢ d⇢
⇢
H0 1
(inverse radial dependence)
E0 ⇢
b) Repeat part a, but use the spherical coordinate form of ( ù H = J: Again, for a constant directed magnetic field, the curl in spherical coordinates reduces to:
(ùH=
H
H
1 d(H sin ✓)
1 d(rH )
ar *
a✓ =
cot ✓ ar * 0 a✓
r sin ✓
d✓
r dr
r
r
Now use a✓ = a⇢ cos ✓ * az sin ✓ and ar = a⇢ sin ✓ + az cos ✓ to obtain
⇧
H0 ⌅
cot ✓(a⇢ sin ✓ + az cos ✓) * a⇢ cos ✓ + az sin ✓
r 4
5
H0 cos2 ✓
H0
H
=
+ sin ✓ az =
az = 0 az
r
sin ✓
r sin ✓
⇢
(ùH=
....thus leading to the same result as in part a.
135
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7.24. Infinitely-long filamentary conductors are located in the y = 0 plane at x = n meters where n =
0, ±1, ±2, … Each carries 1 A in the az direction.
a) Find H on the y axis. As a help,
ÿ
…
n=1
y
=
y2 + n 2
⇡
1
⇡
*
+ 2⇡y
2 2y e * 1
We can begin by determining the field on the y axis arising from two wires only, located at
x = ±n. We start from basics, using the Biot-Savart law:
IdL ù aR
4⇡R2
H=
where R = n2 + y2 + z2
a+
=
R
2
for both wires, and where, for the wire at x = +n
*nax + yay + zaz
n 2 + y2 + z 2
a*
R =
and
1_2
nax + yay + zaz
n2 + y2 + z2
1_2
for the wire located at x = *n. Then with IdL = dz az (I = 1) for both wires, the Biot-Savart
construction becomes
H=
ÿ dz a
zù
*ÿ
This evaluates as
⌅
⇧
(*nax + yay + zaz ) + (nax + yay + zaz )
4⇡
3_2
n2 + y2 + z2
1
H=*
⇡
0
y
2
n + y2
=
*yax
2⇡
ÿ
*ÿ
dz
n 2 + y2 + z 2
3_2
1
ax A_m
Now if we include all wire pairs, the result is the superposition of an infinite number of fields of
the above form. Specifically,
ÿ 0
1…
Hnet = *
⇡ n=1
y
2
n + y2
1
ax A_m
Using the given closed form of the series expansion, our final answer is
4
5
1
1
1
Hnet = *
*
+
a A_m
2 2⇡y e2⇡y * 1 x
b) Compare your result of part a to that obtained if the filaments are replaced by a current sheet in
the y = 0 plane that carries surface current density K = 1az A/m: This is found through Eq.
(11):
1
1
1
H = K ù aN = az ù ay = * ax
2
2
2
Our answer of part a approaches this value as y ô ÿ, demonstrating that at large distances, the
parallel wires act like a current sheet. Interestingly, at close-in locations, such that 2⇡y << 1, we
.
may expand e2⇡y = 1 + 2⇡y, leading to the cancellation of the last two terms in the part a result,
.
and again, Hnet = *1_2 ax .
136
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7.25. When x, y, and z are positive and less than 5, a certain magnetic field intensity may be expressed as
H = [x2 yz_(y + 1)]ax + 3x2 z2 ay * [xyz2 _(y + 1)]az . Find the total current in the ax direction that
crosses the strip, x = 2, 1 f y f 4, 3 f z f 4, by a method utilizing:
a) a surface integral: We need to find the current density by taking the curl of the given H. Actually,
since the strip lies parallel to the yz plane, we need only find the x component of the current
density, as only this component will contribute to the requested current. This is
0
1
0
1
)Hz )Hy
xz2
2
Jx = (( ù H)x =
*
=*
+ 6x z ax
)y
)z
(y + 1)2
The current through the strip is then
1
2z2
I = J ax da = *
+ 24z dy dz = *
(y + 1)2
s
3
1
4⇠
⇡
⇠
⇡4
3 2
1
=*
z + 72z dz = * z3 + 36z2 = *259
5
5
3
3
4
40
40
3
*2z2
+ 24zy
(y + 1)
14
1
dz
b) a closed line integral: We integrate counter-clockwise around the strip boundary (using the righthand convention), where the path normal is positive ax . The current is then
1
2(4)z2
dz +
3(2)2 (4)2 dy +
Õ
(4
+
1)
1
3
4
8 3
1 3
3
3
= 108(3) * (4 * 3 ) + 192(1 * 4) * (3 * 4 ) = *259
15
3
I=
H dL =
4
4
3(2)2 (3)2 dy +
*
3
4
*
2(1)z2
dz
(1 + 1)
7.26. Consider a sphere of radius r = 4 centered at (0,0,3). Let S1 be that portion of the spherical surface
that lies above the xy plane. Find îS (( ù H) dS if H = 3⇢ a in cylindrical coordinates: First, the
1
˘
˘
intersection of the sphere with the x-y plane is a disk in the plane of radius ⇢d = 42 * 32 = 7.
The curl of the given field (having a component that varies only with ⇢) is
(ùH=
1 d
⇢H
⇢ d⇢
az =
1 d
3⇢2 az = 6 az
⇢ d⇢
Since this is a constant field, its flux through S1 will simply be the flux through the disk, which in turn
is simply the product of the field with the disk area:
˘
= 6 ù ⇡( 7)2 = 42⇡
Another way to solve the problem is to use Stokes’ theorem and write the flux as
=
Õ
H dL
where the path integral is taken around the disk perimeter. Doing this gives:
˘
˘
= 2⇡⇢d H ⇢d = 2⇡ 7 ù 3 7 = 42⇡
137
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7.27. The magnetic field intensity is given in a certain region of space as
H=
x + 2y
2
ay + az A_m
2
z
z
a) Find ( ù H: For this field, the general curl expression in rectangular coordinates simplifies to
(ùH=*
)Hy
)z
ax +
)Hy
)x
2(x + 2y)
1
ax + 2 az A_m
z3
z
az =
b) Find J: This will be the answer of part a, since ( ù H = J.
c) Use J to find the total current passing through the surface z = 4, 1 < x < 2, 3 < y < 5, in the az
direction: This will be
I=
Û
JÛ
a dx dy =
Ûz=4 z
5
3
2
1
1
dx dy = 1_8 A
42
d) Show that the same result is obtained using the other side of Stokes’ theorem: We take ó H dL
over the square path at z = 4 as defined in part c. This involves two integrals of the y component
of H over the range 3 < y < 5. Integrals over x, to complete the loop, do not exist since there is
no x component of H. We have
I=
Û
HÛ
dL =
Õ Ûz=4
5
3
2 + 2y
dy +
16
3
5
1 + 2y
1
1
dy = (2) * (2) = 1_8 A
16
8
16
138
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7.28. Given H = (3r2 _ sin ✓)a✓ + 54r cos ✓a A/m in free space:
a) find the total current in the a✓ direction through the conical surface ✓ = 20˝ , 0 f
f 2⇡,
0 f r f 5, by whatever side of Stokes’ theorem you like best. I chose the line integral side,
where the integration path is the circular path in around the top edge of the cone, at r = 5.
The path direction is chosen to be clockwise looking down on the xy plane. This, by convention,
leads to the normal from the cone surface that points in the positive a✓ direction (right hand rule).
We find
Õ
2⇡ ⌅
H dL =
(3r2 _ sin ✓)a✓ + 54r cos ✓a
0
⇧
r=5,✓=20
5 sin(20˝ ) d (*a )
= *2⇡(54)(25) cos(20˝ ) sin(20˝ ) = *2.73 ù 103 A
This result means that there is a component of current that enters the cone surface in the *a✓
direction, to which is associated a component of H in the positive a direction.
b) Check the result by using the other side of Stokes’ theorem: We first find the current density
through the curl of the magnetic field, where three of the six terms in the spherical coordinate
formula survive:
0 3 1
1 )
1 )
1 )
3r
2
a =J
(ùH=
54r cos ✓ a✓ +
(54r cos ✓ sin ✓)) ar *
r sin ✓ )✓
r )r
r )r sin ✓
Thus
9r
a
sin ✓
The calculation of the other side of Stokes’ theorem now involves integrating J over the surface
of the cone, where the outward normal is positive a✓ , as defined in part a:
S
J = 54 cot ✓ ar * 108 cos ✓ a✓ +
(( ù H) dS =
5⌧
2⇡
0
=*
0
2⇡
0
54 cot ✓ ar * 108 cos ✓ a✓ +
5
0
9r
a
sin ✓
20˝
a✓ r sin(20˝ ) dr d
108 cos(20˝ ) sin(20˝ )rdrd = *2⇡(54)(25) cos(20˝ ) sin(20˝ )
= *2.73 ù 103 A
139
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7.29. A long straight non-magnetic conductor of 0.2 mm radius carries a uniformly-distributed current
of 2 A dc.
a) Find J within the conductor: Assuming the current is +z directed,
2
J=
az = 1.59 ù 107 az A_m2
⇡(0.2 ù 10*3 )2
b) Use Ampere’s circuital law to find H and B within the conductor: Inside, at radius ⇢, we have
⇢J
2⇡⇢H = ⇡⇢2 J Ÿ H =
a = 7.96 ù 106 ⇢ a A_m
2
Then B = 0 H = (4⇡ ù 10*7 )(7.96 ù 106 )⇢a = 10⇢ a Wb_m2 .
c) Show that ( ù H = J within the conductor: Using the result of part b, we find,
0
1
1 d
1 d 1.59 ù 107 ⇢2
(ùH=
(⇢H ) az =
az = 1.59 ù 107 az A_m2 = J
⇢ d⇢
⇢ d⇢
2
d) Find H and B outside the conductor (note typo in book): Outside, the entire current is enclosed
by a closed path at radius ⇢, and so
I
1
H=
a =
a A_m
2⇡⇢
⇡⇢
Now B =
0H =
0 _(⇡⇢) a
Wb_m2 .
e) Show that ( ù H = J outside the conductor: Here we use H outside the conductor and write:
0
1
1 d
1 d
1
(ùH=
(⇢H ) az =
⇢
az = 0 (as expected)
⇢ d⇢
⇢ d⇢
⇡⇢
7.30. (an inversion of Problem 8.20). A solid nonmagnetic conductor of circular cross-section has a radius
of 2mm. The conductor is inhomogeneous, with = 106 (1 + 106 ⇢2 ) S/m. If the conductor is 1m in
length and has a voltage of 1mV between its ends, find:
a) H inside: With current along the cylinder length (along az , and with symmetry, H will be directed only. We find E = (V0 _d)az = 10*3 az V_m. Then J = E = 103 (1 + 106 ⇢2 )az A_m2 .
Next we apply Ampere’s circuital law to a circular path of radius ⇢, centered on the z axis and
normal to the axis:
Õ
H dL = 2⇡⇢H =
Thus
103
H =
⇢
J dS =
S
⇢
0
5
3
Finally, H = 500⇢(1 + 5 ù 10 ⇢ )a
2⇡
⇢
0
0
103 (1 + 106 (⇢® )2 )az az ⇢® d⇢® d
4
5
103 ⇢2 106 4
⇢ + 10 (⇢ ) d⇢ =
+
⇢
⇢
2
4
®
6
® 3
®
A_m (0 < ⇢ < 2mm).
b) the total magnetic flux inside the conductor: With field in the direction, a plane normal to B
will be that in the region 0 < ⇢ < 2 mm, 0 < z < 1 m. The flux will be
=
S
B dS =
2ù10*3
1
0
0
0
500⇢ + 2.5 ù 108 ⇢3 d⇢dz = 8⇡ ù 10*10 Wb = 2.5 nWb
140
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7.31. The cylindrical shell defined by 1 cm < ⇢ < 1.4 cm consists of a non-magnetic conducting material
and carries a total current of 50 A in the az direction. Find the total magnetic flux crossing the plane
= 0, 0 < z < 1:
a) 0 < ⇢ < 1.2 cm: We first need to find J, H, and B: The current density will be:
J=
50
⇡[(1.4 ù 10*2 )2 * (1.0 ù 10*2 )2 ]
az = 1.66 ù 105 az A_m2
Next we find H at radius ⇢ between 1.0 and 1.4 cm, by applying Ampere’s circuital law, and
noting that the current density is zero at radii less than 1 cm:
2⇡⇢H = Iencl =
2⇡
⇢
10*2
0
1.66 ù 105 ⇢® d⇢® d
Ÿ H = 8.30 ù 104
Then B =
(⇢2 * 10*4 )
A_m (10*2 m < ⇢ < 1.4 ù 10*2 m)
⇢
0 H, or
B = 0.104
(⇢2 * 10*4 )
a Wb_m2
⇢
Now,
5
10*4
B dS =
0.104 ⇢ *
d⇢ dz
a =
⇢
0
10*2
4
5
⇠ ⇡
(1.2 ù 10*2 )2 * 10*4
1.2
= 0.104
* 10*4 ln
= 3.92 ù 10*7 Wb = 0.392 Wb
2
1.0
1
1.2ù10*2
4
b) 1.0 cm < ⇢ < 1.4 cm (note typo in book): This is part a over again, except we change the upper
limit of the radial integration:
4
5
10*4
B dS =
0.104 ⇢ *
d⇢ dz
b =
⇢
0
10*2
4
⇠ ⇡5
(1.4 ù 10*2 )2 * 10*4
1.4
*4
= 0.104
* 10 ln
= 1.49 ù 10*6 Wb = 1.49 Wb
2
1.0
1
1.4ù10*2
c) 1.4 cm < ⇢ < 20 cm: This is entirely outside the current distribution, so we need B there: We
modify the Ampere’s circuital law result of part a to find:
Bout = 0.104
[(1.4 ù 10*2 )2 * 10*4 ]
10*5
a =
a Wb_m2
⇢
⇢
We now find
c =
1
0
20ù10*2
1.4ù10*2
⇠ ⇡
10*5
20
d⇢ dz = 10*5 ln
= 2.7 ù 10*5 Wb = 27 Wb
⇢
1.4
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7.32. The free space region defined by 1 < z < 4 cm and 2 < ⇢ < 3 cm is a toroid of rectangular crosssection. Let the surface at ⇢ = 3 cm carry a surface current K = 2az kA/m.
a) Specify the current densities on the surfaces at ⇢ = 2 cm, z = 1cm, and z = 4cm. All surfaces
must carry equal currents. With this requirement, we find: K(⇢ = 2) = *3 az kA_m. Next, the
current densities on the z = 1 and z = 4 surfaces must transistion between the current density
values at ⇢ = 2 and ⇢ = 3. Knowing the the radial current density will vary as 1_⇢, we find
K(z = 1) = (60_⇢)a⇢ A_m with ⇢ in meters. Similarly, K(z = 4) = *(60_⇢)a⇢ A_m.
b) Find H everywhere: Outside the toroid, H = 0. Inside, we apply Ampere’s circuital law in the
manner of Problem 8.14:
Õ
H dL = 2⇡⇢H =
2⇡
0
K(⇢ = 2) az (2 ù 10*2 ) d
2⇡(3000)(.02)
Ÿ H=*
a = *60_⇢ a A_m (inside)
⇢
c) Calculate the total flux within the toriod: We have B = *(60 0 _⇢)a Wb_m2 . Then
=
.04
.03
.01
.02
*60 0
a
⇢
(*a ) d⇢ dz = (.03)(60) 0 ln
⇠ ⇡
3
= 0.92 Wb
2
7.33. Use an expansion in rectangular coordinates to show that the curl of the gradient of any scalar field G
is identically equal to zero. We begin with
(G =
and
)G
)G
)G
a +
a +
a
)x x )y y )z z
4
0 15
⇠ ⇡
⌧ ⇠ ⇡
⇠ ⇡
) )G
) )G
) )G
) )G
( ù (G =
*
ax +
*
ay
)y )z
)z )y
)z )x
)x )z
4 0 1
⇠ ⇡5
) )G
) )G
+
*
az = 0 for any G
)x )y
)y )x
142
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7.34. A filamentary conductor on the z axis carries a current of 16A in the az direction, a conducting shell
at ⇢ = 6 carries a total current of 12A in the *az direction, and another shell at ⇢ = 10 carries a total
current of 4A in the *az direction.
a) Find H for 0 < ⇢ < 12: Ampere’s circuital law states that ó H dL = Iencl , where the line
integral and current direction are related in the usual way through the right hand rule. Therefore,
if I is in the positive z direction, H is in the a direction. We proceed as follows:
0 < ⇢ < 6 : 2⇡⇢H = 16 Ÿ H = 16_(2⇡⇢)a
6 < ⇢ < 10 : 2⇡⇢H = 16 * 12 Ÿ H = 4_(2⇡⇢)a
⇢ > 10 : 2⇡⇢H = 16 * 12 * 4 = 0 Ÿ H = 0
b) Plot H vs. ⇢:
c) Find the total flux
=
1
0
6
1
crossing the surface 1 < ⇢ < 7, 0 < z < 1: This will be
16 0
d⇢ dz +
2⇡⇢
1
0
7
6
⇧
4 0
2 ⌅
d⇢ dz = 0 4 ln 6 + ln(7_6) = 5.9 Wb
2⇡⇢
⇡
143
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7.35. A current sheet, K = 20 az A/m, is located at ⇢ = 2, and a second sheet, K = *10 az A/m is located
at ⇢ = 4.
a.) Let Vm = 0 at P (⇢ = 3, = 0, z = 5) and place a barrier at
= ⇡. Find Vm (⇢, , z) for
*⇡ <
< ⇡: Since the current is cylindrically-symmetric, we know that H = I_(2⇡⇢) a ,
where I is the current enclosed, equal in this case to 2⇡(2)K = 80⇡ A. Thus, using the result of
Section 8.6, we find
I
80⇡
Vm = *
=*
= *40 A
2⇡
2⇡
which is valid over the region 2 < ⇢ < 4, *⇡ < < ⇡, and *ÿ < z < ÿ. For ⇢ > 4, the outer
current contributes, leading to a total enclosed current of
Inet = 2⇡(2)(20) * 2⇡(4)(10) = 0
With zero enclosed current, H = 0, and the magnetic potential is zero as well.
b) Let A = 0 at P and find A(⇢, , z) for 2 < ⇢ < 4: Again, we know that H = H (⇢), since
the current is cylindrically symmetric. With the current only in the z direction, and again using
symmmetry, we expect only a z component of A which varies only with ⇢. We can then write:
(ùA=*
Thus
dAz
I
a =B= 0 a
d⇢
2⇡⇢
dAz
I
I
=* 0
Ÿ Az = * 0 ln(⇢) + C
d⇢
2⇡⇢
2⇡
We require that Az = 0 at ⇢ = 3. Therefore C = [( 0 I)_(2⇡)] ln(3), Then, with I = 80⇡, we
finally obtain
A=*
0 (80⇡)
2⇡
0 1
3
az Wb_m
[ln(⇢) * ln(3)] az = 40 0 ln
⇢
7.36. Let A = (3y * z)ax + 2xzay Wb/m in a certain region of free space.
a) Show that ( A = 0:
( A=
)
)
(3y * z) + 2xz = 0
)x
)y
b) At P (2, *1, 3), find A, B, H, and J: First AP = *6ax + 12ay . Then, using the curl formula in
cartesian coordinates,
B = ( ù A = *2xax * ay + (2z * 3)az Ÿ BP = *4ax * ay + 3az Wb_m2
Now
HP = (1_ 0 )BP = *3.2 ù 106 ax * 8.0 ù 105 ay + 2.4 ù 106 az A_m
Then J = ( ù H = (1_ 0 )( ù B = 0, as the curl formula in cartesian coordinates shows.
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7.37. Let N = 1000, I = 0.8 A, ⇢0 = 2 cm, and a = 0.8 cm for the toroid shown in Fig. 8.12b. Find Vm in
the interior of the toroid if Vm = 0 at ⇢ = 2.5 cm, = 0.3⇡. Keep within the range 0 < < 2⇡:
Well-within the toroid, we have
H=
NI
1 dVm
a = *(Vm = *
a
2⇡⇢
⇢ d
Thus
Vm = *
Then,
0=*
or C = 120. Finally
NI
+C
2⇡
1000(0.8)(0.3⇡)
+C
2⇡
⌧
400
Vm = 120 *
⇡
A (0 <
< 2⇡)
7.38. A square filamentary differential current loop, dL on a side, is centered at the origin in the z = 0 plane
in free space. The current I flows generally in the a direction.
a) Assuming that r >> dL, and following a method similar to that in Sec. 4.7, show that
dA =
0 I(dL)
2 sin ✓
4⇡r2
a
We begin with the expression for the differential vector potential, Eq. (48):
dA =
0 IdL
4⇡R
where in our case, we have four differential elements. The net vector potential at some distant
point will consist of the vector sum of the four individual potentials. Referring to the figures
below, the net potential at the indicated point is initially constructed as:
dA =
0 IdL
4⇡
4
0
ax
1
1
*
R2 R1
1
0
+ ay
1
1
*
R3 R4
15
The challenge is to determine the four distances, R1 through R4 , in terms of spherical coordinates,
r, ✓, and , thus referencing the four potentials to a common origin. This is where the treatment in
Sec. 4.7 is useful, although it is more complicated here because the problem is three-dimensional.
The diagram for the y-displaced elements is shown in the left-hand figure. The three distance
lines, r, R1 , and R2 , are approximately parallel because the observation point is in the far zone.
Therefore, beyond the blue line segment that crosses the three lines, the lengths are essentially
equal. The difference in lengths between r and R1 , for example, is the length of the line segment
along r, from the origin to the blue line. This length will be the projection of the distance vector
dL_2 ay along ar , or dL_2 ay ar . As a look ahead, this principle is discussed with illustrations
in Sec. 14.5, which handles antenna arrays of two elements.
145
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b) h = a_4,
7.38
c) h = a_2:
a) (continued). We now may write:
⌧
.
dL
R1 = r *
a a
2 y r
and
⌧
.
dL
R2 = r +
a a
2 y r
Referring to the right-hand figure and applying similar reasoning there leads to
⌧
.
dL
R3 = r *
a a
2 x r
and
⌧
.
dL
R4 = r +
a a
2 x r
where we know that ax ar = sin ✓ cos and ay ar = sin ✓ sin
relations into the original expression for dA:
We now substitute all these
⇡*1 ⇠
⇡*1 1
dL
dL
dA =
r+
sin ✓ sin
* r*
sin ✓ sin
ax
4⇡
2
2
0⇠
1
5
⇡*1 ⇠
⇡*1
dL
dL
+
r*
sin ✓ cos
* r+
sin ✓ cos
ay
2
2
40⇠
⇡*1 ⇠
⇡*1 1
dL
dL
0 IdL
1+
=
sin ✓ sin
* 1*
sin ✓ sin
ax
4⇡r
2r
2r
0⇠
⇡*1 ⇠
⇡*1 1 5
dL
dL
+
1*
sin ✓ cos
* 1+
sin ✓ cos
ay
2r
2r
0 IdL
40⇠
Now, since dL_r << 1, this simplifies to
⌧⇠⇠
⇡ ⇠
⇡⇡
dL
dL
1*
sin ✓ sin
* 1+
sin ✓ sin
ax
4⇡r
2r
2r
⇠⇠
⇡ ⇠
⇡⇡
dL
dL
+
1+
sin ✓ cos
* 1*
sin ✓ cos
ay
2r
2r
⇧
I(dL)2 sin ✓ ⌅
= 0
* sin ax + cos ay
2
4⇡r
2 sin ✓
I(dL)
= 0
a
4⇡r2
.
dA =
0 IdL
146
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7.38
b) Show that
I(dL)2
2 cos ✓ ar + sin ✓ a✓
4⇡r3
Using the part a expression, we construct dH = (1_ 0 )( ù dA, where in this case, we have a
-directed dA that varies with r and ✓. The curl expression in spherical coordinates reduces to:
1 )(dA sin ✓)
1 )(rdA)
( ù dA =
ar *
a✓
r sin ✓
)✓
r )r
or
0
1
0
1
1
) I(dL)2 sin2 ✓
1 ) I(dL)2 sin ✓
( ù dA =
ar *
a✓
r sin ✓ )✓
r )r
4⇡r
4⇡r2
dH =
=
I(dL)2
2 cos ✓ ar + sin ✓ a✓
4⇡r3
7.39. Planar current sheets of K = 30az A/m and *30az A/m are located in free space at x = 0.2 and
x = *0.2 respectively. For the region *0.2 < x < 0.2:
a) Find H: Since we have parallel current sheets carrying equal and opposite currents, we use Eq.
(12), H = K ù aN , where aN is the unit normal directed into the region between currents, and
where either one of the two currents are used. Choosing the sheet at x = 0.2, we find
H = 30az ù *ax = *30ay A_m
b) Obtain and expression for Vm if Vm = 0 at P (0.1, 0.2, 0.3): Use
dV
H = *30ay = *(Vm = * m ay
dy
So
dVm
= 30 Ÿ Vm = 30y + C1
dy
Then
0 = 30(0.2) + C1 Ÿ C1 = *6 Ÿ Vm = 30y * 6 A
c) Find B: B =
2
0 H = *30 0 ay Wb_m .
d) Obtain an expression for A if A = 0 at P : We expect A to be z-directed (with the current), and
so from ( ù A = B, where B is y-directed, we set up
dA
* z = *30 0 Ÿ Az = 30 0 x + C2
dx
Then 0 = 30 0 (0.1) + C2 Ÿ C2 = *3 0 . So finally A = 0 (30x * 3)az Wb_m.
7.40. Show that the line integral of the vector potential A about any closed path is equal to the magnetic flux
enclosed by the path, or ó A dL = î B dS.
We use the fact that B = ( ù A, and substitute this into the desired relation to find
A dL =
( ù A dS
Õ
This is just a statement of Stokes’ theorem (already proved), so we are done.
147
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7.41
a) Demonstrate the principle of Problem 7.40 using the vector potential in a coaxial cable, as given
in Eq. (66): The principle in question is
Õ
A dL =
s
B dS
which we now apply to the situation shown below.
outer
conductor
inner
conductor
2
x
I
I
z1 B 3
4
a
b radius
The above figure shows a cross section plane of a coaxial line (at constant ), where the z axis is
shown as the vertical dashed line. Conductors are shown at radii a and b, which carry equal and
opposite currents, I. Between conductors, the -directed B field is given as
B=
0I
2⇡⇢
a
In addition, A was derived in the development that lead to Eq. (66), and is:
0 1
b
0I
A=
ln
az
2⇡
⇢
in which the boundary condition of A = 0 at ⇢ = b was used. The closed path integral is taken
about the rectangle as shown, where path 3 is coincident with the outer conductor. The integral
will evaluate as
A dL = A1 z + A2 (b * ⇢) + A3 z * A4 (b * ⇢)
Õ
The only surviving term is A1 z because A3 is zero by definition; further, the horizontal components, A2 and A4 , are both orthogonal to the path directions, and thus would be zero (even if
they were not, their contributions would cancel anyway). So we are left with
0 1
b
0I
A dL = A1 z =
ln
z
Õ
2⇡
⇢
Then
B dS =
z
0
b
⇢
0I
d⇢® dz =
2⇡⇢®
0 1
b
ln
z
2⇡
⇢
0I
thus completing the demonstration.
148
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7.41
b) Use the principle to find the vector magnetic potential in a long solenoid of radius a, having n
turns per meter of length, in which each turn carries current I. Assume a long coil, so that B
is approximately 0 nI az Wb_m2 , uniform over the coil cross section: We can assume that A is
circular (in the direction of the current) and is a function of ⇢ only. So
Õ
or
A dL = 2⇡⇢A =
A=
As a check,
(ùA=
0 nI⇡⇢
2⇡
s
B dS =
a
2
0 nI⇡⇢
A
1 d(⇢A )
az =
⇢ d⇢
0 nI az
7.42. Show that (2 (1_R12 ) = *(1 (1_R12 ) = R21 _R312 . First
0
(2
1
R12
1
⌅
⇧*1_2
= (2 (x2 * x1 )2 + (y2 * y1 )2 + (z2 * z1 )2
4
5
*R12
R21
1 2(x2 * x1 )ax + 2(y2 * y1 )ay + 2(z2 * z1 )az
=*
=
=
2 [(x2 * x1 )2 + (y2 * y1 )2 + (z2 * z1 )2 ]3_2
R3
R3
12
12
Also note that (1 (1_R12 ) would give the same result, but of opposite sign.
149
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7.43. Compute the vector magnetic potential within the outer conductor for the coaxial line whose vector
magnetic potential is shown in Fig. 8.20 if the outer radius of the outer conductor is 7a. Select the
proper zero reference and sketch the results on the figure: We do this by first finding B within the outer
conductor and then “uncurling” the result to find A. With *z-directed current I in the outer conductor,
the current density is
I
I
Jout = *
az = *
az
2
2
⇡(7a) * ⇡(5a)
24⇡a2
Since current I flows in both conductors, but in opposite directions, Ampere’s circuital law inside the
outer conductor gives:
2⇡⇢H = I *
2⇡
0
⇢
I
⇢® d⇢® d
2
5a 24⇡a
4
5
49a2 * ⇢2
I
Ÿ H =
2⇡⇢
24a2
Now, with B = 0 H, we note that ( ù A will have a component only, and from the direction and
symmetry of the current, we expect A to be z-directed, and to vary only with ⇢. Therefore
(ùA=*
and so
dAz
a =
d⇢
0H
4
5
2
2
dAz
0 I 49a * ⇢
=*
d⇢
2⇡⇢
24a2
Then by direct integration,
Az =
* 0 I(49)
d⇢ +
48⇡⇢
I
0 I⇢
d⇢ + C = 0
2
96⇡
48⇡a
4 2
5
⇢
* 98 ln ⇢ + C
a2
As per Fig. 8.20, we establish a zero reference at ⇢ = 5a, enabling the evaluation of the integration
constant:
I
C = * 0 [25 * 98 ln(5a)]
96⇡
Finally,
40 2
1
0 15
⇢
5a
0I
Az =
* 25 + 98 ln
Wb_m
2
96⇡
⇢
a
A plot of this continues the plot of Fig. 8.20, in which the curve goes negative at ⇢ = 5a, and then
approaches a minimum of *.09 0 I_⇡ at ⇢ = 7a, at which point the slope becomes zero.
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7.44. By expanding Eq.(58), Sec. 8.7 in cartesian coordinates, show that (59) is correct. Eq. (58) can be
rewritten as
(2 A = ((( A) * ( ù ( ù A
We begin with
( A=
)Ax )Ay )Az
+
+
)x
)y
)z
Then the x component of ((( A) is
[((( A)]x =
Now
0
(ùA=
)Az )Ay
*
)y
)z
1
) 2 Ay ) 2 Az
+
+
)x)y )x)z
)x2
) 2 Ax
0
ax +
)Ax )Az
*
)z
)x
1
0
ay +
)Ay
)Ax
*
)x
)y
1
az
and the x component of ( ù ( ù A is
[( ù ( ù A]x =
) 2 Ay
)x)y
*
) 2 Ax
)y2
*
) 2 Ax
)z2
+
) 2 Az
)z)y
Then, using the underlined results
[((( A) * ( ù ( ù A]x =
) 2 Ax
)x
+
2
) 2 Ax
)y2
+
) 2 Ax
)z2
= (2 Ax
Similar results will be found for the other two components, leading to
((( A) * ( ù ( ù A = (2 Ax ax + (2 Ay ay + (2 Az az í (2 A
QED
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CHAPTER 8 – 9th Edition
8.1. A point charge, Q = *0.3 C and m = 3 ù 10*16 kg, is moving through the field E = 30 az V/m. Use
Eq. (1) and Newton’s laws to develop the appropriate differential equations and solve them, subject to
the initial conditions at t = 0: v = 3 ù 105 ax m/s at the origin. At t = 3 s, find:
a) the position P (x, y, z) of the charge: The force on the charge is given by F = qE, and Newton’s
second law becomes:
F = ma = m
d 2z
= qE = (*0.3 ù 10*6 )(30 az )
dt2
describing motion of the charge in the z direction. The initial velocity in x is constant, and so no
force is applied in that direction. We integrate once:
qE
dz
= vz =
t + C1
dt
m
The initial velocity along z, vz (0) is zero, and so C1 = 0. Integrating a second time yields the
z coordinate:
qE 2
z=
t + C2
2m
The charge lies at the origin at t = 0, and so C2 = 0. Introducing the given values, we find
z=
(*0.3 ù 10*6 )(30) 2
t = *1.5 ù 1010 t2 m
*16
2 ù 3 ù 10
At t = 3 s, z = *(1.5 ù 1010 )(3 ù 10*6 )2 = *.135 cm. Now, considering the initial constant
velocity in x, the charge in 3 s attains an x coordinate of x = vt = (3 ù 105 )(3 ù 10*6 ) = .90 m.
In summary, at t = 3 s we have P (x, y, z) = (.90, 0, *.135).
b) the velocity, v: After the first integration in part a, we find
vz =
qE
t = *(3 ù 1010 )(3 ù 10*6 ) = *9 ù 104 m_s
m
Including the intial x-directed velocity, we finally obtain v = 3 ù 105 ax * 9 ù 104 az m_s.
c) the kinetic energy of the charge: Have
1
1
K.E. = mv2 = (3 ù 10*16 )(1.13 ù 105 )2 = 1.5 ù 10*5 J
2
2
152
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8.2. Compare the magnitudes of the electric and magnetic forces on an electron that has attained a velocity
of 107 m/s. Assume an electric field intensity of 105 V/m, and a magnetic flux density associated
with that of the Earth’s magnetic field in temperate latitudes, 0.5 gauss. We use the Lorentz Law,
F = Fe + Fm = q(E + v ù B), where B = 0.5 G = 5.0 ù 10*5 T. We find
Fe  = (1.6 ù 10*19 C)(105 V_m) = 1.6 ù 10*14 N
Fm  = (1.6 ù 10*19 C)(107 m_s)(5.0 ù 10*5 T) = 8.0 ù 10*17 N = 0.005Fe 
8.3. A point charge for which Q = 2 ù 10*16 C and m = 5 ù 10*26 kg is moving in the combined fields
E = 100ax * 200ay + 300az V/m and B = *3ax + 2ay * az mT. If the charge velocity at t = 0 is
v(0) = (2ax * 3ay * 4az ) ù 105 m/s:
a) give the unit vector showing the direction in which the charge is accelerating at t = 0: Use
F(t = 0) = q[E + (v(0) ù B)], where
v(0) ù B = (2ax * 3ay * 4az )105 ù (*3ax + 2ay * az )10*3 = 1100ax + 1400ay * 500az
So the force in newtons becomes
F(0) = (2ù10*16 )[(100+1100)ax +(1400*200)ay +(300* 500)az ] = 4ù10*14 [6ax +6ay *az ]
The unit vector that gives the acceleration direction is found from the force to be
aF =
6ax + 6ay * az
= .70ax + .70ay * .12az
˘
73
b) find the kinetic energy of the charge at t = 0:
1
1
K.E. = mv(0)2 = (5 ù 10*26 kg)(5.39 ù 105 m_s)2 = 7.25 ù 10*15 J = 7.25 f J
2
2
8.4. Show that a charged particle in a uniform magnetic field describes a circular orbit with an orbital period
that is independent of the radius. Find the relationship between the angular velocity and magnetic flux
density for an electron (the cyclotron frequency).
A circular orbit can be established if the magnetic force on the particle is balanced by the centripital force associated with the circular path. We assume a circular path of radius R, in which
B = B0 az is normal to the plane of the path. Then, with particle angular velocity ⌦, the velocity
is v = R⌦ a . The magnetic force is then Fm = qv ù B = qR⌦ a ù B0 az = qR⌦B0 a⇢ . This
force will be negative (pulling the particle toward the center of the path) if the charge is positive
and motion is in the *a direction, or if the charge is negative, and motion is in positive a . In
either case, the centripital force must counteract the magnetic force. Assuming particle mass m,
the force balance equation is qR⌦B0 = m⌦2 R, from which ⌦ = qB0 _m. The revolution period
is T = 2⇡_⌦ = 2⇡m_(qB0 ), which is independent of R. For an electron, we have q = 1.6 ù 10*9
C, and m = 9.1 ù 1031 kg. The cyclotron frequency is therefore
⌦c =
q
B = 1.76 ù 1011 B0 s*1
m 0
153
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8.5. A cylindrical shell of radius a is oriented along the z axis. The cylinder carries surface current of
density K = K0 az A/m, where K0 is a constant. Magnetic flux density B = *B0 a⇢ is applied at the
surface over length d. Find the torque on the cylinder: The torque will be the sum of all differential
torques acting at all points that comprise the surface. We have dT = R ù dF, where R = aa⇢ , and
dF = Kda ù B = K0 az ad dz ù *B0 a⇢ = *K0 B0 ad dza
So
dT = aa⇢ ù *K0 B0 ad dza = *a2 K0 B0 d dzaz
Summing up all contributions:
T=
dT =
2⇡
d
0
0
*a2 K0 B0 d dz = *2⇡a2 dK0 B0 az N * m
8.6. Show that the differential work in moving a current element IdL through a distance dl in a magnetic
field B is the negative of that done in moving the element Idl through a distance dL in the same field:
The two differential work quantities are written as:
dW = (IdL ù B) dl
and
dW ® = (Idl ù B) dL
We now apply the vector identity, Eq.(A.6), Appendix A: (A ù B) C = (B ù C) A, and write:
(IdL ù B) dl = (B ù dl) IdL = *(Idl ù B) dL
QED
8.7. A conducting strip of infinite length lies in the xy plane with its length oriented along the x axis, and
where *b_2 < y < b_2 defines its width along y. Current I1 flows down the strip in the positive x
direction and is uniformly distributed over the width. Above the strip and parallel to it at z = d is
an infinitely long current filament that carries current I2 in the positive x direction. Find the force of
attraction between the two currents per unit length in x. Assume d << b: Under the latter assumption,
the B field generated by the strip at the location of the filament will be B(z = d) = * 0 K_2 ay (from
Eq. (11), Chapter 7), where K = I1 _b ax . Then the differential force on the filament will be
dF = I2 dL ù B = I2 dx ax ù
* 0 I1
* 0 I1 I2
ay =
az dx
2b
2b
The force per unit length is then
F=
1
0
I I
* 0 I1 I2
* 0 1 2 az dx =
az N_m
2b
2b
where again, d << b.
154
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8.8. Two conducting strips, having infinite length in the z direction, lie in the xz plane. One occupies the
region d_2 < x < b + d_2 and carries surface current density K = K0 az ; the other is situated at
*(b + d_2) < x < *d_2 and carries surface current density *K0 az .
a) Find the force per unit length in z that tends to separate the two strips:
We begin by evaluating the magnetic field arising from the left-hand strip (in the region x < 0)
at any location on the x axis. Because the source strip is infinite in z, this field will not depend
on z and will be valid at any location in the x-z plane. We use the Biot-Savart law and find the
field at a fixed point x0 on the x axis. The Biot-Savart law reads:
K ù aR
da
2
s 4⇡R
H(x0 ) =
where the integral is taken over the left strip area, and where
˘ R is the distance from point (x, z)
on the strip to the fixed observation point, x0 . Thus R = (x * x0 )2 + z2 , and
*(x * x0 ) ax * z az
aR = ˘
(x * x0 )2 + z2
so that
H(x0 ) =
ÿ
*ÿ
⌅
⇧
*K0 az ù *(x * x0 ) ax * z az
dx dz
⌅
⇧3_2
*( d2 +b)
4⇡ (x * x0 )2 + z2
* d2
Taking the cross product leaves only a y component:
H(x0 ) =
ÿ
*ÿ
K0 ay (x * x0 )
* d2
*( d2 +b) 4⇡
⌅
⇧3_2 dx dz
(x * x0 )2 + z2
It is easiest to evaluate the z integral first, leading to
K
H0 = 0 ay
4⇡
* d2
K0
Ûÿ
a
Û dx =
˘
2⇡ y
*( d +b) (x * x ) (x * x )2 + z2 Û*ÿ
z
0
2
0
Evaluate the x integral to find:
d
K
K
Û*
H0 = 0 ay ln(x * x0 )Û 2d
= * 0 ln
Û
*(
+b)
2⇡
2⇡
2
Ld
2
+ b + x0
d
+ x0
2
* d2
*( d2 +b)
dx
(x * x0 )
M
ay A_m
Now the force acting on the right-hand strip per unit length is
F=
s
K ù B da
where K is the surface current density in the right-hand strip, and B is the magnetic flux density
( 0 H) arising from the left strip, evaluated within the right strip area (over which the integral is
taken). Over a unit lengh in z, the force integral is written:
Ld
M
1
(b+ d2 )
+ b + x0
* 0 K0
2
F=
k0 az ù
ln
ay dx0 dz
d
d
2⇡
0
+ x0
2
2
155
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8.8
a) (continued)
The z integration yields a factor of 1, the cross product gives an x-directed force, and we can
rewrite the expression as:
F=
=
2
0 K0
2⇡
2
0 K0
2⇡
(b+ d2 ) ⌧
ax
ax
d
2
⌧⇠
⇠
ln
⇡
⇠
⇡
d
d
+ b + x0 * ln
+ x0 dx0
2
2
⇡ ⇠
⇡
⇠
⇡ ⇠
⇡
(b+ d2 )
d
d
d
d
+ b + x0 ln
+ b + x0 * x0 *
+ x0 ln
+ x0 + x0 d
2
2
2
2
2
Evaluating this result over the integration limits and then simplifying results in the following
expression, which is one of many ways of writing the result:
F=
2
0 dK0
2⇡
L
ax
⇠
⇡
2b
1+
ln
d
H
1 + 2b
d
1 + db
I
⇠
b
* ln 1 +
d
M
⇡
b) let b approach zero while maintaining constant current, I = K0 b, and show that the force per
unit length approaches 0 I 2 _(2⇡d) N/m.
As b gets small, so does the ratio b_d. We may then write:
4
⇠ ⇡2 5
1 .
b
b
= 1* +
b
d
d
1+
d
The force expression now becomes:
F=
2
0 dK0
2⇡
ax
4⇠
5
⇡ 4⇠
⇡0
⇠ ⇡2 15
⌧
2b
2b
b
b
b
1+
ln 1 +
1* +
* ln 1 +
d
d
d
d
d
The product in the natural log function is expanded:
F=
2
0 dK0
2⇡
4⇠
5
5
⇡ 4
⌧
2b
b b2 2b 2b2 2b2
b
ax 1 +
ln 1 * + 2 +
* 2 + 2 * ln 1 +
d
d d
d
d
d
d
All terms in the natural log functions involve 1 + f (b_d) where f (b_d) << 1. Therefore, we
.
may expand the log functions in power series, keeping only the first terms: i.e., ln(1 + f ) = f , if
f << 1. With this simplification, the force becomes:
.
F=
2
0 dK0
2⇡
ax
4⇠
2b
1+
d
⇡0
b b2 2b3
*
+ 3
d d2
d
1
b
*
d
5
Expanding and simplifying, the final result is
.
F=
0 (bK0 )
2⇡d
2
ax N_m
⇠
b
<< 1
d
⇡
where the term, 4b4 _d 4 , has been neglected.
156
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8.9. A current of *100az A/m flows on the conducting cylinder ⇢ = 5 mm and +500az A/m is present
on the conducting cylinder ⇢ = 1 mm. Find the magnitude of the total force acting to split the outer
cylinder apart along its length: The differential force acting on the outer cylinder arising from the field
of the inner cylinder is dF = Kouter ù B, where B is the field from the inner cylinder, evaluated at the
outer cylinder location:
2⇡(1)(500) 0
B=
a = 100 0 a T
2⇡(5)
Thus dF = *100az ù 100 0 a = 104 0 a⇢ N_m2 . We wish to find the force acting to split the outer
cylinder, which means we need to evaluate the net force in one cartesian direction on one half of the
cylinder. We choose the “upper” half (0 < < ⇡), and integrate the y component of dF over this
range, and over a unit length in the z direction:
Fy =
1
0
⇡
0
⇡
104 0 a⇢ ay (5 ù 10*3 ) d dz =
0
50 0 sin d = 100 0 = 4⇡ ù 10*5 N_m
Note that we did not include the “self force” arising from the outer cylinder’s B field on itself. Since
the outer cylinder is a two-dimensional current sheet, its field exists only just outside the cylinder,
and so no force exists. If this cylinder possessed a finite thickness, then we would need to include
its self-force, since there would be an interior field and a volume current density that would spatially
overlap.
8.10. A planar transmission line consists of two conducting planes of width b separated d m in air, carrying
equal and opposite currents of I A. If b >> d, find the force of repulsion per meter of length between
the two conductors.
Take the current in the top plate in the positive z direction, and so the bottom plate current is
directed along negative z. Furthermore, the bottom plate is at y = 0, and the top plate is at
y = d. The magnetic field stength at the bottom plate arising from the current in the top plate is
H = K_2 ax A/m, where the top plate surface current density is K = I_b az A/m. Now the force
per unit length on the bottom plate is
F=
1
0
b
0
Kb ù Bb dS
where Kb is the surface current density on the bottom plate, and Bb is the magnetic flux density
arising from the top plate current, evaluated at the bottom plate location. We obtain
F=
1
0
b
0
I
I2
I
* az ù 0 ax dS = * 0 ay N_m
b
2b
2b
157
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8.11.
a) Use Eq. (14), Sec. 8.3, to show that the force of attraction per unit length between two filamentary
conductors in free space with currents I1 az at x = 0, y = d_2, and I2 az at x = 0, y = *d_2, is
0 I1 I2 _(2⇡d): The force on I2 is given by
L
M
I1 I2
aR12 ù dL1
F2 = 0
ù dL2
4⇡ Õ Õ
R2
12
Let z1 indicate
the z coordinate along I1 , and z2 indicate the z coordinate along I2 . We then have
˘
R12 = (z2 * z1 )2 + d 2 and
(z2 * z1 )az * day
aR12 = ˘
(z2 * z1 )2 + d 2
Also, dL1 = dz1 az and dL2 = dz2 az The “inside” integral becomes:
ÿ
[(z2 * z1 )az * day ] ù dz1 az
*d dz1 ax
aR12 ù dL1
=
=
2
2
2
1.5
2
2 1.5
Õ
Õ
[(z2 * z1 ) + d ]
R
*ÿ [(z2 * z1 ) + d ]
12
The force expression now becomes
4 ÿ
5
d dz1 dz2 ay
*d dz1 ax
I1 I2
I1 I2 1 ÿ
F2 = 0
ù
dz
a
=
2
z
0
2
2
1.5
4⇡ Õ
4⇡ 0 *ÿ [(z2 * z1 )2 + d 2 ]1.5
*ÿ [(z2 * z1 ) + d ]
Note that the “outside” integral is taken over a unit length of current I2 . Evaluating, obtain,
F2 =
0
I1 I2 d ay
4⇡d 2
(2)
as expected.
1
0
dz2 =
0 I1 I2
2⇡d
ay N_m
b) Show how a simpler method can be used to check your result: We use dF2 = I2 dL2 ù B12 , where
the field from current 1 at the location of current 2 is
I
B12 = 0 1 ax T
2⇡d
so over a unit length of I2 , we obtain
I
I I
F2 = I2 az ù 0 1 ax = 0 1 2 ay N_m
2⇡d
2⇡d
This second method is really just the first over again, since we recognize the inside integral of the
first method as the Biot-Savart law, used to find the field from current 1 at the current 2 location.
8.12. Two circular wire rings are parallel to each other, share the same axis, are of radius a, and are separated
by distance d, where d << a. Each ring carries current I. Find the approximate force of attraction
and indicate the relative orientations of the currents.
With the loops very close to each other, the primary force on each “segment” of current in either
loop can be assumed to arise from the local B field from the immediately adjacent segment in
the other loop. So the problem becomes essentially that of straightening out both loops and
considering them as two parallel wires of length L = 2⇡a. The magnetic induction at points very
.
close to either current loop is approximately that of an infinite straight wire, or B = 0 I_(2⇡d).
The force is therefore found using Eq. (11) to be
F = ILB = I(2⇡a)
0I
2⇡d
=
0 aI
2
d
If the force is attractive, then the currents must be in the same direction.
158
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8.13. An infinitely long current filament is oriented along the x axis and carries current I1 in the positive x
direction. A second
˘ current filament is positioned at z = z0 and carries current I2 along its orientation
in the (ax + ay )_ 2 direction.
a) Find the differential magnetic force at any point on I2 arising from I1 : The magnetic flux density
arising from I1 and at any location on I2 will be:
B=
yaz * z0 ay
0 I1
2⇡(y2 + z20 )1_2 (y2 + z20 )1_2
=
0 I1 (xaz * z0 ay )
2⇡(x2 + z20 )
Wb_m2
where the substitution y = x was used, as this is true along the second current. Now dF =
I2 dL ù B, where dL = dxax + dyay , so that
dF = I2 (dxax + dyay ) ù
0 I1 (xaz * z0 ay )
2⇡(x2 + z20 )
=
⌅
⇧
0 I1 I2
x(a
*
a
)
*
z
a
dx N
x
y
0
z
2⇡(x2 + z20 )
where dx = dy has been used — again true along the second current. Note that the term involving
(ax * ay ) is the torque term about the z axis.
b) Find the net z directed force acting on I2 : This will be
Fz =
ÿ
*ÿ
* 0 I1 I2 z0 dx
* 0 I1 I2 z0 1
=
tan*1
2
2
2⇡
z
2⇡(x + z0 )
0
0
x
z0
1
I I
Ûÿ
=* 0 1 2 N
Û
Û*ÿ
2
Note the independence on z0 . This is because the y component of B at I2 increases linearly with
z0 , while the magnitude of B decreases as 1_z0 . The two dependences exactly cancel for infinite
lines.
159
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8.14. A solenoid is 25cm long, 3cm in diameter, and carries 4 A dc in its 400 turns. Its axis is perpendicular
to a uniform magnetic field of 0.8 Wb_m2 in air. Using an origin at the center of the solenoid, calculate
the torque acting on it.
First, we consider the torque, referenced to the origin, of a single wire loop of radius a in the
plane z = z0 . This is one loop of the solenoid. We will take the applied magnetic flux density as
in the ax direction and write it as B = B0 ax . The differential torque associated with a differential
current element on this loop, referenced to the origin, will be:
dT = R ù dF
where R is the vector directed from the origin to the current element, and is given by R =
z0 az + aa⇢ , and where
dF = IdL ù B = Ia d a ù B0 ax = *IaB0 cos d az
So now
dT = z0 az + aa⇢ ù *IaB0 cos d az = a2 IB0 cos d a
Using a = cos ay * sin ax , the differential torque becomes
dT = a2 IB0 cos
cos ay * sin ax d
Note that there is no dependence on z0 . The net torque on the loop is now
T=
dT =
2⇡
0
a2 IB0 cos2
ay * cos sin ax d = ⇡a2 IB0 ay
But we have 400 identical turns at different z locations (which don’t matter), so the total torque
will be just the above result times 400, or:
T = 400⇡a2 IB0 ay = 400⇡(1.5 ù 10*2 )2 (4)(0.8)ay = 0.91ay N m
8.15. A solid conducting filament extends from x = *b to x = b along the line y = 2, z = 0. This filament
carries a current of 3 A in the ax direction. An infinite filament on the z axis carries 5 A in the az
direction. Obtain an expression for the torque exerted on the finite conductor about an origin located
at (0, 2, 0): The differential force on the wire segment arising from the field from the infinite wire is
dF = 3 dx ax ù
5 0
15 0 cos dx
15 0 x dx
a =*
az = *
az
˘
2⇡⇢
2⇡(x2 + 4)
2⇡ x2 + 4
So now the differential torque about the (0, 2, 0) origin is
dT = RT ù dF = x ax ù *
15 0 x dx
2⇡(x2 + 4)
az =
15 0 x2 dx
2⇡(x2 + 4)
ay
The torque is then
⇠ ⇡ b
15 0 ⌧
*1 x
a
x
*
2
tan
y
2
2⇡
2 *b
*b 2⇡(x + 4)
⌧
⇠ ⇡
b
= (6 ù 10*6 ) b * 2 tan*1
ay N m
2
T=
b
15 0 x2 dx
ay =
160
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8.16. Assume that an electron is describing a circular orbit of radius a about a positively-charged nucleus.
a) By selecting an appropriate current and area, show that the equivalent orbital dipole moment is
ea2 !_2, where ! is the electron’s angular velocity: The current magnitude will be I = Te , where
e is the electron charge and T is the orbital period. The latter is T = 2⇡_!, and so I = e!_(2⇡).
Now the dipole moment magnitude will be m = IA, where A is the loop area. Thus
m=
e! 2 1 2
⇡a = ea ! __
2⇡
2
b) Show that the torque produced by a magnetic field parallel to the plane of the orbit is ea2 !B_2:
With B assumed constant over the loop area, we would have T = m ù B. With B parallel to the
loop plane, m and B are orthogonal, and so T = mB. So, using part a, T = ea2 !B_2.
c) by equating the Coulomb and centrifugal forces, show that ! is (4⇡✏0 me a3 _e2 )*1_2 , where me is
the electron mass: The force balance is written as
e2
= me !2 a Ÿ ! =
4⇡✏0 a2
0
4⇡✏0 me a3
1*1_2
e2
__
d) Find values for the angular velocity, torque, and the orbital magnetic moment for a hydrogen
atom, where a is about 6 ù 10*11 m; let B = 0.5 T: First
4
(1.60 ù 10*19 )2
!=
4⇡(8.85 ù 10*12 )(9.1 ù 10*31 )(6 ù 10*11 )3
Finally,
51_2
= 3.42 ù 1016 rad_s
1
T = (3.42 ù 1016 )(1.60 ù 10*19 )(0.5)(6 ù 10*11 )2 = 4.93 ù 10*24 N m
2
m=
T
= 9.86 ù 10*24 A m2
B
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8.17. The hydrogen atom described in Problem 16 is now subjected to a magnetic field having the same
direction as that of the atom. Show that the forces caused by B result in a decrease of the angular
velocity by eB_(2me ) and a decrease in the orbital moment by e2 a2 B_(4me ). What are these decreases
for the hydrogen atom in parts per million for an external magnetic flux density of 0.5 T? We first write
down all forces on the electron, in which we equate its coulomb force toward the nucleus to the sum
of the centrifugal force and the force associated with the applied B field. With the field applied in the
same direction as that of the atom, this would yield a Lorentz force that is radially outward – in the
same direction as the centrifugal force.
Fe = Fcent + FB Ÿ
e2
= me !2 a + e!aB
2
´Ø¨
4⇡✏0 a
QvB
With B = 0, we solve for ! to find:
v
! = !0 =
e2
4⇡✏0 me a3
Then with B present, we find
!2 =
Therefore
e2
e!B
e!B
*
= !20 *
3
me
me
4⇡✏0 me a
v
! = !0
.
But ! = !0 , and so
e!B .
1* 2
= !0
!0 me
0
.
! = !0 1 *
eB
2!0 me
H
e!B
1*
2!20 me
1
= !0 *
I
eB
__
2me
As for the magnetic moment, we have
0
1
e! 2 1
eB
1
1 e2 a2 B
2 . 1 2
m = IS =
⇡a = !ea = ea !0 *
= !0 ea2 *
__
2⇡
2
2
2me
2
4 me
Finally, for a = 6 ù 10*11 m, B = 0.5 T, we have
!
eB 1 . eB 1
1.60 ù 10*19 ù 0.5
=
=
=
= 1.3 ù 10*6
!
2me ! 2me !0
2 ù 9.1 ù 10*31 ù 3.4 ù 1016
where !0 = 3.4 ù 1016 sec*1 is found from Problem 16. Finally,
m e2 a2 B
2 . eB
=
ù
=
= 1.3 ù 10*6
m
4me
2me !0
!ea2
162
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8.18. Calculate the vector torque on the square loop shown in Fig. 8.16 about an origin at A in the field B,
given:
a) A(0, 0, 0) and B = 100ay mT: The field is uniform and so does not produce any translation of the
loop. Therefore, we may use T = IS ù B about any origin, where I = 0.6 A and S = 16az m2 .
We find T = 0.6(16)az ù 0.100ay = *0.96 ax N*m.
b) A(0, 0, 0) and B = 200ax + 100ay mT: Using the same reasoning as in part a, we find
T = 0.6(16)az ù (0.200ax + 0.100ay ) = *0.96ax + 1.92ay N*m
c) A(1, 2, 3) and B = 200ax + 100ay * 300az mT: We observe two things here: 1) The field is again
uniform and so again the torque is independent of the origin chosen, and 2) The field differs from
that of part b only by the addition of a z component. With S in the z direction, this new component
of B will produce no torque, so the answer is the same as part b, or T = *0.96ax + 1.92ay N*m.
d) A(1, 2, 3) and B = 200ax + 100ay * 300az mT for x g 2 and B = 0 elsewhere: Now, force is
acting only on the y-directed segment at x = +2, so we need to be careful, since translation will
occur. So we must use the given origin. The differential torque acting on the differential wire
segment at location (2,y) is dT = R(y) ù dF, where
dF = IdL ù B = 0.6 dy ay ù [0.2ax + 0.1ay * 0.3az ] = [*0.18ax * 0.12az ] dy
and R(y) = (2, y, 0) * (1, 2, 3) = ax + (y * 2)ay * 3az . We thus find
⌅
⇧
dT = R(y) ù dF = ax + (y * 2)ay * 3az ù [*0.18ax * 0.12az ] dy
⌅
⇧
= *0.12(y * 2)ax + 0.66ay + 0.18(y * 2)az dy
The net torque is now
T=
2⌅
*2
⇧
*0.12(y * 2)ax + 0.66ay + 0.18(y * 2)az dy = 0.96ax + 2.64ay * 1.44az N*m
8.19. Given a material for which
a) H: We use B =
0 (1 +
m = 3.1 and within which B = 0.4yaz T, find:
m )H, or
H=
(1 + 3.1) 0
b)
= (1 + 3.1) 0 = 5.15 ù 10*6 H_m.
c)
r = (1 + 3.1) = 4.1.
d) M =
0.4yay
= 77.6yaz kA_m
m H = (3.1)(77.6yay ) = 241yaz kA_m
e) J = ( ù H = (dHz )_(dy) ax = 77.6 ax kA_m2 .
f) Jb = ( ù M = (dMz )_(dy) ax = 241 ax kA_m2 .
g) JT = ( ù B_ 0 = 318ax kA_m2 .
163
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8.20. Find H in a material where:
a)
29
3
*30 a
y
r = 4.2, there are 2.7 ù 10 atoms/m , and each atom has a dipole moment of 2.6 ù 10
2
29
*30
A m . Since all dipoles are identical, we may write M = Nm = (2.7 ù 10 )(2.6 ù 10 ay ) =
0.70ay A/m. Then
H=
0.70 ay
M
=
= 0.22 ay A_m
4.2 * 1
r*1
b) M = 270 az A/m and = 2 H_m: Have
H = 270az _(1.59 * 1) = 456 az A_m.
c)
r =
_ 0 = (2 ù 10*6 )_(4⇡ ù 10*7 ) = 1.59. Then
m = 0.7 and B = 2az T: Use
H=
B
(1
+
0
m)
=
2az
= 936 az kA_m
(4⇡ ù 10*7 )(1.7)
d) Find M in a material where bound surface current densities of 12 az A/m and *9 az A/m exist at
⇢ = 0.3 m and ⇢ = 0.4 m, respectively: We use ó M dL = Ib , where, since currents are in the
z direction and are symmetric about the z axis, we chose the path integrals to be circular loops
centered on and normal to z. From the symmetry, M will be -directed and will vary only with
radius. Note first that for ⇢ < 0.3 m, no bound current will be enclosed by a path integral, so we
conclude that M = 0 for ⇢ < 0.3m. At radii between the currents the path integral will enclose
only the inner current so,
Õ
M dL = 2⇡⇢M = 2⇡(0.3)12 Ÿ M =
3.6
a A_m (0.3 < ⇢ < 0.4m)
⇢
Finally, for ⇢ > 0.4 m, the total enclosed bound current is Ib,tot = 2⇡(0.3)(12) * 2⇡(0.4)(9) = 0,
so therefore M = 0 (⇢ > 0.4m).
8.21. Find the magnitude of the magnetization in a material for which:
a) the magnetic flux density is 0.02 Wb_m2 and the magnetic susceptibility is 0.003 (note that this
latter quantity is missing in the original problem statement): From B = 0 (H + M) and from
M = m H, we write
0
1*1
B
1
B
0.02
M=
+1
=
=
= 47.7 A_m
*7 )(334)
(334)
(4⇡
ù
10
0
m
0
b) the magnetic field intensity is 1200 A/m and the relative permeability is 1.005: From B =
(H + M) = 0 r H, we write
0
M = ( r * 1)H = (.005)(1200) = 6.0 A_m
c) there are 7.2 ù 1028 atoms per cubic meter, each having a dipole moment of 4 ù 10*30 A m2
in the same direction, and the magnetic susceptibility is 0.0003: With all dipoles identical the
dipole moment density becomes
M = n m = (7.2 ù 1028 )(4 ù 10*30 ) = 0.288 A_m
164
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8.22. Under some conditions, it is possible to approximate the effects of ferromagnetic materials by assuming
linearity in the relationship of B and H. Let r = 1000 for a certain material of which a cylindrical
wire of radius 1mm is made. If I = 1 A and the current distribution is uniform, find
a) B: We apply Ampere’s circuital law to a circular path of radius ⇢ around the wire axis, and where
⇢ < a:
1000 0 I⇢
⇡⇢2
I⇢
(103 )4⇡ ù 10*7 (1)⇢
I
Ÿ
H
=
Ÿ
B
=
a
=
a
⇡a2
2⇡a2
2⇡a2
2⇡ ù 10*6
= 200⇢ a Wb_m2
2⇡⇢H =
b) H: Using part a, H = B_ r 0 = ⇢_(2⇡) ù 106 a A_m.
c) M:
M = B_ 0 * H =
(2000 * 2)⇢
ù 106 a = 1.59 ù 108 ⇢ a A_m
4⇡
d) J:
J=(ùH=
1 d(⇢H )
az = 3.18 ù 105 az A_m
⇢ d⇢
Jb = ( ù M =
1 d(⇢M )
az = 3.18 ù 108 az A_m2
⇢ d⇢
e) Jb within the wire:
8.23. Calculate values for H , B , and M at ⇢ = c for a coaxial cable with a = 2.5 mm and b = 6 mm
if it carries current I = 12 A in the center conductor, and = 3 H_m for 2.5 < ⇢ < 3.5 mm,
= 5 H_m for 3.5 < ⇢ < 4.5 mm, and = 10 H_m for 4.5 < ⇢ < 6 mm. Compute for:
a) c = 3 mm: Have
H =
12
I
=
= 637 A_m
2⇡⇢ 2⇡(3 ù 10*3 )
Then B = H = (3 ù 10*6 )(637) = 1.91 ù 10*3 Wb_m2 .
Finally, M = (1_ 0 )B * H = 884 A_m.
b) c = 4 mm: Have
H =
I
12
=
= 478 A_m
2⇡⇢ 2⇡(4 ù 10*3 )
Then B = H = (5 ù 10*6 )(478) = 2.39 ù 10*3 Wb_m2 .
Finally, M = (1_ 0 )B * H = 1.42 ù 103 A_m.
c) c = 5 mm: Have
H =
I
12
=
= 382 A_m
2⇡⇢ 2⇡(5 ù 10*3 )
Then B = H = (10 ù 10*6 )(382) = 3.82 ù 10*3 Wb_m2 .
Finally, M = (1_ 0 )B * H = 2.66 ù 103 A_m.
165
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8.24. Two current sheets, K0 ay A/m at z = 0 and *K0 ay A/m at z = d, are separated by an inhomogeneous
material for which r = az + 1, where a is a constant.
a) Find expressions for H and B in the material: The z variation in the permeability leaves the H
field unaffected, and so we may find this using Ampere’s circuital law. This is done in Chapter
7, culminating in Eq. (12) there. Applying this to the conductor in the z = 0 plane, we find
H = K ù an = K0 ay ù az = K0 ax A_m
b) find the total flux that crosses a 1m2 area on the yz plane: Because the permeability varies with z,
the flux will depend on the location and dimensions of the 1m2 area. Choose a rectangle located
in the range 0 < y < y1 , and z1 < z < z2 , where we require that (z2 * z1 )y1 = 1. Therefore,
y1 = 1_(z2 * z1 ). The flux through this area is now
m =
=
s
H dS =
0 K0
⌧
1_(z2 *z1 )
z2
z1
0
0 K0 (az + 1) ax
a 2
(z * z21 ) + (z2 * z1 ) =
(z2 * z1 ) 2 2
0 K0
⌧
ax dy dz =
z2
0 K0
(z2 * z1 )
z1
(az + 1) dz
a
(z + z1 ) + 1 Wb_m2
2 2
8.25. A conducting filament at z = 0 carries 12 A in the az direction. Let
1 < ⇢ < 2 cm, and r = 1 for ⇢ > 2 cm. Find
r = 1 for ⇢ < 1 cm,
r = 6 for
a) H everywhere: This result will depend on the current and not the materials, and is:
I
1.91
H=
a =
A_m (0 < ⇢ < ÿ)
2⇡⇢
⇢
b) B everywhere: We use B =
B(⇢ < 1 cm) = (1) 0
r 0 H to find:
(1.91_⇢) = (2.4 ù 10*6 _⇢)a
T
B(1 < ⇢ < 2 cm) = (6) 0 (1.91_⇢) = (1.4 ù 10*5 _⇢)a T
B(⇢ > 2 cm) = (1) 0 (1.91_⇢) = (2.4 ù 10*6 _⇢)a T where ⇢ is in meters.
8.26. A long solenoid has a radius of 3cm, 5,000 turns/m, and carries current I = 0.25 A. The region
0 < ⇢ < a within the solenoid has r = 5, while r = 1 for a < ⇢ < 3 cm. Determine a so that
a) a total flux of 10 Wb is present: First, the magnetic flux density in the coil is written in general
as B = nI az Wb_m2 . Using b = 0.03m as the outer radius, the total flux in the coil becomes
2⇡
a
2⇡
b
B dS =
5 0 nI⇢ d⇢ d +
s
0
0
a
⌅
⇧
⌅ 0
⇧
= 2⇡ 0 nI 5a2 + (b2 * a2 ) = 2⇡ 0 nI 4a2 + b2
m =
0 nI ⇢ d⇢ d
Substituting the given numbers, we have
m = 2⇡(4⇡ ù 10
*7
⌅
⇧
)(5000)(0.25) 4a2 + 0.032 = 10*5 Wb (as required)
Solve for a to find a = 5.3 mm.
b) Find a so that the flux is equally-divided between the regions 0 < ⇢ < a and a < ⇢ < 3 cm:
Using the expression for the flux in part a, we set
b
3
5a2 = b2 * a2 Ÿ a = ˘ = ˘ = 1.22 cm
6
6
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8.27. Let r1 = 2 in region 1, defined by 2x + 3y * 4z > 1, while
In region 1, H1 = 50ax * 30ay + 20az A/m. Find:
r2 = 5 in region 2 where 2x + 3y * 4z < 1.
a) HN1 (normal component of H1 at the boundary): We first need a unit vector normal to the surface,
found through
aN =
2ax + 3ay * 4az
( (2x + 3y * 4z)
=
= .37ax + .56ay * .74az
˘
( (2x + 3y * 4z)
29
Since this vector is found through the gradient, it will point in the direction of increasing values
of 2x + 3y * 4z, and so will be directed into region 1. Thus we write aN = aN21 . The normal
component of H1 will now be:
HN1 = (H1 aN21 )aN21
⌅
⇧
= (50ax * 30ay + 20az ) (.37ax + .56ay * .74az ) (.37ax + .56ay * .74az )
= *4.83ax * 7.24ay + 9.66az A_m
b) HT 1 (tangential component of H1 at the boundary):
HT 1 = H1 * HN1
= (50ax * 30ay + 20az ) * (*4.83ax * 7.24ay + 9.66az )
= 54.83ax * 22.76ay + 10.34az A_m
c) HT 2 (tangential component of H2 at the boundary): Since tangential components of H are continuous across a boundary between two media of different permeabilities, we have
HT 2 = HT 1 = 54.83ax * 22.76ay + 10.34az A_m
d) HN2 (normal component of H2 at the boundary): Since normal components of B are continuous
across a boundary between media of different permeabilities, we write 1 HN1 = 2 HN2 or
HN2 =
r1
R2
2
HN1 = (*4.83ax * 7.24ay + 9.66az ) = *1.93ax * 2.90ay + 3.86az A_m
5
e) ✓1 , the angle between H1 and aN21 : This will be
4
5
50ax * 30ay + 20az
H1
cos ✓1 =
a
=
(.37ax + .56ay * .74az ) = *0.21
H1  N21
(502 + 302 + 202 )1_2
Therefore ✓1 = cos*1 (*.21) = 102˝ .
f) ✓2 , the angle between H2 and aN21 : First,
H2 = HT 2 + HN2 = (54.83ax * 22.76ay + 10.34az ) + (*1.93ax * 2.90ay + 3.86az )
= 52.90ax * 25.66ay + 14.20az A_m
4
5
52.90ax * 25.66ay + 14.20az
H2
cos ✓2 =
a
=
(.37ax + .56ay * .74az ) = *0.09
H2  N21
60.49
Therefore ✓2 = cos*1 (*.09) = 95˝ .
167
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8.28. For values of B below the knee on the magnetization curve for silicon steel, approximate the curve by
a straight line with = 5 mH/m. The core shown in Fig. 8.17 has areas of 1.6 cm2 and lengths of 10
cm in each outer leg, and an area of 2.5 cm2 and a length of 3 cm in the central leg. A coil of 1200
turns carrying 12 mA is placed around the central leg. Find B in the:
a) center leg: We use mmf =
Rc =
R, where, in the central leg,
Lin
3 ù 10*2
=
= 2.4 ù 104 H
*3
*4
Ain
(5 ù 10 )(2.5 ù 10 )
In each outer leg, the reluctance is
Ro =
Lout
10 ù 10*2
=
= 1.25 ù 105 H
*3
*4
Aout
(5 ù 10 )(1.6 ù 10 )
The magnetic circuit is formed by the center leg in series with the parallel combination of the
two outer legs. The total reluctance seen at the coil location is RT = Rc + (1_2)Ro = 8.65 ù 104
H. We now have
mmf
14.4
=
=
= 1.66 ù 10*4 Wb
RT
8.65 ù 104
The flux density in the center leg is now
B=
A
=
1.66 ù 10*4
= 0.666 T
2.5 ù 10*4
b) center leg, if a 0.3-mm air gap is present in the center leg: The air gap reluctance adds to the total
reluctance already calculated, where
Rair =
0.3 ù 10*3
= 9.55 ù 105 H
*7
*4
(4⇡ ù 10 )(2.5 ù 10 )
Now the total reluctance is Rnet = RT + Rair = 8.56 ù 104 + 9.55 ù 105 = 1.04 ù 106 . The flux
in the center leg is now
14.4
=
= 1.38 ù 10*5 Wb
1.04 ù 106
and
1.38 ù 10*5
B=
= 55.3 mT
2.5 ù 10*4
168
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8.29. In Problem 8.28, the linear approximation suggested in the statement of the problem leads to a flux
density of 0.666 T in the center leg. Using this value of B and the magnetization curve for silicon steel,
.
what current is required in the 1200-turn coil? With B = 0.666 T, we read Hin = 120 A t_m in Fig.
8.11. The flux in the center leg is = 0.666(2.5 ù 10*4 ) = 1.66 ù 10*4 Wb. This divides equally in
the two outer legs, so that the flux density in each outer leg is
⇠ ⇡
1 1.66 ù 10*4
= 0.52 Wb_m2
2 1.6 ù 10*4
.
Using Fig. 8.11 with this result, we find Hout = 90 A t_m We now use
Bout =
Õ
H dL = NI
to find
I=
(120)(3 ù 10*2 ) + (90)(10 ù 10*2 )
1
Hin Lin + Hout Lout =
= 10.5 mA
N
1200
169
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8.30. A rectangular core has fixed permeability r >> 1, a square cross-section of dimensions a ù a, and
has centerline dimensions around its perimeter of b and d. Coils 1 and 2, having turn numbers N1
and N2 , are wound on the core. Consider a selected core cross-sectional plane as lying within the xy
plane, such that the surface is defined by 0 < x < a, 0 < y < a.
a) With current I1 in coil 1, use Ampere’s circuital law to find the magnetic flux density as a function
of position over the core cross-section: Along the midline of the core (at which x = d_2), the
path integral for H in Ampere’s law becomes
Õ
H dL = (2b + 2d)H
At all other points in the core interior, but off the midline, the path integral becomes
Õ
H dL = [2(d + a * 2x) + 2(b + a * 2x)] H = Iencl = N1 I1
The flux density magnitudes are therefore
B11 = B12 = H =
r 0 N1 I 1
2(d + b + 2a * 4x)
in which we are assuming no y variation.
b) Integrate your result of part a to determine the total magnetic flux within the core: This will be
the integral of B over the core cross-section:
m =
B dS =
a
a
r 0 N1 I1
2(d + b + 2a * 4x)
1
d + b + 2a
=
N I a ln
Wb
8 r 0 1 1
d + b * 2a
s
0
⌧
0
dx dy = *
1
Ûa
r 0 N1 I1 a ln [d + b + 2a * 4x] ÛÛ0
8
c) Find the self-inductance of coil 1:
L11 =
⌧
N1 B11
1
d + b + 2a
2
=
H
r 0 N1 a ln
I1
8
d + b * 2a
d) find the mutual inductance between coils 1 and 2.
M12 = M =
⌧
N2 B12
1
d + b + 2a
=
H
r 0 N1 N2 a ln
I1
8
d + b * 2a
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8.31. A toroid is constructed of a magnetic material having a cross-sectional area of 2.5 cm2 and an effective
length of 8 cm. There is also a short air gap 0.25 mm length and an effective area of 2.8 cm2 . An mmf
of 200 A t is applied to the magnetic circuit. Calculate the total flux in the toroid if:
a) the magnetic material is assumed to have infinite permeability: In this case the core reluctance,
Rc = l_( A), is zero, leaving only the gap reluctance. This is
Rg =
Now
0.25 ù 10*3
d
=
= 7.1 ù 105 H
(4⇡ ù 10*7 )(2.5 ù 10*4 )
0 Ag
=
mmf
200
=
= 2.8 ù 10*4 Wb
Rg
7.1 ù 105
b) the magnetic material is assumed to be linear with r = 1000: Now the core reluctance is no
longer zero, but
8 ù 10*2
Rc =
= 2.6 ù 105 H
*7
*4
(1000)(4⇡ ù 10 )(2.5 ù 10 )
The flux is then
=
mmf
200
=
= 2.1 ù 10*4 Wb
Rc + Rg
9.7 ù 105
c) the magnetic material is silicon steel: In this case we use the magnetization curve, Fig. 8.11, and
employ an iterative process to arrive at the final answer. We can begin with the value of found
in part a, assuming infinite permeability: (1) = 2.8 ù 10*4 Wb. The flux density in the core
is then Bc(1) = (2.8 ù 10*4 )_(2.5 ù 10*4 ) = 1.1 Wb_m2 . From Fig. 8.11, this corresponds to
.
magnetic field strength Hc(1) = 270 A/m. We check this by applying Ampere’s circuital law to
the magnetic circuit:
Õ
H dL = Hc(1) Lc + Hg(1) d
where Hc(1) Lc = (270)(8 ù 10*2 ) = 22, and where Hg(1) d =
199. But we require that
Õ
(1) R
*4
5
g = (2.8 ù 10 )(7.1 ù 10 ) =
H dL = 200 A t
whereas the actual result in this first calculation is 199 + 22 = 221, which is too high. So, for a
second trial, we reduce B to Bc(2) = 1 Wb_m2 . This yields Hc(2) = 200 A/m from Fig. 8.11, and
thus (2) = 2.5 ù 10*4 Wb. Now
Õ
H dL = Hc(2) Lc +
(2)
Rg = 200(8 ù 10*2 ) + (2.5 ù 10*4 )(7.1 ù 105 ) = 194
This is less than 200, meaning that the actual flux is slightly higher than 2.5 ù 10*4 Wb.
I will leave the answer at that, considering the lack of fine resolution in Fig. 8.11.
171
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8.32.
a) Find an expression for the magnetic energy stored per unit length in a coaxial transmission line
consisting of conducting sleeves of negligible thickness, having radii a and b. A medium of
relative permeability r fills the region between conductors. Assume current I flows in both
conductors, in opposite directions.
Within the coax, the magnetic field is H = I_(2⇡⇢) a . The energy density is then
I2
1
wm = B H = r 20 2 J_m3
2
8⇡ ⇢
The energy per unit length in z is therefore
Wm =
v
wm dv =
1
0
2⇡
b
0
a
2
r 0I
⇢ d⇢ d
8⇡ 2 ⇢2
dz =
r 0I
4⇡
2
ln
⇠ ⇡
b
J_m
a
b) Obtain the inductance, L, per unit length of line by equating the energy to (1_2)LI 2 .
L=
2Wm
I2
=
r 0
2⇡
ln
⇠ ⇡
b
H_m
a
8.33. A toroidal core has a square cross section, 2.5 cm < ⇢ < 3.5 cm, *0.5 cm < z < 0.5 cm. The upper
half of the toroid, 0 < z < 0.5 cm, is constructed of a linear material for which r = 10, while the
lower half, *0.5 cm < z < 0, has r = 20. An mmf of 150 A t establishes a flux in the a direction.
For z > 0, find:
a) H (⇢): Ampere’s circuital law gives:
2⇡⇢H = NI = 150 Ÿ H =
b) B (⇢): We use B =
c)
r 0H
150
= 23.9_⇢ A_m
2⇡⇢
= (10)(4⇡ ù 10*7 )(23.9_⇢) = 3.0 ù 10*4 _⇢ Wb_m2 .
z>0 : This will be
z>0 =
B dS =
= 5.0 ù 10*7 Wb
.005
0
.035
.025
⇠
⇡
3.0 ù 10*4
.035
d⇢dz = (.005)(3.0 ù 10*4 ) ln
⇢
.025
d) Repeat for z < 0: First, the magnetic field strength will be the same as in part a, since the
calculation is material-independent. Thus H = 23.9_⇢ A_m. Next, B is modified only by the
new permeability, which is twice the value used in part a: Thus B = 6.0 ù 10*4 _⇢ Wb_m2 .
Finally, since B is twice that of part a, the flux will be increased by the same factor, since the
area of integration for z < 0 is the same. Thus z<0 = 1.0 ù 10*6 Wb.
e) Find total : This will be the sum of the values found for z < 0 and z > 0, or
1.5 ù 10*6 Wb.
total
=
172
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8.34. Determine the energy stored per unit length in the internal magnetic field of an infinitely-long straight
wire of radius a, carrying uniform current I.
We begin with H = I⇢_(2⇡a2 ) a , and find the integral of the energy density over the unit length
in z:
1
2⇡
a
2 2
2
1
0⇢ I
0I
2
We =
H
dv
=
⇢
d⇢
d
dz
=
J_m
0
16⇡
8⇡ 2 a4
vol 2
0
0
0
8.35. The cones ✓ = 21˝ and ✓ = 159˝ are conducting surfaces and carry total currents of 40 A, as shown
in Fig. 8.17. The currents return on a spherical conducting surface of 0.25 m radius.
a) Find H in the region 0 < r < 0.25, 21˝ < ✓ < 159˝ , 0 < < 2⇡: We can apply Ampere’s
circuital law and take advantage of symmetry. We expect to see H in the a direction and it would
be constant at a given distance from the z axis. We thus perform the line integral of H over a
circle, centered on the z axis, and parallel to the xy plane:
Õ
H dL =
2⇡
0
H a
r sin ✓a d = Iencl. = 40 A
Assuming that H is constant over the integration path, we take it outside the integral and solve:
H =
40
20
Ÿ H=
a A_m
2⇡r sin ✓
⇡r sin ✓
b) How much energy is stored in this region? This will be
2⇡
159
.25
200 0
100 0
1
2
2
H
=
r
sin
✓
dr
d✓
d
=
0
2
2 2
⇡
v 2
0
21˝
4
5 0 ⇡ r sin ✓
100 0
tan(159_2)
=
ln
= 1.35 ù 10*4 J
⇡
tan(21_2)
˝
WH =
159˝
21˝
d✓
sin ✓
c) Find the inductance of the cone-sphere configuration: The inductance is that offered at the origin
between the vertices of the cone: Again, the magnetic flux density is B = 20 0 _(⇡r sin ✓). We
integrate this over the cross-sectional area defined by 0 < r < 0.25 and 21˝ < ✓ < 159˝ , to find
the total flux:
4
5
159˝
0.25
20 0
5 0
5
tan(159_2)
=
r dr d✓ =
ln
= 0 (3.37) = 6.74 ù 10*6 Wb
⇡r
sin
✓
⇡
tan(21_2)
⇡
˝
21
0
Now L =
_I = 6.74 ù 10*6 _40 = 0.17 H.
Second method: Use the energy computation in part b, and write
L=
2WH
2(1.35 ù 10*4 )
=
= 0.17 H
I2
(40)2
173
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8.36. The dimensions of the outer conductor of a coaxial cable are b and c, where c > b. Assuming = 0 ,
find the magnetic energy stored per unit length in the region b < ⇢ < c for a uniformly-distributed
total current I flowing in opposite directions in the inner and outer conductors.
We first need to find the magnetic field inside the outer conductor volume. Ampere’s circuital
law is applied to a circular path of radius ⇢, where b < ⇢ < c. This encloses the entire center
conductor current (assumed in the positive z direction), plus that part of the *z-directed outer
conductor current that lies inside ⇢. We obtain:
4 2
5
4 2
5
⇢ * b2
c * ⇢2
2⇡⇢H = I * I 2
=I 2
c * b2
c * b2
So that
4 2
5
c * ⇢2
I
H=
a A_m (b < ⇢ < c)
2⇡⇢ c 2 * b2
The energy within the outer conductor is now
4 2
5
1
2⇡
c
2
1
c
0I
2
2
2
Wm =
* 2c + ⇢ ⇢ d⇢ d , dz
0 H dv =
2 2
2 2 ⇢2
vol 2
0
0
b 8⇡ (c * b )
⌧
2
1
0I
=
ln(c_b) * (1 * b2 _c 2 ) + (1 * b4 _c 4 ) J
2
2
2
4
4⇡(1 * b _c )
8.37. A toroid has known core reluctance R. Two windings, having N1 and N2 turns are present. Find
a) the self inductances of coils 1 and 2: In general, we have NI =
L1 =
m R, and L = N
m _I. So
2
N2
N1 (N1 I1 _R) N1
=
and likewise L2 = 2
I1
R
R
b) the mutual inductance between the two coils:
M=
N2 (N1 I1 _R) N1 (N2 I2 _R) N1 N2
=
=
I1
I2
R
174
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8.38. A toroidal core has a rectangular cross section defined by the surfaces ⇢ = 2 cm, ⇢ = 3 cm, z = 4 cm,
and z = 4.5 cm. The core material has a relative permeability of 80. The core is wound with two coils
having N1 = 1000 and N2 = 2500 turns of wire.
a) Find the self inductances of the two coils: First we apply Ampere’s circuital law to a circular loop
of radius ⇢ in the interior of the toroid, and in the a direction. Doing this for coil 1 (carrying
current I1 ), we find:
Õ
H dL = 2⇡⇢H = N1 I1 Ÿ H =
N1 I1
2⇡⇢
The flux in the toroid is then the integral over the cross section of B:
11 =
B dL =
.045
.04
.03
r 0 N1 I1
2⇡⇢
.02
⇠ ⇡
N I
.03
d⇢ dz = (.005) r 0 1 1 ln
2⇡
.02
The self linkage is then given by ⇤11 = N1 , and the inductance is
L11 =
N1 11
(.005)(80)(4⇡ ù 10*7 )(1000)2
=
ln(1.5) = 32 mH
I1
2⇡
Performing the same procedure for coil 2, we find:
L22 =
N2 22
(.005)(80)(4⇡ ù 10*7 )(2500)2
=
ln(1.5) = 203 mH
I2
2⇡
b) Find the mutual inductance between the two coils: Because the coils share the same core, we
have 12 = 11 and 21 = 22 . Then
⇤12 = N2
12
and ⇤21 = N1
21
and thus
L12 =
N2 12
N
(.005)(80)(4⇡ ù 10*7 )(1000)(2500)
= L21 = 1 21 =
ln(1.5) = 80 mH
I1
I2
2⇡
175
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8.39. Conducting planes in air at z = 0 and z = d carry surface currents of ±K0 ax A_m.
a) Find the energy stored in the magnetic field per unit length (0 < x < 1) in a width w (0 < y < w):
First, assuming current flows in the +ax direction in the sheet at z = d, and in *ax in the sheet
at z = 0, we find that both currents together yield H = K0 ay for 0 < z < d and zero elsewhere.
The stored energy within the specified volume will be:
d
1
2
0 H dv =
2
v
WH =
0
1
w
0
0
1
1
2
2
0 K0 dx dy dz = wd 0 K0 J_m
2
2
b) Calculate the inductance per unit length of this transmission line from WH = (1_2)LI 2 , where
I is the total current in a width w in either conductor: We have I = wK0 , and so
L=
2 wd
I2 2
2
0 K0 =
2 dw
0d
2
K
=
H_m
0
0
w
w2 K02 2
c) Calculate the total flux passing through the rectangle 0 < x < 1, 0 < z < d, in the plane y = 0,
and from this result again find the inductance per unit length:
=
1
d
0
0
Then
0 Hay ay dx dz =
L=
I
0 dK0
=
wK0
1
d
0
=
0 K0 dx dy =
0
0d
w
0 dK0
H_m
8.40. A coaxial cable has conductor radii a and b, where a < b. Material of permeability r ë 1 exists in
the region a < ⇢ < c, while the region c < ⇢ < b is air-filled. Find an expression for the inductance
per unit length.
In both regions, the magnetic field will be H = I_(2⇡⇢) a A/m. So the flux per unit length
between conductors will be the sum of the fluxes in both regions. We integrate over a plane
surface of constant , unit length in z, and between radii a and b:
m =
s
1
c
r 0I
1
b
0I
d⇢ dz
2⇡⇢
2⇡⇢
0
a
0
c
⇠ ⇡
⇠ ⇡
I⌧
c
b
= 0
ln
+
ln
Wb_m
r
2⇡
a
c
B dS =
d⇢ dz +
The inductance per unit length is then L = m _I (with one turn), or
⌧
⇠ ⇡
⇠ ⇡
c
b
L = 0 r ln
+ ln
H_m
2⇡
a
c
176
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8.41. A rectangular coil is composed of 150 turns of a filamentary conductor. Find the mutual inductance
in free space between this coil and an infinite straight filament on the z axis if the four corners of the
coil are located at
a) (0,1,0), (0,3,0), (0,3,1), and (0,1,1): In this case the coil lies in the yz plane. If we assume that
the filament current is in the +az direction, then the B field from the filament penetrates the coil
in the *ax direction (normal to the loop plane). The flux through the loop will thus be
=
1
3
0
1
* 0I
I
ax (*ax ) dy dz = 0 ln 3
2⇡y
2⇡
The mutual inductance is then
M=
N
I
=
150 0
ln 3 = 33 H
2⇡
b) (1,1,0), (1,3,0), (1,3,1), and (1,1,1): Now the coil lies in the x = 1 plane, and the field from the
filament penetrates in a direction that is not normal to the plane of the coil. We write the B field
from the filament at the coil location as
B=
0 Ia
˘
2⇡ y2 + 1
The flux through the coil is now
=
=
1
0
1
0
3
1
3
0 Ia
0 I sin
(*ax ) dy dz =
dy dz
˘
˘
1 2⇡ y2 + 1
0
1 2⇡ y2 + 1
3
I
Û3
0 Iy
dy dz = 0 ln(y2 + 1)Û = (1.6 ù 10*7 )I
2
Û1
2⇡
1 2⇡(y + 1)
The mutual inductance is then
M=
N
I
= (150)(1.6 ù 10*7 ) = 24 H
8.42. Find the mutual inductance between two filaments forming circular rings of radii a and a, where
a << a. The field should be determined by approximate methods. The rings are coplanar and
concentric.
We use the result of Problem 7.4, which asks for the magnetic field on the z axis, arising from a
circular current loop of radius a. That result is specialized for z = 0, leading to:
Hcirc =
2⇡ Iad
0
a ù (*a⇢ )
4⇡a2
=
2⇡
0
Id az
I
=
a A_m
4⇡a
2a z
We now approximate that field as constant over a circular area of radius
linkage (for the single turn) as
.
m = ⇡(
a)2 Bouter =
0 I⇡(
2a
a)2
Ÿ M=
m
I
=
a, and write the flux
0 ⇡(
2a
a)2
177
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8.43.
a) Use energy relationships to show that the internal inductance of a nonmagnetic cylindrical wire
of radius a carrying a uniformly-distributed current I is 0 _(8⇡) H_m. We first find the magnetic
field inside the conductor, then calculate the energy stored there. From Ampere’s circuital law:
2⇡⇢H =
Now
1
1
2
0 H dv =
v 2
WH =
⇡⇢2
I⇢
I Ÿ H =
A_m
2
⇡a
2⇡a2
2⇡
0
a
0
0
Now, with WH = (1_2)LI 2 , we find Lint =
2 2
0I ⇢
⇢ d⇢ d
8⇡ 2 a4
0I
dz =
2
16⇡
J_m
0 _(8⇡) as expected.
b) Find the internal inductance if the portion of the conductor for which ⇢ < c < a is removed: The
hollowed-out conductor still carries current I, so Ampere’s circuital law now reads:
4 2
5
⇡(⇢2 * c 2 )
⇢ * c2
I
2⇡⇢H =
Ÿ H =
A_m
2⇡⇢ a2 * c 2
⇡(a2 * c 2 )
and the energy is now
WH =
=
1
0
2⇡
0
0
I2
a
c
2 2
2 2
0 I (⇢ * c )
⇢ d⇢ d
8⇡ 2 ⇢2 (a2 * c 2 )2
⌧
4⇡(a2 * c 2 )2
2
0I
dz =
4⇡(a2 * c 2 )2
⇠ ⇡
1 4
a
(a * c 4 ) * c 2 (a2 * c 2 ) + c 4 ln
J_m
4
c
a4
c
5
C4
⇢ * 2c ⇢ +
d⇢
⇢
3
2
The internal inductance is then
4 4
5
2 2
4
4
2WH
0 a * 4a c + 3c + 4c ln(a_c)
Lint =
=
H_m
8⇡
I2
(a2 * c 2 )2
8.44. Show that the external inductance per unit length of a two-wire transmission line carrying equal and
opposite currents is approximately ( _⇡) ln(d_a) H/m, where a is the radius of each wire and d is the
center-to-center wire spacing. On what basis is the approximation valid?
Suppose that one line is positioned along the z axis, with the other in the x-z plane at x = d.
With equal and opposite currents, I, and with the z axis wire current in the positive z direction,
the magnetic flux density in the x-z plane (arising from both currents) will be
B(x) =
I
a Wb_m2 (a < x < d * a)
⇡x y
The flux per unit length in z between conductors is now:
m =
s
B dS =
1
0
(d*a)
a
Now, the external inductance will be L =
.
L=
I
I ⇠d * a⇡
ay ay dx dz =
ln
⇡x
⇡
a
m _I, which becomes
⇡
ln
⇠ ⇡
d
H_m
a
under the assumption that a << d.
178
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8.45
a) Beginning with the definition of the scalar magnetic potential, H = *(Vm and using Eq. (25),
show that in a region having permanent magnetization M and zero current, a Poisson equation
for magnetic potential can be developed:
(2 Vm =
*⇢m
0
where ⇢m = * 0 ( M is an equivalent magnetic charge density. What happens when M is
uniform? Eq. (25) can be written as
H=
B
0
* M = *(Vm
Taking the divergence, we find
*( (Vm = *(2 Vm =
1
0
( B*( M
Noting that ( B = 0, we obtain the desired result if we define ⇢m = * 0 ( M.
If M is uniform, its divergence is zero, and hence so is the equivalent charge density. What results
is (2 Vm = 0, or a form of Laplace’s equation.
b) Using the definition of the vector magnetic potential, B = ( ù A, and Eq. (25), show that when
a permanent magnetization exists, and with zero current:
(ù(ùA=
0 Jeq
where Jeq = ( ù M, in analogy to Eq. (57) in Chapter 7: Take
(ùB=(ù(ùA=(ù
0 (H + M) =
0( ù M =
0 Jeq
where H = 0 because we have zero current.
179
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CHAPTER 9 – 9th Edition
9.1. In Fig. 9.4, let B = 0.2 cos 120⇡t T, and assume that the conductor joining the two ends of the resistor
is perfect. It may be assumed that the magnetic field produced by I(t) is negligible. Find:
a) Vab (t): Since B is constant over the loop area, the flux is = ⇡(0.15)2 B = 1.41 ù 10*2 cos 120⇡t
Wb. Now, emf = Vba (t) = *d _dt = (120⇡)(1.41 ù 10*2 ) sin 120⇡t. Then Vab (t) = *Vba (t) =
*5.33 sin 120⇡t V.
b) I(t) = Vba (t)_R = 5.33 sin(120⇡t)_250 = 21.3 sin(120⇡t) mA
9.2. In the example described by Fig. 9.1, replace the constant magnetic flux density by the time-varying
quantity B = B0 sin !t az . Assume that v is constant and that the displacement y of the bar is zero at
t = 0. Find the emf at any time, t.
The magnetic flux through the loop area is
m =
s
B dS =
vt
0
d
0
B0 sin !t (az az ) dx dy = B0 v t d sin !t
Then the emf is
emf =
Õ
E dL = *
d m
= *B0 d v [sin !t + !t cos !t] V
dt
9.3. Given H = 300 az cos(3 ù 108 t * y) A/m in free space, find the emf developed in the general a
direction about the closed path having corners at
a) (0,0,0), (1,0,0), (1,1,0), and (0,1,0): The magnetic flux will be:
1
1
300 0 cos(3 ù 108 t * y) dx dy = 300 0 sin(3 ù 108 t * y)10
⌅
⇧
= 300 0 sin(3 ù 108 t * 1) * sin(3 ù 108 t) Wb
=
0
0
Then
⌅
⇧
d
= *300(3 ù 108 )(4⇡ ù 10*7 ) cos(3 ù 108 t * 1) * cos(3 ù 108 t)
dt
⌅
⇧
= *1.13 ù 105 cos(3 ù 108 t * 1) * cos(3 ù 108 t) V
emf = *
b) corners at (0,0,0), (2⇡,0,0), (2⇡,2⇡,0), (0,2⇡,0): In this case, the flux is
= 2⇡ ù 300 0 sin(3 ù 108 t * y)2⇡
=0
0
The emf is therefore 0.
180
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9.4. A rectangular loop of wire containing a high-resistance voltmeter has corners initially at (a_2, b_2, 0),
(*a_2, b_2, 0), (*a_2, *b_2, 0), and (a_2, *b_2, 0). The loop begins to rotate about the x axis at
constant angular velocity !, with the first-named corner moving in the az direction at t = 0. Assume a
uniform magnetic flux density B = B0 az . Determine the induced emf in the rotating loop and specify
the direction of the current.
The magnetic flux though the loop is found (as usual) through
m =
s
B dS, where S = n da
Because the loop is rotating, the direction of the normal, n, changing, and is in this case given by
n = cos !t az * sin !t ay
Therefore,
m =
b_2
a_2
*b_2
*a_2
B0 a z
cos !t az * sin !t ay dx dy = abB0 cos !t
The integral is taken over the entire loop area (regardless of its immediate orientation). The
important result is that the component of B that is normal to the loop area is varying sinusoidally,
and so it is fine to think of the B field itself rotating about the x axis in the opposite direction
while the loop is stationary. Now the emf is
emf =
Õ
E dL = *
d m
= ab !B0 sin !t V
dt
The direction of the current is the same as the direction of E in the emf expression. It is easiest
to picture this by considering the B field rotating and the loop fixed. By convention, dL will be
counter-clockwise when looking down on the loop from the upper half-space (in the opposite
direction of the normal vector to the plane). The current will be counter-clockwise whenever the
emf is positive, and will be clockwise whenever the emf is negative.
9.5. The location of the sliding bar in Fig. 9.5 is given by x = 5t + 2t3 , and the separation of the two rails
is 20 cm. Let B = 0.8x2 az T. Find the voltmeter reading at:
a) t = 0.4 s: The flux through the loop will be
=
0.2
0
x
0
0.8(x® )2 dx® dy =
0.16 3 0.16
x =
(5t + 2t3 )3 Wb
3
3
Then
emf = *
d
0.16
=
(3)(5t + 2t3 )2 (5 + 6t2 ) = *(0.16)[5(.4) + 2(.4)3 ]2 [5 + 6(.4)2 ] = *4.32 V
dt
3
b) x = 0.6 m: Have 0.6 = 5t + 2t3 , from which we find t = 0.1193. Thus
emf = *(0.16)[5(.1193) + 2(.1193)3 ]2 [5 + 6(.1193)2 ] = *.293 V
181
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9.6. Let the wire loop of Problem 9.4 be stationary in its t = 0 position and find the induced emf that results
from a magnetic flux density given by B(y, t) = B0 cos(!t * y) az , where ! and are constants.
We begin by finding the net magnetic flux through the loop:
m =
=
s
B dS =
b_2
a_2
*b_2
*a_2
B0 cos(!t * y) az az dx dy
⇧
B0 a ⌅
sin(!t + b_2) * sin(!t * b_2)
Now the emf is
emf =
Õ
E dL = *
⇧
d m
B a! ⌅
=* 0
cos(!t + b_2) * cos(!t * b_2)
dt
Using the trig identity, cos(a ± b) = cos a cos b - sin a sin b, we may write the above result as
!
emf = +2B0 a sin(!t) sin( b_2) V
9.7. The rails in Fig. 9.7 each have a resistance of 2.2 ⌦_m. The bar moves to the right at a constant speed
of 9 m/s in a uniform magnetic field of 0.8 T. Find I(t), 0 < t < 1 s, if the bar is at x = 2 m at t = 0
and
a) a 0.3 ⌦ resistor is present across the left end with the right end open-circuited: The flux in the
left-hand closed loop is
l = B ù area = (0.8)(0.2)(2 + 9t)
Then, emf l = *d l _dt = *(0.16)(9) = *1.44 V. With the bar in motion, the loop resistance is
increasing with time, and is given by Rl (t) = 0.3 + 2[2.2(2 + 9t)]. The current is now
Il (t) =
emf l
*1.44
=
A
Rl (t) 9.1 + 39.6t
Note that the sign of the current indicates that it is flowing in the direction opposite that shown
in the figure.
b) Repeat part a, but with a resistor of 0.3 ⌦ across each end: In this case, there will be a contribution
to the current from the right loop, which is now closed. The flux in the right loop, whose area
decreases with time, is
r = (0.8)(0.2)[(16 * 2) * 9t]
and emf r = *d r _dt = (0.16)(9) = 1.44 V. The resistance of the right loop is Rr (t) =
0.3 + 2[2.2(14 * 9t)], and so the contribution to the current from the right loop will be
Ir (t) =
*1.44
A
61.9 * 39.6t
The minus sign has been inserted because again the current must flow in the opposite direction
as that indicated in the figure, with the flux decreasing with time. The total current is found by
adding the part a result, or
⌧
1
1
IT (t) = *1.44
+
A
61.9 * 39.6t 9.1 + 39.6t
182
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9.8. A perfectly-conducting filament is formed into a circular ring of radius a. At one point a resistance
R is inserted into the circuit, and at another a battery of voltage V0 is inserted. Assume that the loop
current itself produces negligible magnetic field.
a) Apply Faraday’s law, Eq. (4), evaluating each side of the equation carefully and independently to
show the equality: With no B field present, and no time variation, the right-hand side of Faraday’s
law is zero, and so therefore
Õ
E dL = 0
This is just a statement of Kirchoff’s voltage law around the loop, stating that the battery voltage
is equal and opposite to the resistor voltage.
b) Repeat part a, assuming the battery removed, the ring closed again, and a linearly-increasing B
field applied in a direction normal to the loop surface: The situation now becomes the same as
that shown in Fig. 9.4, except the loop radius is now a, and the resistor value is not specified.
Consider the loop as in the x-y plane with the positive z axis directed out of the page. The a
direction is thus counter-clockwise around the loop. The B field (out of the page as shown) can
be written as B(t) = B0 t az . With the normal to the loop specified as az , the direction of dL is, by
the right hand convention, a . Since the wire is perfectly-conducting, the only voltage appears
across the resistor, and is given as VR . Faraday’s law becomes
Õ
E dL = VR = *
d m
d
=*
dt
dt
s
B0 t az az da = *⇡a2 B0
This indicates that the resistor voltage, VR = ⇡a2 B0 , has polarity such that the positive terminal
is at point a in the figure, while the negative terminal is at point b. Current flows in the clockwise
direction, and is given in magnitude by I = ⇡a2 B0 _R.
9.9. A square filamentary loop of wire is 25 cm on a side and has a resistance of 125 ⌦ per meter length.
The loop lies in the z = 0 plane with its corners at (0, 0, 0), (0.25, 0, 0), (0.25, 0.25, 0), and (0, 0.25, 0)
at t = 0. The loop is moving with velocity vy = 50 m/s in the field Bz = 8 cos(1.5 ù 108 t * 0.5x) T.
Develop a function of time which expresses the ohmic power being delivered to the loop: First, since
the field does not vary with y, the loop motion in the y direction does not produce any time-varying
flux, and so this motion is immaterial. We can evaluate the flux at the original loop position to obtain:
.25
.25
8 ù 10*6 cos(1.5 ù 108 t * 0.5x) dx dy
0
0
⌅
⇧
= *(4 ù 10*6 ) sin(1.5 ù 108 t * 0.13) * sin(1.5 ù 108 t) Wb
(t) =
⌅
⇧
Now, emf = V (t) = *d _dt = 6.0 ù 102 cos(1.5 ù 108 t * 0.13) * cos(1.5 ù 108 t) , The total loop
resistance is R = 125(0.25 + 0.25 + 0.25 + 0.25) = 125 ⌦. Then the ohmic power is
P (t) =
⌅
⇧2
V 2 (t)
= 2.9 ù 103 cos(1.5 ù 108 t * 0.13) * cos(1.5 ù 108 t) Watts
R
183
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9.10
a) Show that the ratio of the amplitudes of the conduction current density and the displacement
current density is _!✏ for the applied field E = Em cos !t. Assume = 0 . First, D =
✏E = ✏Em cos !t. Then the displacement current density is )D_)t = *!✏Em sin !t. Second,
Jc = E = Em cos !t. Using these results we find Jc _Jd  = _!✏.
b) What is the amplitude ratio if the applied field is E = Em e*t_⌧ , where ⌧ is real? As before, find
D = ✏E = ✏Em e*t_⌧ , and so Jd = )D_)t = *(✏_⌧)Em e*t_⌧ . Also, Jc = Em e*t_⌧ . Finally,
Jc _Jd  = ⌧_✏.
9.11. Let the internal dimension of a coaxial capacitor be a = 1.2 cm, b = 4 cm, and l = 40 cm. The
homogeneous material inside the capacitor has the parameters ✏ = 10*11 F/m, = 10*5 H/m, and
= 10*5 S/m. If the electric field intensity is E = (106 _⇢) cos(105 t)a⇢ V/m, find:
a) J: Use
0
J= E=
10
⇢
1
cos(105 t)a⇢ A_m2
b) the total conduction current, Ic , through the capacitor: Have
Ic =
J dS = 2⇡⇢lJ = 20⇡l cos(105 t) = 8⇡ cos(105 t) A
c) the total displacement current, Id , through the capacitor: First find
Jd =
Now
(105 )(10*11 )(106 )
)D
)
1
= (✏E) = *
sin(105 t)a⇢ = * sin(105 t) A_m
)t
)t
⇢
⇢
Id = 2⇡⇢lJd = *2⇡l sin(105 t) = *0.8⇡ sin(105 t) A
d) the ratio of the amplitude of Id to that of Ic , the quality factor of the capacitor: This will be
Id  0.8
=
= 0.1
Ic 
8
184
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9.12. The magnetic flux density B = B0 cos(!t) cos(k0 z) ay Wb_m2 exists in free space. B0 and k0 are
constants, and J = 0. Find:
a) the displacement current density: In free space, and with zero conduction current, Ampere’s Law
in point form becomes
)E
)D
(ùH=
= ✏0
)t
)t
The given field field is y-directed and varies spatially only with z. So the above becomes
(ùH=(ù
B
0
=*
1 )By
)D k0 B0
ax =
=
cos(!t) sin(k0 z) ax A_m2
)z
)t
0
0
b) the electric field intensity: We write
E=
1
✏0
k B
)D
dt + C = 0 0 sin(k0 z) ax
)t
0 ✏0
cos(!t)dt =
k0 B0
sin(k0 z) sin(!t) ax
! 0 ✏0
where the integration constant, C, is set to zero because no steady electric field component is
expected with a purely time-varying H.
c) k0 : We find this by requiring consistency in applying both of the Maxwell curl equations. The
Faraday Law in point form reads
(ùE=*
)B
Ÿ B=*
)t
( ù E dt
where again the integration constant is set to zero. Using the result for E in part b, and noting
that it is x directed and spatially varies only with z, the curl simplifies to a single term, and the
above equation becomes:
B=*
=
( ù E dt = *
k20 B0
!2 0 ✏0
)Ex
a dt =
)z y
*k20 B0
! 0 ✏0
cos(k0 z) sin(!t)ay dt
cos(k0 z) cos(!t) ay
Requiring this result to be the same as the given field leads to the identification:
˘
*1
k0 = !
0 ✏0 m .
185
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9.13. In free space it is known that E = E0 _r sin ✓ cos(!t * k0 r) a✓ = E✓ a✓ . Show that a component of
H in the a direction arises, where H = E✓ (✏0 _ 0 )1_2 . Do this by applying Eqs. (20) and (21) and
requiring consistency:
Start with Eq. (20) in free space:
(ùE=* 0
where
( ù E✓ (r, ✓) a✓ =
)H
)t
)H
1 )(rE✓ )
1
a = * E0 k0 sin ✓ sin(!t * k0 r)a = * 0
a
r )r
r
)t
After integrating the above result over time, we find
H=H a =
E0 k0
sin ✓ cos(!t * k0 r)a
! 0r
Now, use Eq. (21) in free space:
( ù H = ✏0
where
)E
)t
1 )(H sin ✓)
1 )(rH )
ar *
a✓
r sin ✓
)✓
r )r
E0 k2
2E k
= 2 0 0 2 cos ✓ sin(!t * k0 r)ar + 2 0 2 sin ✓ cos(!t * k0 r)a✓
! 0 ✏0 r
! 0 ✏0 r
˘
The second term in this field will be equal to the given field when k0 = !
0 ✏0 . With this
substitution, the component of H becomes:
u
✏0 E0
H =
sin ✓ cos(!t * k0 r) Q.E.D
0 r
( ù H (r, ✓)a =
186
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9.14. A voltage source, V0 sin !t, is connected between two concentric conducting spheres, r = a and r = b,
b > a, where the region between them is a material for which ✏ = ✏r ✏0 , = 0 , and = 0. Find the
total displacement current through the dielectric and compare it with the source current as determined
from the capacitance (Sec. 6.3) and circuit analysis methods: First, solving Laplace’s equation, we
find the voltage between spheres (see Eq. 39, Chapter 6):
V (t) =
Then
E = *(V =
Now
(1_r) * (1_b)
V sin !t
(1_a) * (1_b) 0
V0 sin !t
ar
2
r (1_a * 1_b)
Jd =
Ÿ D=
✏r ✏0 V0 sin !t
r2 (1_a * 1_b)
ar
)D ✏r ✏0 !V0 cos !t
= 2
ar
)t
r (1_a * 1_b)
The displacement current is then
Id = 4⇡r2 Jd =
4⇡✏r ✏0 !V0 cos !t
dV
=C
(1_a * 1_b)
dt
where, from Eq. 6, Chapter 6,
C=
4⇡✏r ✏0
(1_a * 1_b)
9.15. Use each of Maxwell’s equations in point form to obtain as much information as possible about
a) H if E = 0: In this case we have
( ù H = J (the source of H is conduction current only)
)B
= 0 Ÿ B is constant
)t
⇢v = 0 (no charge exists)
( B = 0 (always true anyway)
As J is non-zero, but E = 0, the current-carrying medium would be a perfect conductor.
b) E if H = 0: Here we have
( ù E = 0 (E is constant in time and conservative)
)D
)D
=0 Ÿ J=
=0
)t
)t
( D = ⇢v (E arises f rom f ree charge only)
(ùH=J+
( B = 0 (irrelevant as B = 0 anyway)
187
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9.16. Derive the continuity equation from Maxwell’s equations: First, take the divergence of both sides of
Ampere’s circuital law:
)⇢
)
( (ùH=( J+ ( D=( J+ v =0
´≠≠Ø≠≠¨
)t
)t
0
where we have used ( D = ⇢v , another Maxwell equation.
9.17. The electric field intensity in the region 0 < x < 5, 0 < y < ⇡_12, 0 < z < 0.06 m in free space is
given by E = C sin(12y) sin(az) cos(2 ù 1010 t) ax V/m. Beginning with the ( ù E relationship, use
Maxwell’s equations to find a numerical value for a, if it is known that a is greater than zero: In this
case we find
)Ex
)Ez
ay *
a
)z
)y z
⌅
⇧
)B
= C a sin(12y) cos(az)ay * 12 cos(12y) sin(az)az cos(2 ù 1010 t) = *
)t
(ùE=
Then
H=*
=*
1
0
( ù E dt + C1
⌅
⇧
C
a sin(12y) cos(az)ay * 12 cos(12y) sin(az)az sin(2 ù 1010 t) A_m
10
0 (2 ù 10
where the integration constant, C1 = 0, since there are no initial conditions. Using this result, we now
find
4
5
)Hz )Hy
C(144 + a2 )
)D
(ùH=
*
ax = *
sin(12y) sin(az) sin(2 ù 1010 t) ax =
10 )
)y
)z
)t
(2
ù
10
0
Now
E=
D
=
✏0
1
( ù H dt + C2 =
✏0
C(144 + a2 )
sin(12y) sin(az) cos(2 ù 1010 t) ax
10 )2
✏
(2
ù
10
0 0
where C2 = 0. This field must be the same as the original field as stated, and so we require that
C(144 + a2 )
=1
10 2
0 ✏0 (2 ù 10 )
Using
8 *2
0 ✏0 = (3 ù 10 ) , we find
4
(2 ù 1010 )2
a=
* 144
(3 ù 108 )2
51_2
= 66 m*1
188
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9.18. The parallel plate transmission line shown in Fig. 9.7 has dimensions b = 4 cm and d = 8 mm, while
the medium between plates is characterized by r = 1, ✏r = 20, and = 0. Neglect fields outside the
dielectric. Given the field H = 5 cos(109 t * z)ay A/m, use Maxwell’s equations to help find:
a)
, if
> 0: Take
(ùH=*
So
)z
ax = *5 sin(109 t * z)ax = 20✏0
)E
)t
*5
sin(109 t * z)ax dt =
cos(109 t * z)ax
20✏0
(4 ù 109 )✏0
E=
Then
(ùE=
So that
)Hy
2
)Ex
)H
ay =
sin(109 t * z)ay = * 0
)z
)t
(4 ù 109 )✏0
2
* 2
9
sin(10
t
*
z)a
dt
=
cos(109 t * z)
x
9
18
(4 ù 10 ) 0 ✏0
(4 ù 10 ) 0 ✏0
H=
= 5 cos(109 t * z)ay
where the last equality is required to maintain consistency. Therefore
2
(4 ù 1018 ) 0 ✏0
=5 Ÿ
b) the displacement current density at z = 0: Since
= 14.9 m*1
= 0, we have
9
( ù H = Jd = *5 sin(10 t * z) = *74.5 sin(109 t * 14.9z)ax
= *74.5 sin(109 t)ax A_m at z = 0
c) the total displacement current crossing the surface x = 0.5d, 0 < y < b, and 0 < z < 0.1 m in
the ax direction. We evaluate the flux integral of Jd over the given cross section:
Id = *74.5b
0.1
0
⌅
⇧
sin(109 t * 14.9z) ax ax dz = 0.20 cos(109 t * 1.49) * cos(109 t) A
9.19. In the first section of this chapter, Faraday’s law was used to show that the field E = * 12 kB0 ⇢ekt a
results from the changing magnetic field B = B0 ekt az .
a) Show that these fields do not satisfy Maxwell’s other curl equation: Note that B as stated is
constant with position, and so will have zero curl. The electric field, however, varies with time,
and so (ùH = )D
would have a zero left-hand side and a non-zero right-hand side. The equation
)t
is thus not valid with these fields.
b) If we let B0 = 1 T and k = 106 s*1 , we are establishing a fairly large magnetic flux density in 1
s. Use the ( ù H equation to show that the rate at which Bz should (but does not) change with
⇢ is only about 5 ù 10*6 T/m in free space at t = 0: Assuming that B varies with ⇢, we write
)H
1 dB0 kt
)E
1
( ù H = * za = *
e = ✏0
= * ✏0 k2 B0 ⇢ekt
)⇢
d⇢
)t
2
0
Thus
dB0
1012 (1)⇢
1
2
=
✏
k
⇢B
=
= 5.6 ù 10*6 ⇢
0
d⇢
2 0 0
2(3 ù 108 )2
which is near the stated value if ⇢ is on the order of 1m.
189
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9.20. Given Maxwell’s equations in point form, assume that all fields vary as est and write the equations
without explicitly involving time: Write all fields in the general form A(r, t) = A0 (r)est , where r is a
position vector in any coordinate system. Maxwell’s equations become:
( ù E0 (r) est = *
( ù H0 (r) est = J0 (r)est +
)
B (r) est = *sB0 (r) est
)t 0
)
D (r) est = J0 (r)est + sD0 (r) est
)t 0
( D0 (r) est = ⇢0 (r) est
( B0 (r) est = 0
In all cases, the est terms divide out, leaving:
( ù E0 (r) = *sB0 (r)
( ù H0 (r) = J0 (r) + sD0 (r)
( D0 (r) = ⇢0 (r)
( B0 (r) = 0
9.21.
a) Show that under static field conditions, Eq. (55) reduces to Ampere’s circuital law. First use the
definition of the vector Laplacian:
(2 A = *( ù ( ù A + ((( A) = * J
which is Eq. (55) with the time derivative set to zero. We also note that ( A = 0 in steady state
(from Eq. (54)). Now, since B = ( ù A, (55) becomes
*( ù B = * J Ÿ ( ù H = J
b) Show that Eq. (51) becomes Faraday’s law when taking the curl: Doing this gives
( ù E = *( ù (V *
)
(ùA
)t
The curl of the gradient is identially zero, and ( ù A = B. We are left with
( ù E = *)B_)t
190
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9.22. In a sourceless medium, in which J = 0 and ⇢v = 0, assume a rectangular coordinate system in which
E and H are functions only of z and t. The medium has permittivity ✏ and permeability .
a) If E = Ex ax and H = Hy ay , begin with Maxwell’s equations and determine the second order
partial differential equation that Ex must satisfy.
First use
)Hy
)Ex
)B
Ÿ
ay = *
a
)t
)z
)t y
in which case, the curl has dictated the direction that H must lie in. Similarly, use the other
Maxwell curl equation to find
(ùE=*
(ùH=
)Hy
)E
)D
Ÿ *
ax = ✏ x ax
)t
)z
)t
Now, differentiate the first equation with respect to z, and the second equation with respect to t:
) 2 Ex
)z2
Combining these two, we find
=*
) 2 Hy
)t)z
) 2 Ex
)z2
and
= ✏
) 2 Hy
)z)t
= *✏
) 2 Ex
)t2
) 2 Ex
)t2
b) Show that Ex = E0 cos(!t * z) is a solution of that equation for a particular value of : Substituting, we find
) 2 Ex
)z2
= * 2 E0 cos(!t * z) and
✏
) 2 Ex
)t2
= *!2 ✏E0 cos(!t * z)
These two will be equal provided the constant multipliers of cos(!t * z) are equal.
c) Find as a function of given parameters. Equating the two constants in part b, we find
˘
=!
✏.
191
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9.23. Referring to the parallel plate transmission line of Fig. 9.7, the electric field between plates is given
as E(z, t) = *E0 cos(!t * z)ay . Find the vector potential A(y, z, t) if it is given that A(0, z, t) = 0.
Assume b >> d, so that x variation is negligible:
Start with
(ùE=*
from which
H=
* E0
0
)Ey
)z
ax = + E0 sin(!t * z)ax = * 0
sin(!t * z)ax dt + C = +
)H
)t
E0
cos(!t * z)ax
! 0
where the integration constant is zero. Next we have
E0
( ù A = B = 0H =
cos(!t * z)ax =
!
0
)Az )Ay
*
)y
)z
1
ax
where the last term is the x component of the curl of A, within which we expect Ay = 0 because
the current will be entirely z directed. The equation to solve is thus
)Az
E0
E0
=
cos(!t * z) Ÿ A =
y cos(!t * z)az
)y
!
!
which is seen to satisfy the boundary condition at y = 0.
192
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9.24. A vector potential is given as A = A0 cos(!t * kz) ay . a) Assuming as many components as possible
are zero, find H, E, and V ;
With A y-directed only, and varying spatially only with z, we find
H=
1
(ùA=*
kA
1 )Ay
ax = * 0 sin(!t * kz) ax A_m
)z
Now, in a lossless medium we will have zero conductivity, so that the point form of Ampere’s
circuital law involves only the displacement current term:
(ùH=
)D
)E
=✏
)t
)t
Using the magnetic field as found above, we find
(ùH=
Now,
or
)Hx
k 2 A0
k 2 A0
)E
ay =
cos(!t * kz) ay = ✏
Ÿ E=
sin(!t * kz) ay V_m
)z
)t
! ✏
⌧
)A
)A
Ÿ (V = *
+E
)t
)t
4
5
k2
)V
(V = A0 ! 1 * 2
sin(!t * kz) ay =
a
)y y
! ✏
Integrating over y we find
E = *(V *
4
5
k2
V = A0 ! y 1 * 2
sin(!t * kz) + C
! ✏
˘
where C, the integration constant, can be taken as zero. In part b, it will be shown that k = !
✏,
which means that V = 0.
b) Specify k in terms of A0 , !, and the constants of the lossless medium, ✏ and . Use the other
Maxwell curl equation:
)B
)H
(ùE=*
=*
)t
)t
so that
k3 A
)H
1
1 )Ey
=* (ùE=
ax = * 2 0 cos(!t * kz) ax
)t
)z
! ✏
Integrate over t (and set the integration constant to zero) and require the result to be consistant
with part a:
k3 A
kA
H = * 2 20 sin(!t * kz) ax = * 0 sin(!t * kz) ax
! ✏
´≠≠≠≠≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠≠≠≠≠¨
f rom part a
We identify
˘
k=!
✏
193
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9.25. In a region where r = ✏r = 1 and = 0, the retarded potentials are given by V = x(z * ct) V and
˘
A = x[(z_c) * t]az Wb/m, where c = 1_
0 ✏0 .
a) Show that ( A = * ✏()V _)t):
First,
˘
)Az
x
= =x
0 ✏0
)z
c
( A=
Second,
)V
x
= *cx = * ˘
)t
0 ✏0
so we observe that ( A = * 0 ✏0 ()V _)t) in free space, implying that the given statement would
hold true in general media.
b) Find B, H, E, and D:
Use
⇠
⇡
)Ax
z
ay = t *
a T
)x
c y
B=(ùA=*
Then
Now,
Then
H=
E = *(V *
B
0
=
1
⇠
0
t*
⇡
z
a A_m
c y
)A
= *(z * ct)ax * xaz + xaz = (ct * z)ax V_m
)t
D = ✏0 E = ✏0 (ct * z)ax C_m2
c) Show that these results satisfy Maxwell’s equations if J and ⇢v are zero:
i. ( D = ( ✏0 (ct * z)ax = 0
ii. ( B = ( (t * z_c)ay = 0
iii.
(ùH=*
)Hy
)z
ax =
which we require to equal )D_)t:
)D
= ✏0 cax =
)t
iv.
(ùE=
which we require to equal *)B_)t:
1
0c
ax =
u
✏0
0
u
✏0
0
ax
ax
)Ex
a = *ay
)z y
)B
= ay
)t
So all four Maxwell equations are satisfied.
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9.26. Write Maxwell’s equations in point form in terms of E and H as they apply to a sourceless medium,
where J and ⇢v are both zero. Replace ✏ by , by ✏, E by H, and H by *E, and show that the
equations are unchanged. This is a more general expression of the duality principle in circuit theory.
Maxwell’s equations in sourceless media can be written as:
)H
)t
(1)
)E
)t
( ✏E = 0
(2)
(3)
H=0
(4)
(ùE=*
(ùH=✏
(
In making the above substitutions, we find that (1) converts to (2), (2) converts to (1), and (3)
and (4) convert to each other.
195
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CHAPTER 10 – 9th Edition
10.1. The parameters of a certain transmission line, operating at ! = 6.0 ù 108 rad/s are L = 0.350 H_m,
C = 40.0 pF/m, G = 0, and R = 15.0 ⌦/m. Evaluate ↵, , , and Z0 .
We use
=
=
˘
˘
ZY =
˘
(R + j!L)(G + j!C)
[15 + j(6 ù 108 )(0.350 ù 10*6 )][0 + j(6 ù 108 )(40 ù 10*12 )]
= 0.080 + j2.25 m*1 = ↵ + j
Therefore, ↵ = 0.080 Np_m,
u
Z0 =
v
Z
=
Y
= 2.25 rad_m, and
R + j!L
=
G + j!C
v
= 2⇡_ = 2.80 m. Finally,
15 + j2.1 ù 102
= 93.6 * j3.34 ⌦ = 93.67⁄ * 0.036 rad
0 + j2.4 ù 10*2
10.2. A sinusoidal wave on a transmission line is specified by voltage and current in phasor form:
Vs (z) = V0 e↵z ej z
and Is (z) = I0 e↵z ej z ej
where V0 and I0 are both real.
a) In which direction does this wave propagate and why? Propagation is in the backward z direction, because of the factor e+j z .
b) It is found that ↵ = 0, Z0 = 50 ⌦, and the wave velocity is vp = 2.5ù108 m/s, with ! = 108 s*1 .
Evaluate R, G, L, C, , and : First, the fact that ↵ = 0 means that the line is lossless,
˘ from
which we immediately conclude that R = G = 0. As this is true it follows that Z0 = L_C
˘
and vp = 1_ LC, from which
C=
Then
1
1
=
= 8.0 ù 10*11 F = 80 pF
Z0 vp
50(2.5 ù 108 )
L = CZ02 = (8.0 ù 10*11 )(50)2 = 2.0 ù 10*7 = 0.20 H
Now,
=
vp
f
=
2⇡vp
!
=
2⇡(2.5 ù 108 )
= 15.7 m
108
Finally, the current phase is found through
I0 ej =
Since V0 , I0 , and Z0 are all real, it follows that
V0
Z0
= 0.
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10.3. A voltage pulse propagates within a lossless transmission line of characteristic impedance Z0 = 50
ohms. The pulse is gaussian in shape, having voltage envelope given by
2
2
V (t) = V0 e*t _(2T )
where V0 = 10 V and T = 20 ns. The pulse is incident on a 100-ohm load at the far end of the line.
Determine the energy in Joules that is dissipated by the load.
The pulse is treated as a modulated dc voltage, so that the power over its envelope will be Pin (t) =
V (t)I(t) = V 2 (t)_Z0 . The power transferred to the load is PL (t) = Pin (t)(1 *  2 ), where =
(100 * 50)_(100 + 50) = 1_3. We find:
PL (t) =
⇡
(10)2 *t2 _T 2 ⇠
1
16 2 2
e
1*
= e*t _T W
50
9
9
The energy dissipated by the load is the time integral of the above, or
W =
ÿ
˘
16 *t2 _T 2
16
e
dt = (20 ù 10*9 ) ⇡ = 63 nJ
9
*ÿ 9
10.4. A sinusoidal voltage wave of amplitude V0 , frequency !, and phase constant, , propagates in the forward z direction toward the open load end in a lossless transmission line of characteristic impedance
Z0 . At the end, the wave totally reflects with zero phase shift, and the reflected wave now interferes
with the incident wave to yield a standing wave pattern over the line length (as per Example 10.1). Determine the standing wave pattern for the current in the line. Express the result in real instantaneous
form and simplify.
In phasor form, the forward and backward waves are:
VsT (z) = V0 e*j z + V0 ej z
The current is found from the voltage by dividing by Z0 (while incorporating the proper sign
for forward and backward waves):
IsT (z) =
V0 *j z
V
V
2V
e
* 0 ej z = * 0 ej z * e*j z = *j 0 sin( z)
Z0
Z0
Z0
Z0
The real instantaneous current is now
I(z, t) = Re IsT (z)e
=
j!t
h
i
n 2V0
n
= Re l*j
sin( z)[cos(!t) + j sin(!t)]m
Z
´≠≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠≠¨ n
0
n
j
k
ej!t
2V0
sin( z) sin(!t)
Z0
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10.5. Two voltage waves of equal amplitude, V0 , and frequency, !, propagate in the forward z direction in a
lossy transmission line having attenuation coefficient ↵, and characteristic impedance, Z0 = Z0 ej
One wave is phase-shifted from the other by radians.
a) Find an expression for the net voltage wave formed by the superposition of the two voltages.
Your result should be a single wave function in real instantaneous form: To do this, express
both given waves in phasor form: The total phasor voltage will be
VsT (z) = V0 e*↵z e*j z + V0 e*↵z e*j z ej
= V0 e*↵z e*j z 1 + ej
= V0 e*↵z e*j z ej _2 e*j _2 + ej _2
= 2V0 cos( _2)ej _2 e*↵z e*j z
In real instantaneous form, this becomes:
V(z, t) = Re VsT (z)ej!t = 2V0 cos( _2)e*↵z cos(!t * z + _2) V
b) Find an expression for the net current in the line, again in the form of a single wave function:
Since the net voltage wave propagates in the forward z direction, we find the current just by
dividing the phasor voltage of part a by Z0 ej :
IsT (z) =
So that
2V0
cos( _2)ej( _2* ) e*↵z e*j z
Z0 
I(z, t) = Re IsT (z)ej!t =
2V0 *↵z
e cos( _2) cos(!t * z + _2 * ) A
Z0 
c) Find an expression for the average power in the line. This is found using the phasor expressions
for the net voltage and current:
1
<
P = Re VsT IsT
2
=
2V02
Z0 
cos2 ( _2) cos e*2↵z W
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10.6. A 50-ohm load is attached to a 50m section of the transmission line of Problem 10.1, and a 100-W
signal is fed to the input end of the line.
a) Evaluate the distributed line loss in dB/m: From Problem 10.1 (or from the answer in Appendix
F) we have ↵ = 0.080 Np/m. Then
Loss[dB_m] = 8.69↵ = 8.69(0.080) = 0.70 dB_m
b) Evaluate the reflection coefficient at the load: We need the characteristic impedance of the
line. Again, in solving Problem 10.1 (or looking up the answer in the appendix), we have
Z0 = 93.6 * j3.34 ohms. The reflection coefficient is
L =
ZL * Z0
50 * (93.6 * j3.34)
=
= *0.304 + j0.016 = 0.304⁄177˝
ZL + Z0
50 + (93.6 * j3.34)
c) Evaluate the power that is dissipated by the load resistor: This will be
⌅
⇧
Pd = 100W ù e*2↵L ù (1 *  L 2 ) = 100 e*2(0.080)(50) 1 * (0.304)2 = 0.30 W = 30 mW
d) What power drop in dB does the dissipated power in the load represent when compared to the
original input power? This we find as a positive number through
4 5
⌧
P
100
Pd [dB] = 10 log10 in = 10 log10
= 35.2 dB
Pd
0.030
e) On partial reflection from the load, how much power returns to the input and what dB drop does
this represent when compared to the original 100-W input power? After one round trip plus a
reflection at the load, the power returning to the input is expressed as
Pout = Pin ù e*2↵(2L) ù  L 2 = 100 e*200(0.080) (0.304)2 = 1.04 ù 10*6 W = 1.04 mW
As a decibel reduction from the original input power, this becomes
4
5
4
5
Pin
100
Pout [dB] = 10 log10
= 10 log10
= 79.8 dB
Pout
1.04 ù 10*6
10.7. A transmitter and receiver are connected using a cascaded pair of transmission lines. At the operating
frequency, Line 1 has a measured loss of 0.1 dB/m, and Line 2 is rated at 0.2 dB/m. The link is
composed of 40m of Line 1, joined to 25m of Line 2. At the joint, a splice loss of 2 dB is measured.
If the transmitted power is 100mW, what is the received power?
The total loss in the link in dB is 40(0.1) + 25(0.2) + 2 = 11 dB. Then the received power is
Pr = 100mW ù 10*0.1(11) = 7.9 mW.
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10.8. An absolute measure of power is the dBm scale,⌅in which power⇧ is specified in decibels relative to
one milliwatt. Specifically, P (dBm) = 10 log10 P (mW)_1 mW . Suppose that a receiver is rated
as having a sensitivity of *20 dBm, indicating the mimimum power that it must receive in order to
adequately interpret the transmitted electronic data. Suppose this receiver is at the load end of a 50ohm transmission line having 100-m length and loss rating of 0.09 dB/m. The receiver impedance is
75 ohms, and so is not matched to the line. What is the minimum required input power to the line in
a) dBm, b) mW?
Method 1 – using decibels: The total loss in dB will be the sum of the transit loss in the line and
the loss arising from partial transmission into the load. The latter will be
0
1
1
Lossload [dB] = 10 log10
1 *  L 2
where
L = (75 * 50)_(75 + 50) = 0.20. So
0
Lossload = 10 log10
The transit loss will be
Losstrans = 10 log10
⇠
1
e*2↵L
⇡
1
1 * (0.20)2
1
= 0.18 dB
= (0.09 dB_m)(100 m) = 9.0 dB
The total loss in dB is then Losstot = 9.0 + 0.18 = 9.2 dB. The minimum required input power
is now
Pin [dBm] = *20 dBm + 9.2 dB = *10.8 dBm
In milliwatts, this is
Pin [mW] = 10*1.08 = 8.3 ù 10*2 mW = 83 W
Method 2 – using loss factors: The 0.09 dB/m line loss corresponds to an exponential voltage
attenuation coefficient of ↵ = 0.09_8.69 = 1.04 ù 10*2 Np/m. Now, the power dropped at the
load will be
⌅
⇧⌅
⇧
Pload = Pin e*2↵L (1 *  L 2 ) = Pin exp *2(1.04 ù 10*2 )(100) 1 * (0.2)2 = 0.12Pin
Since the minimum power at the load of -20 dBm in mW is 10*2 , the minimum input power
will be
10*2
Pin [mW] =
= 8.3 ù 10*2 mW = 83 W as before
0.12
In dBm this is Pin = 10 log10 8.3 ù 10*2 = *10.8 dBm.
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10.9. A 100-m transmission line is used to propagate a signal from a transmitter to a receiver unit whose
input impedance is 50 ohms. The transmitter is capable of launching 12 dBm average power onto
the input end of the line (see Problem 10.8 for the definition of dBm). The line is lossy, and has
characteristic impedance, Z0 = 75 + j10 ohms. The loss coefficient for the line is A = 0.05 dB/m.
Find the power at the receiver in both dBm and mW: First we find the reflection coefficient at the
load:
Z * Z0
50 * 75 * j10 *(25 + j10)
= L
=
=
ZL + Z0
50 + 75 + j10
125 + j10
Then the fraction of the power just before the load that is reflected from it will be:
0
10
1
*(25 + j10)
*(25 * j10)
2
  =
= 0.046
125 + j10
125 * j10
and the fraction of the power just in front of the load that is transferred to the load is 1 *  2 =
1 * 0.046 = 0.954. The loss this represents in dB is then
0
1
⇠
⇡
1
1
10 log10
=
10
log
= 0.205 dB
10
0.954
1 *  2
The net loss from input to load in dB is now:
net loss = A L + 0.205 = 0.05(100) + 0.205 = 5.2 dB
The power that enters the receiver is then
Pr [dBm] = Pin [dBm] * net loss[dB] = 12.0 * 5.2 = 6.8 dBm
In mW, this is:
Pr [mW] = 100.68 = 4.8 mW
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10.10. Two lossless transmission lines having different characteristic impedances are to be joined end-toend. The impedances are Z01 = 100 ohms and Z03 = 25 ohms. The operating frequency is 1 GHz.
a) Find the required characteristic impedance, Z02 , of a quarter-wave section to be inserted between the two, which will impedance-match the joint, thus allowing
˘transmission
˘ total power
through the three lines: The required inpedance will be Z02 = Z01 Z03 = (100)(25) =
50 ohms.
b) The capacitance per unit length of the intermediate line is found to be 100 pF/m. Find the
shortest length in meters of this line that is needed to satisfy the impedance-matching condition:
For the lossless intermediate line,
v
˘
L2
2
Z02 =
Ÿ L2 = C2 Z02
Then 2 = ! L2 C2 = 2⇡f C2 Z02
C2
The line length at _4 (the shortest length that will work) is then
0 1
1 2⇡
1
1
2
l2 =
=
=
=
= 0.05 m
9
4
4
4f C2 Z02
(4 ù 10 )(10*10 )(50)
2
c) With the three-segment setup as found in parts a and b, the frequency is now doubled to 2 GHz.
Find the input impedance at the Line 1-to-Line 2 junction, seen by waves incident from Line 1:
With the frequency doubled, the wavelength is cut in half, which means that the intermediate
section is now a half-wavelength long. In that case, the input impedance is just the impedance
of the far line, or Zin = Z03 = 25 ohms.
d) Under the conditions of part c, and with power incident from Line 1, evaluate the standing wave
ratio that will be measured in Line 1, and the fraction of the incident power from Line 1 that is
reflected and propagates back to the Line 1 input. The reflection coefficient at the junction is
in = (25 * 100)_(25 + 100) = *3_5. So the VSWR = (1 + 3_5)_(1 * 3_5) = 4. The fraction
of the power reflected at the junction is  2 = (3_5)2 = 0.36, or 36%.
202
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10.11. Two voltage waves of equal amplitude, V0 , and which have differing frequencies, ! and 3! (with corresponding phase constants, and 3 ) propagate in the forward z direction in a lossless transmission
line. Find an expression for the net voltage wave formed by the superposition of the given voltages.
To do this, express both given waves in complex instantaneous form. Combine the two algebraically,
and convert the result back to real instantaneous form to produce a single wave function composed
of the product of two cosines. Plot your result as a function of z at t = 0.
The total complex instantaneous voltage will be:
VcT (z, t) = V0 ej!t e*j z + V0 ej3!t e*j3 z
= V0 ej2!t e*j2 z e*j!t e+j z + e+j!t e*j z
= 2V0 ej2!t e*2j z cos (!t * z)
In real instantaneous form, this becomes:
V(z, t) = Re VcT (z, t) = 2V0 cos(!t * z) cos(2!t * 2 z) V
This function (with normalized amplitude) is plotted below as a function of z.
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10.12. In a circuit in which a sinusoidal voltage source drives its internal impedance in series with a load
impedance, it is known that maximum power transfer to the load occurs when the source and load
impedances form a complex conjugate pair. Suppose the source (with its internal impedance) now
drives a complex load of impedance ZL = RL + jXL that has been moved to the end of a lossless
transmission line of length l having characteristic impedance Z0 . If the source impedance is Zg =
Rg + jXg , write an equation that can be solved for the required line length, l, such that the displaced
load will receive the maximum power.
The condition of maximum power transfer will be met if the input impedance to the line is the
conjugate of the internal impedance. Using Eq. (98), we write
4
5
(RL + jXL ) cos( l) + jZ0 sin( l)
Zin = Z0
= Rg * jXg
Z0 cos( l) + j(RL + jXL ) sin( l)
This is the equation that we have to solve for l – assuming that such a solution exists. To find
out, we need to work with the equation a little. Multiplying both sides by the denominator of
the left side gives
Z0 (RL + jXL ) cos( l) + jZ02 sin( l) = (Rg * jXg )[Z0 cos( l) + j(RL + jXL ) sin( l)]
We next separate the equation by equating the real parts of both sides and the imaginary parts
of both sides, giving
(RL * Rg ) cos( l) = *
and
(XL + Xg ) cos( l) =
(Rg XL + RL Xg )
Z0
Rg RL * Xg XL * Z02
Z0
sin( l) (real parts)
sin( l) (imaginary parts)
Using the two equations, we find two conditions on the tangent of l:
tan( l) =
Z0 (Rg * RL )
Rg XL + RL Xg
=
Z0 (XL + Xg )
Rg RL * Xg XL * Z02
For a viable solution to exist for l, both equalities must be satisfied, thus limiting the possible
choices of the two impedances.
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10.13. The skin effect mechanism in transmission lines is responsible for the increase with frequency of of the
line resistance per meter, R. Specifically, resistance scales as the square root of frequency f according
to R = A0 f 1_2 , where A0 is a constant. Consider a low loss line in which the attenuation coefficient
is approximated by Eq. (54a), the phase constant by the first term in (54b), and the characteristic
impedance by Eq. (24). The conductance per unit length, G, is zero. The 50-ohm line has a measured
power loss of 10.0 dB over a 100-m length at f = 100 MHz.
a) Find the value of A0 and the line resistance per meter at 100 MHz: The approximations to be
used are
u
u
. 1
.
. ˘
C
L
↵= R
Z0 =
and
= ! LC
2
L
C
Let l be the line length. We then may write:
u
1 ˘
C
8.69 ˘ 8 100
10 dB = 8.69↵l = 8.69 A0 f
l=
A 10
= 8.69 ù 104 A0
2
L
2 0
Z0
So A0 = (10_8.69) ù 10*4 = 1.15 ù 10*4
˘
and therefore at 100 MHz, R = 1.15 ù 10*4 108 = 1.15 ohms_m
b) At the line input, a 10-W power transmitter is attached. At the far end of the line, a 100-ohm
load impedance˘is attached. How much power is dissipated by the load at frequency 400 MHz?
As ↵ scales as f , changing f from 100 to 400 MHz doubles ↵ as well as the dB loss per unit
length. So the new one-way transit loss at 400 MHz is 20dB instead of 10. At the load, the
reflection coefficient is
100 * 50 1
=
=
100 + 50 3
and so the fraction of power incident on the load that is transferred to it is 1 *  2 = 8_9. This
represents a dB reflective loss of 10 log10 (9_8) = 0.5 dB. So now the total dB power loss from
transmitter to load is 20 + 0.5 = 20.5 dB. The power delivered to (dissipated by) the load is thus
PL = 10W ù 10*2.05 = 89 mW
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10.14. A lossless transmission line having characteristic impedance Z0 = 50 ohms is driven by a source
at the input end that consists of the series combination of a 10-V sinusoidal generator and a 50-ohm
resistor. The line is one-quarter wavelength long. At the other end of the line, a load impedance,
ZL = 50 * j50 ohms is attached.
a) Evaluate the input impedance to the line seen by the voltage source-resistor combination: For
a quarter-wave section,
Z02
Zin =
=
ZL
(50)2
= 25 + j25 ohms
50 * j50
b) Evaluate the power that is dissipated by the load: This will be the same as the power dissipated
by Zin , assuming we replace the line-load section by a lumped element of impedance Zin . The
voltage across Zin will be
4
5
Zin
25 + j25
Vin = Vs0
= 10
= 4 + j2
Zg + Zin
50 + 25 + j25
The power will be
1
Pin = PL = Re
2
<
Vin Vin<
=
<
Zin
1
= Re
2
<
(4 + j2)(4 * j2)
25 * j25
=
= 0.2 W
c) Evaluate the voltage amplitude that appears across the load: The phasor voltage at any point in
the line is given by the sum of forward and backward waves:
Vs (z) = V0+ e*j z + V0* e+j z
where V0* =
+
L V0 , and where
L =
ZL * Z0
50 * j50 * 50
=
= 0.2 * j0.4
ZL + Z0
50 * j50 + 50
By our convention, the load is located at z = 0. The voltage at the line input, Vin , is therefore
given by the above voltage expression evaluated at z = *l, where l = * _4. Thus l = ⇡_2,
and
⌅
⇧
Vin = Vs (*l) = V0+ ej l + L e*j l = V0+ [j + (0.2 * j0.4)(*j)] = V0+ (*0.4 + j0.8)
Using Vin from part b, we have
V0+ =
Now, the voltage at the load will be
VL = V0+ (1 +
As a check,
1
PL = Re
2
<
L) =
VL VL<
ZL<
(4 + j2)
(*0.4 + j0.8)
(4 + j2)
(1 + 0.2 * j0.4) = *2 * j6 V
(*0.4 + j0.8)
=
1
= Re
2
<
(*2 * j6)(*2 + j6)
50 + j50
=
= 0.2 W
which is in agreement with part b.
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10.15. For the transmission line represented in Fig. 10.29, find Vs,out if f =:
a) 60 Hz: At this frequency,
=
!
2⇡ ù 60
=
= 1.9 ù 10*6 rad_m So l = (1.9 ù 10*6 )(80) = 1.5 ù 10*4 << 1
vp
(2_3)(3 ù 108 )
.
The line is thus essentially a lumped circuit, where Zin = ZL = 80 ⌦. Therefore
Vs,out = 120
⌧
80
= 104 V
12 + 80
b) 500 kHz: In this case
=
Now
2⇡ ù 5 ù 105
= 1.57 ù 10*2 rad_s So l = 1.57 ù 10*2 (80) = 1.26 rad
2 ù 108
4
5
80 cos(1.26) + j50 sin(1.26)
Zin = 50
= 33.17 * j9.57 = 34.5⁄ * .28
50 cos(1.26) + j80 sin(1.26)
The equivalent circuit is now the voltage source driving the series combination of Zin and the
12 ohm resistor. The voltage across Zin is thus
4
5
4
5
Zin
33.17 * j9.57
Vin = 120
= 120
= 89.5 * j6.46 = 89.7⁄ * .071
12 + Zin
12 + 33.17 * j9.57
The voltage at the line input is now the sum of the forward and backward-propagating waves
just to the right of the input. We reference the load at z = 0, and so the input is located at
z = *80 m. In general we write Vin = V0+ e*j z + V0* ej z , where
V0* =
At z = *80 m we thus have
⌧
3
Vin = V0+ ej1.26 + e*j1.26
13
80 * 50 +
+
V0 =
L V0 =
80 + 50
Ÿ V0+ =
3 +
V
13 0
89.5 * j6.46
ej1.26 + (3_13)e*j1.26
= 42.7 * j100 V
Now
Vs,out = V0+ (1 +
L ) = (42.7 * j100)(1 + 3_(13)) = 134⁄ * 1.17 rad = 52.6 * j123 V
As a check, we can evaluate the average power reaching the load:
2
1 Vs,out 
1 (134)2
Pavg,L =
=
= 112 W
2 RL
2 80
This must be the same power that occurs at the input impedance:
1
1
<
Pavg,in = Re Vin Iin
= Re {(89.5 * j6.46)(2.54 + j0.54)} = 112 W
2
2
where Iin = Vin _Zin = (89.5 * j6.46)_(33.17 * j9.57) = 2.54 + j0.54.
207
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
10.16. A 100-⌦ lossless transmission line is connected to a second line of 40-⌦ impedance, whose length is
_4. The other end of the short line is terminated by a 25-⌦ resistor. A sinusoidal wave (of frequency
f ) having 50 W average power is incident from the 100-⌦ line.
a) Evaluate the input impedance to the quarter-wave line: For the quarter-wave section,
Zin =
2
Z02
=
ZL
(40)2
= 64 ohms
25
b) Determine the steady state power that is dissipated by the resistor: This will be the same as the
power dropped across a lumped element of impedance Zin at the junction, which replaces the
termimated 40-ohm line. The reflection coefficient at the junction is
in =
Zin * Z01
64 * 100
9
=
=*
Zin + Z01
64 + 100
41
The dissipated power there is then
2
Pin = PL = 50 1 *  in 
0
⇠ ⇡2 1
9
= 50 1 *
= 47.6 W
41
c) Now suppose the operating frequency is lowered to one-half its original value. Determine the
® , for this case: Halving the frequency doubles the wavelength, so that
new input impedance, Zin
now the 40-ohm section is of length l = _8. l is now ⇡_4, and the input impedance, from
Eq. (98) is:
4
5
25 cos(⇡_4) + j40 sin(⇡_4)
®
Zin = 40
= 36.0 + j17.5 ohms
40 cos(⇡_4) + j25 sin(⇡_4)
d) For the new frequency, calculate the power in watts that returns to the input end of the line after
reflection: The new reflection coefficient is
®
in =
® *Z
Zin
01
® +Z
Zin
01
=
36.0 + j17.5 * 100
= *0.447 + j0.186
36.0 + j17.5 + 100
The reflected power (all of which returns to the input) is
Pref = 50 ®in 2 = 50(0.234) = 11.7 W
208
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
10.17. Determine the average power absorbed by each resistor in Fig. 10.30: The problem is made easier by
first converting the current source/100 ohm resistor combination to its Thevenin equivalent. This is
a 50⁄0 V voltage source in series with the 100 ohm resistor. The next step is to determine the input
impedance of the 2.6 length line, terminated by the 25 ohm resistor: We use l = (2⇡_ )(2.6 ) =
16.33 rad. This value, modulo 2⇡ is (by subtracting 2⇡ twice) 3.77 rad. Now
4
5
25 cos(3.77) + j50 sin(3.77)
Zin = 50
= 33.7 + j24.0
50 cos(3.77) + j25 sin(3.77)
The equivalent circuit now consists of the series combination of 50 V source, 100 ohm resistor, and
Zin , as calculated above. The current in this circuit will be
I=
50
= 0.368⁄ * .178
100 + 33.7 + j24.0
The power dissipated by the 25 ohm resistor is the same as the power dissipated by the real part of
Zin , or
1
1
P25 = P33.7 = I2 R = (.368)2 (33.7) = 2.28 W
2
2
To find the power dissipated by the 100 ohm resistor, we need to return to the Norton configuration,
with the original current source in parallel with the 100 ohm resistor, and in parallel with Zin . The
voltage across the 100 ohm resistor will be the same as that across Zin , or
V = IZin = (.368⁄ * .178)(33.7 + j24.0) = 15.2⁄0.44. The power dissipated by the 100 ohm
resistor is now
1 V 2
1 (15.2)2
P100 =
=
= 1.16 W
2 R
2 100
209
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
10.18 The line shown in Fig. 10.31 is lossless. Find s on both sections 1 and 2: For section 2, we consider
the propagation of one forward and one backward wave, comprising the superposition of all reflected
waves from both ends of the section. The ratio of the backward to the forward wave amplitude is
given by the reflection coefficient at the load, which is
50 * j100 * 50
*j
1
=
= (1 * j)
50 * j100 + 50 1 * j
2
˘
˘
Then  L  = (1_2) (1 * j)(1 + j) = 1_ 2. Finally
L =
˘
1 +  L  1 + 1_ 2
s2 =
=
˘ = 5.83
1 *  L  1 * 1_ 2
For section 1, we need the reflection coefficient at the junction (location of the 100 ⌦ resistor) seen
by waves incident from section 1: We first need the input impedance of the .2 length of section 2:
4
5
4
5
(50 * j100) cos( 2 l) + j50 sin( 2 l)
(1 * j2)(0.309) + j0.951
Zin2 = 50
= 50
50 cos( 2 l) + j(50 * j100) sin( 2 l)
0.309 + j(1 * j2)(0.951)
= 8.63 + j3.82 = 9.44⁄0.42 rad
Now, this impedance is in parallel with the 100⌦ resistor, leading to a net junction impedance
found by
1
1
1
=
+
Ÿ ZinT = 8.06 + j3.23 = 8.69⁄0.38 rad
ZinT
100 8.63 + j3.82
The reflection coefficient will be
j =
ZinT * 50
= *0.717 + j0.096 = 0.723⁄3.0 rad
ZinT + 50
and the standing wave ratio is s1 = (1 + 0.723)_(1 * 0.723) = 6.22.
210
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
10.19. A lossless transmission line is 50 cm in length and operating at a frequency of 100 MHz. The line
parameters are L = 0.2 H_m and C = 80 pF/m. The line is terminated by a short circuit at z = 0,
and there is a load, ZL = 50 + j20 ohms across the line at location z = *20 cm. What average power
is delivered to ZL if the input voltage is 100⁄0 V? With the given capacitance and inductance, we
find
u
u
L
2 ù 10*7
Z0 =
=
= 50 ⌦
C
8 ù 10*11
and
1
1
vp = ˘
=˘
= 2.5 ù 108 m_s
(2 ù 10*7 )(9 ù 10*11 )
LC
Now = !_vp = (2⇡ ù 108 )_(2.5 ù 108 ) = 2.5 rad/s. We then find the input impedance to the
shorted line section of length 20 cm (putting this impedance at the location of ZL , so we can combine
them): We have l = (2.5)(0.2) = 0.50, and so, using the input impedance formula with a zero
load impedance, we find Zin1 = j50 tan(0.50) = j27.4 ohms. Now, at the location of ZL , the
net impedance there is the parallel combination of ZL and Zin1 : Znet = (50 + j20)(j27.4) =
7.93 + j19.9. We now transform this impedance to the line input, 30 cm to the left, obtaining (with
l = (2.5)(.3) = 0.75):
4
5
(7.93 + j19.9) cos(.75) + j50 sin(.75)
Zin2 = 50
= 35.9 + j98.0 = 104.3⁄1.22
50 cos(.75) + j(7.93 + j19.9) sin(.75)
The power delivered to ZL is the same as the power delivered to Zin2 : The current magnitude is
I = (100)_(104.3) = 0.96 A. So finally,
1
1
P = I2 R = (0.96)2 (35.9) = 16.5 W
2
2
211
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
10.20
a) Determine s on the transmission line of Fig. 10.32. Note that the dielectric is air: The reflection
coefficient at the load is
L =
40 + j30 * 50
= j0.333 = 0.333⁄1.57 rad
40 + j30 + 50
Then s =
1 + .333
= 2.0
1 * .333
b) Find the input impedance: With the length of the line at 2.7 , we have l = (2⇡)(2.7) = 16.96
rad. The input impedance is then
4
5
4
5
(40 + j30) cos(16.96) + j50 sin(16.96)
*1.236 * j5.682
Zin = 50
= 50
= 61.8 * j37.5 ⌦
50 cos(16.96) + j(40 + j30) sin(16.96)
1.308 * j3.804
c) If !L = 10 ⌦, find Is : The source drives a total impedance given by Znet = 20 + j!L + Zin =
20 + j10 + 61.8 * j37.5 = 81.8 * j27.5. The current is now Is = 100_(81.8 * j27.5) =
1.10 + j0.37 A.
d) What value of L will produce a maximum value for Is  at ! = 1 Grad/s? To achieve this,
the imaginary part of the total impedance of part c must be reduced to zero (so we need an
inductor). The inductor impedance must be equal to negative the imaginary part of the line
input impedance, or !L = 37.5, so that L = 37.5_! = 37.5 nH. Continuing, for this value of
L, calculate the average power:
e) supplied by the source: Ps = (1_2)Re{Vs Is< } = (1_2)(100)(1.10) = 55.0 W.
f) delivered to ZL = 40 + j30 ⌦: The power delivered to the load will be the same as the power
delivered to the input impedance. We write
1
1
PL = Re{Zin }Is 2 = (61.8)[(1.10 + j.37)(1.10 * j.37)] = 41.6 W
2
2
10.21. A lossless line having an air dielectric has a characteristic impedance of 400 ⌦. The line is operating
at 200 MHz and Zin = 200 * j200 ⌦. Use analytic methods or the Smith chart (or both) to find: (a)
s; (b) ZL if the line is 1 m long; (c) the distance from the load to the nearest voltage maximum: I will
first use the analytic approach. Using normalized impedances, Eq. (13) becomes
4
5 4
5
Zin
zL cos( L) + j sin( L)
zL + j tan( L)
zin =
=
=
Z0
cos( L) + jzL sin( L)
1 + jzL tan( L)
Solve for zL :
4
zin * j tan( L)
zL =
1 * jzin tan( L)
5
where, with = c_f = 3 ù 108 _2 ù 108 = 1.50 m, we find L = (2⇡)(1)_(1.50) = 4.19, and so
tan( L) = 1.73. Also, zin = (200 * j200)_400 = 0.5 * j0.5. So
zL =
0.5 * j0.5 * j1.73
= 2.61 + j0.174
1 * j(0.5 * j0.5)(1.73)
Finally, ZL = zL (400) = 1.04 ù 103 + j69.8 ⌦. Next
=
ZL * Z0
6.42 ù 102 + j69.8
= .446 + j2.68 ù 10*2 = .447⁄6.0 ù 10*2 rad
=
ZL + Z0
1.44 ù 103 + j69.8
212
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
10.21. (continued) Now
s=
1 +   1 + .447
=
= 2.62
1 *   1 * .447
=*
(6.0 ù 10*2 )(1.50)
= *7.2 ù 10*3 m = *7.2 mm
4⇡
Finally
zmax = *
2
=*
4⇡
We next solve the problem using the Smith chart. Referring to the figure below, we first locate and
mark the normalized input impedance, zin = 0.5 * j0.5. A line drawn from the origin through
this point intersects the outer chart boundary at the position 0.0881 on the wavelengths toward
load (WTL) scale. With a wavelength of 1.5 m, the 1 meter line is 0.6667 wavelengths long. On
the WTL scale, we add 0.6667 , or equivalently, 0.1667 (since 0.5 is once around the chart),
obtaining (0.0881 + 0.1667) ) = 0.2548 , which is the position of the load. A straight line is now
drawn from the origin though the 0.2548 position. A compass is then used to measure the distance
between the origin and zin . With this distance set, the compass is then used to scribe off the same
distance from the origin to the load impedance, along the line between the origin and the 0.2548
position. That point is the normalized load impedance, which is read to be zL = 2.6 + j0.18. Thus
ZL = zL (400) = 1040+j72. This is in reasonable agreement with the analytic result of 1040+j69.8.
The difference in imaginary parts arises from uncertainty in reading the chart in that region.
In transforming from the input to the load positions, we cross the r > 1 real axis of the chart at r=2.6.
This is close to the value of the VSWR, as we found earlier. We also see that the r > 1 real axis (at
which the first Vmax occurs) is a distance of 0.0048 (marked as .005 on the chart) in front of the
load. The actual distance is zmax = *0.0048(1.5) m = *0.0072 m = *7.2 mm.
Problem 10.21
213
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10.22. A lossless 50-ohm line is terminated by an unknown load impedance. A VSWR of 5.0 is measured,
and the first voltage minimum occurs at a 0.10 wavelengths in front of the load. Using the Smith
chart, find
a) The load impedance: Referring to the Smith chart section below, first mark the VSWR on the
positive real axis and set the compass to that length. The voltage maximum will be at that
location, and will have normalized impedance equal to the the VSWR, or 5.0. The voltage
minimum will be found halfway around the chart, on the negative real axis. Set the compass
to the distance of r = VSWR = 5 from the origin, and mark off that distance on the negative
real axis, to find r = 1_VSWR = 0.20 there (shown on the chart as the point farthest to the
left). Now, from r = 0.2, move counter-clockwise (toward the load) by a distance of 0.10
wavelengths (using the wavelengths toward load scale). The angled line is drawn from the origin
through the 0.10 mark on the scale. Use the compass (set to the VSWR length) to scribe the
point on the angled line that is labeled zL . We identify that as the normalized load impedance,
zL = 0.30 * j0.68. The load impedance is then ZL = 50zL = 15.0 * j34.0 ohms
b) The magnitude and phase of the reflection coefficient: The magnitude of L can be found by
measuring the compass span on the linear “Ref. coeff. E or I” scale on the bottom of the chart.
Set the compass point at the center position, and then scribe on the scale to the left to find
 L  = 0.67. The phase is the angle of the line from the positive real axis, which is read from
the “angle of reflection coefficient” scale as = *108˝ . In summary, L = 0.67⁄ * 108˝
c) The shortest length of line necessary to achieve an entirely resistive input impedance: In moving
toward the generator from the load, we look for the first real axis crossing. This occurs simply
at the Vmin location, and so we identify the shortest length as just 0.10 .
45
1.4
1.2
1.0
50
0.9
55
0.8
1.6
(+
jX
/Z
20
N
75
T
3.0
EC
OM
4
PO
NE
0.6
4.0
TA
NC
0.8
VE
R
15
5.0
1.0
EA
C
1.0
C TI
85
0.47
160
0.2
10
IN D
U
0.6
90
25
0.4
0.8
10
0.1
0.49
0.4
ER A
TO
R
80
>
0.0
0.4
6
15
0
0.3
G EN
2.0
65
0.5
0.0
6
0.4
4
70
0
5
14
0.4
5
0.0
0.4
A RD
0.1
8
0.3
2
50
20
50
20
10
55.0
44.0
33.0
22.0
11.8
11.4
AN
PT
1.0
CT
DU
IN
,O
o)
2.0
1.8
1.6
ITI
VE
1.0
0.9
1.2
0.36
-90
0.12
0.13
0.38
0.37
0.11
-10
1000
0.39
-1110
00..1
RE
A
CT
AN
0.0
9
-1
20
0.0
8
-1
40
(-j
-70
T
6
EN
0.0
N
0.6
AC
0.7
0.14
-80
-4
0
-75
R
1.4
0.15
0.35
0
-70
5
-4
4
-5
6
0.3
0.88
0.1
-35
-55
3
CA P
-60
7
0.3
-60
PO
5
0.0
Z
X/
0.1
OM
<
CE
US
ES
IV
0.22
-30
2
CE
C
0
-65 .5
0.3
1
4
9
0.4
0.1
0.3
)
o
jB/ Y
0.4
0
8
0.1
0
-5
-25
-90
(CE
0.6
-20
4
0.0
0
-15 -80
0.8
5
0.4
0.3
-4
0.10 lambda WTL
0.2
0.4
3.0
0.47
1.0
4.0
9
6
0.4
-15
0.2
j0.68
0.28
zL
0.2
1
-30
0.3
5.0
-160
-85
0.8
0.22
WA
11.6
11.0
11.2
00.9
00.8
00.7
00.6
00.5
00.4
00.3
00.2
0.1
± 180
10
0.6
-20
0.2
G TH
VSWR = 5
0.4
-10
AD <
A RD LO
S TO W
-170
00.2
0.1
EN
VEL
50
20
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE
ONDUCTANCE COMPONENT (G/Yo)
50
0.0
0.2
0.48
0.25
0.26
0.24
0.27
0.23
0.25
0.24
0.26
0.23
O
0.27
REFLEC TIO N C EFFIC IEN T IN D EG
REES
LE O F
AN G
SSIO N C O EFFIC IEN T IN
TRA N SM I
D EG R
LE O F
EES
AN G
0.48
7
3
20
TO W
0.2
0.1
0.3
30
0.28
G TH S
60
0.22
170
4
1
0.2
9
0.2
30
> W A VELEN
CI
)
0.2
0.0
PA
0.3
35
0.3
0.49
CA
R
NC
0.1
6
70
40
,O
TA
EP
SC
SU
VE
TI
Yo
jB/
E (+
0.35
40
1
0.3
o)
120
0.15
0.36
80
9
0.1
3
0.4
0
13
2
0.4
0.14
0.37
90
1.8
7
0.0
0.6 60
0
110
1
0.4
0.7
0
.08
0.38
0.39
100
0.4
0.13
0.12
0.11
0.1
.09
0.0
7
-1
30
0.4
3
0.4
2
1
0.4
00..4
214
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
10.22. (alternate version) A lossless 50-ohm line is terminated by an unknown load impedance. A VSWR
of 5.0 is measured, and the first voltage maximum occurs at a 0.10 wavelengths in front of the load.
Using the Smith chart, find
a) The load impedance: Referring to the Smith chart section below, first mark the VSWR on the
positive real axis and set the compass to that length. The voltage maximum will be at that
location, and will have normalized impedance equal to the the VSWR, or 5.0. Now, move
toward the load by a distance of 0.10 wavelengths (using the wavelengths toward load scale).
The angled line is drawn from the origin through the 0.35 mark on the scale. Use the compass
(set to the VSWR length) to scribe the point on the dashed line that is labeled zL . We identify
that as the normalized load impedance, zL = 0.54 + j1.23. The load impedance is then ZL =
50zL = 27 + j62 ohms
b) The magnitude and phase of the reflection coefficient: The magnitude of L can be found by
measuring the compass span on the linear “Ref. coeff. E or I” scale on the bottom of the chart.
Set the compass point at the center position, and then scribe on the scale to the left to find
 L  = 0.67. The phase is the angle of the line from the positive real axis, which is read from
the “angle of reflection coefficient” scale as = +72˝ . In summary, L = 0.67⁄ + 72˝
c) The shortest length of line necessary to achieve an entirely resistive input impedance: In moving
toward the generator from the load, we look for the first real axis crossing. This occurs simply
at the Vmax location, and so we identify the shortest length as just 0.10 .
45
11.4
1.2
1.0
50
0.9
0.8
55
1.6
0.7
2.0
(+
jX
/Z
T
N
75
NE
PO
OM
4
0.88
EC
>
4.0
TA
NC
80
R
TO
20
3.0
0.6
VE
R
1.0
5.0
10
IN D
U
0.6
.6
10
0.1
0.49
00.4
20
)
/ Yo
(-jB
50
20
10
55.0
44.0
33.0
22.0
11.8
11.4
CE
1.0
CE
US
ES
IV
CT
DU
IN
,O
o)
2.0
1.8
1.6
CA P
1.4
1.2
1.0
0.8
0.9
0.14
-80
-4
0
0.36
5
-4
0.15
0.35
0
-70
-5
4
0.3
6
-90
0.12
0.13
0.38
0.37
0.11
-100
0.39
-55
-35
0.1
AC
ITI
VE
-110
0.1
RE
A
CT
AN
-60
-60
0.7
3
0.3
7
-75
R
0.2
-30
0.1
0.0
9
-12
0
0.0
8
40
(-j
-70
T
-1
EN
6
N
0.6
2
PO
5
0.0
Z
X/
0.3
OM
0
-65 .5
0.3
1
CE
C
0.0
9
4
0.1
0.4
0.4
0
8
0.1
0
-5
-25
<
AN
PT
0.6
-20
3.0
5
0.3
-4
0.4
0.2
0.4
0.8
9
6
0.4
0.2
4
0.0
0
-15 -80
1.0
0.47
0.28
0.22
0.2
1
-30
0.3
4.0
-85
0.8
-15
-90
11.6
11.0
11.2
0.9
0.8
0.7
0.6
0.4
0.5
0.3
0.2
0.1
± 180
0.6
-20
0.2
5.0
D LO A D <
OWAR
-170
VSWR = 5
0.4
10
0.48
0.2
0.1
-10
HS T
N GT
50
50
0.0
00.2
.22
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
E
VEL
WA
-160
0.25
0.26
0.24
0.27
0.25
0.24
0.26
0.23
O
0.27
REFLEC TIO N C EFFIC IEN T IN D EG
REES
LE O F
AN G
SSIO N C O EFFIC IEN T IN
TRA N SM I
D EG R
LE O F
EES
AN G
90
15
0.8
A RD
C TI
85
0.2
160
0.47
EA
C
1.0
20
0.0
0.4
6
15
0
0.3
ER A
1.8
65
0.5
6
0.0
0.4
4
70
0
5
14
0.4
5
0.0
0.4
G EN
zL = 0.54 +j1.23
0.4
0.1
8
0.3
2
50
25
0.23
0.48
0.1
7
3
20
TO W
0.2
0.3
30
0.28
G TH S
60
0.22
170
4
1
0.2
9
0.2
30
> W A VELEN
0.3
35
0.2
0.0
PA
0.1
6
70
40
Yo )
0.3
0.49
CA
R
AN
0.3355
40
,O
T
EP
SC
SU
VE
TI
CI
/
(+jB
CE
0.35 lambda WTL
0.1155
0.36
80
1
0.3
o)
120
0.14
0.37
90
9
0.1
3
0.4
0
13
2
0.4
0.6 60
7
0.0
110
1
0.4
8
0.0
0.38
0.39
100
0.4
0.13
0.12
0.11
0.1
9
0.0
0.0
7
-1
30
0.4
3
0.4
2
0.4
1
0.4
215
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
10.23. The normalized load on a lossless transmission line is zL = 2 + j1. Let
Smith chart to find:
= 20 m Make use of the
a) the shortest distance from the load to the point at which zin = rin + j0, where rin > 1 (not
greater than 0 as stated): Referring to the figure below, we start by marking the given zL on the
chart and drawing a line from the origin through this point to the outer boundary. On the WTG
scale, we read the zL location as 0.213 . Moving from here toward the generator, we cross
the positive R axis (at which the impedance is purely real and greater than 1) at 0.250 . The
distance is then (0.250 * 0.213) = 0.037 from the load. With = 20 m, the actual distance
is 20(0.037) = 0.74 m.
b) Find zin at the point found in part a: Using a compass, we set its radius at the distance between
the origin and zL . We then scribe this distance along the real axis to find zin = rin = 2.61.
Problem 10.23
c) The line is cut at this point and the portion containing zL is thrown away. A resistor r = rin of
part a is connected across the line. What is s on the remainder of the line? This will be just s
for the line as it was before. As we know, s will be the positive real axis value of the normalized
impedance, or s = 2.61.
d) What is the shortest distance from this resistor to a point at which zin = 2 + j1? This would
return us to the original point, requiring a complete circle around the chart (one-half wavelength
distance). The distance from the resistor will therefore be: d = 0.500 * 0.037 = 0.463 .
With = 20 m, the actual distance would be 20(0.463) = 9.26 m.
216
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10.24. With the aid of the Smith chart, plot a curve of Zin  vs. l for the transmission line shown in Fig.
10.33. Cover the range 0 < l_ < 0.25. The required input impedance is that at the actual line input
(to the left of the two 20⌦ resistors. The input to the line section occurs just to the right of the 20⌦
resistors, and the input impedance there we first find with the Smith chart. This impedance is in series
with the two 20⌦ resistors, so we add 40⌦ to the calculated impedance from the Smith chart to find
the net line input impedance. To begin, the 20⌦ load resistor represents a normalized impedance of
zl = 0.4, which we mark on the chart (see below). Then, using a compass, draw a circle beginning
at zL and progressing clockwise to the positive real axis. The circle traces the locus of zin values for
line lengths over the range 0 < l < _4.
Problem 10.24
On the chart, radial lines are drawn at positions corresponding to .025 increments on the WTG scale.
The intersections of the lines and the circle give a total of 11 zin values. To these we add normalized
impedance of 40_50 = 0.8 to add the effect of the 40⌦ resistors and obtain the normalized impedance
at the line input. The magnitudes of these values are then found, and the results are multiplied by
50⌦. The table below summarizes the results.
l_
zinl (to right of 40⌦)
zin = zinl + 0.8
Zin  = 50zin 
0
0.40
1.20
60
.025
0.41 + j.13
1.21 + j.13
61
.050
0.43 + j.27
1.23 + j.27
63
.075
0.48 + j.41
1.28 + j.41
67
.100
0.56 + j.57
1.36 + j.57
74
.125
0.68 + j.73
1.48 + j.73
83
.150
0.90 + j.90
1.70 + j.90
96
.175
1.20 + j1.05
2.00 + j1.05
113
.200
1.65 + j1.05
2.45 + j1.05
134
.225
2.2 + j.7
3.0 + j.7
154
.250
2.5
3.3
165
217
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
10.24. (continued) As a check, the line input input impedance can be found analytically through
4
5
4
5
20 cos(2⇡l_ ) + j50 sin(2⇡l_ )
60 cos(2⇡l_ ) + j66 sin(2⇡l_ )
Zin = 40 + 50
= 50
50 cos(2⇡l_ ) + j20 sin(2⇡l_ )
50 cos(2⇡l_ ) + j20 sin(2⇡l_ )
from which
L
Zin  = 50
36 cos2 (2⇡l_ ) + 43.6 sin2 (2⇡l_ )
M1_2
25 cos2 (2⇡l_ ) + 4 sin2 (2⇡l_ )
This function is plotted below along with the results obtained from the Smith chart. A fairly good
comparison is obtained.
218
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10.25. A 300-ohm transmission line is short-circuited at z = 0. A voltage maximum, V max = 10 V, is
found at z = *25 cm, and the minimum voltage, V min = 0, is found at z = *50 cm. Use the Smith
chart to find ZL (with the short circuit replaced by the load) if the voltage readings are:
a) V max = 12 V at z = *5 cm, and V min = 5 V: First, we know that the maximum and minimum
voltages are spaced by _4. Since this distance is given as 25 cm, we see that = 100 cm = 1 m.
Thus the maximum voltage location is 5_100 = 0.05 in front of the load. The standing wave
ratio is s = V max _V min = 12_5 = 2.4. We mark this on the positive real axis of the chart
(see next page). The load position is now 0.05 wavelengths toward the load from the V max
position, or at 0.30 on the WTL scale. A line is drawn from the origin through this point on the
chart, as shown. We next set the compass to the distance between the origin and the z = r = 2.4
point on the real axis. We then scribe this same distance along the line drawn through the .30
position. The intersection is the value of zL , which we read as zL = 1.65 + j.97. The actual
load impedance is then ZL = 300zL = 495 + j290 ⌦.
b) V max = 17 V at z = *20 cm, and V min = 0. In this case the standing wave ratio is
infinite, which puts the starting point on the r ô ÿ point on the chart. The distance of 20 cm
corresponds to 20_100 = 0.20 , placing the load position at 0.45 on the WTL scale. A line is
drawn from the origin through this location on the chart. An infinite standing wave ratio places
us on the outer boundary of the chart, so we read zL = j0.327 at the 0.45 WTL position. Thus
.
ZL = j300(0.327) = j98 ⌦.
Problem 10.25
219
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10.26. A 75-ohm lossless line is of length 1.2 . It is terminated by an unknown load impedance. The input
end of the 75-ohm line is attached to the load end of a lossless 50-ohm line. A VSWR of 4 is measured
on the 50-ohm line (not the 75-ohm line), on which the first voltage minimum occurs at a distance of
0.15 in front of the junction between the two lines. Use the Smith chart to find the unknown load
impedance.
First, mark the VSWR on the positive real axis, which gives the magnitude of as determined
on the 50-ohm line. Next, scribe this length on the negative real axis to find r = 0.25 there.
The starting point is thus r = 0.25, x = 0, which is the location of the first voltage minimum.
From there, move toward the load by 0.15 wavelengths to reach the junction position, and note
the normalized impedance there, marked as zL1 = 0.64 * j1.16. This is the normalized load
impedance at the junction, as seen by the 50-ohm line.
The next step is to re-normalize zL1 to the 75-ohm line to find zin2 . This will be
50
= 0.43 * j0.77
75
zin2 = zL1
which is marked on the chart as shown, and the chart location for this impedance is seen to be
0.114 (w.t.l.). Now, this point is translated toward the load by 1.2 (equivalent to 0.2 ) to obtain
the normalized load impedance, zL2 = 1.23 + j1.60, marked on the chart. The load impedance
is thus
ZL = 75(1.23 + j1.60) = 92 + j120 ohms
70
45
50
1.4
1.2
0.9
55
1.6
0.7
1.0
2.0
1.8
65
0.5
06
0.
0.
44
0
(+
jX
/Z
14
5
N
75
T
0.4
NE
PO
EC
OM
4
NC
TA
1.0
EA
C
VE
R
C TI
50
20
10
5.0
44.0
33.0
22.0
11.4
11.6
11.0
11.2
00.9
00.8
00.7
00.6
00.5
00.4
00.3
0.2
0.49
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT
ONENT (G/Yo)
50
0.1
50
11.8
TO
IN D
U
0.2
0.1
± 180
0.4
20
.22
00.2
VSWR = 4
0.44
10
/ Yo
(-jB
)
0.66
1.0
N
TA
P
CE
US
ES
IV
CT
,O
o)
2.0
1.8
0.38
CE
0.6
1.6
0.12
0.13
0.37
00.1
.111
-10
100
0.7
14
1.
- 0
-90
CT
AN
-12
0
0.0
8
-70
(-j
06
0.
T
40
IN
1.2
0.36
10
1.0
0.14
-80
-4
0
5
-4
0...11155
35
0..335
00.99
-770
0
-5
4
0.8
6
0.1
0.3
-35
-55
0.15 lambda WTL
3
0.3
CA
RE
A
-60
7
0.1
-60
P AC
ITI
VE
EN
-1
R
0.2
-30
N
0
-65 .5
8
0.1
0
-5
2
PO
5
0.0
Z
X/
0.3
CO
M
-75
DU
0.
31
44
0.
0.4
0.
19
-25
5
0.6
3.0
-20
0
0.0
7
-1
30
0.4
3
0.4
2
0.114 lambda WTL
0.3
-4
4
0.
0.2
0.4
j1.16
9
6
0.4
0..88
4
0.0
0
-15 -80
1.0
zL1
0.2
j0.777
0.2
1
zin2
-30
0.3
0.28
<
8
0.22
0.47
CE
0.
-20
0.2
4.0
R
80
>
0.6
10
1st Vmin location
r = 0.25
0.1
-85
10
-15
85
0.47
5.0
-10
160
0.2
90
15
8
A RD
4.0
1.0
0.
ER A
zL2 = 1.23 +
j1.60
0.8
5.0
0.0
0.4
6
15
0
0.3
G EN
20
3.0
0.6
20
0.0
5
0.
4
0.48
25
0.4
0.25
0.26
0.24
0.27
0.23
0.25
0.24
0.26
0.23
O
0.27
REFLEC TIO N C EFFIC IEN T IN D EG
REES
LE O F
AN G
SSIO N C O EFFIC IEN T IN
TRA N SM I
D EG R
LE O F
EES
AN G
TO W
0.3
2
50
20
G TH S
0.11
8
30
0.28
170
0.2
0.22
0.49
0.48
AD <
A RD LO
S TO W
G TH
0
N
-17
E
VEL
WA
-90
-160
0.1
7
0.314 lambda
0.3 WTL
60
3
)
1
0.2
9
0.2
30
> W A VELEN
C
0.3
4
35
0.2
0.0
PA
C
0.1
6
70
0.3
0.0
CA
R
AN
0.35
40
40
,O
PT
CE
US
ES
IV
IT
Yo
jB/
E (+
0.15
0.36
80
31
0.
o)
120
0.14
0.37
90
19
0.
3
0.4
0
13
2
0.4
0.6 60
7
0.0
1
0.4
8
0.0
110
0.8
0
0.38
0.39
100
0.4
0.13
0.12
0.11
0.1
.09
0.39
-110
0.1
0.0
9
0.4
1
0.4
220
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
10.27. The characteristic admittance (Y0 = 1_Z0 ) of a lossless transmission line is 20 mS. The line is
terminated in a load YL = 40 * j20 mS. Make use of the Smith chart to find:
a) s: We first find the normalized load admittance, which is yL = YL _Y0 = 2 * j1. This is plotted
on the Smith chart below. We then set on the compass the distance between yL and the origin.
The same distance is then scribed along the positive real axis, and the value of s is read as 2.6.
b) Yin if l = 0.15 : First we draw a line from the origin through zL and note its intersection with
the WTG scale on the chart outer boundary. We note a reading on that scale of about 0.287 .
To this we add 0.15 , obtaining about 0.437 , which we then mark on the chart (0.287 is not
the precise value, but I have added 0.15 to that mark to obtain the point shown on the chart that
is near to 0.437 . This “eyeballing” method increases the accuracy a little). A line drawn from
the 0.437 position on the WTG scale to the origin passes through the input admittance. Using
the compass, we scribe the distance found in part a across this line to find yin = 0.56 * j0.35,
or Yin = 20yin = 11 * j7.0 mS.
c) the distance in wavelengths from YL to the nearest voltage maximum: On the admittance chart,
the Vmax position is on the negative r axis. This is at the zero position on the WTL scale. The
load is at the approximate 0.213 point on the WTL scale, so this distance is the one we want.
Problem 10.27
221
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10.28. The wavelength on a certain lossless line is 10cm. If the normalized input impedance is zin = 1 + j2,
use the Smith chart to determine:
a) s: We begin by marking zin on the chart (see below), and setting the compass at its distance
from the origin. We then use the compass at that setting to scribe a mark on the positive real
axis, noting the value there of s = 5.8.
b) zL , if the length of the line is 12 cm: First, use a straight edge to draw a line from the origin
through zin , and through the outer scale. We read the input location as slightly more than 0.312
on the WTL scale (this additional distance beyond the .312 mark is not measured, but is instead
used to add a similar distance when the impedance is transformed). The line length of 12cm
corresponds to 1.2 wavelengths. Thus, to transform to the load, we go counter-clockwise twice
around the chart, plus 0.2 , finally arriving at (again) slightly more than 0.012 on the WTL
scale. A line is drawn to the origin from that position, and the compass (with its previous setting)
is scribed through the line. The intersection is the normalized load impedance, which we read
as zL = 0.173 * j0.078.
c) xL , if zL = 2 + jxL , where xL > 0. For this, use the compass at its original setting to scribe
through the r = 2 circle in the upper half plane. At that point we read xL = 2.62.
Problem 10.28
222
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10.29. A standing wave ratio of 2.5 exists on a lossless 60 ⌦ line. Probe measurements locate a voltage
minimum on the line whose location is marked by a small scratch on the line. When the load is
replaced by a short circuit, the minima are 25 cm apart, and one minimum is located at a point 7 cm
toward the source from the scratch. Find ZL : We note first that the 25 cm separation between minima
imply a wavelength of twice that, or = 50 cm. Suppose that the scratch locates the first voltage
minimum. With the short in place, the first minimum occurs at the load, and the second at 25 cm in
front of the load. The effect of replacing the short with the load is to move the minimum at 25 cm to
a new location 7 cm toward the load, or at 18 cm. This is a possible location for the scratch, which
would otherwise occur at multiples of a half-wavelength farther away from that point, toward the
generator. Our assumed scratch position will be 18 cm or 18_50 = 0.36 wavelengths from the load.
Using the Smith chart (see below) we first draw a line from the origin through the 0.36 point on the
wavelengths toward load scale. We set the compass to the length corresponding to the s = r = 2.5
point on the chart, and then scribe this distance through the straight line. We read zL = 0.79+j0.825,
from which ZL = 47.4 + j49.5 ⌦. As a check, I will do the problem analytically. First, we use
4
5
4(18)
1
zmin = *18 cm = * ( + ⇡) Ÿ
=
* 1 ⇡ = 1.382 rad = 79.2˝
2
50
Now
s * 1 2.5 * 1
=
= 0.4286
s + 1 2.5 + 1
and so L = 0.4286⁄1.382. Using this, we find
 L =
zL =
1+
1*
L
L
= 0.798 + j0.823
and thus ZL = zL (60) = 47.8 + j49.3 ⌦.
Problem 10.29
223
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10.30. A 2-wire line, constructed of lossless wire of circular cross-section is gradually flared into a coupling
loop that looks like an egg beater. At the point X, indicated by the arrow in Fig. 10.34, a short circuit
is placed across the line. A probe is moved along the line and indicates that the first voltage minimum
to the left of X is 16cm from X. With the short circuit removed, a voltage minimum is found 5cm
to the left of X, and a voltage maximum is located that is 3 times voltage of the minimum. Use the
Smith chart to determine:
a) f : No Smith chart is needed to find f , since we know that the first voltage minimum in front
of a short circuit is one-half wavelength away. Therefore, = 2(16) = 32cm, and (assuming an
air-filled line), f = c_ = 3 ù 108 _0.32 = 0.938 GHz.
b) s: Again, no Smith chart is needed, since s is the ratio of the maximum to the minimum voltage
amplitudes. Since we are given that Vmax = 3Vmin , we find s = 3.
c) the normalized input impedance of the egg beater as seen looking the right at point X: Now
we need the chart. From the figure below, s = 3 is marked on the positive real axis, which
determines the compass radius setting. This point is then transformed, using the compass, to
the negative real axis, which corresponds to the location of a voltage minimum. Since the first
Vmin is 5cm in front of X, this corresponds to (5_32) = 0.1563 to the left of X. On the chart,
we now move this distance from the Vmin location toward the load, using the WTL scale. A
line is drawn from the origin through the 0.1563 mark on the WTL scale, and the compass
is used to scribe the original radius through this line. The intersection is the normalized input
impedance, which is read as zin = 0.86 * j1.06.
Problem 10.30
224
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10.31. In order to compare the relative sharpness of the maxima and minima of a standing wave, assume a
load zL = 4 + j0 is located at z = 0. Let V min = 1 and = 1 m. Determine the width of the
a) minimum, where V  < 1.1: We begin with the general phasor voltage in the line:
V (z) = V + (e*j z + ej z )
With zL = 4+j0, we recognize the real part as the standing wave ratio. Since the load impedance
is real, the reflection coefficient is also real, and so we write
= =
s*1 4*1
=
= 0.6
s+1 4+1
The voltage magnitude is then
˘
⌅
⇧1_2
V (z)V < (z) = V + (e*j z + ej z )(ej z + e*j z )
⌅
⇧1_2
= V + 1 + 2 cos(2 z) + 2
V (z) =
Note that with cos(2 z) = ±1, we obtain V  = V + (1 ± ) as expected. With s = 4 and
with V min = 1, we find V max = 4. Then with = 0.6, it follows that V + = 2.5. The net
expression for V (z) is then
˘
V (z) = 2.5 1.36 + 1.2 cos(2 z)
To find the width in z of the voltage minimum, defined as V  < 1.1, we set V (z) = 1.1 and
solve for z: We find
⇠
1.1
2.5
⇡2
= 1.36 + 1.2 cos(2 z) Ÿ 2 z = cos*1 (*0.9726)
Thus 2 z = 2.904. At this stage, we note the the V min point will occur at 2 z = ⇡. We
therefore compute the range, z, over which V  < 1.1 through the equation:
2 ( z) = 2(⇡ * 2.904) Ÿ
where
z=
⇡ * 2.904
= 0.0378 m = 3.8 cm
2⇡_1
= 1 m has been used.
b) Determine the width of the maximum, where V  > 4_1.1: We use the same equation for V (z),
which in this case reads:
˘
4_1.1 = 2.5 1.36 + 1.2 cos(2 z) Ÿ cos(2 z) = 0.6298
Since the maximum corresponds to 2 z = 0, we find the range through
2
z = 2 cos*1 (0.6298) Ÿ
z=
0.8896
= 0.142 m = 14.2 cm
2⇡_1
225
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10.32. In Fig. 10.7, let ZL = 250 ohms, Z0 = 50 ohms, find the shortest attachment distance d and the
shortest length, d1 of a short-circuited stub line that will provide a perfect match on the main line to
the left of the stub. Express all answers in wavelengths.
The first step is to mark the normalized load admittance on the chart. This will be yL = 1_zL =
50_250 = 0.20. Its location is noted as 0.0 on the wavelengths toward generator (WTG) scale. Next,
from the load, move clockwise (toward generator) until the admittance real part is unity. The first
instance of this is at the point yin1 = 1 + j1.8, as shown. Moving farther, the second instance is at the
point yin2 = 1.0 * j1.8. The distance in wavelenghs between yL and yin1 is noted on the WTG scale,
or da = 0.183 . The distance in wavelenghs between yL and yin2 is again noted on the WTG scale,
or db = 0.317 . These are the two possible attachment points for the shorted stub. The shortest of
these is da = 0.183 . The corresponding stub length is found by transforming from the short circuit
(load) position on the stub, Psc toward generator until a normalized admittance of ys = bs = *j1.8
occurs. This is marked on the chart as the point ys1 , located at 0.331 (WTG). The (shortest) stub
length is thus d1a = (0.331 * 0.250) = 0.81 .
226
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10.33. In Fig. 10.17, let ZL = 100 + j150 ⌦, and Z0 = 100 ⌦.
(a) a) Find the shortest length, d1 , of a short-circuited stub, and the shortest distance d that it may
be located from the load to provide a perfect match on the main line to the left of the stub.
Express both answers in wavelengths:
The Smith chart construction is shown below. First we find zL = (100+j150)_100 = 1.0+
j1.5 and plot it on the chart. Next, we find yL = 1_zL by transforming this point halfway
around the chart, where we read yL = 0.308 * j0.462. This point is to be transformed to a
location at which the real part of the normalized admittance is unity. In this problem, this
happens to occur at precisely the original load impedance position. The attachment point
thus lies one quarter-wavelength in front of the load (d = 0.25 ), where the normalized
admittance is yB = 1.0 + j1.5. We now need a stub input normalized susceptance that
will exactly cancel the imaginary part of yB , or ys = *j1.5. This value occurs on the
lower perimeter of the chart as shown, and is located at 0.344 on the WTG scale. The
distance of this point from the short circuit point, Psc (representing the stub load end) is
thus d1 = (0.344 * 0.250) = 0.094 .
70
0
45
1.2
1.0
50
0.9
55
1.6
2.0
20
T
N
75
EC
OM
4
PO
NE
.6
NC
0.8
TA
4.0
15
5.0
1.0
EA
C
1.0
VE
R
0.47
C TI
85
160
0.2
IN D
U
A RD
3.0
0
zL = yB = 1.00
+j1.50
10
0.8
0.6
10
0.1
0.49
0.4
ER A
TO
R
80
>
0.0
0.4
6
15
0
0.3
G EN
0.6 60
(+
jX
/Z
5
14
0.4
5
0.0
0.4
0.48
25
0.4
0.25
25
0.26
0.24
0.27
0.23
0.25
0.2
25
25
0.24
0.26
0.23
O
0.27
REFLEC TIO N C EFFIC IEN T IN D EG
REES
LE O F
AN G
SSIO N C O EFFIC IEN T IN
TRA N SM I
D EG R
LE O F
EES
AN G
90
0.3
2
50
20
20
50
20
10
5.0
4.0
3.0
2.0
1.8
1.4
Psc (0.25 lambda WTG)
0.44
10
jB/
E (-
Yo )
0.66
1.0
AN
PT
CT
IN
,O
o)
1.8
ITI
VE
-90
0.12
0.13
0.38
0.37
0.11
-100
0.39
-110
0.1
RE
A
CT
AN
0.0
9
-12
0
0.0
8
40
(-j
-70
T
6
EN
0
-65 .5
2.0
AC
N
0.6
1.6
1.4
1.0
1.2
0.36
0.9
0.14
-80
-4
0
5
-4
0.15
0.35
0
-70
-5
4
0.3
6
0.11
-35
0.8
7
3
CA P
-55
0.1
0.3
-75
R
0.2
0.7
0.344 lambda WTG
(stub input location)
-60
-60
-30
2
PO
-1
Z
X/
0.3
OM
<
CE
US
ES
IV
DU
0.3
1
CE
C
0.0
9
4
0.1
0.4
0.4
0
8
0.1
0
-5
-25
5
0.0
0.3
-4
5
0.4
0.6
3.0
0.2
0.4
-20
4
0.0
0
-15 -80
0.8
0.47
1.0
4.0
9
6
0.4
-15
0.2
0.2
1
-30
0.3
0.28
j0.462
5.0
C
8
0.8
yL
0.22
0.2
-10
-85
1.6
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
± 180
1.2
0.22
0.2
0.1
-20
0.49
0.48
D<
RD LO A
TO W A
TH S
-170
EN G
VEL
WA
-90
-160
50
20
0.1
0.2
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
50
TO W
0.1
8
30
0.28
0.0
7
3
0.22
G TH S
0.2
0.1
0.3
1
0.2
9
0.2
30
170
C
60
0.2
> W A VELEN
PA
)
0.3
0.0
CA
R
0.3
4
35
40
,O
0.1
6
70
1
0.3
o)
P
CE
US
ES
IV
IT
C
0.35
40
1.8
65
0
0.4
4
13
N
TA
Yo
jB/
E (+
0.15
0.36
80
9
0.1
0.0
6
0
.43
0.5
0
120
0.14
0.37
90
1.4
2
0.4
0.7
8
0.0
.07
110
1
0.4
0.8
0
0.38
0.39
100
0.4
0.13
0.12
0.11
0.1
.09
0.0
7
-1
30
0.4
3
0.4
2
0.4
1
0.4
b) Repeat for an open-circuited stub:
In this case, everything is the same, except for the load-end position of the stub, which now
occurs at the Poc point on the chart (diametrically opposite Psc ). The stub attachment point,
d, is the same as before. The stub length, found by transforming from Poc to the stub input,
is now one quarter-wavelength longer than the previous result, or d1® = (0.094 + 0.250) =
0.344 (or just simply the reading on the WTG scale). Note that this choice, while keeping
d at the minimum value, does not do so for d1 . Do you see why?
227
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10.34. The lossless line shown in Fig. 10.35 is operating with = 100cm. If d1 = 10cm, d = 25cm, and the
line is matched to the left of the stub, what is ZL ? For the line to be matched, it is required that the
sum of the normalized input admittances of the shorted stub and the main line at the point where the
stub is connected be unity. So the input susceptances of the two lines must cancel. To find the stub
input susceptance, use the Smith chart to transform the short circuit point 0.1 toward the generator,
and read the input value as bs = *1.37 (note that the stub length is one-tenth of a wavelength). The
main line input admittance must now be yin = 1 + j1.37. This line is one-quarter wavelength long,
so the normalized load impedance is equal to the normalized input admittance. Thus zL = 1 + j1.37,
so that ZL = 300zL = 300 + j411 ⌦.
Problem 10.34
228
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10.35. A load, ZL = 25 + j75 ⌦, is located at z = 0 on a lossless two-wire line for which Z0 = 50 ⌦ and
v = c.
a) If f = 300 MHz, find the shortest distance d (z = *d) at which the input admittance has a
real part equal to 1_Z0 and a negative imaginary part: The Smith chart construction is shown
below. We begin by calculating zL = (25 + j75)_50 = 0.5 + j1.5, which we then locate on the
chart. Next, this point is transformed by rotation halfway around the chart to find yL = 1_zL =
0.20 * j0.60, which is located at 0.088 on the WTL scale. This point is then transformed
toward the generator until it intersects the g = 1 circle (shown highlighted) with a negative
imaginary part. This occurs at point yin = 1.0 * j2.23, located at 0.308 on the WTG scale.
The total distance between load and input is then d = (0.088 + 0.308) = 0.396 . At 300 MHz,
and with v = c, the wavelength is = 1 m. Thus the distance is d = 0.396 m = 39.6 cm.
b) What value of capacitance C should be connected across the line at that point to provide unity
standing wave ratio on the remaining portion of the line? To cancel the input normalized susceptance of -2.23, we need a capacitive normalized susceptance of +2.23. We therefore write
!C =
2.23
2.23
Ÿ C=
= 2.4 ù 10*11 F = 24 pF
Z0
(50)(2⇡ ù 3 ù 108 )
Problem 10.35
229
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10.36. The two-wire lines shown in Fig. 10.36 are all lossless and have Z0 = 200 ⌦. Find d and the
shortest possible value for d1 to provide a matched load if = 100cm. In this case, we have a
series combination of the loaded line section and the shorted stub, so we use impedances and the
Smith chart as an impedance diagram. The requirement for matching is that the total normalized
impedance at the junction (consisting of the sum of the input impedances to the stub and main loaded
section) is unity. First, we find zL = 100_200 = 0.5 and mark this on the chart (see below). We
then transform this point toward the generator until we reach the r = 1 circle. This happens at two
possible points, indicated as zin1 = 1 + j.71 and zin2 = 1 * j.71. The stub input impedance must
cancel the imaginary part of the loaded section input impedance, or zins = ±j.71. The shortest stub
length that accomplishes this is found by transforming the short circuit point on the chart to the point
zins = +j0.71, which yields a stub length of d1 = .098 = 9.8 cm. The length of the loaded section is
then found by transforming zL = 0.5 to the point zin2 = 1 * j.71, so that zins + zin2 = 1, as required.
This transformation distance is d = 0.347 = 37.7 cm.
Problem 10.36
230
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10.37. In the transmission line of Fig. 10.20, Rg = Z0 = 50 ⌦, and RL = 25 ⌦. Determine and plot the
voltage at the load resistor and the current in the battery as functions of time by constructing appropriate voltage and current reflection diagrams: Referring to the figure, closing the switch launches a
voltage wave whose value is given by Eq. (119):
V1+ =
V 0 Z0
50
1
=
V0 = V0
R g + Z0
100
2
Now, L = (25 * 50)_(25 + 50) = *1_3. So on reflection from the load, the reflected wave is of value
V1* = *V0 _6. On returning to the input end, the reflection coefficient there is zero, and so all is still.
The voltage reflection diagram would be that shown in Fig. 10.21a, except that no waves are present
after time t = 2l_v. Likewise, the current reflection diagram is that of Fig. 10.22a, except, again, no
waves exist after t = l_v. The voltage at the load will be VL = V1+ (1 + L ) = V0 _3 for times beyond
l_v. The current through the battery is initially IB = V1+ _Z0 = V0 _100 for times (0 < t < 2l_v).
When the reflected wave from the load returns to the input end (at time t = 2l_v), the reflected wave
current, I1* = V0 _300, adds to the original current to give IB = V0 _75 A for (t > 2l_v).
10.38. Repeat Problem 37, with Z0 = 50⌦, and RL = Rg = 25⌦. Carry out the analysis for the time period
0 < t < 8l_v. At the generator end, we have g = *1_3. At the load end, we have L = *1_3
as before. The initial wave is of magnitude V + = (2_3)V0 . Using these values, voltage and current
reflection diagrams are constructed, and are shown below:
231
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10.38. (continued) From the diagrams, voltage and current plots are constructed. First, the load voltage is
found by adding voltages along the right side of the voltage diagram at the indicated times. Second,
the current through the battery is found by adding currents along the left side of the current reflection
diagram. Both plots are shown below, where currents and voltages are expressed to three significant
figures. The steady state values, VL = 0.5V and IB = 0.02A, are expected as t ô ÿ.
10.39. In the transmission line of Fig. 10.20, Z0 = 50 ⌦ and RL = Rg = 25 ⌦. The switch is closed at t = 0
and is opened again at time t = l_4v, thus creating a rectangular voltage pulse in the line. Construct
an appropriate voltage reflection diagram for this case and use it to make a plot of the voltage at the
load resistor as a function of time for 0 < t < 8l_v (note that the effect of opening the switch is to
initiate a second voltage wave, whose value is such that it leaves a net current of zero in its wake):
The value of the initial voltage wave, formed by closing the switch, will be
V+ =
Z0
50
2
V0 =
V0 = V0
R g + Z0
25 + 50
3
On opening the switch, a second wave, V +® , is generated which leaves a net current behind it of zero.
This means that V +® = *V + = *(2_3)V0 . Note also that when the switch is opened, the reflection
coefficient at the generator end of the line becomes unity. The reflection coefficient at the load end is
L = (25 * 50)_(25 + 50) = *(1_3). The reflection diagram is now constructed in the usual manner,
and is shown on the next page. The path of the second wave as it reflects from either end is shown
in dashed lines, and is a replica of the first wave path, displaced later in time by l_(4v).a All values
for the second wave after each reflection are equal but of opposite sign to the immediately preceding
first wave values. The load voltage as a function of time is found by accumulating voltage values as
they are read moving up along the right hand boundary of the chart. The resulting function, plotted
just below the reflection diagram, is found to be a sequence of pulses that alternate signs. The pulse
amplitudes are calculated as follows:
232
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10.39. (continued)
l
5l
<t<
:
v
4v
3l
13l
<t<
:
v
4v
5l
21l
<t<
:
v
4v
7l
29l
<t<
:
v
4v
⇠
⇡
1
V1 = 1 *
V + = 0.44 V0
3
⇠
⇡
1
1
V2 = * 1 *
V + = *0.15 V0
3
3
⇠ ⇡2 ⇠
⇡
1
1
V3 =
1*
V + = 0.049 V0
3
3
⇠ ⇡3 ⇠
⇡
1
1
V4 = *
1*
V + = *0.017 V0
3
3
233
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10.40. In the charged line of Fig. 10.25, the characteristic impedance is Z0 = 100⌦, and Rg = 300⌦. The
line is charged to initial voltage V0 = 160 V, and the switch is closed at t = 0. Determine and plot
the voltage and current through the resistor for time 0 < t < 8l_v (four round trips). This problem
accompanies Example 11.12 as the other special case of the basic charged line problem, in which
now Rg > Z0 . On closing the switch, the initial voltage wave is
V + = *V0
Z0
100
= *160
= *40 V
Rg + Z0
400
Now, with g = 1_2 and L = 1, the voltage and current reflection diagrams are constructed as
shown below. Plots of the voltage and current at the resistor are then found by accumulating values
from the left sides of the two charts, producing the plots as shown.
234
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10.41. In the transmission line of Fig. 10.37, the switch is located midway down the line, and is closed at
t = 0. Construct a voltage reflection diagram for this case, where RL = Z0 . Plot the load resistor
voltage as a function of time: With the left half of the line charged to V0 , closing the switch initiates
(at the switch location) two voltage waves: The first is of value *V0 _2 and propagates toward the
left; the second is of value V0 _2 and propagates toward the right. The backward wave reflects at the
battery with g = *1. No reflection occurs at the load end, since the load is matched to the line. The
reflection diagram and load voltage plot are shown below. The results are summarized as follows:
l
: VL = 0
2v
V
l
3l
<t<
: VL = 0
2v
2v
2
3l
t>
: VL = V0
2v
0<t<
235
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10.42. A simple frozen wave generator is shown in Fig. 10.38. Both switches are closed simultaneously
at t = 0. Construct an appropriate voltage reflection diagram for the case in which RL = Z0 .
Determine and plot the load voltage as a function of time: Closing the switches sets up a total of four
voltage waves as shown in the diagram below. Note that the first and second waves from the left are
of magnitude V0 , since in fact we are superimposing voltage waves from the *V0 and +V0 charged
sections acting alone. The reflection diagram is drawn and is used to construct the load voltage with
time by accumulating voltages up the right hand vertical axis.
236
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10.43. In Fig. 10.39, RL = Z0 and Rg = Z0 _3. The switch is closed at t = 0. Determine and plot as
functions of time: a) the voltage across RL ; b) the voltage across Rg ; c) the current through the
battery.
With the switch at the opposite end from the battery, the entire line is initially charged to V0 . So, on
closing the switch, the initial wave propagates backward, originating at the switch, and is of value
V * = *V0 Z0 _(RL + Z0 ) = *V0 _2. On reaching the left end, the wave reflects with reflection
coefficient G = (RG * Z0 )_(RG + Z0 ) = *1_2. The reflected wave returns to the switch end, sees
a matched load there, and there is no further reflection. The resulting voltage and current reflection
diagrams are shown below.
237
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10.43 (continued) The load voltage is read from the right side of the voltage diagram, and is plotted as VL
below. The voltage across the entire left end is read from the left side of the voltage diagram, and
is plotted as VT G below. The voltage across the resistor, RG , will be VG = VT G * V0 (choosing the
resistor voltage polarity as positive), which leads to the VG plot below. Finally, the battery current is
read from the left side of the current diagram, and is plotted as IB below.
238
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CHAPTER 11 – 9th Edition
11.1. Show that Exs = Aejk0 z+ is a solution to the vector Helmholtz equation, Sec. 11.1, Eq. (30), for
˘
k0 = !
and A: We take
0 ✏0 and any
d2
Aejk0 z+ = (jk0 )2 Aejk0 z+ = *k20 Exs
dz2
11.2. A 20-GHz uniform plane wave propagates in the forward z direction in a lossless medium for which
✏r = r = 3. Assume a real electric field amplitude, E0 , and take the wave as polarized in the x
direction. Find:
˘
˘
8
a) vp : vp = 1_
✏ = c_
r ✏r = c_3 = 1.0 ù 10 m_s.
b)
: In this lossless medium, k =
c)
:
= !_vp = (4⇡ ù 1010 )_(1.0 ù 108 ) = 1.26 ù 103 m*1 .
= 2⇡_ = 5.0 ù 10*3 m = 5.0 mm.
d) Es : With real amplitude E0 , forward z travel, and x polarization, we write
Es = E0 exp(*j z)ax = E0 exp(*j1.26 ù 103 z) ax V_m.
˘
˘
e) Hs : First, the intrinsic impedance of the medium is ⌘ =
_✏ = ⌘0
r _✏r = ⌘0 = 377 ohms.
Then Hs = (E0 _⌘0 ) exp(*j z) ay = (E0 _377) exp(*j1.26 ù 103 z) ay A_m.
f) < S >= (1_2)Re Es ù H<s = (E02 _754) az W_m2
11.3. An H field in free space is given as H(x, t) = 10 cos(108 t * x)ay A/m. Find
a)
: Since we have a uniform plane wave,
= 108 _(3 ù 108 ) = 0.33 rad_m.
b)
: We know
= !_c, where we identify ! = 108 sec*1 . Thus
= 2⇡_ = 18.9 m.
c) E(x, t) at P (0.1, 0.2, 0.3) at t = 1 ns: Use E(x, t) = *⌘0 H(x, t) = *(377)(10) cos(108 t *
x) = *3.77 ù 103 cos(108 t * x). The vector direction of E will be *az , since we require that
S = E ù H, where S is x-directed. At the given point, the relevant coordinate is x = 0.1. Using
this, along with t = 10*9 sec, we finally obtain
E(x, t) = *3.77 ù 103 cos[(108 )(10*9 ) * (0.33)(0.1)]az = *3.77 ù 103 cos(6.7 ù 10*2 )az
= *3.76 ù 103 az V_m
11.4. Small antennas have low efficiencies (as will be seen in Chapter 14) and the efficiency increases
with size up to the point at which a critical dimension of the antenna is an appreciable fraction of a
wavelength, say _8.
a) An antenna is that is 12cm long is operated in air at 1 MHz. What fraction of a wavelength long
is it? The free space wavelength will be
air =
3.0 ù 108 m_s
c
0.12
=
= 300 m, so that the f raction =
= 4.0 ù 10*4
f
300
106 s*1
b) The same antenna is embedded in a ferrite material for which ✏r = 20 and r = 2, 000. What
fraction of a wavelength is it now?
300
0.12
air
=˘
= 1.5m Ÿ f raction =
= 0.08
f errite = ˘
1.5
(20)(2000)
r ✏r
239
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11.5. Consider two x-polarized light waves that counter-propagate along the z axis. The wave traveling in
the forward z direction is of frequency !2 ; the backward z-directed wave is of frequency !1 , where
!1 < !2 . Both frequencies are very slightly detuned on either side of their mean frequency !0 , such
that !0 * !1 = !2 * !0 << !0 . Using the complex field forms, construct the expression for the total
electric field, and from this find the light intensity distribution (proportional to EE < ). Your answer
should be in the form of a “standing wave” that in fact moves along the z axis. Find an expression
for the velocity of the standing wave pattern in terms of given or known parameters. Does the pattern
move in the forward or backward z direction?
The fact that the frequencies differ means that the wavenumbers will differ as well. If we assume that
the medium is free space, we would have k2 = !2 _c and k1 = !1 _c. Assuming equal amplitudes,
E0 , the total complex field, representing the sum of the forward and backward waves is
⌅
⇧
EcT (z) = E0 exp(*jk2 z) exp(j!2 t) + exp(+jk1 z) exp(j!1 t) ax
Now define the mean frequency and wavenumber:
!0 =
!2 + !1
2
and
km =
k2 + k1
2
then define k = k2 * k1 and ! = !2 * !1 , from which
k2 = km +
k
,
2
k1 = km *
k
,
2
!2 = !0 +
!
,
2
and
!1 = !0 *
!
2
The total complex field can now be expressed as
⌅
⇧
EcT (z) = E0 e*j kz_2 ej!0 t e*jkm z ej !t_2 + e+jkm z e*j !t_2 ax
⇠
⇡
!
= 2E0 e*j kz_2 ej!0 t cos
t * km z ax
2
The power density (light intensity) is now proportional to
⇠
⇡
!
<
EcT EcT
= 4E02 cos2
t * km z
2
The intensity pattern moves in the forward z direction at velocity v = !_2km = c !_2!0 .
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11.6. A uniform plane wave has electric field Es = (Ey0 ay * Ez0 az ) e*↵x e*j x V_m. The intrinsic
impedance of the medium is given as ⌘ = ⌘ ej , where is a constant phase.
a) Describe the wave polarization and state the direction of propagation: The wave is linearly
polarized in the y-z plane, and propagates in the forward x direction (from the e*j x factor).
b) Find Hs : Each component of Es , when crossed into its companion component of H<s , must give
a vector in the positive-x direction of travel. Using this rule, we find
4
5 4
5
Ey
Ey0
Ez
Ez0
Hs =
a +
a =
a +
a e*↵x e*j e*j x A_m
⌘ z
⌘ y
⌘ z ⌘ y
⌅
⇧
c) Find E(x, t) and H(x, t): E(x, t) = Re Es ej!t = Ey0 ay * Ez0 az e*↵x cos(!t * x)
H(x, t) = Re Hs ej!t =
⇧
1 ⌅
Ey0 az + Ez0 ay e*↵x cos(!t * x * )
⌘
where all amplitudes are assumed real.
d) Find < S > in W_m2 :
⇠
⇡
1
1
2
2
< S >= Re Es ù H<s =
Ey0
+ Ez0
e*2↵x cos ax W_m2
2
2⌘
e) Find the time-average power in watts that is intercepted by an antenna of rectangular crosssection, having width w and height h, suspended parallel to the yz plane, and at a distance d
from the wave source. This will be
⇠
⇡
1
2
2
P =
< S > dS =  < S > x=d ù area =
(wh) Ey0
+ Ez0
e*2↵d cos W
2⌘
plate
241
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11.7. Express in terms of the penetration depth, , the distance into a lossy medium at which the wave
power has attenuated by
a) 1 dB: The decibel loss is given by
Loss (dB) = 8.69↵L =
8.69L
Ÿ L1 =
8.69
= 0.12
b) 3 dB: This would be L3 = 3L1 = 0.35
c) 30 dB: L30 = 30L1 = 3.45 .
11.8. An electric field in free space is given in spherical coordinates as Es (r) = E0 (r)e*jkr a✓ V/m.
a) find Hs (r) assuming uniform plane wave behavior: Knowing that the cross product of Es with
the complex conjugate of the phasor Hs field must give a vector in the direction of propagation,
we obtain,
E (r)
Hs (r) = 0 e*jkr a A_m
⌘0
b) Find < S >: This will be
E 2 (r)
1
< S >= Re Es ù H<s = 0 ar W_m2
2
2⌘0
c) Express the average outward power in watts through a closed spherical shell of radius r, centered
at the origin: The power will be (in this case) just the product of the power density magnitude
in part b with the sphere area, or
E 2 (r)
P = 4⇡r2 0
W
2⌘0
where E0 (r) is assumed real.
d) Establish the required functional form of E0 (r) that will enable the power flow in part c to be
independent of radius: Evidently this condition is met when E0 (r) ◊ 1_r
242
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11.9. An example of a nonuniform plane wave is a surface or evanescent wave, an example of which in
phasor form is shown here, exhibiting diminishing amplitude with x while propagating in the forward
z direction:
Es (x, z) = E0 e*↵x e*j z ay
Such fields form part of the mode structure in dielectric waveguides, as will be explored in
Chapter 13.
a) Assuming free space conditions, use Eq. (23) to find the associated phasor magnetic field,
Hs (x, z) (note that there will be two components). Use ( ù Es = *j! 0 Hs , which leads to
0
1
)Ey
)Ey
j
j
j
Hs =
( ù Ey ay =
az *
ax =
*↵Ey az + j Ey ax
! 0
! 0 )x
)z
! 0
where the subscript s has been dropped everywhere, with the understanding that that all fields
are in phasor form. Finally,
H(x, z) = *
E0
! 0
ax + j↵az e*↵x e*j z
b) Find ↵ as a function of , !, ✏0 , and 0 , such that all of Maxwell’s equations are satisfied by the
electric and magnetic fields: With a source-free medium, we first of all have ( E = ( H = 0,
which by inspection are seen to be satisfied by the above fields. The ( ù E = *j! 0 H equation
is of course automatically satisfied. This leaves ( ù H = j!✏0 E. Substituting the known fields,
we find
0
1
)Hx )Hz
(ùH=
*
ay = j!✏0 Ey ay
)z
)x
or:
jE0 2
* ↵ 2 e*↵x e*j z = j!✏0 E0 e*↵x e*j z
! 0
which simplifies to
2
* ↵ 2 = !2 0 ✏0 Ÿ ↵ =
t
2 * !2
0 ✏0
11.10. In a medium characterized by intrinsic impedance ⌘ = ⌘ej , a linearly-polarized plane wave propagates, with magnetic field given as Hs = H0y ay + H0z az e*↵x e*j x . Find:
a) Es : Requiring orthogonal components of Es for each component of Hs , we find
⌅
⇧
Es = ⌘ H0z ay * H0y az e*↵x e*j x ej
⌅
⇧
b) E(x, t) = Re {Es ej!t } = ⌘ H0z ay * H0y az e*↵x cos(!t * x + ).
⌅
⇧
c) H(x, t) = Re {Hs ej!t } = H0y ay + H0z az e*↵x cos(!t * x).
⌧
2 + H 2 e*2↵x cos a W_m2
d) < S >= 12 Re{Es ù H<s } = 12 ⌘ H0y
x
0z
243
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11.11. A uniform plane wave at frequency f = 100 MHz propagates in a material having conductivity
= 3.0 S/m and dielectric constant ✏r® = 8.00. The wave carries electric field amplitude E0 = 100
V/m.
a) Calculate the loss tangent and determine whether the medium would qualify as a good dielectric
or a good conductor.
!✏ ®
=
3.0
(2⇡ ù 108 )(8.00)(8.85 ù 10*12 )
= 67.5
This falls within the good conductor approximation.
b) Calculate ↵, , and ⌘. Assuming a good conductor, we have:
˘
.
. ˘
↵ = = ⇡f 0 = ⇡(108 )(4⇡ ù 10*7 )(3) = 34.4 m*1
Then
. 1+j
(1 + j)↵ (1 + j)(34.4)
⌘=
(Eq. (85)) =
=
= 11.5(1 + j) ohms
3.0
c) Assuming forward z propagation and x polarization, write the electric field in phasor form:
Es = 100e*↵z e*j z ax
where ↵ and
are as found in part b.
d) Write the magnetic field in phasor form:
Hs =
100
e*↵z e*j z ay = 6.15 e*j0.79 e*↵z e*j z ay
11.5(1 + j)
e) Write the time-average Poynting vector, < S > in W_m2 .
1
1
< S >= Re Es ù H<s = (100)(6.15)e*2↵z cos(0.79) az = 216 e*2↵z az W_m2
2
2
f) Find the 6-dB material thickness; i.e., the propagation distance over which the wave power drops
to 25% of its initial value.
z(25%) =
6
6
=
= 2.0 ù 10*2 m = 2.0 cm
8.69↵ (8.69)(34.4)
g) What dB power drop corresponds to one penetration depth, ? In general, the decibel power
loss is given by 8.69↵z. So, when z = = 1_↵, we have 8.69 dB.
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11.12. Repeat Problem 11.11, except that the wave now propagates in fresh water, having conductivity
10*3 S/m, dielectric constant ✏r® = 80.0, and permeability 0 :
=
a) Calculate the loss tangent and determine whether the medium would qualify as a good dielectric
or a good conductor:
!✏
=
®
10*3
= 0.002
(2⇡ ù 108 )(80.0)(8.85 ù 10*12 )
This falls within the good dielectric approximation.
b) Calculate ↵, , and ⌘. Assuming a good dielectric, we have:
u
⌘0
.
377(10*3 )
0
↵=
= ˘
=
= 0.0211 Np_m
˘
2 ✏0
2 ✏r®
2 80
t
8˘
. t
® = !
® = 2⇡ ù 10
=!
✏
✏
✏
80 = 18.7 rad_m
0 0 r
r
c
3 ù 108
v
.
377
0
⌘=
= ˘ = 42.2 ohms
®
✏r ✏0
80
c) Assuming forward z propagation and x polarization, write the electric field in phasor form:
Es = 100e*↵z e*j z ax
where ↵ and
are as found in part b.
d) Write the magnetic field in phasor form:
Hs =
100 *↵z *j z
e e
ay = 2.37e*↵z e*j z ay
42.2
e) Write the time-average Poynting vector, < S > in W_m2 .
1
1
< S >= Re Es ù H<s = (100)(2.37)e*2↵z az = 118 e*2↵z az W_m2
2
2
f) Find the 6-dB material thickness; i.e., the propagation distance over which the wave power drops
to 25% of its initial value.
z(25%) =
6
6
=
= 32.7 m
8.69↵ (8.69)(0.0211)
g) What dB power drop corresponds to one penetration depth, ? In general, the decibel power
loss is given by 8.69↵z. So, when z = = 1_↵, we have 8.69 dB.
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11.13. Let jk = 0.2+j1.5 m*1 and ⌘ = 450+j60 ⌦ for a uniform plane wave propagating in the az direction.
If ! = 300 Mrad/s, find , ✏ ® , and ✏ ®® : We begin with
u
1
⌘=
= 450 + j60
˘
®
✏
1 * j(✏ ®® _✏ ® )
and
Then
and
˘
˘
jk = j!
✏ ® 1 * j(✏ ®® _✏ ® ) = 0.2 + j1.5
⌘⌘ < =
˘
✏®
1
= (450 + j60)(450 * j60) = 2.06 ù 105
(1)
1 + (✏ ®® _✏ ® )2 = (0.2 + j1.5)(0.2 * j1.5) = 2.29
(2)
1 + (✏ ®® _✏ ® )2
(jk)(jk)< = !2 ✏ ®
t
Taking the ratio of (2) to (1),
(jk)(jk)<
2.29
= !2 (✏ ® )2 1 + (✏ ®® _✏ ® )2 =
= 1.11 ù 10*5
<
⌘⌘
2.06 ù 105
Then with ! = 3 ù 108 ,
(✏ ® )2 =
1.11 ù 10*5
1.23 ù 10*22
=
(3 ù 108 )2 1 + (✏ ®® _✏ ® )2
1 + (✏ ®® _✏ ® )2
(3)
Now, we use Eqs. (35) and (36). Squaring these and taking their ratio gives
˘
1 + (✏ ®® _✏ ® )2
(0.2)2
↵2
=
=
˘
2
(1.5)2
1 + (✏ ®® _✏ ® )2
We solve this to find ✏ ®® _✏ ® = 0.271. Substituting this result into (3) gives ✏ ® = 1.07 ù 10*11 F/m.
Since ✏ ®® _✏ ® = 0.271, we then find ✏ ®® = 2.90 ù 10*12 F/m. Finally, using these results in either (1)
or (2) we find = 2.28 ù 10*6 H/m. Summary: = 2.28 ù 10*6 H_m,
✏ ® = 1.07 ù 10*11 F_m, and ✏ ®® = 2.90 ù 10*12 F_m.
246
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11.14. A certain nonmagnetic material has the material constants ✏r® = 2 and ✏ ®® _✏ ® = 4 ù 10*4 at ! = 1.5
Grad/s. Find the distance a uniform plane wave can propagate through the material before:
a) it is attenuated by 1 Np: First, ✏ ®® = (4 ù 104 )(2)(8.854 ù 10*12 ) = 7.1 ù 10*15 F/m. Then, since
✏ ®® _✏ ® << 1, we use the approximate form for ↵, given by Eq. (51) (written in terms of ✏ ®® ):
u
. !✏ ®®
(1.5 ù 109 )(7.1 ù 10*15 ) 377
*3
↵=
=
˘ = 1.42 ù 10 Np_m
2
✏®
2
2
The required distance is now z1 = (1.42 ù 10*3 )*1 = 706 m
b) the power level is reduced by one-half: The governing relation is e*2↵z1_2 = 1_2, or z1_2 =
ln 2_2↵ = ln 2_2(1.42 ù 10* 3) = 244 m.
c) the phase shifts 360˝ : This distance is defined as
˘ one wavelength, where
˘
= (2⇡c)_(! ✏r® ) = [2⇡(3 ù 108 )]_[(1.5 ù 109 ) 2] = 0.89 m.
= 2⇡_
11.15. A 10 GHz radar signal may be represented as a uniform plane wave in a sufficiently small region.
Calculate the wavelength in centimeters and the attenuation in nepers per meter if the wave is propagating in a non-magnetic material for which
a) ✏r® = 1 and ✏r®® = 0: In a non-magnetic material, we would have:
v
1_2
u
0 ®® 12
c
® b
✏r
0 ✏0 ✏r f
↵=!
1+
* 1g
g
2 f
✏r®
d
e
and
v
1_2
u
0 ®® 12
c
® b
✏
✏
✏
0 0 r f
r
=!
1+
+ 1g
g
2 f
✏r®
d
e
˘
®
®®
With the given values of ✏r and ✏r , it is clear that
= !
0 ✏0 = !_c, and so
= 2⇡_ = 2⇡c_! = 3 ù 1010 _1010 = 3 cm. It is also clear that ↵ = 0.
. ˘
b) ✏r® = 1.04 and ✏r®® = 9.00 ù 10*4 : In this case ✏r®® _✏r® << 1, and so = ! ✏r® _c = 2.13 cm*1 .
Thus = 2⇡_ = 2.95 cm. Then
˘
u
®®
!✏r®®
. !✏ ®®
0 ✏0
! ✏r
2⇡ ù 1010 (9.00 ù 10*4 )
↵=
=
=
=
˘
˘
˘
2
✏®
2
2c ✏ ®
2 ù 3 ù 108
✏r®
1.04
r
*2
= 9.24 ù 10 Np_m
c) ✏r® = 2.5 and ✏r®® = 7.2: Using the above formulas, we obtain
v
1_2
˘
b
c
⇠ ⇡2
2⇡ ù 1010 2.5 f
7.2
=
1+
+ 1g = 4.71 cm*1
˘
f
g
2.5
10
(3 ù 10 ) 2 d
e
and so
= 2⇡_ = 1.33 cm. Then
v
1_2
˘
b
c
⇠ ⇡2
2.5 f
7.2
↵=
1+
* 1g = 335 Np_m
˘
f
g
2.5
8
(3 ù 10 ) 2 d
e
2⇡ ù 1010
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11.16. Consider the power dissipation term, î E Jdv in Poynting’s theorem (Eq.(70)). This gives the power
lost to heat within a volume into which electromagnetic waves enter. The term pd = E J is thus the
power dissipation per unit volume in W_m3 . Following the same reasoning that resulted in Eq.(77),
the time-average power dissipation per volume will be < pd >= (1_2)Re Es J<s .
a) Show that in a conducting medium, through which a uniform plane wave of amplitude E0 propagates in the forward z direction, < pd >= ( _2)E0 2 e*2↵z : Begin with the phasor expression
for the electric field, assuming complex amplitude E0 , and x-polarization:
Es = E0 e*↵z e*j z ax V_m2
Then
Js = Es = E0 e*↵z e*j z ax A_m2
So that
< pd >= (1_2)Re E0 e*↵z e*j z ax
E0< e*↵z e+j z ax = ( _2)E0 2 e*2↵z
b) Confirm this result for the special case of a good conductor by using the left hand side of
Eq. (70), and consider a very small volume. In a good conductor, the intrinsic impedance
is, from Eq. (85), ⌘c = (1 + j)_( ), where the skin depth, = 1_↵. The magnetic field phasor
is then
E
Hs = s ay =
E e*↵z e*j z ay A_m
⌘c
(1 + j)↵ 0
The time-average Poynting vector is then
1
< S >= Re Es ù H<s =
E 2 e*2↵z az W_m2
2
4↵ 0
Now, consider a rectangular volume of side lengths, x, y, and z, all of which are very
small. As the wave passes through this volume in the forward z direction, the power dissipated
will be the difference between the power at entry (at z = 0), and the power that exits the volume
.
(at z = z). With small z, we may approximate e*2↵z = 1 * 2↵z, and the dissipated power in
the volume becomes
⌧
⌧
Pd = Pin * Pout =
E0 2 x y *
E 2 (1 * 2↵ z) x y = E0 2 ( x y z)
4↵
4↵ 0
2
This is just the result of part a, evaluated at z = 0 and multiplied by the volume. The relation
becomes exact as z ô 0, in which case < pd >ô ( _2)E0 2 .
It is also possible to show the relation by using Eq. (69) (which involves taking the divergence
of < S >), or by removing the restriction of a small volume and evaluating the integrals in
Eq. (70) without approximations. Either method is straightforward.
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11.17. Consider a long solid cylindrical wire having uniform conductivity . The wire is oriented along the
z axis, and has radius a and length L. Constant voltage V0 is applied between the two ends.
a. Write the electric field intensity, E, in terms of V0 and L. Assume a positive z directed field:
Would have E = V0 _L az .
b. Using Ampere’s circuital law, find the magnetic field intensity, H, inside the wire as a function
of ⇢.
Õ
J nda Ÿ 2⇡⇢H = ⇡⇢2 J = ⇡⇢2 V0 _L
H dL =
where J = E = ( V0 _L) az . So
⇢V0
a A_m
2L
H=
(0 < ⇢ < a)
c. Find the Poynting vector S = E ù H:
V0
⇢V0
az ù
a
L
2L
S=
=
* ⇢V02
2L2
a⇢ W_m2
d. Evaluate the left hand side of Poynting’s theorem by integrating the Poynting vector over the
wire surface to find the power transferred into the volume:
*
Õ
S n da = *
0
⇢V02
L *
2⇡
2L2
0
a⇢ a⇢ ad dz =
⇡a2
V02 = V02 _R W
L
where the wire resistance is R = L_(⇡a2 ) ohms.
e. Evaluate the right hand side of Poynting’s theorem, and thus verify that the theorem is satisfied
in this situation: In the absence of time variation, all time derivatives are zero, and Poynting’s
theorem is simplified to read:
*
Õs
S n da =
v
E J dv
The right hand side evaluates as
v
E J dv =
2⇡
L
0
0
V02
a
0
L
⇢ d⇢ d dz =
2
⇡a2
V02 = V02 _R
L
which is the same as the left side evaluation, found in part d. So the theorem works as it should!
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11.18. Given, a 100MHz uniform plane wave in a medium known to be a good dielectric. The phasor electric
field is Es = 4e*0.5z e*j20z ax V/m. Not stated in the problem is the permeability, which we take to be
0 . Determine:
a) ✏ ® : As a first step, it is useful to see just how much of a good dielectric we have. We use the
good dielectric approximations, Eqs. (60a) and (60b), with = !✏ ®® . Using these, we take the
ratio, _↵, to find
˘
⇧
⌅
0 ®1
0 1
!
✏ ® 1 + (1_8)(✏ ®® _✏ ® )2
20
✏
1 ✏ ®®
=
=
= 2 ®® +
˘
↵ 0.5
✏
4 ✏®
(!✏ ®® _2)
_✏ ®
This becomes the quadratic equation:
0 ®® 12
0 ®® 1
✏
✏
*
160
+8=0
✏®
✏®
The solution to the quadratic is ✏ ®® _✏ ® = 0.05, which means that we can neglect the second
˘
. ˘ ®
term in Eq. (60b), so that = !
✏ = (!_c) ✏r® . With the given frequency of 100 MHz,
˘
and with = 0 , we find ✏r® = 20(3_2⇡) = 9.55, so that ✏r® = 91.3, and finally ✏ ® = ✏r® ✏0 =
8.1 ù 10*10 F_m.
b) ✏ ®® : Using Eq. (60a), the set up is
u
u
2(0.5)
!✏ ®®
✏®
10*8 ˘
®®
↵ = 0.5 =
Ÿ
✏
=
=
91.3 = 4.0 ù 10*11 F_m
2
✏®
2⇡(377)
2⇡ ù 108
c) ⌘: Using Eq. (62b), we find
u 4
0 15
.
1 ✏ ®®
377
⌘=
1
+
j
=˘
(1 + j.025) = (39.5 + j0.99) ohms
✏®
2 ✏®
91.3
d) Hs : This will be a y-directed field, and will be
Hs =
Es
4
ay =
e*0.5z e*j20z ay = 0.101e*0.5z e*j20z e*j0.025 ay A_m
⌘
(39.5 + j0.99)
e) < S >: Using the given field and the result of part d, obtain
(0.101)(4) *2(0.5)z
1
< S >= Re{Es ù H<s } =
e
cos(0.025) az = 0.202e*z az W_m2
2
2
f) the power in watts that is incident on a rectangular surface measuring 20m x 30m at z = 10m:
At 10m, the power density is < S >= 0.202e*10 = 9.2 ù 10*6 W_m2 . The incident power on
the given area is then P = 9.2 ù 10*6 ù (20)(30) = 5.5 mW.
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11.19. Perfectly-conducting cylinders with radii of 8 mm and 20 mm are coaxial. The region between the
cylinders is filled with a perfect dielectric for which ✏ = 10*9 _4⇡ F/m and r = 1. If E in this region
is (500_⇢) cos(!t * 4z)a⇢ V/m, find:
a) !, with the help of Maxwell’s equations in cylindrical coordinates: We use the two curl equations, beginning with ( ù E = *)B_)t, where in this case,
)E⇢
(ùE=
So
)z
a =
)B
2000
sin(!t * 4z)a = *
a
⇢
)t
2000
2000
sin(!t * 4z)dt =
cos(!t * 4z) T
⇢
!⇢
B =
Then
H =
B
0
=
2000
cos(!t * 4z) A_m
(4⇡ ù 10*7 )!⇢
We next use ( ù H = )D_)t, where in this case
(ùH=*
)H
)z
a⇢ +
1 )(⇢H )
az
⇢ )⇢
where the second term on the right hand side becomes zero when substituting our H . So
(ùH=*
)H
)z
a⇢ = *
)D⇢
8000
sin(!t
*
4z)a
=
a
⇢
)t ⇢
(4⇡ ù 10*7 )!⇢
And
D⇢ =
*
8000
8000
sin(!t * 4z)dt =
cos(!t * 4z) C_m2
(4⇡ ù 10*7 )!⇢
(4⇡ ù 10*7 )!2 ⇢
Finally, using the given ✏,
D⇢
E⇢ =
✏
=
8000
cos(!t * 4z) V_m
(10*16 )!2 ⇢
This must be the same as the given field, so we require
8000
500
=
Ÿ ! = 4 ù 108 rad_s
*16
2
⇢
(10 )! ⇢
b) H(⇢, z, t): From part a, we have
H(⇢, z, t) =
2000
4.0
cos(!t * 4z)a =
cos(4 ù 108 t * 4z)a A_m
*7
⇢
(4⇡ ù 10 )!⇢
c) S(⇢, , z): This will be
S(⇢, , z) = E ù H =
=
500
4.0
cos(4 ù 108 t * 4z)a⇢ ù
cos(4 ù 108 t * 4z)a
⇢
⇢
2.0 ù 10*3
cos2 (4 ù 108 t * 4z)az W_m2
⇢2
251
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11.19
d) the average power passing through every cross-section 8 < ⇢ < 20 mm, 0 < < 2⇡. Using
the result of part c, we find < S >= (1.0 ù 103 )_⇢2 az W_m2 . The power through the given
cross-section is now
P=
2⇡
0
.020
.008
⇠ ⇡
1.0 ù 103
20
3
⇢
d⇢
d
=
2⇡
ù
10
ln
= 5.7 kW
2
8
⇢
11.20. Voltage breakdown in air at standard temperature and pressure occurs at an electric field strength
of approximately 3 ù 106 V/m. This becomes an issue in some high-power optical experiments,
in which tight focusing of light may be necessary. Estimate the lightwave power in watts that can
be focused into a cylindrical beam of 10 m radius before breakdown occurs. Assume uniform plane
wave behavior (although this assumption will produce an answer that is higher than the actual number
by as much as a factor of 2, depending on the actual beam shape).
The power density in the beam in free space can be found as a special case of Eq. (76) (with
⌘ = ⌘0 , ✓⌘ = ↵ = 0):
<S>=
E02
2⌘0
=
(3 ù 106 )2
= 1.2 ù 1010 W_m2
2(377)
To avoid breakdown, the power in a 10- m radius cylinder is then bounded by
P < (1.2 ù 1010 )(⇡ ù (10*5 )2 ) = 3.75 W
252
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11.21. The cylindrical shell, 1 cm < ⇢ < 1.2 cm, is composed of a conducting material for which = 106
S/m. The external and internal regions are non-conducting. Let H = 2000 A/m at ⇢ = 1.2 cm.
a) Find H everywhere: Use Ampere’s circuital law, which states:
Õ
H dL = 2⇡⇢(2000) = 2⇡(1.2 ù 10*2 )(2000) = 48⇡ A = Iencl
Then in this case
J=
I
48
a =
a = 1.09 ù 106 az A_m2
Area z (1.44 * 1.00) ù 10*4 z
With this result we again use Ampere’s circuital law to find H everywhere within the shell as a
function of ⇢ (in meters):
H 1 (⇢) =
1
2⇡⇢
2⇡
0
⇢
.01
1.09 ù 106 ⇢ d⇢ d =
54.5 4 2
(10 ⇢ * 1) A_m (.01 < ⇢ < .012)
⇢
Outside the shell, we would have
H 2 (⇢) =
48⇡
= 24_⇢ A_m (⇢ > .012)
2⇡⇢
Inside the shell (⇢ < .01 m), H = 0 since there is no enclosed current.
b) Find E everywhere: We use
E=
J
=
1.09 ù 106
az = 1.09 az V_m
106
which is valid, presumeably, outside as well as inside the shell.
c) Find S everywhere: Use
P = E ù H = 1.09 az ù
=*
54.5 4 2
(10 ⇢ * 1) a
⇢
59.4 4 2
(10 ⇢ * 1) a⇢ W_m2 (.01 < ⇢ < .012 m)
⇢
Outside the shell,
S = 1.09 az ù
24
26
a = * a⇢ W_m2 (⇢ > .012 m)
⇢
⇢
253
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11.22. The inner and outer dimensions of a copper coaxial transmission line are 2 and 7 mm, respectively.
Both conductors have thicknesses much greater than . The dielectric is lossless and the operating
frequency is 400 MHz. Calculate the resistance per meter length of the:
a) inner conductor: First
1
=˘
⇡f
=˘
1
⇡(4 ù 108 )(4⇡ ù 10*7 )(5.8 ù 107 )
= 3.3 ù 10*6 m = 3.3 m
Now, using (90) with a unit length, we find
Rin =
1
2⇡a
=
1
2⇡(2 ù 10*3 )(5.8 ù 107 )(3.3 ù 10*6 )
= 0.42 ohms_m
b) outer conductor: Again, (90) applies but with a different conductor radius. Thus
a
2
Rout = Rin = (0.42) = 0.12 ohms_m
b
7
c) transmission line: Since the two resistances found above are in series, the line resistance is their
sum, or R = Rin + Rout = 0.54 ohms_m.
11.23. A hollow tubular conductor is constructed from a type of brass having a conductivity of 1.2 ù 107
S/m. The inner and outer radii are 9 mm and 10 mm respectively. Calculate the resistance per meter
length at a frequency of
a) dc: In this case the current density is uniform over the entire tube cross-section. We write:
R(dc) =
L
1
= 1.4 ù 10*3 ⌦_m
=
7
A (1.2 ù 10 )⇡(.012 * .0092 )
b) 20 MHz: Now the skin effect will limit the effective cross-section. At 20 MHz, the skin depth
is
(20MHz) = [⇡f 0 ]*1_2 = [⇡(20 ù 106 )(4⇡ ù 10*7 )(1.2 ù 107 )]*1_2 = 3.25 ù 10*5 m
This is much less than the outer radius of the tube. Therefore we can approximate the resistance
using the formula:
R(20MHz) =
L
1
1
=
=
= 4.1 ù 10*2 ⌦_m
7
A 2⇡b
(1.2 ù 10 )(2⇡(.01))(3.25 ù 10*5 )
c) 2 GHz: Using the same formula as in part b, we find the skin depth at 2 GHz to be = 3.25ù10*6
m. The resistance (using the other formula) is R(2GHz) = 4.1 ù 10*1 ⌦_m.
254
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11.24
a) Most microwave ovens operate at 2.45 GHz. Assume that = 1.2 ù 106 S/m and
the stainless steel interior, and find the depth of penetration:
1
=˘
⇡f
=˘
1
⇡(2.45 ù 109 )(4⇡ ù 10*7 )(1.2 ù 106 )
r = 500 for
= 9.28 ù 10*6 m = 9.28 m
b) Let Es = 50⁄ 0˝ V/m at the surface of the conductor, and plot a curve of the amplitude of Es vs.
the angle of Es as the field propagates
into the stainless steel: Since the conductivity is high, we
.
. ˘
use (82) to write ↵ = = ⇡f
= 1_ . So, assuming that the direction into the conductor
is z, the depth-dependent field is written as
Es (z) = 50e*↵z e*j z = 50e*z_ e*jz_ = 50 exp(*z_9.28) exp(*j z_9.28 )
´≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠¨
´Ø¨
amplitude
angle
where z is in microns. Therefore, the plot of amplitude versus angle is simply a plot of e*x versus
x, where x = z_9.28; the starting amplitude is 50 and the 1_e amplitude (at z = 9.28 m) is
18.4.
11.25. A good conductor is planar in form and carries a uniform plane wave that has a wavelength of 0.3
mm and a velocity of 3 ù 105 m/s. Assuming the conductor is non-magnetic, determine the frequency
and the conductivity: First, we use
f=
v
Ÿ
=
=
3 ù 105
= 109 Hz = 1 GHz
3 ù 10*4
Next, for a good conductor,
=
2⇡
=˘
1
⇡f
4⇡
2f
=
4⇡
= 1.1 ù 105 S_m
(9 ù 10*8 )(109 )(4⇡ ù 10*7 )
255
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11.26. The dimensions of a certain coaxial transmission line are a = 0.8mm and b = 4mm. The outer
conductor thickness is 0.6mm, and all conductors have = 1.6 ù 107 S/m.
a) Find R, the resistance per unit length, at an operating frequency of 2.4 GHz: First
=˘
1
⇡f
1
=˘
= 2.57 ù 10*6 m = 2.57 m
8
*7
7
⇡(2.4 ù 10 )(4⇡ ù 10 )(1.6 ù 10 )
Then, using (90) with a unit length, we find
Rin =
1
2⇡a
=
1
2⇡(0.8 ù 10*3 )(1.6 ù 107 )(2.57 ù 10*6 )
= 4.84 ohms_m
The outer conductor resistance is then found from the inner through
a
0.8
Rout = Rin =
(4.84) = 0.97 ohms_m
b
4
The net resistance per length is then the sum, R = Rin + Rout = 5.81 ohms_m.
b) Use information from Secs. 6.3 and 8.10 to find C and L, the capacitance and inductance per
unit length, respectively. The coax is air-filled. From those sections, we find (in free space)
C=
L=
c) Find ↵ and if ↵ + j
expression, we find
↵ = Re
and
= Im
2⇡✏0
2⇡(8.854 ù 10*12 )
=
= 3.46 ù 10*11 F_m
ln(b_a)
ln(4_.8)
0
ln(b_a) =
2⇡
=
$˘
˘
4⇡ ù 10*7
ln(4_.8) = 3.22 ù 10*7 H_m
2⇡
j!C(R + j!L): Taking real and imaginary parts of the given
Lu
M1_2
˘
⇠
⇡
! LC
R 2
j!C(R + j!L) = ˘
1+
*1
!L
2
$˘
%
Lu
M1_2
˘
⇠
⇡
! LC
R 2
j!C(R + j!L) = ˘
1+
+1
!L
2
%
These can be found by writing out
$˘
%
˘
↵ = Re
j!C(R + j!L) = (1_2) j!C(R + j!L) + c.c.
where c.c denotes the complex conjugate. The result is squared, terms collected, and the
square root taken. Now, using the values of R, C, and L found in parts a and b, we find
↵ = 3.0 ù 10*2 Np_m and = 50.3 rad_m.
256
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11.27. The planar surface at z = 0 is a brass-Teflon interface. Use data available in Appendix C to evaluate
the following ratios for a uniform plane wave having ! = 4 ù 1010 rad/s:
a) ↵Tef _↵brass : From the appendix we find ✏ ®® _✏ ® = .0003 for Teflon, making the material a good
dielectric. Also, for Teflon, ✏r® = 2.1. For brass, we find = 1.5ù107 S/m, making brass a good
conductor at the stated frequency. For a good dielectric (Teflon) we use the approximations:
u
0 ®® 1 ⇠ ⇡ ˘
0 1 t
.
✏
1
1 ✏ ®® !
®
↵=
=
!
✏ =
✏r®
2 ✏®
✏®
2
2 ✏® c
4
0 15
t
. ˘ ®
. ˘
1 ✏ ®®
!
=!
✏ 1+
=
!
✏®
=
✏r®
8 ✏®
c
For brass (good conductor) we have
u ⇠ ⇡
.
. ˘
1
↵ = = ⇡f brass = ⇡
(4 ù 1010 )(4⇡ ù 10*7 )(1.5 ù 107 ) = 6.14 ù 105 m*1
2⇡
Now
˘
˘
1_2 ✏ ®® _✏ ® (!_c) ✏r®
↵Tef
(1_2)(.0003)(4 ù 1010 _3 ù 108 ) 2.1
=
=
= 4.7 ù 10*8
˘
5
↵brass
6.14
ù
10
⇡f brass
b)
(2⇡_ Tef )
=
=
(2⇡_
brass
brass )
Tef
c)
brass
Tef
˘
c ⇡f brass
(3 ù 108 )(6.14 ù 105 )
=
=
= 3.2 ù 103
t
˘
(4 ù 1010 ) 2.1
! ✏r® Tef
vTef
(!_ Tef )
=
=
vbrass
(!_ brass )
brass
Tef
= 3.2 ù 103 as before
11.28. A uniform plane wave in free space has electric field given by Es = 10e*j x az + 15e*j x ay V/m.
a) Describe the wave polarization: Since the two components have a fixed phase difference (in
this case zero) with respect to time and position, the wave has linear polarization, with the field
vector in the yz plane at angle = tan*1 (10_15) = 33.7˝ to the y axis.
b) Find Hs : With propagation in forward x, we would have
Hs =
*10 *j x
15 *j x
e
ay +
e
az A_m = *26.5e*j x ay + 39.8e*j x az mA_m
377
377
c) determine the average power density in the wave in W_m2 : Use
4
5
(15)2
1
1 (10)2
Pavg = Re Es ù H<s =
ax +
ax = 0.43ax W_m2 or Pavg = 0.43 W_m2
2
2 377
377
257
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11.29. Consider a left-circularly polarized wave in free space that propagates in the forward z direction. The
electric field is given by the appropriate form of Eq. (100).
a) Determine the magnetic field phasor, Hs :
We begin, using (100), with Es = E0 (ax + jay )e*j z . We find the two components of Hs
separately, using the two components of Es . Specifically, the x component of Es is associated
with a y component of Hs , and the y component of Es is associated with a negative x component
of Hs . The result is
E
Hs = 0 ay * jax e*j z
⌘0
b) Determine an expression for the average power density in the wave in W_m2 by direct application of Eq. (77): We have
0
1
E0
1
1
<
*j z
+j z
Pz,avg = Re(Es ù Hs ) = Re E0 (ax + jay )e
ù
(a * jax )e
2
2
⌘0 y
=
E02
⌘0
az W_m2 (assuming E0 is real)
11.30. In an anisotropic medium, permittivity varies with electric field direction, and is a property seen in
most crystals. Consider a uniform plane wave propagating in the z direction in such a medium, and
which enters the material with equal field components along the x and y axes. The field phasor will
take the form:
Es (z) = E0 (ax + ay ej z ) e*j z
where
= x * y is the difference in phase constants for waves that are linearly-polarized in the x
and y directions. Find distances into the material (in terms of
) at which the field is:
a) Linearly-polarized: We want the x and y components to be in phase, so therefore
m⇡
zlin = m⇡ Ÿ zlin =
, (m = 1, 2, 3...)
b) Circularly-polarized: In this case, we want the two field components to be in quadrature phase,
such that the total field is of the form, Es = E0 (ax ± jay )e*j z . Therefore,
zcirc =
(2n + 1)⇡
(2n + 1)⇡
Ÿ zcirc =
, (n = 0, 1, 2, 3...)
2
2
c) Assume intrinsic impedance ⌘ that is approximately constant with field orientation and find Hs
and < S >: Magnetic field is found by looking at the individual components:
Hs (z) =
E0
ay * ax ej
⌘
z
e*j z and
E2
1
< S >= Re Es ù H<s = 0 az W_m2
2
⌘
where it is assumed that E0 is real. ⌘ is real because the medium is evidently lossless.
258
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11.31. A linearly-polarized uniform plane wave, propagating in the forward z direction, is input to a lossless anisotropic material, in which the dielectric constant encountered by waves polarized along
y (✏ry ) differs from that seen by waves polarized along x (✏rx ). Suppose ✏rx = 2.15, ✏ry = 2.10,
and the wave electric field at input is polarized at 45˝ to the positive x and y axes. Assume free space
wavelength .
a) Determine the shortest length of the material such that the wave as it emerges from the output
end is circularly polarized: With the input field at 45˝ , the x and y components are of equal
magnitude, and circular polarization will result if the phase difference between the components
is ⇡_2. Our requirement over length L is thus x L * y L = ⇡_2, or
L=
⇡
2( x *
y)
=
⇡c
˘
˘
2!( ✏rx * ✏ry )
With the given values, we find,
L=
(58.3)⇡c
= 58.3 = 14.6
2!
4
b) Will the output wave be right- or left-circularly-polarized? With the dielectric constant greater
for x-polarized waves, the x component will lag the y component in time at the output. The
field can thus be written as E = E0 (ay * jax ), which is lef t circular polarization.
11.32. Suppose that the length of the medium of Problem 11.31 is made to be twice that as determined in
the problem. Describe the polarization of the output wave in this case: With the length doubled, a
phase shift of ⇡ radians develops between the two components. At the input, we can write the field
as Es (0) = E0 (ax + ay ). After propagating through length L, we would have,
Es (L) = E0 [e*j x L ax + e*j y L ay ] = E0 e*j x L [ax + e*j( y * x )L ay ]
where ( y * x )L = *⇡ (since x > y ), and so Es (L) = E0 e*j x L [ax * ay ]. With the reversal of the
y component, the wave polarization is rotated by 90˝ , but is still linear polarization.
11.33. Given a wave for which Es = 15e*j z ax + 18e*j z ej ay V/m, propagating in a medium characterized
by complex intrinsic impedance, ⌘.
a) Find Hs : With the wave propagating in the forward z direction, we find:
Hs =
⇧
1⌅
*18ej ax + 15ay e*j z A_m
⌘
b) Determine the average power density in W_m2 : We find
<
=
< =
(15)2 (18)2
1
1
1
Pz,avg = Re Es ù H<s = Re
+
=
275
Re
W_m2
2
2
⌘<
⌘<
⌘<
259
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11.34. Given the general elliptically-polarized wave as per Eq. (93):
Es = [Ex0 ax + Ey0 ej ay ]e*j z
a) Show, using methods similar to those of Example 11.7, that a linearly polarized wave results
when superimposing the given field and a phase-shifted field of the form:
Es = [Ex0 ax + Ey0 e*j ay ]e*j z ej
where
is a constant: Adding the two fields gives
⌅
Es,tot = Ex0 1 + ej ax + Ey0 ej + e*j ej
⇧
ay e*j z
c
b
f
g
= fEx0 ej _2 e*j _2 + ej _2 ax + Ey0 ej _2 e*j _2 ej + e*j ej _2 ay g e*j z
f
´≠≠≠≠≠≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠≠≠≠≠≠¨ g
´≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠¨
f
g
2 cos( _2)
2 cos( * _2)
d
e
⌅
⇧ j _2 *j z
This simplifies to Es,tot = 2 Ex0 cos( _2)ax + Ey0 cos( * _2)ay e e
, which is
linearly polarized.
b) Find in terms of such that the resultant wave is polarized along x: By inspecting the part a
result, we achieve a zero y component when 2 * = ⇡ (or odd multiples of ⇡).
260
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CHAPTER 12 – 9th Edition
+
+
12.1. A uniform plane wave in air, Ex1
= Ex10
cos(1010 t * z) V/m, is normally-incident on a copper
surface at z = 0. What percentage of the incident power density is transmitted into the copper? We
need to find the reflection coefficient. The intrinsic impedance of copper (a good conductor) is
v
u
u
j!
!
1010 (4⇡ ù 107 )
⌘c =
= (1 + j)
= (1 + j)
= (1 + j)(.0104)
2
2(5.8 ù 107 )
Note that the accuracy here is questionable, since we know the conductivity to only two significant
figures. We nevertheless proceed: Using ⌘0 = 376.7288 ohms, we write
=
⌘c * ⌘0
.0104 * 376.7288 + j.0104
=
= *.9999 + j.0001
⌘c + ⌘0
.0104 + 376.7288 + j.0104
Now  2 = .9999, and so the transmitted power fraction is 1 *  2 = .0001, or about 0.01% is
transmitted.
®® = 0, and
12.2. The plane z = 0 defines the boundary between two dielectrics. For z < 0, ✏r1 = 9, ✏r1
+
®
®®
1 = 0 . For z > 0, ✏r2 = 3, ✏r2 = 0, and 2 = 0 . Let Ex1 = 10 cos(!t * 15z) V/m and find
a) !: We have
1.5 ù 109 s*1 .
t
t
t
˘
®
®
®
8
= !
0 ✏1 = ! ✏r1 _c = 15. So ! = 15c_ ✏r1 = 15 ù (3 ù 10 )_ 9 =
t
t
˘
® = ⌘ _ ✏ ® = 377_ 9 = 126 ohms. Next we apply
b) < S+
>:
First
we
need
⌘
=
_✏
1
0 1
0
1
r1
Eq. (76), Chapter 11, to evaluate the Poynting vector (with no loss and consequently with no
phase difference between electric and magnetic fields). We find < S+
>= (1_2)E1 2 _⌘1 az =
1
(1_2)(10)2 _126 az = 0.40 az W_m2 .
c) < S*
>: First, we need to evaluate the reflection coefficient:
1
t
t
t
t
˘
˘
® * ⌘ _ ✏®
® *
®
⌘
_
✏
✏
✏r2
0
0
r2
r1
r1
⌘2 * ⌘1
9* 3
=
=
=t
=˘
t
t
t
˘ = 0.27
⌘2 + ⌘1
® + ⌘ _ ✏®
® +
®
9
+
3
⌘0 _ ✏r2
✏
✏
0
r1
r1
r2
Then < S*
>= * 2 < S+
>= *(0.27)2 (0.40) az = *0.03 az W_m2 .
1
1
d) < S+
>: This will be the remaining power, propagating in the forward z direction, or < S+
>=
2
2
0.37 az W_m2 .
261
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12.3. A uniform plane wave in region 1 is normally-incident on the planar boundary separating regions 1
® = 3 and ✏ ® = 3 , find the ratio ✏ ® _✏ ® if 20% of the energy in the
and 2. If ✏1®® = ✏2®® = 0, while ✏r1
r1
r2
r2
r2 r1
incident wave is reflected at the boundary. There are two possible answers. First, since  2 = .20,
and since both permittivities and permeabilities are real, = ±0.447. we then set up
t
t
® )*⌘
® )
⌘
(
_✏
( r1 _✏r1
0
r2 r2
0
⌘2 * ⌘1
= ±0.447 =
= t
t
⌘2 + ⌘1
® )+⌘
® )
⌘0 ( r2 _✏r2
( r1 _✏r1
0
t
t
3
3
( r2 _ r2
) * ( r1 _ r1
)
* r2
=t
= r1
t
3
3
r1 + r2
( r2 _ r2
) + ( r1 _ r1
)
Therefore
®
✏r2
1 - 0.447
=
= (0.382, 2.62) Ÿ ® =
1 ± 0.447
✏r1
r1
r2
0
r2
13
= (0.056, 17.9)
r1
12.4. A 10-MHz uniform plane wave having an initial average power density of 5W_m2 is normally® = 5, and
incident from free space onto the surface of a lossy material in which ✏2®® _✏2® = 0.05, ✏r2
2 = 0 . Calculate the distance into the lossy medium at which the transmitted wave power density
is down by 10dB from the initial 5W_m2 :
First, since ✏2®® _✏2® = 0.05 << 1, we recognize region 2 as a good dielectric. Its intrinsic
impedance is therefore approximated well by Eq. (62b), Chapter 11:
M
v L
®®
✏
1
377
0
⌘2 =
1 + j 2® = ˘ [1 + j0.025]
2 ✏2
✏2®
5
The reflection coefficient encountered by the incident wave from region 1 is therefore
˘
˘
⌘2 * ⌘1
(377_ 5)[1 + j.025] * 377 (1 * 5) + j.025
=
=
=
= *0.383 + j0.011
˘
˘
⌘2 + ⌘1
(377_ 5)[1 + j.025] + 377 (1 + 5) + j.025
The fraction of the incident power that is reflected is then  2 = 0.147, and thus the fraction
of the power that is transmitted into region 2 is 1 *  2 = 0.853. Still using the good dielectric
approximation, the attenuation coefficient in region 2 is found from Eq. (60a), Chapter 11:
®® v
. !✏2
377
0
↵=
= (2⇡ ù 107 )(0.05 ù 5 ù 8.854 ù 10*12 ) ˘ = 2.34 ù 10*2 Np_m
2
✏2®
2 5
Now, the power that propagates into region 2 is expressed in terms of the incident power through
*2
< S2 > (z) = 5(1 *  2 )e*2↵z = 5(.853)e*2(2.34ù10 )z = 0.5 W_m2
in which the last equality indicates a factor of ten reduction from the incident power, as occurs
for a 10 dB loss. Solve for z to obtain
z=
ln(8.53)
= 45.8 m
2(2.34 ù 10*2 )
262
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12.5. The region z < 0 is characterized by ✏r® = r = 1 and ✏r®® = 0. The total E field here is given as the
sum of the two uniform plane waves, Es = 150e*j10z ax + (50⁄20˝ )ej10z ax V/m.
a) What is the operating frequency? In free space,
! = 3 ù 109 s*1 , or f = !_2⇡ = 4.7 ù 108 Hz.
= k0 = 10 = !_c = !_3 ù 108 . Thus,
b) Specify the intrinsic impedance of the region z > 0 that would provide the appropriate reflected
wave: Use
˝
⌘ * ⌘0
E
˝
50ej20
1
= r =
= ej20 = 0.31 + j0.11 =
Einc
150
3
⌘ + ⌘0
Now
⌘ = ⌘0
⇠
1+
1*
⇡
0
= 377
1 + 0.31 + j0.11
1 * 0.31 * j0.31
1
= 691 + j177 ⌦
c) At what value of z (*10 cm < z < 0) is the total electric field intensity a maximum amplitude?
We found the phase of the reflection coefficient to be = 20˝ = .349rad, and we use
zmax =
*
2
=
*.349
= *0.017 m = *1.7 cm
20
12.6. In the beam-steering prism of Example 12.8, suppose the anti-reflective coatings are removed, leaving
bare glass-to-air interfaces. Calcluate the ratio of the prism output power to the input power, assuming
a single transit.
In making the transit, the light encounters two interfaces at normal incidence, at which loss will
occur. The reflection coefficient at the front surface (air to glass) is
f =
⌘g * ⌘0
⌘g + ⌘0
=
⌘0 _ng * ⌘0
⌘0 _ng + ⌘0
=
1 * ng
1 + ng
Taking the glass index, ng , as 1.45, we find f = *0.18. The interface on exit from the prism
is glass to air, and so the reflection coefficient there will be equal and opposite to f ; i.e.,
b = * f.
Now, the wave power that makes it through (assuming total reflection at the 45˝ interface)
will be
Pout = Pin (1 *  f 2 )(1 *  b 2 ) = Pin (1 * 0.182 )2 = 0.93Pin
So we have 93% net transmission.
263
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12.7. The semi-infinite regions z < 0 and z > 1 m are free space. For 0 < z < 1 m, ✏r® = 4, r = 1, and
✏r®® = 0. A uniform plane wave with ! = 4 ù 108 rad/s is travelling in the az direction toward the
interface at z = 0.
a) Find the standing wave ratio in each of the three regions: First we find the phase constant in the
middle region,
˘
! ✏r®
2(4 ù 108 )
=
= 2.67 rad_m
2 =
c
3 ù 108
Then, with the middle layer ˘
thickness of 1 m, 2 d = 2.67 rad. Also, the intrinsic impedance of
the middle layer is ⌘2 = ⌘0 _ ✏r® = ⌘0 _2. We now find the input impedance:
4
⌘in = ⌘2
5
4
5
⌘0 cos( 2 d) + j⌘2 sin( 2 d)
377 2 cos(2.67) + j sin(2.67)
=
= 231 + j141
⌘2 cos( 2 d) + j⌘0 sin( 2 d)
2 cos(2.67) + j2 sin(2.67)
Now, at the first interface,
12 =
⌘in * ⌘0
231 + j141 * 377
=
= *.176 + j.273 = .325⁄123˝
⌘in + ⌘0
231 + j141 + 377
The standing wave ratio measured in region 1 is thus
s1 =
1 +  12  1 + 0.325
=
= 1.96
1 *  12  1 * 0.325
In region 2 the standing wave ratio is found by considering the reflection coefficient for waves
incident from region 2 on the second interface:
23 =
⌘0 * ⌘0 _2 1 * 1_2 1
=
=
⌘0 + ⌘0 _2 1 + 1_2 3
Then
s2 =
1 + 1_3
=2
1 * 1_3
Finally, s3 = 1, since no reflected waves exist in region 3.
b) Find the location of the maximum E for z < 0 that is nearest to z = 0. We note that the phase
of 12 is = 123˝ = 2.15 rad. Thus
zmax =
*
2
=
*2.15
= *.81 m
2(4_3)
12.8. A wave starts at point a, propagates 1m through a lossy dielectric rated at ↵dB = 0.1dB/cm, reflects
at normal incidence at a boundary at which = 0.3 + j0.4, and then returns to point a. Calculate
the ratio of the final power to the incident power after this round trip: Final power, Pf , and incident
power, Pi , are related through
Pf = Pi 10*0.1↵dB L  2 10*0.1↵dB L Ÿ
Pf
Pi
= 0.3 + j0.42 10*0.2(0.1)100 = 2.5 ù 10*3
264
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12.9. Region 1, z < 0, and region 2, z > 0, are both perfect dielectrics ( = 0 , ✏ ®® = 0). A uniform plane
wave traveling in the az direction has a radian frequency of 3 ù 1010 rad/s. Its wavelengths in the two
regions are 1 = 5 cm and 2 = 3 cm. What percentage of the energy incident on the boundary is
a) reflected; We first note that
®
✏r1
=
0
2⇡c
1!
12
® _✏ ® = ( _ )2 . Then with
Therefore ✏r1
2
1
r2
=
=
2*
2+
1
1
=
® *⌘
1_✏r2
0
t
0
2⇡c
2!
12
0 in both regions, we find
t
® _✏ ® * 1
✏r1
r2
⌘2 * ⌘1
( _ )*1
=
= t
=t
= 2 1
t
⌘2 + ⌘1
( 2_ 1) + 1
® +⌘
®
® _✏ ® + 1
⌘0 1_✏r2
1_✏r1
✏r1
0
r2
⌘0
t
®
and ✏r2
=
®
1_✏r1
3*5
1
=*
3+5
4
The fraction of the incident energy that is reflected is then  2 = 1_16 = 6.25 ù 10*2 .
b) transmitted? We use part a and find the transmitted fraction to be
1 *  2 = 15_16 = 0.938.
c) What is the standing wave ratio in region 1? Use
s=
1 +   1 + 1_4 5
=
= = 1.67
1 *   1 * 1_4 3
12.10. In Fig. 12.1, let region 2 be free space, while
®®
®
®
r1 = 1, ✏r1 = 0, and ✏r1 is unknown. Find ✏r1 if
a) the amplitude of E*
is one-half that of E+
: Since region 2 is free space, the reflection coeffi1
1
cient is
t
t
®
® *1
*
⌘
*
⌘
_
✏
✏r1
0
0
E1  ⌘0 * ⌘1
r1
1
®
= + =
=
=t
=
Ÿ ✏r1
= 9.
t
2
E1  ⌘0 + ⌘1
®
®
⌘0 + ⌘0 _ ✏r1
✏r1 + 1
b) < S*
> is one-half of < S+
>: This time
1
1
Ût ®
Û2
Û ✏ * 1Û
Û
Û
r1
Û = 1 Ÿ ✏ ® = 34
 2 = ÛÛ t
Û
r1
2
Û ✏® + 1 Û
Û
Û
r1
Û
Û
c) E1 min is one-half E1 max : Use
t
E1 max
1+ 
=s=
=2 Ÿ  =
E1 min
1* 
® *1
✏r1
1
®
= =t
Ÿ ✏r1
=4
3
® +1
✏r1
265
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12.11. A 150 MHz uniform plane wave in normally-incident from air onto a material whose intrinsic
impedance is unknown. Measurements yield a standing wave ratio of 3 and the appearance of an
electric field minimum at 0.3 wavelengths in front of the interface. Determine the impedance of the
unknown material: First, the field minimum is used to find the phase of the reflection coefficient,
where
1
zmin = * ( + ⇡) = *0.3 Ÿ
= 0.2⇡
2
where
= 2⇡_ has been used. Next,
 =
So we now have
s*1 3*1 1
=
=
s+1 3+1 2
= 0.5ej0.2⇡ =
We solve for ⌘u to find
⌘u * ⌘0
⌘u + ⌘0
⌘u = ⌘0 (1.70 + j1.33) = 641 + j501 ⌦
12.12. A 50MHz uniform plane wave is normally incident from air onto the surface of a calm ocean.
For seawater, = 4 S/m, and ✏r® = 78.
a) Determine the fractions of the incident power that are reflected and transmitted: First we find
the loss tangent:
4
=
= 18.4
!✏ ®
2⇡(50 ù 106 )(78)(8.854 ù 10*12 )
This value is sufficiently greater than 1 to enable seawater to be considered a good conductor
at 50MHz.
Then, using the approximation (Eq. 85, Chapter 11), the intrinsic impedance is
˘
⌘s = ⇡f _ (1 + j), and the reflection coefficient becomes
˘
=˘
where
⇡f _ (1 + j) * ⌘0
⇡f _ (1 + j) + ⌘0
˘
˘
⇡f _ = ⇡(50 ù 106 )(4⇡ ù 10*7 )_4 = 7.0. The fraction of the power reflected is
˘
[
⇡f _ * ⌘0 ]2 + ⇡f _
Pr
[7.0 * 377]2 + 49.0
=  2 = ˘
=
= 0.93
Pi
[7.0 + 377]2 + 49.0
[ ⇡f _ + ⌘0 ]2 + ⇡f _
The transmitted fraction is then
Pt
= 1 *  2 = 1 * 0.93 = 0.07
Pi
b) Qualitatively, how will these answers change (if at all) as the frequency is increased? Within
the limits of our good conductor approximation (loss tangent greater than about ten), the reflected power fraction, using the formula derived in part a, is found to decrease with increasing
frequency. The transmitted power fraction thus increases.
266
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12.13. A right-circularly-polarized plane wave is normally incident from air onto a semi-infinite slab of plexiglas (✏r® = 3.45, ✏r®® = 0). Calculate the fractions of the incident power that are reflected and transmitted. Also, describe the polarizations
of the reflected and transmitted waves. First, the impedance
˘
of the plexiglas will be ⌘ = ⌘0 _ 3.45 = 203 ⌦. Then
=
203 * 377
= *0.30
203 + 377
The reflected power fraction is thus  2 = 0.09. The total electric field in the plane of the interface
must rotate in the same direction as the incident field, in order to continually satisfy the boundary
condition of tangential electric field continuity across the interface. Therefore, the reflected wave will
have to be lef t circularly polarized in order to make this happen. The transmitted power fraction is
now 1 *  2 = 0.91. The transmitted field will be right circularly polarized (as the incident field)
for the same reasons.
12.14. A left-circularly-polarized plane wave is normally-incident onto the surface of a perfect conductor.
a) Construct the superposition of the incident and reflected waves in phasor form: Assume positive
z travel for the incident electric field. Then, with reflection coefficient, = *1, the incident
and reflected fields will add to give the total field:
Etot = Ei + Er = E0 (ax + jay )e*j z * E0 (ax + jay )e+j z
b
c
f
g
⌅
⇧
= E0 f e*j z * ej z ax + j e*j z * ej z ay g = 2E0 sin( z) ay * jax
f´≠≠≠≠≠≠Ø≠≠≠≠≠≠¨
´≠≠≠≠≠≠Ø≠≠≠≠≠≠¨ g
f
g
*2j sin( z)
d *2j sin( z)
e
b) Determine the real instantaneous form of the result of part a:
⌅
⇧
E(z, t) = Re Etot ej!t = 2E0 sin( z) cos(!t)ay + sin(!t)ax
c) Describe the wave that is formed: This is a standing wave exhibiting circular polarization in
time. At each location along the z axis, the field vector rotates clockwise in the xy plane, and
has amplitude (constant with time) given by 2E0 sin( z).
267
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12.15. Sulfur hexafluoride (SF6 ) is a high-density gas that has refractive index, ns = 1.8 at a specified
pressure, temperature, and wavelength. Consider the retro-reflecting prism shown in Fig. 12.16, that
is immersed in SF6 . Light enters through a quarter-wave antireflective coating and then totally reflects
from the back surfaces of the glass. In principle, the beam should experience zero loss at the design
wavelength (Pout = Pin ).
a) Determine the minimum required value of the glass refractive index, ng , so that the interior
beam will totally reflect: We set the critical angle of total reflection equal to 45˝ , which gives
sin ✓c =
˘
ns
1
= sin(45˝ ) = ˘ Ÿ ng = ns 2 = 2.55
ng
2
b) Knowing ng , find the required refractive index of the quarter-wave film, nf : For a quarter-wave
section, we know that the film intrinsic impedance will be
˘
˘
˘
⌘f = ⌘s ⌘g Ÿ nf = ns ng = (1.80)(2.55) = 2.14
c) With the SF6 gas evacuated from the chamber, and with the glass and film values as previously
found, find the ratio, Pout _Pin . Assume very slight misalignment, so that the long beam path
through the prism is not re-traced by reflected waves. The beam loses power at the two normalincidence boundaries, whereas the back reflections at 45˝ will still be lossless, as that angle is
now greater than ✓c with the reduced surrounding index. At the first normal incidence boundary
(from air to film to glass), the input intrinsic impedance is
H I
⌘f2
⌘02 _n2f
ng
⌘in1 =
=
= ⌘0
⌘g
⌘0 _ng
n2f
At the second normal incidence boundary at the prism exit (glass to film to air), the input intrinsic impedance is
H I
⌘f2
⌘02 _n2f
1
⌘in2 =
=
= ⌘0
⌘0
⌘0
n2f
The reflection coefficients at the two boundaries will be
ng * n2f
⌘in1 * ⌘0
2.55 * (2.14)2
=
=
=
= *0.285
1
⌘in1 + ⌘0
2.55 + (2.14)2
ng + n2
f
2 =
⌘in2 * ⌘g
⌘in2 + ⌘g
=
ng * n2f
ng + n2f
=
1
The power ratio will be:
Pout
= 1 *  1 2
Pin
1 *  2 2 = 1 * (0.285)2
2
= 0.845
268
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12.16. In Fig. 12.5, let regions 2 and 3 both be of quarter-wave thickness. Region 4 is glass, having refractive
index, n4 = 1.45; region 1 is air.
a) Find ⌘in,b : Since region 3 is a quarter-wave layer,
⌘in,b =
3 lb = ⇡_2, and (47) reduces to
⌘32
⌘4
b) Find ⌘in,a : Again, with region 2 of quarter-wave thickness,
⌘in,a =
⌘22
⌘in,b
=
2 la = ⇡_2 and (48) becomes
⌘22 ⌘4
⌘32
c) Specify a relation between the four intrinsic impedances that will enable total transmission of
waves incident from the left into region 4: At the front surface, we need to have a zero reflection
coefficient, so the input impedance there must match that of free space:
⌘in,a = ⌘0 Ÿ ⌘22 ⌘4 = ⌘32 ⌘0
d) Specify refractive index values for regions 2 and 3 that will accomplish the condition of part c:
We can rewrite the part c result as
H 2I0 1 H 2I
⌘0
⌘0
n23
⌘0
=
⌘
Ÿ
n
=
0
4
⌘4
n22
n23
n22
So any combination of indices that satisfy this result will work. One combination, for example,
would be n2 = 1.10 and n3 = 1.33. It is better to have the indices ascending (or descending)
monotonically in value from layer to layer because the high transmission feature is then less
sensitive to changes in wavelength (as an exercise for fun, show this).
e) Find the fraction of incident power transmitted if the two layers were of half-wave thickness
instead of quarter-wave: For any half-wave layer, we know that the input impedance is equal to
that of the load. Therefore, ⌘in,b = ⌘in,a = ⌘4 . The reflection coefficient at the front surface is
therefore
⌘in,a * ⌘0
⌘ * ⌘0
1 * n4
1 * 1.45
= 4
=
=
= *0.184
in =
⌘in,a + ⌘0
⌘4 + ⌘0
1 + n4
1 + 1.45
The transmitted power fraction is then
Pt
= 1 *  in 2 = 1 * (0.184)2 = 0.97
Pin
269
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12.17. A uniform plane wave in free space is normally-incident onto a dense dielectric plate of thickness
_4, having refractive index n. Find the required value of n such that exactly half the incident power
is reflected (and half transmitted). Remember that n > 1.
In this problem, ⌘1 = ⌘3 = ⌘0 , and the quarter-wave section input impedance is therefore
⌘in =
⌘22
⌘3
=
⌘02 _n2
⌘0
=
⌘0
n2
The reflection coefficient at the front surface is then
in =
⌘in * ⌘0
1 * n2
=
⌘in + ⌘0
1 + n2
For half-power reflection, we require that  in 2 = 0.5, or
greater than 1, we choose the minus sign option and write:
1 * n2
1
= *˘ Ÿ n =
1 + n2
2
˘
=
±1_
2. Since n must be
in
L˘
M1_2
2+1
= 2.41
˘
2*1
12.18. A uniform plane wave is normally-incident onto a slab of glass (n = 1.45) whose back surface is in
contact with a perfect conductor. Determine the reflective phase shift at the front surface of the glass
if the glass thickness is: (a) _2; (b) _4; (c) _8.
With region 3 being a perfect conductor, ⌘3 = 0, and Eq. (36) gives the input impedance to the
structure as ⌘in = j⌘2 tan l. The reflection coefficient is then
⌘22 tan2 l * ⌘02 + j2⌘0 ⌘2 tan l
⌘in * ⌘0
j⌘2 tan l * ⌘0
=
=
=
=
⌘in + ⌘0
j⌘2 tan l + ⌘0
⌘22 tan2 l + ⌘02
r+j i
where the last equality occurs by multiplying the numerator and denominator of the middle term
by the complex conjugate of its denominator. The reflective phase is now
L
M
0 1
4
5
2⌘
⌘
tan
l
i
2 0
*1
*1
*1 (2.90) tan l
= tan
= tan
= tan
tan l * 2.10
⌘ 2 tan2 l * ⌘ 2
r
2
0
where ⌘2 = ⌘0 _1.45 has been used. We can now evaluate the phase shift for the three given
cases. First, when l = _2, l = ⇡, and thus ( _2) = 0. Next, when l = _4, l = ⇡_2,
and
( _4) ô tan*1 [2.90] = 71˝
as l ô _4. Finally, when l = _8, l = ⇡_4, and
⌧
2.90
( _8) = tan*1
= *69.2˝ (or 291˝ )
1 * 2.10
270
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12.19. You are given four slabs of lossless dielectric, all with the same intrinsic impedance, ⌘, known to
be different from that of free space. The thickness of each slab is _4, where is the wavelength
as measured in the slab material. The slabs are to be positioned parallel to one another, and the
combination lies in the path of a uniform plane wave, normally-incident. The slabs are to be arranged
such that the air spaces between them are either zero, one-quarter wavelength, or one-half wavelength
in thickness. Specify an arrangement of slabs and air spaces such that
a) the wave is totally transmitted through the stack: In this case, we look for a combination
of half-wave sections. Let the inter-slab distances be d1 , d2 , and d3 (from left to right). Two
possibilities are i.) d1 = d2 = d3 = 0, thus creating a single section of thickness , or ii.)
d1 = d3 = 0, d2 = _2, thus yielding two half-wave sections separated by a half-wavelength.
b) the stack presents the highest reflectivity to the incident wave: The best choice here is to make
d1 = d2 = d3 = _4. Thus every thickness is one-quarter wavelength. The impedances transform as follows: First, the input impedance at the front surface of the last slab (slab 4) is
⌘in,1 = ⌘ 2 _⌘0 . We transform this back to the back surface of slab 3, moving through a distance of _4 in free space: ⌘in,2 = ⌘02 _⌘in,1 = ⌘03 _⌘ 2 . We next transform this impedance to
the front surface of slab 3, producing ⌘in,3 = ⌘ 2 _⌘in,2 = ⌘ 4 _⌘03 . We continue in this manner
until reaching the front surface of slab 1, where we find ⌘in,7 = ⌘ 8 _⌘07 . Assuming ⌘ < ⌘0 , the
ratio ⌘ n _⌘0n*1 becomes smaller as n increases (as the number of slabs increases). The reflection
coefficient for waves incident on the front slab thus gets close to unity, and approaches 1 as the
number of slabs approaches infinity.
271
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12.20. The 50MHz plane wave of Problem 12.12 is incident onto the ocean surface at an angle to the normal
of 60˝ . Determine the fractions of the incident power that are reflected and transmitted for
a) s polarization: To review Problem 12, we first we find the loss tangent:
!✏ ®
=
4
2⇡(50 ù 106 )(78)(8.854 ù 10*12 )
= 18.4
This value is sufficiently greater than 1 to enable seawater to be considered a good conductor
at 50MHz. Then, using
˘ the approximation (Eq. 85, Chapter 11), and with = 0 , the intrinsic
impedance is ⌘s = ⇡f _ (1 + j) = 7.0(1 + j). Next we need the angle of refraction, which
means that we need to know the refractive index of seawater at 50MHz. For a uniform plane
wave in a good conductor, the phase constant is
u
nsea ! . ˘
.
=
= ⇡f
Ÿ nsea = c
= 26.8
c
4⇡f
Then, using Snell’s law, the angle of refraction is found:
sin ✓2 =
nsea
sin ✓1 = 26.8 sin(60˝ ) Ÿ ✓2 = 1.9˝
n1
.
This angle is small enough so that cos ✓2 = 1. Therefore, for s polarization,
. ⌘s2 * ⌘s1
7.0(1 + j) * 377_ cos 60˝
=
= *0.98 + j0.018 = 0.98⁄179˝
⌘s2 + ⌘s1
7.0(1 + j) + 377_ cos 60˝
s =
The fraction of the power reflected is now  s 2 = 0.96. The fraction transmitted is then 0.04.
b) p polarization: Again, with the refracted angle close to zero, the relection coefficient for
p polarization is
. ⌘p2 * ⌘p1
7.0(1 + j) * 377 cos 60˝
=
= *0.93 + j0.069 = 0.93⁄176˝
⌘p2 + ⌘p1
7.0(1 + j) + 377 cos 60˝
p =
The fraction of the power reflected is now  p 2 = 0.86. The fraction transmitted is then 0.14.
272
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12.21. A right-circularly polarized plane wave in air is incident at Brewster’s angle onto a semi-infinite slab
of plexiglas (✏r® = 3.45, ✏r®® = 0, = 0 ).
a) Determine the fractions of the incident power that are reflected
and transmitted: In plexiglas,
˘
®
®
*1
*1
Brewster’s angle is ✓B = ✓1 = tan (✏r2 _✏r1 ) = tan ( 3.45) = 61.7˝ . Then the angle of
refraction is ✓2 = 90˝ * ✓B (see Example 12.9), or ✓2 = 28.3˝ . With incidence at Brewster’s
angle, all p-polarized power will be transmitted — only s-polarized power will be reflected.
This is found through
⌘2s * ⌘1s
.614⌘0 * 2.11⌘0
=
= *0.549
s =
⌘2s + ⌘1s
.614⌘0 + 2.11⌘0
˝ ) = 2.11⌘ ,
where ⌘1s = ⌘1 sec ✓1 = ⌘0 sec(61.7
0
˘
˝
and ⌘2s = ⌘2 sec ✓2 = (⌘0 _ 3.45) sec(28.3 ) = 0.614⌘0 . Now, the reflected power fraction
is  2 = (*.549)2 = .302. Since the wave is circularly-polarized, the s-polarized component
represents one-half the total incident wave power, and so the fraction of the total power that is
reflected is .302_2 = 0.15, or 15%. The fraction of the incident power that is transmitted is then
the remainder, or 85%.
b) Describe the polarizations of the reflected and transmitted waves: Since all the p-polarized component is transmitted, the reflected wave will be entirely s-polarized (linear). The transmitted
wave, while having all the incident p-polarized power, will have a reduced s-component, and
so this wave will be right-elliptically polarized.
12.22. A dielectric waveguide is shown in Fig. 12.16 with refractive indices as labeled. Incident light enters
the guide at angle from the front surface normal as shown. Once inside, the light totally reflects
at the upper n1 * n2 interface, where n1 > n2 . All subsequent reflections from the upper an lower
boundaries will be total as well, and so the light is confined to the guide. Express, in terms of n1 and
n2 , the maximum value of such that total confinement will occur, with n0 = 1. The quantity sin
is known as the numerical aperture of the guide.
From the illustration we see that 1 maximizes when ✓1 is at its minimum value. This minimum will
be the critical angle for the n1 * n2 interface, where sin ✓c = sin ✓1 = n2 _n1 . Let the refracted angle
to the right of the vertical interface (not shown) be 2 , where n0 sin 1 = n1 sin 2 . Then we see that
˝
2 + ✓1 = 90 , and so sin ✓1 = cos 2 . Now, the numerical aperture becomes
t
t
t
n1
sin 1max =
sin 2 = n1 cos ✓1 = n1 1 * sin2 ✓1 = n1 1 * (n2 _n1 )2 = n21 * n22
n0
1
0t
*1
2
2
Finally, 1max = sin
n1 * n2 is the numerical aperture angle.
12.23. Suppose that 1 in Fig. 12.16 is Brewster’s angle, and that ✓1 is the critical angle. Find n0 in terms of
n1 and n2 : With the incoming ray at Brewster’s angle, the refracted angle of this ray (measured from
the inside normal to the front surface) will be 90˝ * 1 . Therefore, 1 = ✓1 , and thus sin 1 = sin ✓1 .
Thus
t
n1
n2
sin 1 = t
= sin ✓1 =
Ÿ n0 = (n1 _n2 ) n21 * n22
n1
2
2
n0 + n1
Alternatively, we
t could have used the result of Problem 12.22, in which it was found that
sin
1 = (1_n0 )
n21 * n22 , which we then set equal to sin ✓1 = n2 _n1 to get the same result.
273
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12.24. A Brewster prism is designed to pass p-polarized light without any reflective loss. The prism of
Fig. 12.17 is made of glass (n = 1.45), and is in air. Considering the light path shown, determine
the vertex angle, ↵: With entrance and exit rays at Brewster’s angle (to eliminate reflective loss), the
interior ray must be horizontal, or parallel to the bottom surface of the prism. From the geometry,
the angle between the interior ray and the normal to the prism surfaces that it intersects is ↵_2. Since
this angle is also Brewster’s angle, we may write:
H
I
H
I
1
1
*1
*1
↵ = 2 sin
= 2 sin
= 1.21 rad = 69.2˝
˘
˘
2
2
1
+
(1.45)
1+n
12.25. In the Brewster prism of Fig. 12.17, determine for s-polarized light the fraction of the incident power
that is transmitted through the prism, and from this specify the dB insertion loss, defined as 10 log10
of that number:
We use
s = (⌘s2 * ⌘s1 )_(⌘s2 + ⌘s1 ), where
⌘s2 =
⌘ ˘
⌘2
⌘
= ˘2
= 02 1 + n2
cos(✓B2 ) n_ 1 + n2
n
⌘s1 =
˘
⌘1
⌘
= ˘1
= ⌘0 1 + n2
cos(✓B1 ) 1_ 1 + n2
and
Thus, at the first interface, = (1 * n2 )_(1 + n2 ). At the second interface, will be equal but
of opposite sign to the above value. The power transmission coefficient through each interface
is 1 *  2 , so that for both interfaces, we have, with n = 1.45:
L
0 2
1 2 M2
Ptr
n *1
2 2
= 1* 
= 1*
= 0.76
Pinc
n2 + 1
The insertion loss is now
li (dB) = 10 log10 (0.76) = *1.19 dB
274
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12.26. Show how a single block of glass can be used to turn a p-polarized beam of iight through 180˝ , with
the light suffering, in principle, zero reflective loss. The light is incident from air, and the returning
beam (also in air) may be displaced sideways from the incident beam. Specify all pertinent angles
and use n = 1.45 for glass. More than one design is possible here.
The prism below is designed such that light enters at Brewster’s angle, and once inside, is turned
around using total reflection. Using the result of Example 12.9, we find that with glass, ✓B = 55.4˝ ,
which, by the geometry, is also the incident angle for total reflection at the back of the prism. For
this to work, the Brewster angle must be greater than or equal to the critical angle. This is in fact the
case, since ✓c = sin*1 (n2 _n1 ) = sin*1 (1_1.45) = 43.6˝ .
12.27. Using Eq. (79) in Chapter 11 as a starting point, determine the ratio of the group and phase velocities
of an electromagnetic wave in a good conductor. Assume conductivity does not vary with frequency:
In a good conductor:
u
⌧
˘
*1_2
!
d
1 !
= ⇡f
=
ô
=
2
d! 2
2
2
Thus
d!
=
d
0
d
d!
1*1
u
2!
=2
= vg
and
vp =
!
=˘
u
!
!
_2
=
2!
Therefore vg _vp = 2.
275
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12.28. Over a small wavelength range, the refractive index of a certain material varies approximately linearly
.
with wavelength as n( ) = na + nb ( * a ), where na , nb , and a are constants, and where is the
free space wavelength.
a) Show that d_d! = *(2⇡c_!2 )d_d : With as the free space wavelength, we use = 2⇡c_!,
from which d _d! = *2⇡c_!2 . Then d_d! = (d _d!) d_d = *(2⇡c_!2 ) d_d .
b) Using ( ) = 2⇡n_ , determine the wavelength-dependent (or independent) group delay over
a unit distance: This will be
4
5
⌧
⇧
d
1
d 2⇡n( )
2⇡c d 2⇡ ⌅
tg =
=
=
=* 2
na + nb ( * a )
vg
d! d!
! d
⌧
⇧ 2⇡
2⇡c
2⇡ ⌅
= * 2 * 2 na + nb ( * a ) +
nb
! 4
5
2
2⇡n
2⇡nb a
1
=*
* 2a +
= (na * nb a ) s_m
2
2⇡c
c
c) Determine 2 from your result of part b: 2 = d 2 _d!2 !0 . Since the part b result is independent
of wavelength (and of frequency), it follows that 2 = 0.
d) Discuss the implications of these results, if any, on pulse broadening: A wavelength-independent
group delay (leading to zero 2 ) means that there will simply be no pulse broadening at all. All
frequency components arrive simultaneously. This sort of thing happens in most transparent
materials – that is, there will be a certain wavelength, known as the zero dispersion wavelength,
around which the variation of n with is locally linear. Transmitting pulses at this wavelength
will result in no pulse broadening (to first order).
12.29. A T = 5 ps transform-limited pulse propagates in a dispersive channel for which 2 = 10 ps2 _km.
Over ˘
what distance will the pulse spread
˘ to twice its initial width? After propagation, the width is
®
2
2
T = T + ( ⌧) = 2T . Thus ⌧ = 3T , where ⌧ = 2 z_T . Therefore
˘
˘
2
˘
z
3(5 ps)2
3T
2
= 3T or z =
=
= 4.3 km
T
10 ps2 _km
2
12.30. A T = 20 ps transform-limited pulse propagates through 10 km of a dispersive channel for which
2
2 = 12 ps _km. The pulse then propagates through a second 10 km channel for which 2 =
2
*12 ps _km. Describe the pulse at the output of the second channel and give a physical explanation for what happened.
Our theory of pulse spreading will allow for changes in 2 down the length of the channel. In fact,
we may write in general:
L
1
⌧=
(z) dz
T 0 2
Having 2 change sign at the midpoint, yields a zero ⌧, and so the pulse emerges from the output
unchanged! Physically, the pulse acquires a positive linear chirp (frequency increases with time over
the pulse envelope) during the first half of the channel. When 2 switches sign, the pulse begins to
acquire a negative chirp in the second half, which, over an equal distance, will completely eliminate
the chirp acquired during the first half. The pulse, if originally transform-limited at input, will emerge,
again transform-limited, at its original width. More generally, complete dispersion compensation is
achieved using a two-segment channel when 2 L = * 2® L® , assuming dispersion terms of higher
order than 2 do not exist.
276
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CHAPTER 13 – 9th Edition
13.1. The conductors of a coaxial transmission line are copper ( c = 5.8 ù 10*7 S/m) and the dielectric is
polyethylene (✏r® = 2.26, _!✏ ® = 0.0002). If the inner radius of the outer conductor is 4 mm, find
the radius of the inner conductor so that (assuming a lossless line):
a) Z0 = 50 ⌦: Us
1
Z0 =
2⇡
u
˘
⇠ ⇡
⇠ ⇡ 2⇡ ✏ ® (50)
b
b
r
ln
= 50 Ÿ ln
=
= 1.25
✏®
a
a
377
Thus b_a = e1.25 = 3.50, or a = 4_3.50 = 1.142 mm
b) C = 100 pF/m: Begin with
C=
⇠ ⇡
2⇡✏ ®
b
= 10*10 Ÿ ln
= 2⇡(2.26)(8.854 ù 10*2 ) = 1.257
ln(b_a)
a
So b_a = e1.257 = 3.51, or a = 4_3.51 = 1.138 mm.
c) L = 0.2 H_m: Use
L=
0
2⇡
ln
⇠ ⇡
⇠ ⇡ 2⇡(0.2 ù 10*6 )
b
b
= 0.2 ù 10*6 Ÿ ln
=
=1
a
a
4⇡ ù 10*7
Thus b_a = e1 = 2.718, or a = b_2.718 = 1.472 mm.
277
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13.2. Find R, L, C, and G for a coaxial cable with a = 0.25 mm, b = 2.50 mm, c = 3.30 mm, ✏r = 2.0,
7
= 1.0 ù 10*5 S/m, and f = 300 MHz.
r = 1, c = 1.0 ù 10 S/m,
First, we note that the metal is a good conductor, as confirmed by the loss tangent:
c
!✏
=
®
1.0 ù 107
= 3.0 ù 108 >> 1
8
*12
2⇡(3.00 ù 10 )(2)(8.854 ù 10 )
So the skin depth into the metal is
v
u
2
2
=
=
= 9.2 m
8
! c
2⇡(3.00 ù 10 )(4⇡ ù 10*7 )(1.0 ù 107 )
The current can be said to exist in layers of thickness just beneath the inner conductor radius,
a, and just inside the outer conductor at its inner radius, b. The outer conductor far radius, c,
is of no consequence. Both conductors behave essentially as thin films of thickness . We are
thus in the high frequency regime, and the equations in Sec. 13.1.1 apply. The calculations for
the primary constants are as follows:
0
1
⇠
⇡
1
1 1
1
1
1
R=
+
=
+
= 7.6 ⌦_m
2⇡ c a b
2⇡(9.2 ù 10*6 )(1.0 ù 107 ) 0.25 ù 10*3 2.5 ù 10*3
using Eq. (12).
L = Lext =
using Eq. (11).
C=
2⇡
ln
⇠ ⇡
⇠
⇡
b
4⇡ ù 10*7
2.5
=
ln
= 0.46 H_m
a
2⇡
0.25
2⇡(2)(8.854 ù 10*12 )
2⇡✏ ®
=
= 48 pF_m
ln(b_a)
ln(2.5_0.25)
using Eq. (9).
G=
2⇡(1.0 ù 10*5 )
2⇡
=
= 27 S_m
ln(b_a)
ln(2.5_0.25)
using Eq. (10).
278
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13.3. Two aluminum-clad steel conductors are used to construct a two-wire transmission line. Let
7
6
Al = 3.8 ù 10 S/m, St = 5 ù 10 S/m, and St = 100 H_m. The radius of the steel wire is
0.5 in., and the aluminum coating is 0.05 in. thick. The dielectric is air, and the center-to-center wire
separation is 4 in. Find C, L, G, and R for the line at 10 MHz: The first question is whether we are in
the high frequency or low frequency regime. Calculation of the skin depth, , will tell us. We have,
for aluminum,
1
1
=˘
=˘
= 2.58 ù 10*5 m
7
*7
7
⇡f 0 Al
⇡(10 )(4⇡ ù 10 )(3.8 ù 10 )
so we are clearly in the high frequency regime, where uniform current distributions cannot be assumed. Furthermore, the skin depth is considerably less than the aluminum layer thickness, so the
bulk of the current resides in the aluminum, and we may neglect the steel. Assuming solid aluminum
wires of radius a = 0.5 + 0.05 = 0.55 in = 0.014 m, the resistance of the two-wire line is now
R=
1
⇡a
=
Al
1
= 0.023 ⌦_m
⇡(.014)(2.58 ù 10*5 )(3.8 ù 107 )
Next, since the dielectric is air, no leakage will occur from wire to wire, and so G = 0 S_m. Now the
capacitance will be
C=
⇡✏0
cosh*1 (d_2a)
=
⇡ ù 8.85 ù 10*12
= 1.42 ù 10*11 F_m = 14.2 pF_m
cosh*1 (4_(2 ù 0.55))
Finally, the inductance per unit length will be
L=
0
⇡
cosh(d_2a) =
4⇡ ù 10*7
cosh (4_(2 ù 0.55)) = 7.86 ù 10*7 H_m = 0.786 H_m
⇡
279
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13.4. Find R, L, C, and G for a two-wire transmission line in polyethylene at f = 800 MHz. Assume
copper conductors of radius 0.50 mm and separation 0.80 cm. Use ✏r = 2.26 and _(!✏ ® ) = 4.0 ù
10*4 .
From the loss tangent, we find the conductivity of polyethylene:
= (4.0 ù 10*4 )(2⇡ ù 8.00 ù 108 )(2.26)(8.854 ù 10*12 ) = 4.0 ù 10*5 S_m
As polyethylene is a good dielectric, its penetration depth is (using Eq. (60a), Chapter 11):
˘
u
2✏r®
2 2.26
1
2 ✏®
=
=
=
= 199 m
p =
↵p
⌘0
(4.0 ù 10*5 )(377)
Therefore, we can assume field distributions over a any cross-sectional plane to be the same
as those of the lossless line. On the other hand, within the copper conductors (for which
7
c = 5.8 ù 10 ) the skin depth will be (from Eq. (82), Chapter 11):
c =
1
⇡f
=
c
1
⇡(8.00 ù 108 )(4⇡ ù 10*7 )(5.8 ù 107 )
= 2.3 m
As this value is much less than the overall conductor dimensions, the line operates in the high
frequency regime. The primary constants are found using Eqs. (20) - (23). We have
C=
⇡(2.26)(8.854 ù 10*12 )
⇡✏ ®
=
= 22.7 pF_m
cosh*1 (d_2a)
cosh*1 (8)
L = Lext =
⇡
G=
R=
1
⇡a c c
=
cosh*1 (d_2a) =
⇡
cosh*1 (d_2a)
=
4⇡ ù 10*7
cosh*1 (8) = 1.11 H_m
⇡
⇡(4.0 ù 10*5 )
cosh*1 (8)
1
= 46 S_m
⇡(5.0 ù 10*4 )(2.3 ù 10*6 )(5.8 ù 107 )
= 4.8 ohms_m
280
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13.5. Each conductor of a two-wire transmission line has a radius of 0.5mm; their center-to-center distance
is 0.8cm. Let f = 150MHz and assume and c are zero. Find the dielectric constant of the
insulating medium if
a) Z0 = 300 ⌦: Use
v
0
1
t
⇠ ⇡
1
d
120⇡
8
0
*1
*1
®
300 =
cosh
Ÿ
✏r =
cosh
= 1.107 Ÿ ✏r® = 1.23
⇡ ✏r® ✏0
2a
300⇡
2(.5)
b) C = 20 pF/m: Use
20 ù 10*12 =
⇡✏ ®
20 ù 10*12
®
Ÿ
✏
=
cosh*1 (8) = 1.99
r
*1
⇡✏
cosh (d_2a)
0
c) vp = 2.6 ù 108 m/s:
1
1
c
vp = ˘
=˘
=˘
Ÿ ✏r® =
®
®
✏
✏
✏
LC
0 0 r
r
0
3.0 ù 108
2.6 ù 108
12
= 1.33
281
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13.6. Consider an air-filled coaxial transmission line having inner and outer radii a and b. Assume low
loss, such that the attenuation coefficient is given by Eq. (54a) in Chapter 10.
a) Show that the skin effect loss in the line is minimized when b_a = 3.6, yielding a characteristic
impedance of 77 ohms: Start with
u
u
⇠ ⇡
⌘
. 1
.
C
1 R
L
b
↵= R
=
where Z0 =
= 0 ln
2
L 2 Z0
C
2⇡
a
Then the skin effect resistance for the coax line is
⇠
⇡
1
1 1
1
R=
+
=
2⇡ c a b
2⇡b
So
.
↵=
2⇡b
⇡
⇠
c
b
+1
a
⇡
(b_a) + 1
c ⌘0 ln(b_a)
Let x = b_a, which we use as the parameter through which we minimize the function
(x + 1)_ ln x. We have
⌧
ln x * (1 + x)_x
d x+1
=
=0
dx ln x
(ln x)2
whose solution is
x = e(1+x)_x Ÿ x = 3.6
so that
Z0 =
377
ln(3.6) = 77 ohms
2⇡
b) Show that the maximum power that the line can carry before air breakdown occurs is achieved
when b_a = 1.65, yielding a characteristic impedance of 30 ohms (for some historical perspective, see US patent number 2298428): The maximum electric field intensity in the line occurs
at the surface of the inner conductor, and this field must not exceed the breakdown field for air,
Ebd . Between conductors, and with the inner conductor at potential V0 , we have
E(⇢) =
V0,max
V0
a⇢ Ÿ Emax =
a = Ebd
⇢ ln(b_a)
a ln(b_a) ⇢
So the maximum voltage we can apply is V0,max = Ebd a ln(b_a). The current in the line is now
I0,max =
V0,max
Z0
=
Ebd a ln(b_a)
2⇡aEbd
=
(⌘0 _2⇡) ln(b_a)
⌘0
The maximum line power (under dc operation) is
Pmax = V0,max I0,max =
2 ln(b_a)
2⇡a2 Ebd
⌘0
=
2⇡b2 2 ln(b_a)
E
⌘0 bd (b_a)2
Defining x = b_a, we need to maximize the function ln(x)_x2 :
0
1
x * 2x ln(x)
d ln(x)
1
=
= 0 Ÿ ln(x) =
Ÿ x = e1_2 = 1.65
2
4
dx
2
x
x
So
Z0 =
⌘0 ⇠ b ⇡ ⌘0
ln
=
= 30 ohms
2⇡
a
4⇡
282
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13.7. Pertinent dimensions for the transmission line shown in Fig. 13.2 are b = 3 mm, and d = 0.2 mm.
The conductors and the dielectric are non-magnetic.
a) If the characteristic impedance of the line is 15 ⌦, find ✏r® : We use
u
Z0 =
⇠ ⇡
⇠
⇡
d
377 2 .04
®
=
15
Ÿ
✏
=
= 2.8
r
✏® b
15
9
b) Assume copper conductors and operation at 2 ù 108 rad/s. If RC = GL, determine the loss
tangent of the dielectric: For copper, c = 5.8 ù 107 S/m, and the skin depth is
v
u
2
2
=
=
= 1.2 ù 10*5 m
8
! 0 c
(2 ù 10 )(4⇡ ù 10*7 )(5.8 ù 107 )
Then
2
R=
b
c
Now
C=
and
Then, with RC = GL,
Thus
d = (4.4 ù 10
2
(5.8 ù 107 )(1.2 ù 10*5 )(.003)
= 0.98 ⌦_m
✏ ® b (2.8)(8.85 ù 10*12 )(3)
=
= 3.7 ù 10*10 F_m
d
0.2
L=
G=
=
0d
b
=
(4⇡ ù 10*7 )(0.2)
= 8.4 ù 10*8 H_m
3
b
(.98)(3.7 ù 10*10 )
RC
=
= 4.4 ù 10*3 S_m = d
*8
L
d
(8.4 ù 10 )
*3 )(0.2_3) = 2.9 ù 10*4 S_m. The loss tangent is
l.t. =
d
!✏ ®
=
2.9 ù 10*4
= 5.85 ù 10*2
(2 ù 108 )(2.8)(8.85 ù 10*12 )
283
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13.8. A transmission line constructed from perfect conductors and an air dielectric is to have a maximum
dimension of 8mm for its cross-section. The line is to be used at high frequencies. Specify its dimensions if it is:
a) a two-wire line with Z0 = 300 ⌦: With the maximum dimension of 8mm, we have, using (24):
u
⇠
⇡
⇠
⇡
1
8 * 2a
300⇡
*1 8 * 2a
Z0 =
cosh
=
300
Ÿ
=
cosh
= 6.13
⇡ ✏®
2a
2a
120⇡
Solve for a to find a = 0.56 mm. Then d = 8 * 2a = 6.88 mm.
b) a planar line with Z0 = 15 ⌦: In this case our maximum dimension dictates that
So, using (8), we write
u ˘
˘
64 * b2
15
Z0 =
=
15
Ÿ
64 * b2 =
b
✏®
b
377
˘
d 2 + b2 = 8.
Solving, we find b = 7.99 mm and d = 0.32 mm.
c) a 72 ⌦ coax having a zero-thickness outer conductor: With a zero-thickness outer conductor,
we note that the outer radius is b = 8_2 = 4mm. Using (13), we write
u
⇠ ⇡
⇠ ⇡ 2⇡(72)
1
b
b
Z0 =
ln
=
72
Ÿ
ln
=
= 1.20 Ÿ a = be*1.20 = 4e*1.20 = 1.2
®
2⇡ ✏
a
a
120⇡
Summarizing, a = 1.2 mm and b = 4 mm.
13.9. A microstrip line is to be constructed using a lossless dielectric for which ✏r® = 7.0. If the line is to
have a 50-⌦ characteristic impedance, determine:
a) ✏r,ef f : We use Eq. (34) (under the assumption that w_d > 1.3) to find:
⇧*1
⌅
✏r,ef f = 7.0 0.96 + 7.0(0.109 * 0.004 ù 7.0) log10 (10 + 50) * 1
= 5.0
b) w_d: Still under the assumption that w_d > 1.3, we solve Eq. (33) for d_w to find
L
0
1 M1_0.555
✏r® * 1
✏r® + 1 *1
d
= (0.1)
✏r,ef f *
* 0.10
w
2
2
⌅
⇧1_0.555
w
= (0.1) 3.0 (5.0 * 4.0)*1
* 0.10 = 0.624 Ÿ
= 1.60
d
284
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13.10. Two microstrip lines are fabricated end-to-end on a 2-mm thick wafer of lithium niobate (✏r® = 4.8).
Line 1 is of 4mm width; line 2 (unfortunately) has been fabricated with a 5mm width. Determine the
power loss in dB for waves transmitted through the junction.
We first note that w1 _d1 = 2.0 and w2 _d2 = 2.5, so that Eqs. (32) and (33) apply. As the first
step, solve for the effective dielectric constants for the two lines, using (33). For line 1:
✏r1,ef f =
⌧
⇠ ⇡ *0.555
5.8 3.8
1
+
1 + 10
= 3.60
2
2
2
For line 2:
⌧
⇠ ⇡ *0.555
5.8 3.8
1
+
1 + 10
= 3.68
2
2
2.5
We next use Eq. (32) to find the characteristic impedances for the air-filled cases. For line 1:
L
M
u
⇠ ⇡
⇠ ⇡2
1
1
air
Z01
= 60 ln 4
+ 16
+ 2 = 89.6 ohms
2
2
✏r2,ef f =
and for line 2:
b ⇠ ⇡
1
air
Z02
= 60 ln f4
+
f 2.5
d
v
16
⇠
1
2.5
⇡2
c
+ 2g = 79.1 ohms
g
e
The actual line impedances are given by Eq. (31). Using our results, we find
Z air
Z air
89.6
79.1
Z01 = ˘ 01
=˘
= 47.2 ⌦ and Z02 = ˘ 02
=˘
= 41.2 ⌦
✏r1,ef f
✏
r2,ef f
3.60
3.68
The reflection coefficient at the junction is now
=
Z02 * Z01
47.2 * 41.2
=
= 0.068
Z02 + Z01
47.2 + 41.2
The transmission loss in dB is then
PL (dB) = *10 log10 (1 *  2 ) = *10 log10 (0.995) = 0.02 dB
285
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13.11. A parallel-plate waveguide is known to have a cutoff wavelength for the m = 1 TE and TM modes
of c1 = 0.4 cm. The guide is operated at wavelength = 1 mm. How many modes propagate? The
cutoff wavelength for mode m is cm = 2nd_m, where n is the refractive index of the guide interior.
For the first mode, we are given
c1 =
2nd
0.4 0.2
= 0.4 cm Ÿ d =
=
cm
1
2n
n
Now, for mode m to propagate, we require
f
2nd
0.4
0.4 0.4
=
Ÿ mf
=
=4
m
m
0.1
So, accounting for 2 modes (TE and TM) for each value of m, and the single TEM mode, we will
have a total of 9 modes.
13.12. A parallel-plate guide is to be constructed for operation in the TEM mode only over the frequency
range 0 < f < 3 GHz. The dielectric between plates is to be teflon (✏r® = 2.1). Determine the
maximum allowable plate separation, d: We require that f < fc1 , which, using (41), becomes
f<
c
c
3 ù 108
Ÿ dmax =
= ˘
= 3.45 cm
2nd
2nfmax
2 2.1 (3 ù 109 )
13.13. A lossless parallel-plate waveguide is known to propagate the m = 2 TE and TM modes at frequencies
as low as 10GHz. If the plate separation is 1 cm, determine the dielectric constant of the medium
between plates: Use
fc2 =
c
3 ù 1010
=
= 1010 Ÿ n = 3 or ✏r = 9
nd
n(1)
13.14. A d = 1 cm parallel-plate guide is made with glass (n = 1.45) between plates. If the operating
frequency is 32 GHz, which modes will propagate? For a propagating mode, we require f > fcm
Using (41) and the given values, we write
f>
2f nd
2(32 ù 109 )(1.45)(.01)
mc
Ÿ m<
=
= 3.09
2nd
c
3 ù 108
The maximum allowed m in this case is thus 3, and the propagating modes will be TM1 , TE1 , TM2 ,
TE2 , TM3 , and TE3 .
286
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13.15. For the guide of Problem 13.14, and at the 32 GHz frequency, determine the difference between
the group delays of the highest order mode (TE or TM) and the TEM mode. Assume a propagation
distance of 10 cm: From Problem 13.14, we found mmax = 3. The group velocity of a TE or TM mode
for m = 3 is
v
0
1
fc3 2
3(3 ù 1010 )
c
vg3 =
1*
where fc3 =
= 3.1 ù 1010 = 31 GHz
n
f
2(1.45)(1)
Thus
3 ù 1010
vg3 =
1.45
u
1*
⇠
31
32
⇡2
= 5.13 ù 109 cm_s
For the TEM mode (assuming no material dispersion) vg,T EM = c_n = 3 ù 1010 _1.45 = 2.07 ù 1010
cm/s. The group delay difference is now
0
1
0
1
1
1
1
1
tg = z
*
= 10
*
= 1.5 ns
vg3 vg,T EM
5.13 ù 109 2.07 ù 1010
13.16. The cutoff frequency of the m = 1 TE and TM modes in an air-filled parallel-plate guide is known to
be fc1 = 7.5 GHz. The guide is used at wavelength = 1.5 cm. Find the group velocity of the m = 2
TE and TM modes. First we know that fc2 = 2fc1 = 15 GHz. Then f = c_ = 3 ù 108 _.015 = 20
GHz. Now, using (57),
v
u
0
1
⇠ ⇡2
fc2 2
c
c
15
vg2 =
1*
=
1*
= 2 ù 108 m_s
n
f
(1)
20
® = 4.0,
13.17. A parallel-plate guide is partially filled with two lossless dielectrics (Fig. 13.25) where ✏r1
®
✏r2 = 2.1, and d = 1 cm. At a certain frequency, it is found that the TM1 mode propagates through
the guide without suffering any reflective loss at the dielectric interface.
a) Find this frequency: The ray angle is such that the wave is incident on the interface at Brewster’s
angle. In this case
u
2.1
*1
✓B = tan
= 35.9˝
4.0
The ray angle is thus ✓ = 90 * 35.9 = 54.1˝ . The cutoff frequency for the m = 1 mode is
fc1 =
2d
c
t
®
✏r1
=
3 ù 1010
= 7.5 GHz
2(1)(2)
The frequency is thus f = fc1 _ cos ✓ = 7.5_ cos(54.1˝ ) = 12.8 GHz.
b) Is the guide operating at a single TM mode at the frequency found in part a? The cutoff frequency for the next higher mode, TM2 is fc2 = 2fc1 = 15 GHz. The 12.8 GHz operating
frequency is below this, so TM2 will not propagate. So the answer is yes.
287
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13.18. In the guide of Figure 13.25, it is found that m = 1 modes propagating from left to right totally reflect
® .
at the interface, so that no power is transmitted into the region of dielectric constant ✏r2
a) Determine the range of frequencies over which this will occur: For total reflection, the ray angle
measured from the normal to the interface must be greater than or equal to the critical angle,
® _✏ ® )1_2 . The maximum mode ray angle is then ✓
˝
✓c , where sin ✓c = (✏r2
1 max = 90 * ✓c . Now,
r1
using (39), we write
H
I
1
0
1
0
⇡
⇡c
c
˝
*1
*1
*1
90 * ✓c = cos
= cos
=
cos
˘
kmax d
4dfmax
2⇡fmax d 4
Now
v
cos(90 * ✓c ) = sin ✓c =
®
✏r2
®
✏r1
=
c
4dfmax
˘
˘
Therefore fmax = c_(2 2.1d) = (3 ù 108 )_(2 2.1(.01)) = 10.35 GHz. The frequency range
is thus f < 10.35 GHz.
b) Does your part a answer in any way
˘ relate to the cutoff frequency for m = 1 modes in any
region? We note that fmax = c_(2 2.1d) = fc1 in guide 2. To summarize, as frequency is
lowered, the ray angle in guide 1 decreases, which leads to the incident angle at the interface
increasing to eventually reach and surpass the critical angle. At the critical angle, the refracted
angle in guide 2 is 90˝ , which corresponds to a zero degree ray angle in that guide. This defines
the cutoff condition in guide 2. So it would make sense that fmax = fc1 (guide 2).
13.19. A rectangular waveguide has dimensions a = 6 cm and b = 4 cm.
a) Over what range of frequencies will the guide operate single mode? The cutoff frequency for
mode mp is, using Eq. (101):
u
⇠ ⇡2 ⇠ p ⇡2
c
m
fc,mn =
+
2n
a
b
where n is the refractive index of the guide interior. We require that the frequency lie between
the cutoff frequencies of the T E10 and T E01 modes. These will be:
fc10 =
c
3 ù 108
2.5 ù 109
=
=
2na
2n(.06)
n
fc01 =
c
3 ù 108
3.75 ù 109
=
=
2nb
2n(.04)
n
Thus, the range of frequencies for single mode operation is 2.5_n < f < 3.75_n GHz
b) Over what frequency range will the guide support both T E10 and T E01 modes and no others?
We note first that f must be greater than fc01 to support both modes, but must be less than the
cutoff frequency for the next higher order mode. This will be fc11 , given by
v
⇠ ⇡2 ⇠ ⇡2
c
1
1
30c
4.5 ù 109
fc11 =
+
=
=
2n
.06
.04
2n
n
The allowed frequency range is then
3.75
4.5
GHz < f <
GHz
n
n
288
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13.20. Two rectangular waveguides are joined end-to-end. The guides have identical dimensions, where
a = 2b. One guide is air-filled; the other is filled with a lossless dielectric characterized by ✏r® .
a) Determine the maximum allowable value of ✏r® such that single mode operation can be simultaneously ensured in both guides at some frequency: Since a = 2b, the cutoff frequency for any
mode in either guide is written using (101):
u
⇠
⇡
⇠ pc ⇡2
mc 2
fcmp =
+
4nb
2nb
˘
where n = 1 in guide 1 and n = ✏r® in guide 2. We see that, with a = 2b, the next modes (having
the next higher cutoff frequency) above TE10 with be TE20 and TE01 . We also see that in general,
fcmp (guide 2) < fcmp (guide 1). To assure single mode operation in both guides, the operating
frequency must be above cutoff for TE10 in both guides, and below cutoff for the next mode in
both guides. The allowed frequency
range is therefore fc10 (guide 1) < f < fc20 (guide 2). This
˘
leads to c_(2a) < f < c_(a ✏r® ). For this range to be viable, it is required that ✏r® < 4.
b) Write an expression for the frequency range over which single mode operation will occur in
both guides; your answer should be in terms of ✏r® , guide dimensions as needed, and other
known constants: This was already found in part a:
c
c
<f < ˘
2a
✏® a
r
where ✏r® < 4.
13.21. An air-filled rectangular waveguide is to be constructed for single-mode operation at 15 GHz. Specify
the guide dimensions, a and b, such that the design frequency is 10% higher than the cutoff frequency
for the TE10 mode, while being 10% lower than the cutoff frequency for the next higher-order mode:
For an air-filled guide, we have
u
⇠ ⇡2 ⇠ pc ⇡2
mc
fcmp =
+
2a
2b
For TE10 we have fc10 = c_2a, while for the next mode (TE01 ), fc01 = c_2b. Our requirements state
that f = 1.1fc10 = 0.9fc01 . So fc10 = 15_1.1 = 13.6 GHz and fc01 = 15_0.9 = 16.7 GHz. The
guide dimensions will be
a=
c
3 ù 1010
c
3 ù 1010
=
= 1.1 cm and b =
=
= 0.90 cm
9
2fc10
2fc01
2(13.6 ù 10 )
2(16.7 ù 109 )
289
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13.22. Consider the TE11 mode in a rectangular waveguide having width a and height b. Using Eqs. (96d)
and (96e), along with the methods described in Chapter 2, Section 2.6, show that the equation for the
streamlines for the electric field (at z = 0) is
4
5
b
C
*1
y = cos
(0 < x < a, 0 < y < b)
⇡
cos(⇡x_a)
where C is a constant that identifies a given streamline: Using Eqs. (96d) and (96e), we have
Ey
Ex
=
k sin(km x) cos(kp y)
k tan(km x)
dy
=* m
=* m
dx
kp sin(kp y) cos(km x)
kp tan(kp y)
So therefore
*km tan(km x)dx = kp tan(kp y)dy
Next, both sides of the above equation are integrated, and the integration constants are combined and
are written as ln(C):
*km
tan(km x)dx = kp
tan(kp y)dy Ÿ ln cos(km x) = * ln cos(kp y) + ln(C)
Combining terms:
4
ln cos(km x) = * ln
cos(kp y)
C
5
4
C
= + ln
cos(kp y)
5
Ÿ cos(km x) =
C
cos(kp y)
Then, for TE11 , we have kp = ⇡_b and km = ⇡_a. Incorporating these leads to
4
5
4
5
1
C
b
C
*1
*1
y=
cos
= cos
Q.E.D
kp
cos(km x)
⇡
cos(⇡x_a)
290
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13.23
a) Using the relation Pav = (1_2)Re{Es ù H<s }, and Eqs. (106) through (108), show that the
average power density in the TE10 mode in a rectangular waveguide is given by
Pav =
10
2!
E02 sin2 (10 x) az
W_m2
Inspecting (106) through (108), we see that (108) includes a factor of j, and so would lead to
an imaginary part of the total power when the cross product with Ey is taken. Therefore, the
real power in this case is found through the cross product of (106) with the complex conjugate
of (107), or
1
Pav = Re Eys ù H<xs = 10 E02 sin2 (10 x) az
2
2!
W_m2
b) Integrate the result of part a over the guide cross-section 0 < x < a, 0 < y < b, to show that
the power in Watts transmitted down the guide is given as
P =
10 ab
4!
E02 =
ab 2
E sin ✓10 W
4⌘ 0
˘
where ⌘ =
_✏, and ✓10 is the wave angle associated with the TE10 mode. Interpret. First,
the integration:
b
a
10
E02 sin2 (10 x) az az dx dy =
10 ab
E02
2!
4!
0
0
˘
Next, from (54), we have 10 = !
✏ sin ✓10 , which, on substitution, leads to
P =
ab
P = E02 sin ✓10 W with ⌘ =
4⌘
u
✏
The sin ✓10 dependence demonstrates the principle of group velocity as energy velocity (or
power). This was considered in the discussion leading to Eq. (57).
13.24. Show that the group dispersion parameter, d 2 _d!2 , for given mode in a parallel-plate or rectangular
waveguide is given by
⇠ ⇡ 4
⇠ ! ⇡2 5*3_2
d2
n !c 2
c
=*
1*
!c !
!
d!2
where !c is the radian cutoff frequency for the mode in question (note that the first derivative form
was already found, resulting in Eq. (57)). First, taking the reciprocal of (57), we find
4
⇠ ! ⇡2 5*1_2
d
n
c
=
1*
d! c
!
Taking the derivative of this equation with respect to ! leads to
H
I
⇠ ⇡4
⇠ ! ⇡2 5*3_2 2!2
⇠ ⇡ 4
⇠ ! ⇡2 5*3_2
d2
n
1
n !c 2
c
c
c
=
*
1
*
=
*
1
*
2
3
c
2
!
!c
!
!
d!
!
291
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13.25. Consider a transform-limited pulse of center frequency f = 10 GHz and of full-width 2T = 1.0 ns.
The pulse propagates in a lossless single mode rectangular guide which is air-filled and in which the
10 GHz operating frequency is 1.1 times the cutoff frequency of the T E10 mode. Using the result of
Problem 13.24, determine the length of the guide over which
˘ the pulse broadens to twice its initial
®
width: The broadened pulse will have width given by T = T 2 + ( ⌧)2 , where ⌧ = 2 L_T for a
transform limited pulse (assumed gaussian). 2 is the Problem 13.24 result evaluated at the operating
frequency, or
⇠ ⇡2 4
⇠ ⇡2 5*3_2
d2
1
1
1
!=10 GHz = *
1*
2 =
2
10
8
1.1
d!
(2⇡ ù 10 )(3 ù 10 ) 1.1
= 6.1 ù 10*19 s2 _m = 0.61 ns2 _m
Now ⌧ = 0.61L_0.5 = 1.2L ns. For the pulse width to double, we have T ® = 1 ns, and
˘
(.05)2 + (1.2L)2 = 1 Ÿ L = 0.72 m = 72 cm
What simple step can be taken to reduce the amount of pulse broadening in this guide, while maintaining the same initial pulse width? It can be seen that 2 can be reduced by increasing the operating
frequency relative to the cutoff frequency; i.e., operate as far above cutoff as possible, without allowing the next higher-order modes to propagate.
13.26. A symmetric dielectric slab waveguide has a slab thickness d = 10 m, with n1 = 1.48 and
n2 = 1.45. If the operating wavelength t
is = 1.3 m, what modes will propagate? We use the
condition expressed through (141): k0 d n21 * n22 g (m * 1)⇡. Since k0 = 2⇡_ , the condition
becomes
t
2(10) ˘
2d
n21 * n22 g (m * 1) Ÿ
(1.48)2 * (1.45)2 = 4.56 g m * 1
1.3
Therefore, mmax = 5, and we have TE and TM modes for which m = 1, 2, 3, 4, 5 propagating (ten
total).
13.27. A symmetric slab waveguide is known to support only a single pair of TE and TM modes at wavelength = 1.55 m. If the slab thickness is 5 m, what is the maximum value of n1 if n2 = 3.30?
Using (142) we have
v
u
0
12
t
⇠ ⇡2
2⇡d
1.55
2
2
2
n1 * n2 < ⇡ Ÿ n1 <
+ n2 =
+ (3.30)2 = 3.304
2d
2(5)
13.28. In a symmetric slab waveguide, n1 = 1.50, n2 = 1.45, and d = 10 m.
a) What is the phase velocity of the m = 1 TE or TM mode at cutoff? At cutoff, the mode propagates in the slab at the critical angle, which means that the phase velocity will be equal to
that of a plane wave in the upper or lower media of index n2 . Phase velocity will therefore be
vp (cutof f ) = c_n2 = 3 ù 108 _1.45 = 2.07 ù 108 m_s.
b) What is the phase velocity of the m = 2 TE or TM modes at cutoff? The reasoning of part a
applies to all modes, so the answer is the same, or 2.07 ù 108 m_s.
292
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13.29. An asymmetric slab waveguide is shown in Fig. 13.26. In this case, the regions above and below the
slab have unequal refractive indices, where n1 > n3 > n2 .
a) Write, in terms of the appropriate indices, an expression for the minimum possible wave angle,
✓1 , that a guided mode may have: The wave angle must be equal to or greater than the critical
angle of total reflection at both interfaces. The minimum wave angle is thus determined by the
greater of the two critical angles. Since n3 > n2 , we find ✓min = ✓c,13 = sin*1 (n3 _n1 ).
b) Write an expression for the maximum phase velocity a guided mode may have in this structure,
using given or known parameters: We have vp,max = !_ min , where min = n1 k0 sin ✓1,min =
n1 k0 n3 _n1 = n3 k0 . Thus vp,max = !_(n3 k0 ) = c_n3 .
13.30. A step index optical fiber is known to be single mode at wavelengths > 1.2 m. Another fiber is
to be fabricated from the same materials, but is to be single mode at wavelengths > 0.63 m. By
what percentage must the core radius of the new fiber differ from the old one, and should it be larger
or smaller? We use the cutoff condition, given by (159):
t
2⇡a
> c=
n21 * n22
2.405
With reduced, the core radius, a, must also be reduced by the same fraction. Therefore, the percentage reduction required in the core radius will be
%=
1.2 * .63
ù 100 = 47.5%
1.2
13.31. Is the mode field radius greater than or less than the fiber core radius in single-mode step-index fiber?
The answer to this can be found by inspecting Eq. (164). Clearly the mode field radius decreases
with increasing V , so we can look at the extreme case of V = 2.405, which is the upper limit
to single-mode operation. The equation evaluates as
⇢0
1.619
2.879
= 0.65 +
+
= 1.10
3_2
a
(2.405)
(2.405)6
Therefore, ⇢0 is always greater than a within the single-mode regime, V < 2.405.
13.32. The mode field radius of a step-index fiber is measured as 4.5 m at free space wavelength = 1.30
m. If the cutoff wavelength is specified as c = 1.20 m, find the expected mode field radius at
= 1.55 m.
In this problem it is helpful to use the relation V = 2.405( c _ ), and rewrite Eq. (164) to read:
0 13_2
0 16
⇢0
= 0.65 + 0.434
+ 0.015
a
c
c
At = 1.30 m, _ c = 1.08, and at 1.55 m, _ c = 1.29. Using these values, along with
our new equation, we write
4
5
0.65 + 0.434(1.29)3_2 + 0.015(1.29)6
⇢0 (1.55) = 4.5
= 5.3 m
0.65 + 0.434(1.08)3_2 + 0.015(1.08)6
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CHAPTER 14 – 9th Edition
14.1. A short dipole carrying current I0 cos !t in the az direction is located at the origin in free space.
a) If k = 1 rad/m, r = 2 m, ✓ = 45˝ , = 0, and t = 0, give a unit vector in rectangular components
that shows the instantaneous direction of E: In spherical coordinates, the components of E are
given by (13a) and (13b):
0
1
I0 d
1
1
*jkr
Ers =
⌘ cos ✓ e
+
(13a)
2⇡
r2 jkr3
0
1
I0 d
1
1
*jkr jk
E✓s =
⌘ sin ✓ e
+ 2+
(13b)
4⇡
r
r
jkr3
Since we want a unit vector at t = 0, we need only the relative amplitudes of ˘
the two compo˝
nents, but we need the absolute phases. Since ✓ = 45 , sin ✓ = cos ✓ = 1_ 2. Also, with
k = 1 = 2⇡_ , it follows that = 2⇡ m. The above two equations can be simplified by these
substitutions, while dropping all amplitude terms that are common to both. Obtain
0
1
1
1
Ar =
+
e*jr
r2 jr3
0
1
1
1
1
1
A✓ =
j + 2 + 3 e*jr
2
r r
jr
Now with r = 2 m, we obtain
⇡
˝
1
1 *j2 1
*j
e
= (1.12)e*j26.6 e*j2
4
8
4
⇠
⇡
˝
1 1
1
1
A✓ = j + * j
e*j2 = (0.90)ej56.3 e*j2
4 8
16
4
The total vector is now A = Ar ar + A✓ a✓ . We can normalize the vector by first finding the
magnitude:
˘
1˘
A = A A< =
(1.12)2 + (0.90)2 = 0.359
4
Dividing the field vector by this magnitude and converting 2 rad to 114.6˝ , we write the normalized vector as
˝
˝
ANs = 0.780e*j141.2 ar + 0.627e*58.3 a✓
Ar =
⇠
In real instantaneous form, this becomes
AN (t) = Re ANs ej!t = 0.780 cos(!t * 141.2˝ )ar + 0.627 cos(!t * 58.3˝ )a✓
We evaluate this at t = 0 to find
AN (0) = 0.780 cos(141.2˝ )ar + 0.627 cos(58.3˝ )a✓ = *0.608ar + 0.330a✓
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14.1
a) (continued)
˘
Dividing by the magnitude, (0.608)2 + (0.330)2 = 0.692, we obtain the unit vector at t = 0:
aN (0) = *0.879ar + 0.477a✓ . We next convert this to rectangular components:
aNx = aN (0) ax = *0.879 sin ✓ cos
1
= ˘ (*0.879 + 0.477) = *0.284
2
+ 0.477 cos ✓ cos
aNy = aN (0) ay = *0.879 sin ✓ sin
+ 0.477 cos ✓ sin
= 0 since
=0
1
aNz = aN (0) az = *0.879 cos ✓ * 0.477 sin ✓ = ˘ (*0.879 * 0.477) = *0.959
2
The final result is then
aN (0) = *0.284ax * 0.959az
b) What fraction of the total average power is radiated in the belt, 80˝ < ✓ < 100˝ ? We use the
far-zone phasor fields, (22) and (23), with k = 2⇡_ , and first find the average power density:
I 2d 2⌘
1
Pavg = Re[E✓s H <s ] = 0 2 2 sin2 ✓ W_m2
2
8 r
We integrate this over the given belt, an at radius r:
Pbelt =
2⇡
0
100˝ I 2 d 2 ⌘
0
sin2 ✓ r2 sin ✓ d✓ d
2 r2
˝
8
80
=
⇡I02 d 2 ⌘
4 2
100˝
80˝
sin3 ✓ d✓
Evaluating the integral, we find
Pbelt =
⇡I02 d 2 ⌘ ⌧ 1
* cos ✓ sin2 ✓ + 2
3
4 2
100
80
= (0.344)
⇡I02 d 2 ⌘
4 2
The total power is found by performing the same integral over ✓, where 0 < ✓ < 180˝ . Doing
this, it is found that
⇡I 2 d 2 ⌘
Ptot = (1.333) 0 2
4
The fraction of the total power in the belt is then f = 0.344_1.333 = 0.258.
295
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14.2. For the Hertzian dipole, prepare a curve, r vs. ✓ in polar coordinates, showing the locus in the
plane where:
=0
a) The radiation field E✓s  is one-half of its value at r = 104 m, ✓ = ⇡_2: Assuming the far field
approximation, we use (22) with k = 2⇡_ to set up the equation:
E✓s  =
I0 d⌘
I0 d⌘
1
sin ✓ = ù
2 r
2 2 ù 104
Ÿ r = 2 ù 104 sin ✓
b) The average radiated power density, Pr,av , is one-half of its value at r = 104 m, ✓ = ⇡_2. To
find the average power, we use (22) and (23) in
2 2
2 2
˘
1
1 I0 d ⌘ 2
1 1 I0 d ⌘
Pr,av = Re{E✓s H <s } =
sin
✓
=
ù
Ÿ
r
=
2 ù 104 sin ✓
2
2 4 2 r2
2 2 4 2 (108 )
˘
The polar plots for field (r = 2 ù 104 sin ✓) and power (r = 2 ù 104 sin ✓) are shown below.
Both are circles.
296
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14.3. Two short antennas at the origin in free space carry identical currents of 5 cos !t A, one in the az
direction, one in the ay direction. Let = 2⇡ m and d = 0.1 m. Find Es at the distant point:
a) (x = 0, y = 1000, z = 0): This point lies along the axial direction of the ay antenna, so its
contribution to the field will be zero. This leaves the az antenna, and since ✓ = 90˝ , only the
E✓s component will be present (as (136) and (137) show). Since we are in the far zone, (138)
applies. We use ✓ = 90˝ , d = 0.1, = 2⇡, ⌘ = ⌘0 = 120⇡, and r = 1000 to write:
I d⌘
5(0.1)(120⇡) *j1000
Es = E✓s a✓ = j 0 sin ✓e*j2⇡r_ a✓ = j
e
a✓
2 r
4⇡(1000)
= j(1.5 ù 10*2 )e*j1000 a✓ = *j(1.5 ù 10*2 )e*j1000 az V_m
b) (0, 0, 1000): Along the z axis, only the ay antenna will contribute to the field. Since the distance
is the same, we can apply the part a result, modified such the the field direction is in *ay :
Es = *j(1.5 ù 10*2 )e*j1000 ay V_m
c) (1000, 0, 0): Here, both antennas will contribute. Applying the results of parts a and b, we find
Es = *j(1.5 ù 10*2 )(ay + az ).
d) Find E at (1000, 0, 0) at t = 0: This is found through
E(t) = Re Es ej!t = (1.5 ù 10*2 ) sin(!t * 1000)(ay + az )
Evaluating at t = 0, we find
E(0) = (1.5 ù 10*2 )[* sin(1000)](ay + az ) = *(1.24 ù 10*2 )(ay + az ) V_m.
e) Find E at (1000, 0, 0) at t = 0: Taking the magnitude of the part d result, we find E =
1.75 ù 10*2 V_m.
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14.4. Write the Hertzian dipole electric field, whose components are given in Eqs. (15) and (16), in the
near-zone in free space, where kr << 1. In this case, only a single term in each of the two equations
survives, and the phases, r and ✓ , simplify to a single value. Construct the resulting electric field
vector and compare your result to the static dipole result (Eq. (35) (not (36)) in Chapter 4). What
relation must exist between the static dipole charge, Q, and the current amplitude, I0 , so that the two
results are identical?
First, we evaluate the phase terms under the approximation, kr << 1: Using (17b) and (18) we
have
.
.
⇡
⇡
⇡
*1
= *
and ✓ = tan*1 (*ÿ) = *
r = tan (0) *
2
2
2
These are used in (15) and (16). In those equations, the (kr)*2 term is retained in (15) and the
(kr)*4 term is kept in (16). The results are
.
Ers =
.
E✓s =
I0 d
2⇡r2 (kr)
I0 kd
⌘ cos ✓ exp(*j⇡_2) = *j
4⇡r(kr)2
where, in free space
⌘ sin ✓ exp(*j⇡_2) = *j
I0 ⌘d
2⇡kr3
I0 ⌘d
4⇡kr3
cos ✓
sin ✓
˘
⌘
1
0 _✏0
= ˘
=
k !
!✏0
0 ✏0
The field vector can now be constructed as
Es =
⇧
(*jI0 _!)d ⌅
2 cos ✓ ar + sin ✓ a✓
3
4⇡✏0 r
This expression is equivalent to the static dipole field (Eq. (35) in Chapter 4), provided that
I0 = j!Q, which is just a statement that the current is the time derivative of the time-harmonic
charge.
298
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14.5. Consider the term in Eq. (14) (or in Eq. (10)) that gives the 1_r2 dependence in the Hertzian dipole
magnetic field. Assuming this term dominates and that kr << 1, show that the resulting magnetic
field is the same as that found by applying the Biot-Savart law (Eq. (2), Chapter 7) to a current
element of differential length d, oriented along the z axis, and centered at the origin.
We begin by writing Eq. (14) under the condition kr << 1, for which the phase term in (14)
becomes, using Eq. (17a):
exp[*j(kr *
.
)] = exp[*j(kr * tan*1 (kr))] = exp[*j(kr * kr)] = 0
With kr << 1, the 1_(kr)2 term in (14) dominates, and the field becomes:
I d
. I kd 1
H = 0
sin ✓ exp(j0) = 0 2 sin ✓
4⇡r (kr)
4⇡r
Now, using the Biot-Savart law for a short element of assumed differential length, d, we have
H=
I0 dL ù aR
4⇡R2
where, with the element at the origin and oriented along z, we have R = r, aR = ar , and
dL = daz . With these subsitutions:
H=
I0 d (az ù ar )
4⇡r2
=
I0 d
4⇡r2
sin ✓ a
(done)
14.6. Evaluate the time-average Poynting vector, < S >= (1_2)Re Es ù H<s for the Hertzian dipole,
assuming the general case that involves the field components as given by Eqs. (10), (13a), and (13b).
Compare your result to the far-zone case, Eq. (26).
With radial and theta components of E, and with only a phi component of H, the Poynting vector
becomes
$
%
1
< S >= Re E✓s H <s ar * Ers H <s a✓
2
Substituting (10), (13a) and (13b), this becomes:
T0
1
4 2
5
I0 d 2
1
k
k
1
k
1
k
< S > = Re
⌘ sin2 ✓ 2 * j 3 * 4 + j 3 + 4 * j 5 ar
2
4⇡
r
r
r
r
r
r
=
⌧
(I d)2
k
1
1
k
* 0 2 sin ✓ cos ✓ *j 3 * 4 + 4 * j 5 a✓
8⇡
r
r
r
r
In taking the real part, the imaginary terms that do not cancel are removed, other real terms
cancel, and only the first term in the radial component remains. The final result is
< S >=
1
2
0
I0 d k
4⇡r
12
⌘ sin✓ ar W_m2
which we see to be identical to Eq. (26).
299
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14.7. A short current element has d = 0.03 . Calculate the radiation resistance for each of the following
current distributions:
a) Uniform: In this case, (30) applies directly and we find
Rrad = 80⇡ 2
⇠ ⇡2
d
= 80⇡ 2 (.03)2 = 0.711 ⌦
b) Linear: I(z) = I0 (0.5d * z)_0.5d: Here, the average current is 0.5I0 , and so the average
power drops by a factor of 0.25. The radiation resistance therefore is down to one-fourth the
value found in part a, or Rrad = (0.25)(0.711) = 0.178 ⌦.
c) Step: I0 for 0 < z < 0.25d and 0.5I0 for 0.25d < z < 0.5d: In this case the average current
on the wire is 0.75I0 . The radiated power (and radiation resistance) are down to a factor of
(0.75)2 times their values for a uniform current, and so Rrad = (0.75)2 (0.711) = 0.400 ⌦.
14.8. Evaluate the time-average Poynting vector, < S >= (1_2)Re Es ù H<s for the magnetic dipole
antenna in the far-zone, in which all terms of order 1_r2 and 1_r4 are neglected in Eqs. (48), (49),
and (50). Compare your result to the far-zone power density of the Hertzian dipole, Eq. (26). In this
comparison, and assuming equal current amplitudes, what relation between loop radius, a, and dipole
length, d, would result in equal radiated powers from the two devices?
First, neglecting the indicated terms in (48)-(50), only the terms in 1_r survive. We find:
! ⇡a2 I0 k
.
E s = *j 0
sin ✓ exp[*j(kr *
4⇡r
)]
. ! ⇡a2 I0 k 1
H✓s = j 0
sin ✓ exp[*j(kr * ✓ )]
4⇡r
⌘
.
where, in the approximation,
= ✓ = tan*1 (kr). So now
1
<
< S >= (1_2)Re E s ù H<✓s = * Re E s H✓s
ar
2
Using the above field expressions, we find
1
 < S >  = Sr(md) =
2
0
I0 k
4⇡r
12
[! 0 ⇡a2 ]2
1 2
sin ✓
⌘
which we now compare to the power density from the Hertzian dipole, Eq. (26):
0
12
1 I0 k
Sr(Hd) =
d 2 ⌘ sin2 ✓
2 4⇡r
For equal power densities from the two structures, we thus require (assuming free space):
⇠ ⇡
˘
! 0
1
2 2⇡
[! 0 ⇡a2 ]2 = d 2 ⌘ Ÿ d = ⇡a2
= ⇡a2 (!
✏
)
=
⇡a
0 0
⌘
⌘
or
d=
(2⇡a)2
2
(the square of the loop circumference over twice the wavelength)
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14.9. A dipole antenna in free space has a linear current distribution. If the length is 0.02 , what value of
I0 is required to:
a) provide a radiation-field amplitude of 100 mV/m at a distance of one mile, at ✓ = 90˝ : With a
linear current distribution, the peak current, I0 , occurs at the center of the dipole; current decreases linearly to zero at the two ends. The average current is thus I0 _2, and we use
Eq. (138) to write:
I d⌘
I0 (0.02)(120⇡)
E✓  = 0 0 sin(90˝ ) =
= 0.1 Ÿ I0 = 85.4 A
4 r
(4)(5280)(12)(0.0254)
b) radiate a total power of 1 watt? We use
⇠ ⇡⇠
⇡
1
1 2
I0 Rrad
4
2
where the radiation resistance is given by Eq. (140), and where the factor of 1/4 arises from the
average current of I0 _2: We obtain Pavg = 10⇡ 2 I02 (0.02)2 = 1 Ÿ I0 = 5.03 A.
Pavg =
14.10. Show that the chord length in the E-plane plot of Fig. 14.4 is equal to b sin ✓, where b is the circle
diameter.
Referring to the construction below, half the chord length, c_2, is given by
c
b
b
= cos(90 * ✓) = sin ✓ Ÿ c = b sin ✓ (done)
2 2
2
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14.11. A monopole antenna in free space, extending vertically over a perfectly conducting plane, has a linear
current distribution. If the length of the antenna is 0.01 , what value of I0 is required to
a) provide a radiation field amplitude of 100 mV/m at a distance of 1 mi, at ✓ = 90˝ : The image
antenna below the plane provides a radiation pattern that is identical to a dipole antenna of
length 0.02 . The radiation field is thus given by (22) in free space, where k = 2⇡_ , ✓ = 90˝ ,
and with an additional factor of 1_2 included to account for the linear current distribution:
E✓  =
4rE✓ 
4(5289)(12 ù .0254)(100 ù 10*3 )
1 I0 d⌘0
Ÿ I0 =
=
= 85.4 A
2 2 r
(d_ )⌘0
(.02)(377)
b) radiate a total power of 1W: For the monopole over the conducting plane, power is radiated
only over the upper half-space. This reduces the radiation resistance of the equivalent dipole
antenna by a factor of one-half. Additionally, the linear current distribution reduces the radiation
resistance of a dipole having uniform current by a factor of one-fourth. Therefore, Rrad is oneeighth the value obtained from (30), or Rrad = 10⇡ 2 (d_ )2 . The current magnitude is now
˘
4
5
4
51_2
2Pav 1_2
2
2(1)
I0 =
=
=˘
= 7.1 A
2
2
Rrad
10⇡ (d_ )
10 ⇡(.02)
14.12. Find the zeros in ✓ for the E-plane pattern of a dipole antenna for which (using Fig. 14.8 as a guide):
a) l = : We look for zeros in the pattern function, Eq. (59), for which kl = (2⇡_ ) = 2⇡.
4
5 4
5
cos(kl cos ✓) * cos(kl)
cos(2⇡ cos ✓) * cos(2⇡)
F (✓) =
=
sin ✓
sin ✓
Zeros will occur whenever
cos(2⇡ cos ✓) = 1 Ÿ ✓ = (0, 90˝ , 180˝ , 270˝ )
You are encouraged to show (using small argument approximations for cosine and sine) that
F (✓) does approach zero when ✓ approaches zero or 180˝ (despite the sin ✓ term in the
denominator).
b) 2l = 1.3 : In this case kl = (2⇡_ )(1.3 _2) = 1.3⇡. The pattern function becomes
4
5
cos(1.3⇡ cos ✓) * cos(1.3⇡)
F (✓) =
sin ✓
Zeros of this function will occur at ✓ = (0, 57.42˝ , 122.58˝ , 180˝ ).
302
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14.13. The radiation field of a certain short vertical current element is E✓s = (20_r) sin ✓ e*j10⇡r V/m if it
is located at the origin in free space.
a) Find E✓s at P (r = 100, ✓ = 90˝ , = 30˝ ): Substituting these values into the given formula,
find
20
E✓s =
sin(90˝ )e*j10⇡(100) = 0.2e*j1000⇡ V_m
100
b) Find E✓s at P if the vertical element is located at A(0.1, 90˝ , 90˝ ): This places the element on
the y axis at y = 0.1. As a result of moving the antenna from the origin to y = 0.1, the change in
distance to point P is negligible when considering the change in field amplitude, but is not when
considering the change in phase. Consider lines drawn from the origin to P and from y = 0.1
to P . These lines can be considered essentially parallel, and so the difference in their lengths
.
is l = 0.1 sin(30˝ ), with the line from y = 0.1 being shorter by this amount. The construction
and arguments are similar to those used in the discussion of the electric dipole in Sec. 4.7. The
electric field is now the result of part a, modified by including a shorter distance, r, in the phase
term only. We show this as an additional phase factor:
E✓s = 0.2e*j1000⇡ ej10⇡(0.1 sin 30 = 0.2e*j1000⇡ ej0.5⇡ V_m
c) Find E✓s at P if identical elements are located at A(0.1, 90˝ , 90˝ ) and B(0.1, 90˝ , 270˝ ): The
original element of part b is still in place, but a new one has been added at y = *0.1. Again,
constructing a line between B and P , we find, using the same arguments as in part b, that the
length of this line is approximately 0.1 sin(30˝ ) longer than the distance from the origin to P .
The part b result is thus modified to include the contribution from the second element, whose
field will add to that of the first:
E✓s = 0.2e*j1000⇡ ej0.5⇡ + e*j0.5⇡ = 0.2e*j1000⇡ 2 cos(0.5⇡) = 0
The two fields are out of phase at P under the approximations we have used.
14.14. For a dipole antenna of overall length 2l =
half-power beamwidth.
, evaluate the maximum directivity in dB, and the
Dmax is found using Eq. (64) with kl = ⇡, and involves a numerical integration. This I did
using a Mathematica code to find Dmax = 2.41, and in decibels this is 10 log10 (2.41) = 3.82 dB,
with the maximum occurring in the direction ✓ = 90˝ .
With kl = ⇡, the pattern function, Eq. (59), reaches a maximum value of 2 at ✓ ˘
= 90˝ . The
half-power beamwidth is found numerically by setting the pattern function equal to 2 and thus
solving
4
5 ˘
cos(⇡ cos ✓) * cos(⇡)
F (✓) =
= 2
sin ✓
The result is ✓ = 66.1˝ , which means that the deviation angle from the direction along the
maximum is ✓1_2 _2 = 90 * 66.1 = 23.9˝ . The half-power beamwidth is therefore ✓1_2 =
2 ù 23.9 = 47.8˝ .
303
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14.15. For a dipole antenna of overall length 2l = 1.3 , determine the locations in ✓ and the peak intensity
of the sidelobes, expressed as a fraction of the main lobe intensity. Express your result as the sidelobe
level in decibels, given by Ss [dB] = 10 log10 Sr,main _Sr,sidelobe . Again, use Fig. 14.8 as a guide.
Here, kl = (2⇡_ )(1.3 _2) = 1.3⇡. The angular intensity distribution is given by the square
of the pattern function, Eq. (59), with kl = 1.3⇡ substituted:
4
cos(1.3⇡ cos ✓) * cos(1.3⇡)
 < S >  ◊ [F (✓)] =
sin ✓
2
52
To find the maxima, the easiest way is to create a polar plot of the function, as shown in
Fig. 14.8, and then numerically determine the relative intensities of the peaks by searching
for local maxima at locations indicated in the plot. The main lobe occurs at ✓ = 90˝ , and has
value [F (✓)]2 = 2.52. The secondary maxima occur at ✓ = 33.8˝ and 146.2˝ , and both peaks
have values [F (✓)]2 = 0.47. The sidelobe level in decibels is thus
⇠
⇡
2.52
Ss [dB] = 10 log10
= 7.3 dB
0.47
304
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14.16. For a dipole antenna of overall length, 2l = 1.5 :
a) Evaluate the locations in ✓ at which the zeros and maxima in the E-plane pattern occur: In this
case, kl = (2⇡_ )(1.5 _2) = 1.5⇡. The angular intensity distribution is given by the square of
the pattern function, Eq. (59), with kl = 1.3⇡ substituted:
4
cos(1.5⇡ cos ✓) * cos(1.5⇡)
 < S >  ◊ [F (✓)] =
sin ✓
2
52
4
cos(1.5⇡ cos ✓)
=
sin ✓
52
Zeros occur whenever cos(1.5⇡ cos ✓) = 0, which in turn occurs whenever 1.5⇡ cos ✓ is an odd
multiple of ⇡_2 (positive or negative). All zeros over the range 0 f ✓ f ⇡ are found with just
the first two odd multiples, or
1
cos ✓z = ± , ±1 Ÿ ✓z = (0, 70.5˝ , 109.5˝ , 180˝ )
3
The function does indeed zero at ✓ = 0 and 180˝ , despite the presence of the sin ✓ term in the
denominator, as can be demonstrated using small argument approximations in the trig functions
(show this!). The maxima are found numerically, as was done in Problem 14.15. The results
are ✓max,m = (42.6˝ , 137.4˝ ) (main lobes) and ✓max,s = 90˝ (sidelobe). The pattern in plotted
below.
b) Determine the sidelobe level, as per the definition in Problem 14.15: At the main lobe angle
(✓ = 42.6˝ ), [F (✓)]2 evaluates as 1.96. The sidelobe at ✓ = 90˝ gives a value of 1.00. The
sidelobe level in decibles is thus
⇠
⇡
1.96
Ss [dB] = 10 log10
= 2.9 dB
1.00
c) Determine the maximum directivity: This is found evaluating Eq. (64) numerically (which
I did using a Mathematica code), and the result is Dmax = 2.23, or in decibels this becomes
10 log10 (2.23) = 3.5 dB.
305
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14.17. Consider a lossless half-wave dipole in free space, with radiation resistance, Rrad = 73 ohms, and
maximum directivity Dmax = 1.64. If the antenna carries a 1-A current amplitude,
a) How much total power (in watts) is radiated? We use the definition of radiation resistance,
Eq. (29), and write the radiated power:
1
1
Pr = I02 Rrad = (1)2 (73) = 36.5 W
2
2
b) How much power is intercepted by a 1 m2 aperture situated at distance r = 1 km away. The
aperture is on the equatorial plane and squarely faces the antenna. Assume uniform power
density over the aperture: The maximum directivity is given by Dmax = 4⇡Kmax _Pr , where the
maximum radiation intensity, Kmax , is related to the power density in W_m2 at distance r from
the antenna as Sr = Kmax _r2 . The power intercepted by the aperture is then the power density
times the aperture area, or
Prec = Sr ù Area =
Dmax Pr
4⇡r2
ù Area =
1.64(36.5) 2
(1) = 4.77 W
4⇡(103 )2
14.18. Repeat Problem 14.17, but with a full-wave antenna (2l = ). Numerical integrals may be necessary.
In this case, kl = ⇡, and the radiation resistance is found by applying Eq. (65):
Rrad = 60
⇡4
0
cos(⇡ cos ✓) * cos(⇡)
sin ✓
52
sin ✓ d✓ = 199 ohms
where the answer was found by numerical integration. With the same 1-A current amplitude as before,
the total radiated power is now
1
1
Pr = I02 Rrad = (1)2 (199) = 99.5 W
2
2
The maximum directivity is the result of Problem 14.14 (involving a numerical integration in
Eq. (64) with kl = ⇡), and the result was found to be Dmax = 2.41. The received power is now
found as in Problem 14.17:
Prec =
Dmax Pr
4⇡r2
ù Area =
2.41(99.5) 2
(1) = 19.1 W
4⇡(103 )2
306
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14.19. Design a two-element dipole array that will radiate equal intensities in the = 0, ⇡_2, ⇡, and 3⇡_2
directions in the H-plane. Specify the smallest relative current phasing, ⇠, and the smallest element
spacing, d.
The array function is given by Eq. (81), with n = 2:
A2 ( ) =
1 ÛÛ sin( ) ÛÛ
2 ÛÛ sin( _2) ÛÛ
This has periodic maxima occurring at = 0, ±2m⇡, and in the H-plane, = ⇠ + kd cos .
Now, for broadside operation, we have maxima at
= ±⇡_2, at which
= ⇠, so we set
⇠ = 0 to get the array function principal maximum at = 0. Having done this, we now have
= kd cos , which we need to have equal to ±2⇡ when = 0, ⇡, in order to get endfire
operation. This will happen when kd = 2⇡d_ = 2⇡, so we set d = .
14.20. A two-element dipole array is configured to provide zero radiation in the broadside ( = ±90˝ ) and
endfire ( = 0, 180˝ ) directions, but with maxima occurring at angles in between. Consider such a
set-up with = ⇡ at = 0 and = *3⇡ at = ⇡, with both values determined in the H-plane.
a) Verify that these values give zero broadside and endfire radiation: For two elements, the array
function is given by Eq. (81), with n = 2:
A2 ( ) =
This is clearly zero at
= ⇡ and at
1 ÛÛ sin( ) ÛÛ
2 ÛÛ sin( _2) ÛÛ
= *3⇡.
b) Determine the required relative current phase, ⇠: We know that
Setting up the two given conditions, we have:
= ⇠ +kd cos
in the H-plane.
⇡ = ⇠ + kd cos(0) = ⇠ + kd
and
*3⇡ = ⇠ + kd cos(⇡) = ⇠ * kd
Adding these two equations gives ⇠ = *⇡, which means that the contributions from the two
elements will in fact completely cancel in the broadside direction, regardless of the element
spacing.
c) Determine the required element spacing, d: We can now use the first condition, for example,
substituting values that we know:
( = 0) = ⇡ = *⇡ + kd cos(0) = *⇡ + kd
Therefore, kd = 2⇡_ = 2⇡, or d = .
d) Determine the values of at which maxima in the radiation pattern occur: With the values as
found, the array function now becomes
A2 ( ) =
1 ÛÛ sin[⇡(2 cos
Û
2 ÛÛ sin[ ⇡ (2 cos
2
This function maximizes at
* 1)] ÛÛ
Û
* 1)] ÛÛ
= 60˝ , 120˝ , 240˝ , and 300˝ , as found numerically.
307
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14.21. In the two-element endfire array of Example 14.4, consider the effect of varying the operating
frequency, f , away from the original design frequency, f0 , while maintaining the original current
phasing, ⇠ = *⇡_2. Determine the values of at which the maxima occur when the frequency is
changed to
a) f = 1.5f0 : The original phase and length from the example are ⇠ = *⇡_2 and d = 0 _4,
where 0 = c_f0 . Thus kd = (2⇡_ 0 )( 0 _4) = ⇡_2. The H-plane array function in that case
was found to be
4
5
⌧
⌧
⇠ kd
⇡ ⇡
Û
A(⇡_2, )Û = cos
= cos
+
cos
= cos * + cos
Û0
2
2
2
4 4
Now, if we change the frequency to 1.5f0 , while leaving ⇠ fixed (and the antenna physical length
is unchanged), we obtain = 0 _1.5, and kd = (2⇡_ 0 )(1.5)( 0 _4) = 3⇡_4. The new array
function is then:
4
5
⌧
⌧
⇠ kd
⇡ 3⇡
⇡
Û
A(⇡_2, )Û
= cos
+
cos
= cos * +
cos
= cos (3 cos * 2)
Û 0 _1.5
2
2
4
8
8
This function amplitude maximizes when 3 cos
b) f = 2f0 : Now,
=
* 2 = 0, or
= cos*1 (2_3) = ±48.2˝ .
0 _2 and kd = (2⇡_ 0 )(2)( 0 _4) = ⇡. The array function becomes
4
5
⌧
⇠ kd
⇡ ⇡
Û
A(⇡_2, )Û
= cos
+
cos
= cos * + cos
Û 0 _2
2
2
4 2
This function amplitude maximizes when 2 cos
* 1 = 0, or
= cos
⌧
⇡
(2 cos
4
* 1)
= cos*1 (1_2) = ±60.0˝ .
308
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14.22. Revisit Problem 14.21, but with the current phase allowed to vary with frequency (this will automatically occur if the phase difference is established by a simple time delay between the feed currents).
Now, the current phase difference will be ⇠ ® = ⇠ f _f0 , where f0 is the original (design) frequency.
Under this condition, radiation will maximize in the = 0 direction regardless of frequency (show
this). Backward radiation (along = ⇡) will develop however as the frequency is tuned away from
f0 . Derive an expression for the front-to-back ratio, defined as the ratio of the radiation intensities
at = 0 and = ⇡, expressed in decibels. Express this result as a function of the frequency ratio
f _f0 .
With the current phase and wavenumber both changing with frequency, we would have
0 1
0 1
f
f
®
=
⇠+
kd cos
f0
f0
With ⇠ = *⇡_2 and d =
0 _4, The H-plane array function at the original frequency, f0 , is
4
5
⌧
⌧
⇠ kd
⇡
Û
A(⇡_2, )Û = cos
= cos
+
cos
= cos (cos
Ûf0
2
2
2
4
* 1)
The effect on this of changing the frequency is then
4 ®5
4 0
15
40 1
f ⇠ kd
f ⇡
Û
A(⇡_2, )Û = cos
= cos
+
cos
= cos
(cos
Ûf
2
f0 2
2
f0 4
5
* 1)
The condition for the zero argument (which maximizes A) is (cos * 1) = 0 This term is
unaffected by changing the frequency, and so the array function will always maximize in the
= 0 direction.
In the forward direction ( = 0), the value of A2 (proportional to the radiation intensity) is
unity. In the backward direction ( = 180˝ ), we find
0
1
0
1
f ⇡
⇡f
A( = ⇡)2 = cos2 *
= cos2
f0 2
2f0
The front-to-back ratio is then
4
Rf b = 10 log10
L
M
5
A( = 0)2
1
= 10 log10
A( = ⇡)2
cos2 ⇡f _2f0
Evaluate the front-to-back ratio for a) f = 1.5f0 , b) f = 2f0 , c) f = 0.75f0 . Substituting these
values, we find:
5
1
f = 1.5f0 : Rf b = 10 log10
= 3 dB
cos2 (1.5⇡_2)
4
5
1
f = 2f0 : Rf b = 10 log10
= 0 dB
cos2 (⇡)
4
5
1
f = 0.75f0 : Rf b = 10 log10
= 8.3 dB
cos2 (⇡_2.67)
4
309
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14.23. A turnstile antenna consists of two crossed dipole antennas, positioned in this case in the xy plane.
The dipoles are identical, lie along the x and y axes, and are both fed at the origin. Assume that equal
currents are supplied to each antenna, and that a zero phase reference is applied to the x-directed
antenna. determine the relative phase, ⇠, of the y-directed antenna so that the net radiated electric
field as measured on the positive z axis is a) left circularly-polarized; b) linearly polarized along the
45˝ axis between x and y.
When looking at the field along the z axis, the expression for E can be constructed using
Eq. (57), evaluated at ✓ = ⇡_2 (so that we consider the direction normal to the antenna), and
in which r is replaced by z. An additional “array” term includes the two polarization directions
(unit vectors) which are out of phase by ⇠:
E(z, ✓ = ⇡_2) = J
⌅
⇧
I0 ⌘
[1 * cos(kl)] e*jkz ax + ay ej⇠
2⇡z
a) To achieve left circular polarization for propagation in the forward z direction, the vector array
function must be
⌅
⇧ ⌅
⇧
ax + ay ej⇠ = ax + jay Ÿ ⇠ = ⇡_2
b) To achieve 45˝ polarization, the vector array function must be
⌅
⇧ ⌅
⇧
ax + ay ej⇠ = ax ± ay Ÿ ⇠ = 0, ⇡
14.24. Consider a linear endfire array, designed for maximum radiation intensity at = 0, using ⇠ and d
values as suggested in Example 14.5. Determine an expression for the front-to-back ratio (defined in
Problem 14.22) as a function of the number of elements, n, if n is an odd number.
From Example 14.5, we use ⇠ = *⇡_2 and d = _4, so that kd = ⇡_2. Then in the H-plane
= ⇠ + kd cos
=
⇡
(cos
2
* 1)
The array function is then:
4
5
4
5
1 sin(n _2)
1 sin[(n⇡_4)(cos * 1)]
A( ) =
=
n sin( _2)
n sin[(⇡_4)(cos * 1)]
In the forward direction ( = 0), A = 1. In the backward direction ( = ⇡), A = ±1_n, where
n is an odd integer. The front-to-back ratio is therefore:
0
1
A( = 0)2
Rf b = 10 log10
= 10 log10 (n2 ) dB
A( = ⇡)2
in which performance is clearly better as n increases.
310
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14.25. A six-element linear dipole array has element spacing d = _2.
a) Select the appropriate current phasing, ⇠, to achieve maximum radiation along
d = _2, we have kd = ⇡ and in the H-plane,
= ⇠ + kd cos
= ±60˝ : With
= ⇠ + ⇡ cos
We set the principal maximum to occur at
= 0, and with = 60˝ , the condition for this
becomes
⇡
⇡
= 0 = ⇠ + ⇡ cos(60˝ ) = ⇠ +
Ÿ ⇠=*
2
2
b) With the phase set as in part a, evaluate the intensities (relative to the maximum) in the broadside
and endfire directions. With the above result, the array function (Eq. (81)) in the H-plane
becomes
4
5
4
5
sin[3(⇠ + kd cos )]
1
1 sin[(3⇡_2)(2 cos * 1)]
A(*⇡_2, ) =
=
6 sin[(1_2)(⇠ + kd cos )]
6 sin[(⇡_4)(2 cos * 1)]
In the broadside direction ( = 90˝ ) the array function becomes
4
5 ˘
2
1 sin[*3⇡_2]
A(*⇡_2, ⇡_2) =
=
6 sin[*⇡_4]
6
The intensity in this direction (relative to that along
= 60˝ ) is the square of this or 1_18
In the endfire direction ( = 0˝ ) the array function becomes
4
5 ˘
2
1 sin[3⇡_2]
A(*⇡_2, 0) =
=
6 sin[⇡_4]
6
The intensity in this direction (relative to that along = 60˝ ) is the square of this or 1_18, as
before. The same result is found for = 180˝ and for = *90˝ . The intensity pattern is shown
below:
311
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14.26. In a linear endfire array of n elements, a choice of current phasing that improves the directivity is
given by the Hansen-Woodyard condition:
⇠
⇡
2⇡d ⇡
⇠=±
+
n
where the plus or minus sign choices give maximum radiation along = 180˝ and 0˝ respectively.
Applying this phasing may not necessarily lead to unidirectional endfire operation (zero backward
radiation), but will do so with the proper choice of element spacing, d.
a) Determine this required spacing as a function of n and : We first construct the expression for
in the H-plane, using the given current phase expression, and in which the minus sign is chosen:
= ⇠ + kd cos
Note that
=*
2⇡d
*
⇡ 2⇡d
+
cos
n
( = 0) = *⇡_n. The array function, Eq. (81) is:
4
5
4
5
1 sin(n _2)
1 sin[(n⇡d_ )(cos * 1) * ⇡_2]
A(⇠, ) =
=
n sin( _2)
n sin[(⇡d_ )(cos * 1) * (⇡_2n)]
This function is equal to 1 when
function becomes
= 0, regardless of the choice of d. When
4
5
1 sin[(*2n⇡d_ ) * (⇡_2)]
A(⇠, ⇡) =
n sin[(*2⇡d_ ) * (⇡_2n)]
= ⇡, the array
which we require to be zero. This will happen when ( = ⇡) = *⇡. Using the above formula
for , we set up the condition:
⇠
⇡
4⇡d ⇡
n*1
( = ⇡) = *
* = *⇡ Ÿ d =
n
n
4
b) Show that the spacing as found in part a approaches _4 for a large number of elements: It is
readily seen from the part a result that when n ô ÿ, d ô _4.
c) Show that an even number of elements is required: This can be shown by substituting d as found
in part a into the array function at = ⇡. The result is
4
5
1 sin(*n⇡_2)
A(⇠, ⇡) =
n sin(*⇡_2)
We require this result to be zero, which will only happen if n is an even integer. This can be
more quickly seen from the general expression:
4
5
1 sin(n _2)
A(⇠, ) =
n sin( _2)
wherein if
= *⇡, we will get a zero result only if n is even.
312
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14.27. Consider an n-element broadside linear array. Increasing the number of elements has the effect of
narrowing the main beam. Demonstrate this by evaluating the separation in between the zeros on
either side of the principal maximum at = 90˝ . Show that for large n this separation is approximated
.
.
by
= 2 _L, where L = nd is the overall length of the array.
For broadside operation, ⇠ = 0, and, with kd = 2⇡d_ , we have
= ⇠ + kd cos
=
2⇡d
cos
With this condition, the array function, Eq. (81) is:
4
5
4
5
1 sin(n _2)
1 sin[(n⇡d_ ) cos ]
A(⇠, ) =
=
n sin( _2)
n sin[(⇡d_ ) cos ]
Zeros in this function will occur whenever
n⇡d
cos
= ±m⇡
where m is an integer. Now, on either side of the principal maximum at = 0, zeros will occur
at +⇡ and *⇡ (giving a 2⇡ separation), which we can associate with the two angles + and *
respectively, which lie on either side of 90˝ . We can therefore write:
n⇡d
cos
+
* cos
*
= 2⇡
Using a trig identity, this becomes:
⌧
⌧
2n⇡d
1
1
sin ( + + * ) sin ( * *
2
2
´≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠¨
+
) = 2⇡
1
Because we are considering the broadside direction, we have ( + +
first sine term is just unity. We find
⌧
1
sin ( * * + ) =
2
nd
* )_2 = ⇡_2 and so the
Now as n gets large, both sides of the equation become << 1, so that we may approximate the
sine term by just its argument:
.
1 *
( * +) =
2
nd
*
+
Defining
=
* , the above expression becomes
. 2 . 2
=
=
(done)
nd
L
313
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14.28. A large ground-based transmitter radiates 10kW, and communicates with a mobile receiving station
that dissipates 1mW on the matched load of its antenna. The receiver (not having moved) now transmits back to the ground station. If the mobile unit radiates 100W, what power is received (at a matched
load) by the ground station?
We can use Eq. (93) and the fact that Z21 = Z12 to write:
PL2
Z21 2
Z12 2
P
=
=
= L1
Pr1
R11 R22
R11 R22
Pr2
We have
So
PL2
PL1
10*3
*7
=
=
10
=
Pr1
Pr2
104
PL1 = Pr2 ù 10*7 = 100 ù 10*7 = 10*5 = 10 W
14.29. Signals are transmitted at a 1m carrier wavelength between two identical half-wave dipole antennas
spaced by 1km. The antennas are oriented such that they are exactly parallel to each other.
a) If the transmitting antenna radiates 100 watts, how much power is dissipated by a matched load
at the receiving antenna? To find the dissipated power, use Eq. (106):
PL2 =
Pr1 2
D1 (✓1 1 ) D2 (✓2 2 )
(4⇡r)2
Since the antennas face each other squarely, the directivities are both the maximum values for
the two, where for the half-wave dipole, we know that Dmax = 1.64 So
PL2 =
100(1)2
(1.64)2 = 1.7 W
3
2
(4⇡ ù 10 )
b) Suppose the receiving antenna is tilted by 45˝ while the two antennas remain in the same plane.
What is the received power in this case? With the receiving antenna tilted, its directivity is
reduced accordingly. Using Eq. (63), with the help of the results of Example 14.2, we have
Û cos[(⇡_2) cos(45˝ )] Û2
Û = 0.643
D2 (✓2 = 45˝ ) = Dmax ù F (45˝ )2 = 1.64 ù ÛÛ
Û
sin(45˝ )
Û
Û
So now,
PL2 (45˝ ) =
100(1)2
(1.64)(0.643) = 672 nW
(4⇡ ù 103 )2
314
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14.30. A half-wave dipole antenna is known to have a maximum effective area, given as Amax .
a) Write the maximum directivity of this antenna in terms of Amax and wavelength : From
Eq. (104), it follows that
4⇡
Dmax = 2 Amax
b) Express the current amplitude, I0 , needed to radiate total power, Pr , in terms of Pr , Amax , and
: Combining Eqs. (64) and (65), we can write
2Pr
120[F (✓)]2max
I0
Dmax
=
2
where, for a half-wave dipole, [F (✓)]2max = 1. Then, using the part a result, we find
u
I0 =
⇡Pr Amax
15 2
c) At what values of ✓ and will the antenna effective area be equal to Amax ? These will be the
same angles over which the half-wave dipole directivity maximizes, or at which F (✓) for the
antenna reaches its maximum. We know from Example 14.2 that this happens at ✓ = ⇡_2, and
at all values of .
315
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