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 )
= (1.5𝐚𝑥 + 2.5𝐚𝑦 + 0.5𝐚𝑧 )∕2.96
𝐚𝑚𝑝 = √
(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 − 32 𝐵)𝐚𝑦 − (4 + 13 𝐵)𝐚𝑧 | = 10
Expanding, obtain
36 − 8𝐵 + 94 𝐵 2 + 4 − 38 𝐵 + 49 𝐵 2 + 16 + 83 𝐵 + 19 𝐵 2 = 100
or 𝐵 2 − 8𝐵 − 44 = 0. Thus 𝐵 =
√
8± 64−176
2
= 11.75 (taking positive option) and so
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:
A unit vector tangent to this circle in the general increasing 𝑥 direction is 𝐭 = +𝐚𝜙 . Its 𝑥 and
√
𝑦 components are 𝐭𝑥 = 𝐚𝜙 ⋅ 𝐚𝑥 = − sin 𝜙, and 𝐭𝑦 = 𝐚𝜙 ⋅ 𝐚𝑦 = cos 𝜙. At the point ( 3, −1),
√
𝜙 = 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 =
2
14,
a) the dot product:√
First, 𝐀 ⋅ 𝐁 = 2 −
3
+
6
=
5
=
𝐴𝐵
cos
𝜃,
where
𝐴
=
√
◦
2
2
2
and where 𝐵 = 1 + 3 + 2 = 14. Therefore cos 𝜃 = 5∕14, so that 𝜃 = 69.1 .
b) the cross product: Begin with
|𝐚 𝐚 𝐚 |
| 𝑥
𝑦
𝑧|
|
|
𝐀 × 𝐁 = | 2 1 3 | = 11𝐚𝑥 − 𝐚𝑦 − 7𝐚𝑧
|
|
| 1 −3 2 |
|
|
√
√
√
2
2
2
and then |𝐀 × 𝐁|
(√= 11 )+ 1 + 7 = 171. So now, with |𝐀 × 𝐁| = 𝐴𝐵 sin 𝜃 = 171,
find 𝜃 = sin−1
171∕14 = 69.1◦
1.7. Given the field 𝐅 = 𝑥𝐚𝑥 + 𝑦𝐚𝑦 . If 𝐅 ⋅ 𝐆 = 2𝑥𝑦 and 𝐅 × 𝐆 = (𝑥2 − 𝑦2 ) 𝐚𝑧 , find 𝐆:
Let 𝐆 = 𝑔1 𝐚𝑥 + 𝑔2 𝐚𝑦 + 𝑔3 𝐚𝑧 Then 𝐅 ⋅ 𝐆 = 𝑔1 𝑥 + 𝑔2 𝑦 = 2𝑥𝑦, and
|𝐚 𝐚 𝐚 |
| 𝑥 𝑦 𝑧|
|
|
𝐅 × 𝐆 = | 𝑥 𝑦 0 | = 𝑔3 𝑦 𝐚𝑥 − 𝑔3 𝑥 𝐚𝑦 + (𝑔2 𝑥 − 𝑔1 𝑦) 𝐚𝑧 = (𝑥2 − 𝑦2 ) 𝐚𝑧
|
|
|𝑔1 𝑔2 𝑔3 |
|
|
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 × 𝐁 = |𝐀||𝐁| sin 𝜃𝐧. With the given vectors we find
[
]
√ 2𝐚𝑦 + 𝐚𝑧
√
√
𝐀 × 𝐁 = 14𝐚𝑦 + 7𝐚𝑧 = 7 5
= 9 + 4 + 16 4 + 1 + 4 sin 𝜃 𝐧
√
5
⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟
±𝐧
where 𝐧 is identified as shown; we see that 𝐧 can be positive or negative, as sin 𝜃 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 𝜃 = 7 5∕( 29 9) = 0.969. Two values of 𝜃 (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
𝜃
=
3
29 cos 𝜃, or
√
cos 𝜃 = −4∕(3 29) = −0.248 ⇒ 𝜃 = −75.7◦ . Again, the minus sign is not important, as we
care only about the angle magnitude. The main point is that only one 𝜃 value results when using
the dot product, so no ambiguity.
1.9. A field is given as
𝐆=
25
(𝑥𝐚𝑥 + 𝑦𝐚𝑦 )
(𝑥2 + 𝑦2 )
Find:
a) a unit vector in the direction of 𝐆 at 𝑃 (3, 4, −2): Have 𝐆𝑝 = 25∕(9 + 16) × (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 𝜃. So cos 𝜃 =
(0.6, 0.8, 0) ⋅ (1, 0, 0) = 0.6. Thus 𝜃 = 53◦ .
c) the value of the following double integral on the plane 𝑦 = 7:
4
2
∫0 ∫0
4
∫0 ∫0
2
𝐆 ⋅ 𝐚𝑦 𝑑𝑧𝑑𝑥
4
2
4
25
25
350
(𝑥𝐚𝑥 + 𝑦𝐚𝑦 ) ⋅ 𝐚𝑦 𝑑𝑧𝑑𝑥 =
× 7 𝑑𝑧𝑑𝑥 =
𝑑𝑥
∫0 ∫0 𝑥2 + 49
∫0 𝑥2 + 49
𝑥2 + 𝑦2
[
( )
]
4
1
= 350 ×
tan−1
− 0 = 26
7
7
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 𝜃, or 𝑏2 (1 − 1 + 1) = ( 3𝑏)( 3𝑏) cos 𝜃 ⇒
cos 𝜃 = 1∕3 ⇒ 𝜃 = 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
𝜃𝑀 = cos
(
𝐑𝑀𝑁 ⋅ 𝐑𝑀𝑃
|𝐑𝑀𝑁 ||𝐑𝑀𝑃 |
(
)
−1
= cos
0.05
√
√
0.34 0.17
)
= 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
(0.1, 0.2, 0.3) = (0.93, 1.86, 2.79)
𝐆
=
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
(10, −6, 5) = (0.02, 0.25, 0.26)
𝐅 = (0.1, 0.2, 0.3) −
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 × 𝐫2
(5, 25, 55)
=
= (0.08, 0.41, 0.91)
|𝐫1 × 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 × 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 × 𝐫2 | = 30.3
2
d) the area of the triangle defined by the heads of 𝐫1 , 𝐫2 , and 𝐫3 :
1
1
Area = |(𝐫2 − 𝐫1 ) × (𝐫2 − 𝐫3 )| = |(−9, 4, −1) × (−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 × 𝐑𝐴𝑁
(350, −200, 340)
=
= (0.664, −0.379, 0.645)
|𝐑𝐴𝑀 × 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 × 𝐚𝐴𝑁 = (0.664, −0.379, 0.645) × (−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 𝜃 𝐚𝜃 and 𝐇 = (𝐵∕𝑟) sin 𝜃 𝐚𝜙 , where 𝐴 and 𝐵 are constants,
a) evaluate 𝐒 = 𝐄 × 𝐇 and express the result in rectangular coordinates: Begin with
𝐴𝐵 2
sin 𝜃 𝐚𝑟
𝑟2
Now, find the three rectangular components by using:
𝐒=𝐄×𝐇=
𝑆𝑥 = 𝐒 ⋅ 𝐚𝑥 =
𝑥(𝑥2 + 𝑦2 )
𝐴𝐵 2
sin
𝜃
𝐚
⋅
𝐚
=
𝐴𝐵
𝑟
𝑥
𝑟2
(𝑥2 + 𝑦2 + 𝑧2 )5∕2
where we have used 𝐚𝑟 ⋅ 𝐚𝑥 = sin 𝜃 cos 𝜙 with 𝑟 = (𝑥2 + 𝑦2 + 𝑧2 )1∕2 ,
sin 𝜃 = (𝑥2 + 𝑦2 )1∕2 ∕(𝑥2 + 𝑦2 + 𝑧2 )1∕2 and cos 𝜙 = 𝑥∕(𝑥2 + 𝑦2 )1∕2 .
Then
𝑆𝑦 = 𝐒 ⋅ 𝐚𝑦 =
𝑦(𝑥2 + 𝑦2 )
𝐴𝐵 2
sin
𝜃
𝐚
⋅
𝐚
=
𝐴𝐵
𝑟
𝑦
𝑟2
(𝑥2 + 𝑦2 + 𝑧2 )5∕2
where we have used 𝐚𝑟 ⋅ 𝐚𝑦 = sin 𝜃 sin 𝜙 with sin 𝜙 = 𝑦∕(𝑥2 + 𝑦2 )1∕2 .
Finally
𝑆𝑧 = 𝐒 ⋅ 𝐚 𝑧 =
𝑧(𝑥2 + 𝑦2 )
𝐴𝐵 2
sin
𝜃
𝐚
⋅
𝐚
=
𝐴𝐵
𝑟
𝑧
𝑟2
(𝑥2 + 𝑦2 + 𝑧2 )5∕2
where 𝐚𝑟 ⋅ 𝐚𝑧 = cos 𝜃 = 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 = 𝜌2 + 𝑧2 , the given field may
be written:
]
𝐴 [
𝐅= 2
𝑓
𝐚
+
𝑓
𝐚
+
𝑓
𝐚
𝜌
𝜌
𝜙
𝜙
𝑧
𝑧
𝜌 + 𝑧2
Now
𝜌
𝜌
𝑓𝜌 = 𝐚𝑟 ⋅ 𝐚𝜌 = sin 𝜃 = = 2
𝑟
(𝜌 + 𝑧2 )1∕2
𝑓𝜙 = 𝐚𝑟 ⋅ 𝐚𝜙 = 0
𝑧
𝑧
𝑓𝑧 = 𝐚𝑟 ⋅ 𝐚𝑧 = cos 𝜃 = = 2
𝑟
(𝜌 + 𝑧2 )1∕2
Substituting all, we find
]
𝐴 [
𝐅= 2
𝜌𝐚
+
𝑧𝐚
𝜌
𝑧
𝜌 + 𝑧2
b) Transform the given field
√ into rectangular coordinates: The quickest way is to use√the results of
part 𝑎 by using 𝜌 = 𝑥2 + 𝑦2 and 𝐚𝜌 = cos 𝜙 𝐚𝑥 + sin 𝜙 𝐚𝑦 , where cos 𝜙 = 𝑥∕ 𝑥2 + 𝑦2 and
√
sin 𝜙 = 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◦ − 𝛼) = −𝐴𝐵 cos 𝛼 where 𝛼 is the interior angle at the junction of 𝐀 and
𝐁. Using this, we have 𝐶 2 = 𝐴2 + 𝐵 2 − 2𝐴𝐵 cos 𝛼, 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) → 𝐶(𝜌 = 3.61, 𝜙 = 33.7◦ , 𝑧 = −7) and
𝐷(−1, −4, 2) → 𝐷(𝜌 = 4.12, 𝜙 = −104.0◦ , 𝑧 = 2).
Now 𝐑𝐶𝐷 = (−4, −6, 9) and 𝑅𝜌 = 𝐑𝐶𝐷 ⋅ 𝐚𝜌 = −4 cos(33.7) − 6 sin(33.7) = −6.66. Then
𝑅𝜙 = 𝐑𝐶𝐷 ⋅ 𝐚𝜙 = 4 sin(33.7) − 6 cos(33.7) = −2.77. So 𝐑𝐶𝐷 = −6.66𝐚𝜌 − 2.77𝐚𝜙 + 9𝐚𝑧
b) a unit vector at 𝐷 directed toward 𝐶:
𝐑𝐶𝐷 = (4, 6, −9) and 𝑅𝜌 = 𝐑𝐷𝐶 ⋅ 𝐚𝜌 = 4 cos(−104.0) + 6 sin(−104.0) = −6.79. Then 𝑅𝜙 =
𝐑𝐷𝐶 ⋅ 𝐚𝜙 = 4[− sin(−104.0)] + 6 cos(−104.0) = 2.43. So 𝐑𝐷𝐶 = −6.79𝐚𝜌 + 2.43𝐚𝜙 − 9𝐚𝑧
Thus 𝐚𝐷𝐶 = −0.59𝐚𝜌 + 0.21𝐚𝜙 − 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:
𝑎𝜌 = (0.22, 0.87, −0.44) ⋅ 𝐚𝜌 = 0.22 cos(−104.0) + 0.87 sin(−104.0) = −0.90, and
𝑎𝜙 = (0.22, 0.87, −0.44) ⋅ 𝐚𝜙 = 0.22[− sin(−104.0)] + 0.87 cos(−104.0) = 0, so that finally,
𝐚 = −0.90𝐚𝜌 − 0.44𝐚𝑧 .
1.22. A sphere of radius 𝑎, centered at the origin, rotates about the 𝑧 axis at angular velocity Ω 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(𝜃) = Ω𝑎 sin 𝜃 𝐚𝜙 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 Δ𝜃 in radians. Assume Δ𝜃 is small:
Method 1: The velocity difference between the two points can be approximated as
.
𝑑𝑣 |
Δ𝑣 = Δ𝜃
|
𝑑𝜃 |𝜃0
where 𝜃0 is the mean angle between the two points, and where Δ𝜃 is small enough so that 𝑣(𝜃)
can be approximated as a linear function. Using this along with the part 𝑎 result, we find:
.
Δ𝑣 = 𝑎ΩΔ𝜃 cos 𝜃0 m∕s
8
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1.22
b) (continued) Method 2 (harder): Write
Δ𝑣 = 𝑎Ω(sin 𝜃2 − sin 𝜃1 )
where 𝜃1 and 𝜃2 lie on either side of 𝜃0 , and where 𝜃2 − 𝜃1 = Δ𝜃. Using a trig identity, the above
expression becomes
(
)
) .
(
1
1
Δ𝑣 = 2𝑎Ω cos (𝜃2 + 𝜃1 ) sin (𝜃2 − 𝜃1 ) = 𝑎ΩΔ𝜃 cos 𝜃0 m∕s
2
2
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟
.
cos 𝜃0
=Δ𝜃∕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, 𝜃0 = 45◦ (same in spherical and geographic coordinates), Δ𝜃 = 2.0◦ =
0.035 rad, and Ω = 2𝜋∕24 hr = 7.3 × 10−5 rad/s. Using part 𝑏, the velocity difference becomes
√
.
Δ𝑣 = (6.37 × 106 m)(7.3 × 10−5 rad∕s)(0.035 rad)( 2∕2) = 11.5 m∕s
with speed increasing southward.
1.23. The surfaces 𝜌 = 3, 𝜌 = 5, 𝜙 = 100◦ , 𝜙 = 130◦ , 𝑧 = 3, and 𝑧 = 4.5 define a closed surface.
a) Find the enclosed volume:
130◦
4.5
Vol =
∫3
∫100◦ ∫3
5
𝜌 𝑑𝜌 𝑑𝜙 𝑑𝑧 = 6.28
NOTE: The limits on the 𝜙 integration must be converted to radians (as was done here, but not shown).
b) Find the total area of the enclosing surface:
130◦
Area = 2
4.5
+
∫3
5
∫100◦ ∫3
130◦
∫100◦
130◦
4.5
𝜌 𝑑𝜌 𝑑𝜙 +
5 𝑑𝜙 𝑑𝑧 + 2
∫3
∫3
∫100◦
4.5
5
∫3
3 𝑑𝜙 𝑑𝑧
𝑑𝜌 𝑑𝑧 = 20.7
c) Find the total length of the twelve edges of the surfaces:
[ ◦
]
30
30◦
Length = 4 × 1.5 + 4 × 2 + 2 ×
×
2𝜋
×
3
+
×
2𝜋
×
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(𝜌 = 3, 𝜙 = 100◦ , 𝑧 = 3) and B(𝜌 = 5, 𝜙 = 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 𝜙1 and
𝜙2 with the 𝑥 axis respectively.
a) Express each vector in rectangular components; Have 𝐚1 = 𝐴𝑥1 𝐚𝑥 +𝐴𝑦1 𝐚𝑦 , so that 𝐴𝑥1 = 𝐚1 ⋅𝐚𝑥 =
cos 𝜙1 . Then, 𝐴𝑦1 = 𝐚1 ⋅ 𝐚𝑦 = cos(90 − 𝜙1 ) = sin 𝜙1 . Therefore,
𝐚1 = cos 𝜙1 𝐚𝑥 + sin 𝜙1 𝐚𝑦
and similarly, 𝐚2 = cos 𝜙2 𝐚𝑥 + sin 𝜙2 𝐚𝑦
b) take the dot product and verify the trigonometric identity, cos(𝜙1 − 𝜙2 ) = cos 𝜙1 cos 𝜙2 +
sin 𝜙1 sin 𝜙2 : From the definition of the dot product,
𝐚1 ⋅ 𝐚2 = (1)(1) cos(𝜙1 − 𝜙2 )
= (cos 𝜙1 𝐚𝑥 + sin 𝜙1 𝐚𝑦 ) ⋅ (cos 𝜙2 𝐚𝑥 + sin 𝜙2 𝐚𝑦 ) = cos 𝜙1 cos 𝜙2 + sin 𝜙1 sin 𝜙2
c) take the cross product and verify the trigonometric identity sin(𝜙2 − 𝜙1 ) = sin 𝜙2 cos 𝜙1 −
cos 𝜙2 sin 𝜙1 : From the definition of the cross product, and since 𝐚1 and 𝐚2 both lie in the 𝑥𝑦 plane,
| 𝐚
𝐚𝑦
𝐚𝑧 ||
| 𝑥
|
|
𝐚1 × 𝐚2 = (1)(1) sin(𝜙1 − 𝜙2 ) 𝐚𝑧 = |cos 𝜙1 sin 𝜙1 0 |
|
|
|cos 𝜙2 sin 𝜙2 0 |
|
|
[
]
= sin 𝜙2 cos 𝜙1 − cos 𝜙2 sin 𝜙1 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 𝜌 and 𝜙 components to find. First,
]
𝐴 [
𝑥
𝐚
⋅
𝐚
−
𝑦
𝐚
⋅
𝐚
𝑦
𝜌
𝑥
𝜌
𝑥2 + 𝑦2
√
√
where 𝐚𝑦 ⋅ 𝐚𝜌 = sin 𝜙 = 𝑦∕ 𝑥2 + 𝑦2 and 𝐚𝑥 ⋅ 𝐚𝜌 = cos 𝜙 = 𝑥∕ 𝑥2 + 𝑦2 . Thus
𝐻𝜌 = 𝐇 ⋅ 𝐚𝜌 =
𝐻𝜌 =
𝐴
[𝑥𝑦 − 𝑦𝑥] = 0
(𝑥2 + 𝑦2 )3∕2
Then
]
𝐴 [
𝑥
𝐚
⋅
𝐚
−
𝑦
𝐚
⋅
𝐚
𝑦
𝜙
𝑥
𝜙
𝑥2 + 𝑦2
√
√
where 𝐚𝑦 ⋅ 𝐚𝜙 = cos 𝜙 = 𝑥∕ 𝑥2 + 𝑦2 and 𝐚𝑥 ⋅ 𝐚𝜙 = − sin 𝜙 = −𝑦∕ 𝑥2 + 𝑦2 . Thus
𝐻𝜙 = 𝐇 ⋅ 𝐚𝜙 =
𝐻𝜙 =
[ 2
]
𝐴
𝐴
𝐴
2
𝑥
+
𝑦
= 2
=
𝜌
(𝑥2 + 𝑦2 )3∕2
(𝑥 + 𝑦2 )1∕2
Finally
𝐇=
𝐴
𝐚
𝜌 𝜙
10
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1.26. Express the uniform vector field, 𝐅 = 10 𝐚𝑦 in
a) cylindrical components:
𝐹𝜌 = 10 𝐚𝑦 ⋅ 𝐚𝜌 = 10 sin 𝜙, 𝐹𝜙 = 10 𝐚𝑦 ⋅ 𝐚𝜙 = 10 cos 𝜙, and 𝐹𝑧 = 0.
Combining, we obtain 𝐅(𝜌, 𝜙) = 10(sin 𝜙 𝐚𝜌 + cos 𝜙 𝐚𝜙 ).
b) spherical components:
𝐹𝑟 = 10 𝐚𝑦 ⋅ 𝐚𝑟 = 10 sin 𝜃 sin 𝜙; 𝐹𝜃 = 10 𝐚𝑦 ⋅ 𝐚𝜃 = 10 cos 𝜃 sin 𝜙; 𝐹𝜙 = 10 𝐚𝑦 ⋅ 𝐚𝜙 = 10 cos 𝜙.
[
]
Combining, we obtain 𝐅(𝑟, 𝜃, 𝜙) = 10 sin 𝜃 sin 𝜙 𝐚𝑟 + cos 𝜃 sin 𝜙 𝐚𝜃 + cos 𝜙 𝐚𝜙 .
1.27. The dipole field is given in spherical coordinates as:
𝐄=
𝐴
(2 cos 𝜃 𝐚𝑟 + sin 𝜃 𝐚𝜃 )
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 𝐚𝜙 component, the cylindrical
coordinate field components will be 𝐚𝜌 and 𝐚𝑧 only. We look for the surface on which the 𝐚𝜌
component is zero. So we set up:
𝐸𝜌 = 𝐄 ⋅ 𝐚𝜌 =
𝐴
(2 cos 𝜃 𝐚𝑟 ⋅ 𝐚𝜌 + sin 𝜃 𝐚𝜃 ⋅ 𝐚𝜌 ) = 0
𝑟3
⏟⏟⏟
⏟⏟⏟
cos 𝜃
sin 𝜃
From which the condition for zero 𝐸𝜌 is identified:
3 cos 𝜃 sin 𝜃 = 0 ⇒ 𝜃 = 0, 90◦
The 𝜃 = 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 𝜃 = 90◦ . On that surface:
𝐄(𝜃 = 90) =
𝐴
𝐴
𝐚𝜃 = − 3 𝐚𝑧
𝑟3
𝜌
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 𝜃 = 0 solution in part 𝑎, which is the 𝑧
axis. The field on that axis is
2𝐴
𝐄(𝜃 = 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 𝜃 𝐚𝑟 ⋅ 𝐚𝑧 + sin 𝜃 𝐚𝜃 ⋅ 𝐚𝑧 ) = 0
𝑟3
⏟⏟⏟
⏟⏟⏟
cos 𝜃
− sin 𝜃
This tells us that for a zero 𝑧 component, we must have
2 cos2 𝜃 − sin2 𝜃 = 0 ⇒ 3 cos2 𝜃 − 1 = 0 ⇒ 𝜃 = 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.
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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, 𝜃, from
the 𝑥 axis. But this would allow either vector to lie anywhere along a conical surface of angle 𝜃
about the 𝑥 axis. Therefore, 𝐀 can be equal to 𝐁, but not necessarily.
b) 𝐀 × 𝐚𝑥 = 𝐁 × 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 𝜃 to the 𝑥 axis, 𝐁 could lie at 𝜃 or at 180◦ − 𝜃 to give the same cross product. So again,
𝐀 can be equal to 𝐁, but not necessarily.
c) 𝐀 ⋅ 𝐚𝑥 = 𝐁 ⋅ 𝐚𝑥 and 𝐀 × 𝐚𝑥 = 𝐁 × 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, 𝜃, to
𝐚𝑥 . With coplanar vectors, this latter condition might imply that both +𝜃 and −𝜃 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 𝐀×𝐂 = 𝐁×𝐂 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 = 10𝑧𝐚̇ 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 𝜃
where, on the sphere surface, 𝑧 = 𝑎 cos 𝜃. Therefore 𝐹𝑟 (on surface) = 10𝑎 cos2 𝜃.
b) The tangential component is found by projecting 𝐅 into the 𝐚𝜃 or 𝐚𝜙 directions. Note that
𝐚𝑧 ⋅ 𝐚𝜙 = 0, so we are left with the tangential component lying in only the direction of 𝐚𝜃 :
𝐹𝜃 = 10𝑧𝐚𝑧 ⋅ 𝐚𝜃 |𝑟=𝑎 = −10𝑎 cos 𝜃 sin 𝜃
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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.
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Engineering Electromagnetics 9th solution
Engineering Electromagnetics (Chungnam National University)
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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
[
(
) ]
]
𝑞
𝑞 [
1
1
1
𝐚𝑥 + 𝐚𝑦
𝐄𝑜 =
− 2 𝐚𝑦 =
𝐚𝑥 + √
2
4𝜋𝜖0 2
16𝜋𝜖0
( 2)2 2
The magnitude of this field is
(
|𝐄𝑜 | = 𝐄𝑜 ⋅ 𝐄𝑜
)1∕2
√
=
2𝑞
16𝜋𝜖0
◦ line in the first quadrant. Its
To counter this field, the fourth charge must be positioned
√ √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:
)
(
(
)
1
1
𝐄12 =
𝐚𝑥 + 𝐚𝑦 + 𝐚𝑧
nC∕m
√
4𝜋𝜖0 (3)
3
√
in which the distance between charges is 3 m. The force on the -2nC charge is then
𝐅12 = 𝑞2 𝐄12 =
)
(
)
−1 (
−2
𝐚𝑥 + 𝐚𝑦 + 𝐚𝑧
𝐚𝑥 + 𝐚𝑦 + 𝐚𝑧 =
√
10.4 𝜋𝜖0
12 3 𝜋𝜖0
nN
The force on the 1nC charge at the origin is just the opposite of this result, or
𝐅21 =
)
+1 (
𝐚𝑥 + 𝐚𝑦 + 𝐚𝑧
10.4 𝜋𝜖0
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:
[
]
𝐑𝐷𝐴
𝐑𝐵𝐴
(50 × 10−9 )2 𝐑𝐶𝐴
+
+
𝐅=
4𝜋𝜖0
|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
]
[
(50 × 10−9 )2
1
2
1
𝐅=
𝐚𝑥 = 21.5𝐚𝑥 𝜇N
√ + √ +
4𝜋𝜖0
2 2 2 2 8
where distances are in meters.
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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)
⎡
⎤
⎢
⎥
⎥
𝑞 ⎢⎢ 𝐚𝑥 + 𝐚𝑦 + 𝐚𝑧 𝐚𝑦 + 𝐚𝑧 𝐚𝑥 + 𝐚𝑧 𝐚𝑥 + 𝐚𝑦
⎥
+
=
+
+
+
𝐚
+
𝐚
+
𝐚
√
√
√
√
𝑥
𝑦
𝑧
4𝜋𝜖0 𝑎2 ⎢
⏟⏟⏟ ⏟⏟⏟ ⏟⏟⏟⎥
3
3
2
2
2
2
2
2
⎢⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟ ⏟⏟⏟ ⏟⏟⏟ ⏟⏟⏟ (0,𝑎,𝑎)
⎥
(𝑎,𝑎,0)
(𝑎,0,𝑎)
⎢
⎥
(0,0,0)
(𝑎,0,0)
(0,𝑎,0)
(0,0,𝑎)
⎣
⎦
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)
[
]
)
(
) 1.90 𝑞 2 (
𝑞2
1
1
𝐚
+
𝐚
+
𝐚
𝐚
+
𝐚
+
𝐚
+
+
1
=
=
√
√
𝑥
𝑦
𝑧
𝑥
𝑦
𝑧
4𝜋𝜖0 𝑎2 3 3
4𝜋𝜖0 𝑎2
2
in which the magnitude is |𝐅| = 3.29 𝑞 2 ∕(4𝜋𝜖0 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′ )
4𝜋𝜖0 |𝐫 − 𝐫1′ |3
+
𝑞2 (𝐫 − 𝐫2′ )
4𝜋𝜖0 |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:
[
]
−3(𝐚𝑥 + 𝐚𝑦 + 𝐚𝑧 ) −𝑞2 (𝐚𝑥 + 3𝐚𝑦 + 2𝐚𝑧 )
1
𝐄=
+
4𝜋𝜖0 (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.
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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 ′ )
4𝜋𝜖0 |𝐫 − 𝐫 ′ |3
In the case of two charges, we would therefore have
𝐄𝑇 =
𝑞1 (𝐫 − 𝐫1′ )
4𝜋𝜖0 |𝐫 − 𝐫1′ |3
+
𝑞2 (𝐫 − 𝐫2′ )
(1)
4𝜋𝜖0 |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 ,
𝐫 − 𝐫2′ = [𝑧 + (𝑑∕2)]𝐚𝑧 ,
then
|𝐫 − 𝐫1 |3 = [𝑧 − (𝑑∕2)]3
and
|𝐫 − 𝐫2 |3 = [𝑧 + (𝑑∕2)]3
Substitute these results into (1) to obtain:
[
]
𝑞
1
1
𝐄𝑇 (𝑧) =
+
𝐚
4𝜋𝜖0 [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
[
]
𝑥𝐚𝑥 − (𝑑∕2)𝐚𝑧
𝑥𝐚𝑥 + (𝑑∕2)𝐚𝑧
𝑞
+
𝐄𝑇 (𝑥) =
(3)
4𝜋𝜖0 |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
4𝜋𝜖0 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
𝜌, and use 𝐚𝜌 instead of 𝐚𝑥 . The field then becomes:
𝐄𝑇 (𝜌) =
2𝑞𝜌 𝐚𝜌
[
]3∕2
4𝜋𝜖0 𝜌2 + (𝑑∕2)2
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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=
4𝜋𝜖0 |𝐫 − 𝐫+′ |3 4𝜋𝜖0 |𝐫 − 𝐫−′ |3
where 𝐫 = 𝑧𝐚𝑧 , 𝐫+′ = +𝑑∕2 𝐚𝑧 , and 𝐫−′ = −𝑑∕2 𝐚𝑧 . Using these substitutions, we find:
[
𝑞
𝐄=
4𝜋𝜖0
(𝑧 − 𝑑2 )𝐚𝑧
|𝑧 − 𝑑2 |3
−
(𝑧 + 𝑑2 )𝐚𝑧
]
|𝑧 + 𝑑2 |3
b) Evaluate your part 𝑎 result at the origin: We set 𝑧 = 0 in the above result to obtain
𝐄(𝑧 = 0) =
−2𝑞 𝐚𝑧 [ 2 ]2 −2𝑞 𝐚𝑧
=
4𝜋𝜖0 𝑑
𝜋𝜖0 𝑑 2
as expected
c) Find the electric field intensity everywhere on the 𝑥𝑦 plane, expressing your result as a function
of radius 𝜌 in cylindrical coordinates: This will begin with the same initial setup as in part 𝑎,
except now 𝐫 = 𝜌𝐚𝜌 describes the observation point in the 𝑥𝑦 plane. With this change, we have
]
[
𝜌𝐚𝜌 + (𝑑∕2)𝐚𝑧
𝜌𝐚𝜌 − (𝑑∕2)𝐚𝑧
−𝑞𝑑 𝐚𝑧
𝑞
𝐄=
− 2
=
2
2
3∕2
2
3∕2
2
4𝜋𝜖0 [𝜌 + (𝑑∕2) ]
[𝜌 + (𝑑∕2) ]
4𝜋𝜖0 [𝜌 + (𝑑∕2)2 ]3∕2
d) Evaluate your part 𝑐 result at the origin: Setting 𝜌 = 0 in the part 𝑐 expression, we find:
𝐄(𝜌 = 0) =
−2𝑞 𝐚𝑧
𝜋𝜖0 𝑑 2
as in part 𝑏, − and as expected
e) Simplify your part 𝑐 result for the case in which 𝜌 >> 𝑑: With this requirement, we find
. −𝑞𝑑 𝐚𝑧
𝐄(𝜌 >> 𝑑) =
4𝜋𝜖0 𝜌3
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
4𝜋𝜖0 (𝑑 − 𝑥)2
√
from which 𝑄 = 2(𝑑 − 𝑥) 𝜋𝜖0 𝑘𝑥.
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2.8
b) Determine the maximum charge that can be measured in terms of 𝜖0 , 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 𝜋𝜖0 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 × 10−9 𝐑𝐴𝑃
4𝜋𝜖0
|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
]
[
(𝑥 + 1)
100 × 10−9
= 500 V∕m
𝐸𝑥 =
4𝜋𝜖0
[(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 𝜌: 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
)
′
4𝜋𝜖0 |𝐫 − 𝐫𝑙𝑜𝑤𝑒𝑟
|3
−
′
2𝑞(𝐫 − 𝐫𝑚𝑖𝑑𝑑𝑙𝑒
)
′
4𝜋𝜖0 |𝐫 − 𝐫𝑚𝑖𝑑𝑑𝑙𝑒
|3
+
′
𝑞(𝐫 − 𝐫𝑢𝑝𝑝𝑒𝑟
)
′
4𝜋𝜖0 |𝐫 − 𝐫𝑢𝑝𝑝𝑒𝑟
|3
′
′
′
The position vectors will be 𝐫 = 𝜌𝐚𝜌 , 𝐫𝑙𝑜𝑤𝑒𝑟
= −𝑑𝐚𝑧 , 𝐫𝑢𝑝𝑝𝑒𝑟
= +𝑑𝐚𝑧 , and 𝐫𝑚𝑖𝑑𝑑𝑙𝑒
= 0. With these
substitutions, the field expression becomes:
[
]
[
]
𝜌𝐚𝜌 + 𝑑𝐚𝑧
2𝜌𝐚𝜌
𝜌𝐚𝜌 − 𝑑𝐚𝑧
−𝑞𝐚𝜌
𝑞
1
𝐄=
− 3 + 2
=
1−
4𝜋𝜖0 (𝜌2 + 𝑑 2 )3∕2
𝜌
(𝜌 + 𝑑 2 )3∕2
2𝜋𝜖0 𝜌2
(1 + 𝑑 2 ∕𝜌2 )3∕2
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2.10
b) Specialize your part 𝑎 result for large distances, 𝜌 >> 𝑑: Under this condition, we may use the
expansion:
1
(1 +
𝑑 2 ∕𝜌2 )3∕2
(
)−1 (
)−1∕2 . (
)(
)
= 1 + 𝑑 2 ∕𝜌2
1 + 𝑑 2 ∕𝜌2
= 1 − 𝑑 2 ∕𝜌2 1 − 𝑑 2 ∕2𝜌2
Carrying out the product and neglecting the term involving 𝑑 4 ∕𝜌4 , we find:
(
)(
) .
3 𝑑2
1 − 𝑑 2 ∕𝜌2 1 − 𝑑 2 ∕2𝜌2 = 1 −
2 𝜌2
from which
[
]
−3𝑞𝑑 2 𝐚𝜌
. −𝑞𝐚𝜌
3 𝑑2
1
−
1
+
=
𝐄(𝜌 >> 𝑑) =
2 𝜌2
2𝜋𝜖0 𝜌2
4𝜋𝜖0 𝜌4
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
[
]
𝑄0 −2𝐚𝑥 + 𝐚𝑦 − 𝐚𝑧
𝐄𝑃 =
4𝜋𝜖0
61.5
Since the 𝑧 component is of value 1 kV/m, we find 𝑄0 = −4𝜋𝜖0 61.5 × 103 = −1.63 𝜇C.
b) Find 𝐄 at 𝑀(1, 6, 5) in cartesian coordinates: This field will be:
[
]
−1.63 × 10−6 𝐚𝑥 + 6𝐚𝑦 + 5𝐚𝑧
𝐄𝑀 =
4𝜋𝜖0
[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, 𝜌 =
80.54◦ , and 𝑧 = 5. Now
√
1 + 36 = 6.08, 𝜙 = tan−1 (6∕1) =
𝐸𝜌 = 𝐄𝑀 ⋅ 𝐚𝜌 = −30.11 cos 𝜙 − 180.63 sin 𝜙 = −183.12
𝐸𝜙 = 𝐄𝑀 ⋅ 𝐚𝜙 = −30.11(− sin 𝜙) − 180.63 cos 𝜙 = 0 (as expected)
so that 𝐄𝑀 = −183.12𝐚𝜌 − 150.53𝐚𝑧 .
√
d) Find 𝐄 at 𝑀(1, 6, 5) in spherical coordinates: At 𝑀, 𝑟 = 1 + 36 + 25 = 7.87, 𝜙 = 80.54◦ (as
before), and 𝜃 = 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 𝜃 cos 𝜙 − 180.63 sin 𝜃 sin 𝜙 − 150.53 cos 𝜃 = −237.1
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2.12. Electrons are in random motion in a fixed region in space. During any 1𝜇s 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 × 10−19 C, the density becomes
𝜌𝑣 = −
0.27 × 1.602 × 10−19
= −43.3 𝜇C∕m3
10−15
2.13. A uniform volume charge density of 0.2 𝜇C∕m3 is present throughout the spherical shell extending
from 𝑟 = 3 cm to 𝑟 = 5 cm. If 𝜌𝑣 = 0 elsewhere:
a) find the total charge present throughout the shell: This will be
𝑄=
∫0
2𝜋
𝜋
.05
∫0 ∫.03
].05
[
𝑟3
= 8.21 × 10−5 𝜇C = 82.1 pC
0.2 𝑟 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜙 = 4𝜋(0.2)
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
[
]𝑟
𝑟3 1
= 4.105 × 10−5
4𝜋(0.2)
3 .03
Thus
[
3 × 4.105 × 10−5
𝑟1 =
+ (.03)3
0.2 × 4𝜋
]1∕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 𝜌𝑣 = −0.1∕(𝜌2 + 10−8 ) pC∕m3 for 0 < 𝜌 < 3 × 10−4 m, and 𝜌𝑣 = 0 for
𝜌 > 3 × 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 × 10−4 m, and length 1m.
1
𝑞=
∫0 ∫0
2𝜋
∫0
3×10−4
(𝜌2
−0.1
𝜌 𝑑𝜌 𝑑𝜙 𝑑𝑧
+ 10−8 )
From integral tables, this evaluates as
𝑞 = −0.2𝜋
( ) (
) |3×10−4
1
ln 𝜌2 + 10−8 |
= 0.1𝜋 ln(10) = −0.23𝜋 pC∕m
|0
2
b) if the electron velocity is 5×107 m/s, and with one ampere defined as 1C/s, find the beam current:
Current = charge∕m×𝑣 = −0.23𝜋 [pC∕m]×5×107 [m∕s] = −11.5𝜋×106 [pC∕s] = −11.5𝜋 𝜇A
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2.15. A spherical volume having a 2 𝜇m 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)𝜋(2 × 10−6 )3 × 105 = 3.35 × 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
𝜌𝑣,𝑎𝑣𝑔 =
3.35 × 10−12
= 1.24 × 10−4 C∕m3
(0.003)3
(
)
2.16. Within a region of free space, charge density is given as 𝜌𝑣 = 𝜌0 𝑟∕𝑎 cos 𝜃 C∕m3 , where 𝜌0 and 𝑎 are
constants. Find the total charge lying within:
a) the sphere, 𝑟 ≤ 𝑎: This will be
𝑄𝑎 =
∫0
2𝜋
𝜋
∫0 ∫0
𝑎
𝜌0 𝑟
cos 𝜃 𝑟2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜙 = 0
𝑎
It is the integral over 𝜃, performed first, that gives the zero result.
b) the cone, 𝑟 ≤ 𝑎, 0 ≤ 𝜃 ≤ 0.1𝜋:
𝑄𝑏 =
∫0
2𝜋
∫0
0.1𝜋
∫0
𝑎
]
𝜌0 𝑟
𝜌 𝑎3 [
cos 𝜃 𝑟2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜙 = 𝜋 0
1 − cos2 (0.1𝜋) = 0.024𝜋𝜌0 𝑎3
𝑎
4
c) the region, 𝑟 ≤ 𝑎, 0 ≤ 𝜃 ≤ 0.1𝜋, 0 ≤ 𝜙 ≤ 0.2𝜋.
𝑄𝑐 =
∫0
0.2𝜋
∫0
0.1𝜋
∫0
)
(
𝜌0 𝑟
0.2𝜋
cos 𝜃 𝑟2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜙 = 0.024𝜋𝜌0 𝑎3
= 0.0024𝜋𝜌0 𝑎3
𝑎
2𝜋
𝑎
2.17. A length 𝑑 of line charge lies on the 𝑧 axis in free space. The charge density on the line is
𝜌𝐿 = +𝜌0 C/m (0 < 𝑧 < 𝑑∕2) and 𝜌𝐿 = −𝜌0 C/m (−𝑑∕2 < 𝑧 < 0), where 𝜌0 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 𝜌: Begin by constructing the differential field at radius 𝜌 in the 𝑥𝑦 plane that
arises from a point charge 𝑑𝑞 = 𝜌𝐿 𝑑𝑧 on the 𝑧 axis. To do this, use the general expression:
𝑑𝐄 =
𝑑𝑞(𝐫 − 𝐫 ′ )
4𝜋𝜖0 |𝐫 − 𝐫 ′ |3
where in this case, 𝐫 = 𝜌𝐚𝜌 and 𝐫 ′ = 𝑧𝐚𝑧 . With these substitutions, we find
𝑑𝐄 =
±𝜌0 𝑑𝑧(𝜌𝐚𝜌 − 𝑧𝐚𝑧 )
4𝜋𝜖0 (𝜌2 − 𝑧2 )3∕2
where the positive sign applies to the region 𝑧 > 0; the negative sign to 𝑧 < 0. The total field at
radius 𝜌 is then found by integrating 𝑑𝐄 over the total charge length:
0
𝐄=
−𝜌0 𝑑𝑧(𝜌𝐚𝜌 − 𝑧𝐚𝑧 )
∫−𝑑∕2 4𝜋𝜖0 (𝜌2 − 𝑧2 )3∕2
+
∫0
+𝑑∕2
+𝜌0 𝑑𝑧(𝜌𝐚𝜌 − 𝑧𝐚𝑧 )
4𝜋𝜖0 (𝜌2 − 𝑧2 )3∕2
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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:
[
]
𝑑∕2
−𝜌0 𝑧𝐚𝑧 𝑑𝑧
𝜌0 𝐚𝑧
1
|𝑑∕2 −𝜌0 𝐚𝑧
=
1− √
𝐄=2
=
|
∫0
2𝜋𝜖0 𝜌
4𝜋𝜖0 (𝜌2 + 𝑧2 )3∕2
2𝜋𝜖0 (𝜌2 + 𝑧2 )1∕2 |0
1 + (𝑑∕2𝜌)2
b) Simplify your part 𝑎 result for the case in which radius 𝜌 >> 𝑑, and express this result in terms
of charge 𝑞 = 𝜌0 𝑑∕2: At large radii, we can use the binomial expansion to the first two terms:
√
.
1
= 1−
2
1
1 + (𝑑∕2𝜌)2
with which
.
𝐄 =
−𝜌0 𝑑 2
16𝜋𝜖0
𝜌3
𝐚𝑧 =
(
𝑑
2𝜌
)2
−𝑞𝑑
𝐚𝑧
4𝜋𝜖0 𝜌3
where 𝑞 = 𝜌0 𝑑∕2.
2.18.
a) Find 𝐄 in the plane 𝑧 = 0 that is produced by a uniform line charge, 𝜌𝐿 , extending along the 𝑧
axis over the range −𝐿 < 𝑧 < 𝐿 in a cylindrical coordinate system: We find 𝐄 through
𝐿
𝐄=
∫−𝐿
𝜌𝐿 𝑑𝑧(𝐫 − 𝐫 ′ )
4𝜋𝜖0 |𝐫 − 𝐫 ′ |3
where the observation point position vector is 𝐫 = 𝜌𝐚𝜌 (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 ′ = 𝜌𝐚𝜌 − 𝑧𝐚𝑧 , and |𝐫 − 𝐫 ′ | = (𝜌2 + 𝑧2 )1∕2 . These relations are substituted into the integral
to yield:
𝐿 𝜌 𝑑𝑧(𝜌𝐚 − 𝑧𝐚 )
𝜌𝐿 𝜌 𝐚𝜌 𝐿
𝐿
𝜌
𝑧
𝑑𝑧
𝐄=
=
= 𝐸𝜌 𝐚𝜌
∫−𝐿 4𝜋𝜖0 (𝜌2 + 𝑧2 )3∕2
4𝜋𝜖0 ∫−𝐿 (𝜌2 + 𝑧2 )3∕2
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
𝐸𝜌 =
𝜌𝐿 𝜌
𝜌𝐿
𝜌𝐿
𝐿
1
𝑧
|𝐿
=
| =
√
√
√
|
−𝐿
4𝜋𝜖0 𝜌2 𝜌2 + 𝑧2
2𝜋𝜖0 𝜌 𝜌2 + 𝐿2
2𝜋𝜖0 𝜌 1 + (𝜌∕𝐿)2
Note that as 𝐿 → ∞, the expression reduces to the expected field of the infinite line charge in
free space, 𝜌𝐿 ∕(2𝜋𝜖0 𝜌).
b) if the finite line charge is approximated by an infinite line charge (𝐿 → ∞), by what percentage
is 𝐸𝜌 in error if 𝜌 = 0.5𝐿? The percent error in this situation will be
[
]
% error = 1 − √
1
× 100
1 + (𝜌∕𝐿)2
For 𝜌 = 0.5𝐿, this becomes % error = 10.6 %
c) repeat 𝑏 with 𝜌 = 0.1𝐿. For this value, obtain % error = 0.496 %.
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2.19. A line having charge density 𝜌0 |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 = 𝜌𝐚𝜌 . Then, with 𝐫 ′ = 𝑧𝐚𝑧 , we may write
𝓁∕2
𝐄=
∫−𝓁∕2
𝓁∕2 𝜌 |𝑧|𝑑𝑧(𝜌𝐚 − 𝑧𝐚 )
𝜌𝐿 (𝑧)𝑑𝑧(𝐫 − 𝐫 ′ )
0
𝜌
𝑧
=
′
3
2
2
3∕2
∫
4𝜋𝜖0 |𝐫 − 𝐫 |
−𝓁∕2 4𝜋𝜖0 (𝜌 + 𝑧 )
Note that the second term in the integrand (the 𝑧 component) is zero, because of odd parity.
We are left with
𝐄=
𝜌0 𝜌 𝐚𝜌
𝓁∕2
4𝜋𝜖0 ∫−𝓁∕2
2𝜌0 𝜌 𝐚𝜌 𝓁∕2
−𝜌0 𝜌 𝐚𝜌
|𝑧|𝑑𝑧
1
𝑧𝑑𝑧
|𝓁∕2
=
=
|
4𝜋𝜖0 ∫0
2𝜋𝜖0 (𝜌2 + 𝑧2 )1∕2 |0
(𝜌2 + 𝑧2 )3∕2
(𝜌2 + 𝑧2 )3∕2
Evaluating the limits the final result can be written as
[
]
𝜌0 𝐚𝜌
1
𝐄=
1− √
V∕m
2𝜋𝜖0
1 − (𝓁∕2𝜌)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 → ∞) =
𝜌0 𝐚𝜌
2𝜋𝜖0
V∕m
thus exhibiting no radial variation!
2.20. A line charge of uniform charge density 𝜌0 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
𝓁∕2
𝐄=
𝜌0 𝑑𝑧(𝐫 − 𝐫 ′ )
∫−𝓁∕2 4𝜋𝜖0 |𝐫 − 𝐫 ′ |3
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
𝜌0 𝑑𝑧(𝑥𝐚𝑥 − 𝑧𝐚𝑧 ) 𝜌0 𝑥 𝐚𝑥 𝓁∕2
𝑑𝑧
=
= 𝐸𝑥 𝐚𝑥
𝐄=
2
∫
∫−𝓁∕2 4𝜋𝜖0 (𝑥2 + 𝑧2 )3∕2
4𝜋𝜖0 −𝓁∕2 (𝑥 + 𝑧2 )3∕2
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 =
𝜌0 𝑥
𝜌0
𝜌0
𝓁∕2
1
𝑧
|𝓁∕2
=
=
|
√
√
√
4𝜋𝜖0 𝑥2 𝑥2 + 𝑧2 |−𝓁∕2 2𝜋𝜖0 𝑥 𝑥2 + (𝓁∕2)2
2𝜋𝜖0 𝑥 1 + (2𝑥∕𝓁)2
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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 = (𝜌0 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
𝜌20
1
𝑑𝑥 𝐚𝑥
𝐅=
√
∫𝓁∕2 2𝜋𝜖0 𝑥 1 + (2𝑥∕𝓁)2
This can be re-written and then evaluated using integral tables as
[
]3𝓁∕2
√
⎞
2 + (𝓁∕2)2
𝜌20 𝓁 𝐚𝑥 3𝓁∕2
−𝜌20 𝓁 𝐚𝑥 ⎛ 1
𝓁∕2
+
𝑥
𝑑𝑥
⎜
⎟
𝐅=
=
ln
√
⎟
4𝜋𝜖0 ∫𝓁∕2 𝑥 𝑥2 + (𝓁∕2)2
4𝜋𝜖0 ⎜ (𝓁∕2)
𝑥
𝓁∕2 ⎠
⎝
(
√ )⎤
[
⎡
√ ]
0.55𝜌20
−𝜌20 𝐚𝑥 ⎢ (𝓁∕2) 1 + 10 ⎥ 𝜌20 𝐚𝑥
3(1 + 2)
ln ⎢
ln
𝐚 N
=
=
(
√
√ )⎥ =
2𝜋𝜖0
2𝜋𝜖0 𝑥
1 + 10
⎢ 3(𝓁∕2) 1 + 2 ⎥ 2𝜋𝜖0
⎣
⎦
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 +𝜌0 C/m for −𝜋∕2 < 𝜙 < 𝜋∕2, and −𝜌0 C/m for 𝜋∕2 < 𝜙 < 3𝜋∕2.
Find the electric field intensity at the origin:
The field at the origin arising from a differential length 𝑎𝑑𝜙 of charge on the ring will be
𝑑𝐄𝑜 =
±𝜌0 𝑎𝑑𝜙
4𝜋𝜖0 𝑎2
𝐚𝜌
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 𝐚𝜌 = cos 𝜙 𝐚𝑥 + sin 𝜙 𝐚𝑦 (as is required to include all 𝜙 dependence), we have
𝜋∕2
𝐄𝑜 =
∫−𝜋∕2
3𝜋∕2
+𝜌0 𝑑𝜙
−𝜌0 𝑑𝜙
(cos 𝜙 𝐚𝑥 + sin 𝜙 𝐚𝑦 ) +
(cos 𝜙 𝐚𝑥 + sin 𝜙 𝐚𝑦 )
∫𝜋∕2 4𝜋𝜖0 𝑎
4𝜋𝜖0 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:
⎡
⎤
⎢
⎥
−𝜌0 𝐚𝑥
−2𝜌0 ⎢
|𝜋∕2
|𝜋∕2
𝐄𝑜 =
sin 𝜙|
V∕m
𝐚𝑥 − cos 𝜙|
𝐚𝑦 ⎥ =
|−𝜋∕2
|−𝜋∕2 ⎥
4𝜋𝜖0 𝑎 ⎢
𝜋𝜖0 𝑎
⏟⏞⏞⏟⏞⏞⏟ ⎥
⎢⏟⏞⏞⏟⏞⏞⏟
⎣
⎦
2
0
2.22. Two identical uniform sheet charges with 𝜌𝑠 = 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 = −𝜌𝑠 ∕(2𝜖0 ) 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 = 𝜌𝑠 𝑑𝑎𝐄. The force per unit area is
then just 𝐅 = 𝜌𝑠 𝐄 = (100 × 10−9 )(−100 × 10−9 )∕(2𝜖0 ) 𝐚𝑧 = −5.6 × 10−4 𝐚𝑧 N∕m2 .
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2.23. A disk of radius 𝑎 in the 𝑥𝑦 plane carries surface charge of density 𝜌𝑠 = 𝜌0 ∕𝜌, where 𝜌0 is a constant.
Find the electric field strength, 𝐄, everywhere on the 𝑧 axis.
We find the field through
𝐄=
𝜌𝑠 𝑑𝑎(𝐫 − 𝐫 ′ )
∫ ∫ 4𝜋𝜖0 |𝐫 − 𝐫 ′ |3
where the integral is taken over the surface of the disk, and where 𝐫 = 𝑧𝐚𝑧 and 𝐫 ′ = 𝜌𝐚𝜌 . The
integral then becomes
(
)
𝑎 (𝜌 ∕𝜌) 𝜌 𝑑𝜌 𝑑𝜙 𝑧𝐚 − 𝜌𝐚
2𝜋
0
𝑧
𝜌
𝐄=
∫0 ∫0
4𝜋𝜖0 (𝑧2 + 𝜌2 )3∕2
In evaluating this integral, we need to introduce the 𝜙 dependence in 𝐚𝜌 by writing it as 𝐚𝜌 =
cos 𝜙 𝐚𝑥 + sin 𝜙 𝐚𝑦 , where 𝐚𝑥 and 𝐚𝑦 are invariant in their orientation as 𝜙 varies. So the integral
now simplifies to
[
]𝑎
2𝜋𝜌0 𝑧 𝐚𝑧 ∞
2𝜋𝜌0 𝑧 𝐚𝑧
2𝜋𝑎𝜌0 𝐚𝑧
𝜌
𝑑𝜌
𝐄=
=
=
√
4𝜋𝜖0 ∫0 (𝑧2 + 𝜌2 )3∕2
4𝜋𝜖0
4𝜋𝜖0 𝑧2 [1 + (𝑎∕𝑧)2 ]1∕2
𝑧2 𝑧2 + 𝜌2
𝜌=0
2.24.
a) Find the electric field on the 𝑧 axis produced by an annular ring of uniform surface charge density
𝜌𝑠 in free space. The ring occupies the region 𝑧 = 0, 𝑎 ≤ 𝜌 ≤ 𝑏, 0 ≤ 𝜙 ≤ 2𝜋 in cylindrical
coordinates: We find the field through
𝐄=
𝜌𝑠 𝑑𝑎(𝐫 − 𝐫 ′ )
∫ ∫ 4𝜋𝜖0 |𝐫 − 𝐫 ′ |3
where the integral is taken over the surface of the annular ring, and where 𝐫 = 𝑧𝐚𝑧 and 𝐫 ′ = 𝜌𝐚𝜌 .
The integral then becomes
(
)
𝑏 𝜌 𝜌 𝑑𝜌 𝑑𝜙 𝑧𝐚 − 𝜌𝐚
2𝜋
𝑠
𝑧
𝜌
𝐄=
∫0 ∫𝑎
4𝜋𝜖0 (𝑧2 + 𝜌2 )3∕2
In evaluating this integral, we first note that the term involving 𝜌𝐚𝜌 integrates to zero over the
𝜙 integration range of 0 to 2𝜋. This is because we need to introduce the 𝜙 dependence in 𝐚𝜌
by writing it as 𝐚𝜌 = cos 𝜙 𝐚𝑥 + sin 𝜙 𝐚𝑦 , where 𝐚𝑥 and 𝐚𝑦 are invariant in their orientation as 𝜙
varies. So the integral now simplifies to
]𝑏
[
𝜌𝑠 𝑧 𝐚𝑧
2𝜋𝜌𝑠 𝑧 𝐚𝑧 𝑏
𝜌 𝑑𝜌
−1
=
𝐄=
√
4𝜋𝜖0 ∫𝑎 (𝑧2 + 𝜌2 )3∕2
2𝜖0
𝑧2 + 𝜌2 𝑎
]
[
𝜌
1
1
= 𝑠 √
−√
𝐚𝑧
2𝜖0
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 → 0 and 𝑏 → ∞, in which case 𝐄 → 𝜌𝑠 ∕(2𝜖0 ) 𝐚𝑧
as expected.
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2.25. A disk of radius 𝑎 in the 𝑥𝑦 plane carries surface charge of density 𝜌𝑠1 = +𝜌𝑠0 ∕𝜌 C∕m2 for 0 < 𝜙 < 𝜋,
and 𝜌𝑠2 = −𝜌𝑠0 ∕𝜌 C∕m2 for 𝜋 < 𝜙 < 2𝜋, where 𝜌𝑠0 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=
𝜌𝑠 𝑑𝑎(𝐫 − 𝐫 ′ )
∫ ∫𝑑𝑖𝑠𝑘 𝑎𝑟𝑒𝑎 4𝜋𝜖0 |𝐫 − 𝐫 ′ |3
where 𝐫 = 𝑧𝐚𝑧 and 𝐫 ′ = 𝜌𝐚𝜌 . Because the charge is
𝜙-dependent, we need to include the 𝜙 dependence
in all terms. This means that we must use 𝐚𝜌 =
𝐚𝑥 cos 𝜙+𝐚𝑦 sin 𝜙. Then, with |𝐫−𝐫 ′ |3 = (𝑧2 +𝜌2 )3∕2
and with the positive and negative charges accounted
for, the integral is performed piecewise over the two
halves of the disk:
𝜋
𝐄=
+
∫0 ∫0
∫𝜋
2𝜋
𝑎
∫0
𝑎
r-r
r
ρs2
ρs1
r
da
y
x
+(𝜌𝑠0 ∕𝜌)[𝑧𝐚𝑧 − 𝜌 cos 𝜙𝐚𝑥 − 𝜌 sin 𝜙𝐚𝑦 ] 𝜌𝑑𝜌𝑑𝜙
4𝜋𝜖0 (𝑧2 + 𝜌2 )3∕2
−(𝜌𝑠0 ∕𝜌)[𝑧𝐚𝑧 − 𝜌 cos 𝜙𝐚𝑥 − 𝜌 sin 𝜙𝐚𝑦 ] 𝜌𝑑𝜌𝑑𝜙
4𝜋𝜖0 (𝑧2 + 𝜌2 )3∕2
Noting that the 𝑧 components will cancel, and performing the 𝜙 integration first, we find:
[
]
𝑎 − sin 𝜙|𝜋 𝐚 + cos 𝜙|𝜋 𝐚 + sin 𝜙|2𝜋 𝐚 − cos 𝜙|2𝜋 𝐚 𝜌𝑑𝜌
𝜌𝑠0
𝜋 𝑥
𝜋 𝑦
0 𝑥
0 𝑦
𝐄=
4𝜋𝜖0 ∫0
(𝑧2 + 𝜌2 )3∕2
]
[
]𝑎
[
−𝜌
𝐚
−𝜌𝑠0 𝐚𝑦
−4𝜌𝑠0 𝐚𝑦 𝑎
𝜌𝑑𝜌
𝑠0
𝑦
−1
1
=
=
=
1− √
4𝜋𝜖0 ∫0 (𝑧2 + 𝜌2 )3∕2
𝜋𝜖0
𝜋𝜖0 𝑧
(𝑧2 + 𝜌2 )3∕2 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
[
]
−𝜌𝑠0 𝑎2 𝐚𝑦
. −𝜌𝑠0 𝐚𝑦
𝑎2
1−1+ 2 =
𝐄(𝑧 >> 𝑎) =
𝜋𝜖0 𝑧
2𝑧
2𝜋𝜖0 𝑧3
Note the inverse distance cubed dependence that is characteristic of a dipole field.
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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
𝜌𝑠 (𝑟) = 𝜌0 ∕𝑟 C∕m2 in free space. The cone has its vertex at the origin and occupies the region
𝜃 = 𝛼, 0 < 𝑟 < 𝑎, and 0 < 𝜙 < 2𝜋 in spherical coordinates. Differential area in spherical
coordinates is given as 𝑑𝑎 = 𝑟 sin 𝛼𝑑𝑟𝑑𝜙: 𝐄 is found using the general surface integral
𝐄=
𝜌𝑠 𝑑𝑎(𝐫 − 𝐫 ′ )
∫ ∫𝑐𝑜𝑛𝑒 4𝜋𝜖0 |𝐫 − 𝐫 ′ |3
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 𝛼 + 𝑟2 ,
where 𝐚𝑧 ⋅ 𝐚𝑟 = cos 𝛼 has been used. The integral now becomes:
𝐄=
∫0
2𝜋
∫0
𝑎
(𝜌0 ∕𝑟)𝑟 sin 𝛼𝑑𝑟𝑑𝜙(𝑧𝐚𝑧 − 𝑟𝐚𝑟 )
4𝜋𝜖0 (𝑧2 − 2𝑟𝑧 cos 𝛼 + 𝑟2 )3∕2
It is necessary to include the 𝜙 dependence in 𝐚𝑟 .
We substitute 𝐚𝑟 = sin 𝛼 cos 𝜙𝐚𝑥 + sin 𝛼 sin 𝜙𝐚𝑦 + cos 𝛼𝐚𝑧 . The first two terms of this, involving
sin 𝜙 and cos 𝜙, will integrate to zero when taking 𝜙 from 0 to 2𝜋. Only the 𝑧 component survives,
and the integral becomes, after performing the 𝜙 integration:
𝑎
𝜌0 sin 𝛼(𝑧 − 𝑟 cos 𝛼)𝑑𝑟 𝐚𝑧
∫0 4𝜋𝜖0 (𝑧2 − 2𝑟𝑧 cos 𝛼 + 𝑟2 )3∕2
[
]
2𝜋𝜌0 sin 𝛼𝐚𝑧 𝑎
𝑧𝑑𝑟
𝑟 cos 𝛼𝑑𝑟
=
−
∫0 (𝑧2 − 2𝑟𝑧 cos 𝛼 + 𝑟2 )3∕2 (𝑧2 − 2𝑟𝑧 cos 𝛼 + 𝑟2 )3∕2
4𝜋𝜖0
𝐄 = 2𝜋
The two integrals evaluate as
∫0
𝑎
∫0
𝑎
(𝑧2
(𝑟 − 𝑧 cos 𝛼)𝑧
𝑧𝑑𝑟
|𝑎
=
|
2
2
3∕2
− 2𝑟𝑧 cos 𝛼 + 𝑟 )
𝑧2 sin 𝛼(𝑧2 − 2𝑟𝑧 cos 𝛼 + 𝑟2 )1∕2 |0
−(𝑧 − 𝑟 cos 𝛼)𝑧 cos 𝛼
𝑟 cos 𝛼𝑑𝑟
|𝑎
=
|
(𝑧2 − 2𝑟𝑧 cos 𝛼 + 𝑟2 )3∕2
𝑧2 sin2 𝛼(𝑧2 − 2𝑟𝑧 cos 𝛼 + 𝑟2 )1∕2 |0
Combining these, cancelling terms, and rearranging, finally leads to
𝐄=
2𝜋𝑎𝜌0 sin 𝛼 𝐚𝑧
4𝜋𝜖0 𝑧(𝑧2 − 2𝑎𝑧 cos 𝛼 + 𝑎2 )1∕2
b) Find the total charge on the cone: This will be
𝑄𝑐 =
∫ ∫𝑐𝑜𝑛𝑒
𝜌𝑠 𝑑𝑎 =
∫0
2𝜋
∫0
𝑎
𝜌0
𝑟 sin 𝛼 𝑑𝑟𝑑𝜙 = 2𝜋𝑎𝜌0 sin 𝛼
𝑟
c) Specialize your result of part 𝑎 to the case in which 𝛼 = 90◦ , at which the cone flattens to a disk
in the 𝑥𝑦 plane. Compare this result to the answer to Problem 2.23.
When 𝛼 = 90◦ , cos 𝛼 = 0, and the field expression in part 𝑎 can be expressed as:
𝐄(𝛼 = 0) =
2𝜋𝑎𝜌0 𝐚𝑧
4𝜋𝜖0
𝑧2 [1
+ (𝑎∕𝑧)2 ]1∕2
which is the same result as found in Problem 2.23.
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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
4𝜋𝜖0 𝑧2 [1 − (2𝑎∕𝑧) cos 𝛼 + (𝑎∕𝑧)2 ]1∕2
. 𝑄𝑐 𝐚𝑧
=
4𝜋𝜖0 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 𝛼 + (𝑎∕𝑧)2 ]1∕2 = 𝑎∕𝑧 and thus
𝐄=
𝑄𝑐 𝐚𝑧
4𝜋𝜖0
𝑧2 [1
− (2𝑎∕𝑧) cos 𝛼 +
(𝑎∕𝑧)2 ]1∕2
. 𝑄 𝑎𝐚
= 𝑐 𝑧
4𝜋𝜖0 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, 𝜃) = 𝑄𝑑∕(4𝜋𝜖0 𝑟3 ) 2 cos 𝜃𝐚𝑟 + sin 𝜃𝐚𝜃 , 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, 𝜃 = 0, 𝐚𝑟 = 𝐚𝑧 , and 𝑟 = 𝑧. Therefore
𝐅(0, 0, 𝑧) =
𝑞𝑄𝑑 𝐚𝑧
4𝜋𝜖0 𝑧3
N
b) at (0, 𝑦, 0): Here, 𝜃 = 90◦ , 𝐚𝜃 = −𝐚𝑧 , and 𝑟 = 𝑦. The force is
𝐅(0, 𝑦, 0) =
−𝑞𝑄𝑑 𝐚𝑧
4𝜋𝜖0 𝑦3
N
(
)
2.29. If 𝐄 = 20𝑒−5𝑦 cos 5𝑥𝐚𝑥 − sin 5𝑥𝐚𝑦 , find:
a) |𝐄| at 𝑃 (𝜋∕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 ⇒ 𝑑𝑦 = − tan 5𝑥 𝑑𝑥
𝑑𝑥
cos 5𝑥
Thus 𝑦 =
1
5
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 𝐸𝜌 ∕𝐸𝜙 = 𝑑𝜌(𝜌𝑑𝜙). Find the equation of the line passing
through the point (2, 30◦ , 0) for the field 𝐄 = 𝜌 cos 2𝜙 𝐚𝜌 − 𝜌 sin 2𝜙 𝐚𝜙 :
𝐸𝜌
𝐸𝜙
=
𝑑𝜌
−𝜌 cos 2𝜙
𝑑𝜌
=
= − cot 2𝜙 ⇒
= − cot 2𝜙 𝑑𝜙
𝜌𝑑𝜙
𝜌 sin 2𝜙
𝜌
Integrate to obtain
[
𝐶
2 ln 𝜌 = ln sin 2𝜙 + ln 𝐶 = ln
sin 2𝜙
]
⇒ 𝜌2 =
𝐶
sin 2𝜙
√
◦ ) ⇒ 𝐶 = 4 sin 60◦ = 2 3. Finally, the equation for
At the given point, we have
4
=
𝐶∕
sin(60
√
the streamline is 𝜌2 = 2 3∕ sin 2𝜙.
29
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Engineering Electromagnetics 9th solution
Engineering Electromagnetics (Chungnam National University)
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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
4𝜋𝜖0 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 ∕𝜖0 ) 𝐚̂ 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 = 𝜖0 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 =
∮
2
𝐃 ⋅ 𝐧 𝑑𝑎 =
2
2
∫−2 ∫−2
5(2)2 𝐚𝑧 ⋅ 𝐚𝑧 𝑑𝑥 𝑑𝑦 +
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.
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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
4𝜋𝜖0
[𝜌2
+ (𝑑∕2)2 ]3∕2
where 𝜌 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
Ψ𝑒 =
∫ ∫𝑥𝑦
𝐃 ⋅ 𝑑𝐒 where 𝐃 = 𝜖0 𝐄 and 𝑑𝐒 = −𝐚𝑧 𝜌𝑑𝜌𝑑𝜙
Over the entire 𝑥𝑦 plane, the flus is then:
∞
2𝜋
−𝑞𝑑𝐚𝑧 ⋅ (−𝐚𝑧 )
∫0 ∫0 4𝜋[𝜌2 + (𝑑∕2)2 ]3∕2
𝑞𝑑
−1
|∞
=
| =𝑞
2 [𝜌2 + (𝑑∕2)2 ]1∕2 |0
Ψ𝑒 =
𝜌𝑑𝜌𝑑𝜙 =
𝜌𝑑𝜌
2𝜋𝑞𝑑 ∞
4𝜋 ∫0 [𝜌2 + (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 ∕𝜖0 ) 𝐚̂ 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=
∮
𝜖0 𝐄 ⋅ 𝐧 𝑑𝑎 =
∫0
2𝜋
∫0
𝜋
5𝑧3 𝐚𝑧 ⋅ 𝐚𝑟 (3)2 sin 𝜃 𝑑𝜃 𝑑𝜙
where in this case 𝑧 = 3 cos 𝜃 and (in all cases) 𝐚𝑧 ⋅ 𝐚𝑟 = cos 𝜃. These are substituted to yield
𝑄 = 2𝜋
∫0
𝜋
5(3)5 cos4 𝜃 sin 𝜃𝑑𝜃 = −2𝜋(5)(3)5
( )
1
|𝜋
cos5 𝜃 | = 972𝜋
|0
5
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3.5. A volume charge distribution in free space is characterized by the density:
𝑞
exp(−|𝑧|∕𝑑)
2𝐴𝑑
𝜌𝑣 =
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:
∮
Therefore
|𝐷𝑧 | =
𝐃 ⋅ 𝑑𝐒 = 2𝐴1 |𝐷𝑧 | = 𝑄𝑒𝑛𝑐𝑙 = 2𝐴1
∫0
𝑧0
𝑞
exp(−𝑧∕𝑑) 𝑑𝑧
2𝐴𝑑
]
]
𝑞 [
𝑞 [
1 − exp(−𝑧0 ∕𝑑) ⇒ 𝐄(𝑧) = ±
1 − exp(−|𝑧|∕𝑑) 𝐚𝑧 V∕m
2𝐴
2𝜖0 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 𝜌𝑣 = 𝜌0 exists within the region −∞ < 𝑥 < ∞,
−∞ < 𝑦 < ∞, 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:
Φ𝑖𝑛 =
∮
1
𝐃 ⋅ 𝑑𝐒 = 2
1
∫−1 ∫−1
𝑧
𝐷𝑧 𝑑𝑥𝑑𝑦 = 𝑄𝑒𝑛𝑐𝑙 =
1
1
∫−𝑧 ∫−1 ∫−1
𝜌0 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 𝜌0 are constants, leading to
𝐃𝑖𝑛 = 𝜌0 𝑧 𝐚𝑧 C∕m2 (|𝑧| < 𝑑∕2), and therefore 𝐄𝑖𝑛 = (𝜌0 𝑧∕𝜖0 ) 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:
Φ𝑜𝑢𝑡 =
∮
1
𝐃 ⋅ 𝑑𝐒 = 2
1
∫−1 ∫−1
𝑑∕2
𝐷𝑧 𝑑𝑥𝑑𝑦 = 𝑄𝑒𝑛𝑐𝑙 =
1
1
∫−𝑑∕2 ∫−1 ∫−1
𝜌0 𝑑𝑥𝑑𝑦𝑑𝑧′
Solve for 𝐷𝑧 to find the constant values:
{
{
(𝜌0 𝑑∕2𝜖0 ) 𝐚𝑧 (𝑧 > 𝑑∕2)
(𝜌0 𝑑∕2) 𝐚𝑧 (𝑧 > 𝑑∕2)
2
V∕m
C∕m and 𝐄𝑜𝑢𝑡 =
𝐃𝑜𝑢𝑡 =
−(𝜌0 𝑑∕2𝜖0 ) 𝐚𝑧 (𝑧 < 𝑑∕2)
−(𝜌0 𝑑∕2) 𝐚𝑧 (𝑧 < 𝑑∕2)
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3.7. A spherically symmetric charge distribution in free space is characterized by the charge density
𝜌𝑣 =
𝑞𝑏
exp(−𝑏𝑟) C∕m3
2
𝑟
(0 < 𝑟 < ∞)
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:
∮𝑟
𝐃 ⋅ 𝑑𝐒 = 4𝜋𝑟2 𝐷𝑟 =
= 4𝜋𝑞𝑏
∫0
∫𝑒𝑛𝑐𝑙. 𝑣𝑜𝑙
𝜌𝑣 𝑑𝑣 =
∫0
2𝜋
𝜋
𝑟
𝑞𝑏
exp[−𝑏(𝑟′ )2 ] (𝑟′ )2 sin 𝜃𝑑𝑟𝑑𝜃𝑑𝜙
∫0 ∫0 (𝑟′ )2
𝑟
exp[−𝑏(𝑟′ )2 ] = 4𝜋𝑞[1 − exp(−𝑏𝑟)]
Therefore
]
]
[
1 − exp(−𝑏𝑟)
𝑞 1 − exp(−𝑏𝑟)
2
𝐚𝑟 C∕m ⇒ 𝐄(𝑟) =
𝐚𝑟 V∕m
𝐃(𝑟) = 𝑞
𝜖0
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) = 4𝜋𝑞[1 − exp(−𝑏𝑟)]. The total charge
present will be this result evaluated as 𝑟 → ∞, or 𝑄𝑛𝑒𝑡 = 4𝜋𝑞.
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 4𝜋𝐴 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
Φ=
∮
𝐃 ⋅ 𝑑𝐒 =
∫0
2𝜋
∫0
𝜋
𝐴
𝐚 ⋅ 𝐚 𝑟2 sin 𝜃 𝑑𝜃 𝑑𝜙 = 4𝜋𝐴𝑟 = 𝑄𝑒𝑛𝑐𝑙
𝑟 𝑟 𝑟
We see from this that with every increase in 𝑟 by one m, the enclosed charge increases by 4𝜋𝐴
(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 = 4𝜋𝐴𝑟 =
∫0
2𝜋
𝜋
∫0 ∫0
𝑟
𝜌(𝑟′ ) (𝑟′ )2 sin 𝜃 𝑑𝑟′ 𝑑𝜃 𝑑𝜙 = 4𝜋
∫0
𝑟
𝜌(𝑟′ ) (𝑟′ )2 𝑑𝑟′
To obtain the correct enclosed charge, the integrand must be 𝜌(𝑟) = 𝐴∕𝑟2 .
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3.9. A sphere of radius 𝑎 in free space contains charge of density 𝜌𝑣 = 𝜌0 𝑟∕𝑎, where 𝜌0 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
∮𝑠
𝐃𝐼 ⋅ 𝑑𝐒 = 4𝜋𝑟2 𝐷𝐼 = 𝑄𝑒𝑛𝑐𝑙 =
So
𝐷𝐼 =
∫𝑣
𝜌𝑣 𝑑𝑣 = 4𝜋
∫0
𝑟
𝜌0 𝑟′ ′ 2
𝑟4
(𝑟 ) 𝑑𝑟 = 4𝜋𝜌0
𝑎
4𝑎
𝜌 𝑟2
𝜌0 𝑟2
C∕m2 ⇒ 𝐄𝐼 = 0 𝐚𝑟 V∕m (𝑟 < 𝑎)
4𝑎
4𝜖0 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:
∮𝑠
𝐃𝐼𝐼 ⋅ 𝑑𝐒 = 4𝜋𝑟2 𝐷𝐼𝐼 = 𝑄𝑡𝑜𝑡 =
∫𝑣
𝜌𝑣 𝑑𝑣 = 4𝜋
∫0
𝑎
𝜌0 𝑟′ ′ 2
𝑎3
(𝑟 ) 𝑑𝑟 = 4𝜋𝜌0
𝑎
4
which gives
𝐷𝐼𝐼 =
𝜌0 𝑎3
4𝑟2
C∕m2 ⇒ 𝐄𝐼𝐼 =
𝜌0 𝑎3
4𝜖0 𝑟2
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, 𝜌𝑠 , 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:
4𝜋𝑏2 𝜌𝑠 = −𝑄𝑡𝑜𝑡 = −𝜋𝜌0 𝑎3 ⇒ 𝜌𝑠 =
−𝜌0 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:
−𝜌20 𝑎6
−𝜌0 𝑎3 𝜌0 𝑎3
𝐅
|
𝐚𝑟 =
𝐚𝑟 N∕m2
= 𝜌𝑠 𝐄𝐼𝐼 | =
|𝑟=𝑏
𝐴
4𝑏2 4𝜖0 𝑏2
16𝜖0 𝑏4
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3.10. An infinitely long cylindrical dielectric of radius 𝑏 contains charge within its volume of density 𝜌𝑣 =
𝑎𝜌2 , 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 𝜌, concentric with the charge
distribution, having unit length in 𝑧, and where 𝜌 < 𝑏. The outward normal to the surface is 𝐚𝜌 .
∮
1
𝐃 ⋅ 𝐧 𝑑𝑎 =
∫0 ∫0
2𝜋
1
𝐷𝜌 𝐚𝜌 ⋅ 𝐚𝜌 𝜌 𝑑𝜙 𝑑𝑧 = 𝑄𝑒𝑛𝑐𝑙 =
∫0 ∫0
2𝜋
∫0
𝜌
𝑎(𝜌′ )2 𝜌′ 𝑑𝜌′ 𝑑𝜙 𝑑𝑧
in which the dummy variable 𝜌′ must be used in the far-right integral because the upper radial
limit is 𝜌. 𝐷𝜌 is constant over the surface and can be factored outside the integral. Evaluating
both integrals leads to
2𝜋(1)𝜌𝐷𝜌 = 2𝜋𝑎
( )
𝑎𝜌3
𝑎𝜌3
1 4
𝜌 ⇒ 𝐷𝜌 =
𝐚 (𝜌 < 𝑏)
or 𝐄𝑖𝑛 =
4
4
4𝜖0 𝜌
To find the field outside the cylinder, we apply Gauss’ law to a cylinder of radius 𝜌 > 𝑏. The
setup now changes only by the upper radius limit for the charge integral, which is now the charge
radius, 𝑏:
∮
1
𝐃 ⋅ 𝐧 𝑑𝑎 =
∫0 ∫0
2𝜋
1
𝐷𝜌 𝐚𝜌 ⋅ 𝐚𝜌 𝜌 𝑑𝜙 𝑑𝑧 = 𝑄𝑒𝑛𝑐𝑙 =
∫0 ∫0
2𝜋
∫0
𝑏
𝑎𝜌2 𝜌 𝑑𝜌 𝑑𝜙 𝑑𝑧
where the dummy variable is no longer needed. Evaluating as before, the result is
𝐷𝜌 =
𝑎𝑏4
𝑎𝑏4
or 𝐄𝑜𝑢𝑡 =
𝐚 (𝜌 > 𝑏)
4𝜌
4𝜖0 𝜌 𝜌
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, 𝜌𝑣 (𝜌) = 𝜌0 exp[−(𝜌∕𝑏)2 ] C∕m3 , where 𝑏 is a constant.
a) Find the electric field intensity 𝐄 at large radii, 𝜌 >> 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 𝜌 >> 𝑏 and of unit length in 𝑧, that is concentric with the charge.
Also assume that 𝐃 = 𝐷𝜌 𝐚𝜌 is uniform over the surface. Gauss’ Law is thus written:
∮
1
𝐃 ⋅ 𝑑𝐒 =
∫0 ∫0
2𝜋
.
𝐷𝜌 𝜌𝑑𝜙𝑑𝑧 = 2𝜋(1)𝜌𝐷𝜌 = 𝑄𝑒𝑛𝑐𝑙 =
which becomes
.
2𝜋𝜌𝐷𝜌 = 2𝜋𝜌0
∫0
∞
𝜌𝑒−𝜌
2 ∕𝑏2
1
∫0 ∫0
𝑑𝜌 = 2𝜋𝜌0
2𝜋
∫0
∞
2 ∕𝑏2
𝜌0 𝑒−𝜌
𝜌𝑑𝜌𝑑𝜙𝑑𝑧
𝑏2
2
Therefore
. 𝜌 𝑏2
. 𝜌 𝑏2
𝐷𝜌 = 0 and thus 𝐄 = 0 𝐚𝜌 V∕m (𝜌 >> 𝑏)
2𝜌
2𝜖0 𝜌
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3.11
b) Compare your result to the field outside a charged cylinder of radius 𝑏 containing uniform charge
density 𝜌0 : The setup is the same as in part 𝑎, except the charge integral is taken only to radius
𝑏, over the constant charge density:
1
2𝜋𝜌𝐷𝜌 = 𝑄𝑒𝑛𝑐𝑙 =
∫0 ∫0
2𝜋
∫0
𝑏
𝜌0 𝜌𝑑𝜌𝑑𝜙𝑑𝑧 = 𝜋𝜌0 𝑏2 ⇒ 𝐷𝜌 =
𝜌0 𝑏2
2𝜌
and thus 𝐄 is the same as in part 𝑎!
3.12. The sun radiates a total power of about 3.86 × 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 𝜃 = 40◦ and 𝜃 = 30◦ . 12◦ and 27◦
correspond directly to the limits on 𝜙. Since the sun for our purposes is spherically-symmetric,
the flux density emitted by it is 𝐈 = 3.86 × 1026 ∕(4𝜋𝑟2 ) 𝐚𝑟 W∕m2 . The required power is now
found through
27◦
𝑃1 =
=
40◦
∫12◦ ∫30◦
3.86 × 1026
4𝜋
3.86 × 1026
𝐚𝑟 ⋅ 𝐚𝑟 𝑟2 sin 𝜃 𝑑𝜃 𝑑𝜙
4𝜋𝑟2
)
(
[
]
2𝜋
= 8.1 × 1023 W
cos(30◦ ) − cos(40◦ ) (27◦ − 12◦ )
360
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 × 1011 m. Use this distance in the flux density
expression above to obtain
𝐈=
3.86 × 1026
𝐚𝑟 = 1200 𝐚𝑟 W∕m2
4𝜋(1.55 × 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 𝜌𝑠0 , 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 ∶ 4𝜋𝑟2 𝐷𝑟 = 4𝜋(2)2 (20 × 10−9 ) ⇒ 𝐷𝑟 =
80 × 10−9
C∕m2
2
𝑟
So 𝐷𝑟 (𝑟 = 3) = 8.9 × 10−9 C∕m2 .
4 < 𝑟 < 6 ∶ 4𝜋𝑟2 𝐷𝑟 = 4𝜋(2)2 (20 × 10−9 ) + 4𝜋(4)2 (−4 × 10−9 ) ⇒ 𝐷𝑟 =
16 × 10−9
𝑟2
So 𝐷𝑟 (𝑟 = 5) = 6.4 × 10−10 C∕m2 .
b) Determine 𝜌𝑠0 such that 𝐃 = 0 at 𝑟 = 7 m. Since fields will decrease as 1∕𝑟2 , the question could
be re-phrased to ask for 𝜌𝑠0 such that 𝐃 = 0 at all points where 𝑟 > 6 m. In this region, the total
field will be
2
16 × 10−9 𝜌𝑠0 (6)
+
𝐷𝑟 (𝑟 > 6) =
𝑟2
𝑟2
−9
Requiring this to be zero, we find 𝜌𝑠0 = −(4∕9) × 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 𝜃
𝐚𝑟
Watts∕m2
𝐏𝑑 = 𝑃0
2𝜋𝑟2
where 𝜃 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 =
∫0
2𝜋
∫0
𝜋∕2
𝑃0
𝑃
cos2 𝜃
|𝜋∕2 𝑃0
=
𝐚𝑟 ⋅ 𝐚𝑟 𝑟2 sin 𝜃 𝑑𝜃 𝑑𝜙 = − 0 cos3 𝜃 |
2
|0
3
3
2𝜋𝑟
b) Find the cone angle, 𝜃1 , within which half the total power is radiated; i.e., within the range
0 < 𝜃 < 𝜃1 : We perform the same integral as in part 𝑎 except the upper limit for 𝜃 is now 𝜃1 .
The result must be one-half that of part 𝑎, so we write:
(
)
)
𝑃𝑡
𝑃
𝑃
1
|𝜃1 𝑃 (
= 0 = − 0 cos3 𝜃 | = 0 1 − cos3 𝜃1 ⇒ 𝜃1 = cos−1 1∕3 = 37.5◦
|0
2
6
3
3
2
c) An optical detector, having a 1 mm2 cross-sectional area, is positioned at 𝑟 = 1 m and at 𝜃 = 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
2𝜋𝑟2
4𝜋𝑟2
Then the detected power in a 1-mm2 area at 𝑟 = 1 m approximates as
.
. 𝑃
𝑃 [𝑊 ] = 0 × 10−6 = 10−9 ⇒ 𝑃0 = 4𝜋 × 10−3 W
4𝜋
If the originally stated 1mW value is used for the detected power, the answer would have been
4𝜋 kW (!).
3.15. Volume charge density is located as follows: 𝜌𝑣 = 0 for 𝜌 < 1 mm and for 𝜌 > 2 mm, 𝜌𝑣 = 4𝜌 𝜇C∕m3
for 1 < 𝜌 < 2 mm.
a) Calculate the total charge in the region 0 < 𝜌 < 𝜌1 , 0 < 𝑧 < 𝐿, where 1 < 𝜌1 < 2 mm: We find,
𝐿
𝑄=
∫0 ∫0
2𝜋
𝜌1
∫.001
4𝜌 𝜌 𝑑𝜌 𝑑𝜙 𝑑𝑧 =
8𝜋𝐿 3
[𝜌1 − 10−9 ] 𝜇C
3
b) Use Gauss’ law to determine 𝐷𝜌 at 𝜌 = 𝜌1 : Gauss’ law states that 2𝜋𝜌1 𝐿𝐷𝜌 = 𝑄, where 𝑄 is the
result of part 𝑎. So, with 𝜌1 in meters,
𝐷𝜌 (𝜌1 ) =
4(𝜌31 − 10−9 )
3𝜌1
𝜇C∕m2
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3.15
c) Evaluate 𝐷𝜌 at 𝜌 = 0.8 mm, 1.6 mm, and 2.4 mm: At 𝜌 = 0.8 mm, no charge is enclosed by a
cylindrical gaussian surface of that radius, so 𝐷𝜌 (0.8mm) = 0. At 𝜌 = 1.6 mm, we evaluate the
part 𝑏 result at 𝜌1 = 1.6 to obtain:
𝐷𝜌 (1.6mm) =
4[(.0016)3 − (.0010)3 ]
= 3.6 × 10−6 𝜇C∕m2
3(.0016)
At 𝜌 = 2.4, we evaluate the charge integral of part 𝑎 from .001 to .002, and Gauss’ law is written
as
8𝜋𝐿
[(.002)2 − (.001)2 ] 𝜇C
2𝜋𝜌𝐿𝐷𝜌 =
3
from which 𝐷𝜌 (2.4mm) = 3.9 × 10−6 𝜇C∕m2 .
3.16. An electric flux density is given by 𝐃 = 𝐷0 𝐚𝜌 , 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
𝜌𝑣 = ∇ ⋅ 𝐃 =
(𝜌𝐷0 ) = 0 C∕m3
𝜌 𝑑𝜌
𝜌
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:
𝑏
𝑄=
∫0 ∫0
2𝜋
∫0
𝑎
2𝜋
𝑏
𝐷0
𝜌 𝑑𝜌 𝑑𝜙 𝑑𝑧 =
𝐷0 𝐚𝜌 ⋅ 𝐚𝜌 𝑎 𝑑𝜙 𝑑𝑧 = 2𝜋𝑎𝑏𝐷0 C
∫0 ∫0
𝜌
3.17. In a region having spherical symmetry, volume charge is distributed according to:
𝜌𝑣 (𝑟) = 𝜌0
sin(𝜋𝑟∕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
𝜌𝑣 𝑑𝑣 = 4𝜋
∫0
𝑟0
𝜌0
[
]
sin(𝜋𝑟∕𝑎) 2
𝑟
𝑑𝑟
=
4𝑎𝜌
1
−
cos(𝜋𝑟
∕𝑎)
C
0
0
𝑟2
This will be zero when cos(𝜋𝑟0 ∕𝑎) = +1 or when 𝑟0 = 2𝑚𝑎 where 𝑚 = 1, 2, 3....
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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
∮𝑠
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 Φ = ∇ ⋅ 𝐃Δ𝑣,
where in this case ∇ ⋅ 𝐃 = (𝜕∕𝜕𝑥)𝑥 = 1 C∕m3 . Thus
. 4
Φ = 𝜋(.003)3 (1) = 1.13 × 10−7 C = 113 nC
3
3.20. A radial electric field distribution in free space is given in spherical coordinates as:
𝑟𝜌0
𝐚
(𝑟 ≤ 𝑎)
𝐄1 =
3𝜖0 𝑟
(2𝑎3 − 𝑟3 )𝜌0
𝐚𝑟
(𝑎 ≤ 𝑟 ≤ 𝑏)
𝐄2 =
3𝜖0 𝑟2
(2𝑎3 − 𝑏3 )𝜌0
𝐄3 =
𝐚𝑟
(𝑟 ≥ 𝑏)
3𝜖0 𝑟2
where 𝜌0 , 𝑎, and 𝑏 are constants.
a) Determine the volume charge density in the entire region (0 ≤ 𝑟 ≤ ∞) by appropriate use of
∇ ⋅ 𝐃 = 𝜌𝑣 . We find 𝜌𝑣 by taking the divergence of 𝐃 in all three regions, where 𝐃 = 𝜖0 𝐄. As 𝐃
has only a radial component, the divergences become:
( 3 )
1 𝑑 𝑟 𝜌0
1 𝑑 ( 2 )
𝑟 𝐷1 = 2
= 𝜌0
(𝑟 ≤ 𝑎)
𝜌𝑣1 = ∇ ⋅ 𝐃1 = 2
3
𝑟 𝑑𝑟
𝑟 𝑑𝑟
(
)
1 𝑑 1 3
1 𝑑 ( 2 )
(2𝑎 − 𝑟3 )𝜌0 = −𝜌0
𝑟 𝐷2 = 2
𝜌𝑣2 = 2
(𝑎 ≤ 𝑟 ≤ 𝑏)
𝑟 𝑑𝑟
𝑟 𝑑𝑟 3
(
)
1 𝑑 1 3
1 𝑑 ( 2 )
(2𝑎 − 𝑏3 )𝜌0 = 0
𝑟 𝐷3 = 2
(𝑟 ≥ 𝑏)
𝜌𝑣3 = 2
𝑟 𝑑𝑟
𝑟 𝑑𝑟 3
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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=
∫0
2𝜋
𝜋
∫0 ∫0
𝑎
𝜌0 𝑟2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜙 −
∫0
2𝜋
𝜋
∫0 ∫𝑎
𝑏
)
4 (
𝜌0 𝑟2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜙 = 𝜋 2𝑎3 − 𝑏3 𝜌0
3
3.21. In a region exhibiting spherical symmetry, electric flux density is found to be
𝐃1 = 𝜌0 𝑟∕3 𝐚𝑟 (0 < 𝑟 < 𝑎), 𝐃2 = 0 (𝑎 < 𝑟 < 𝑏), and 𝐃3 = (𝑎3 𝜌0 )∕(3𝑟2 ) 𝐚𝑟 (𝑟 > 𝑏).
a) Find the charge configuration that would produce the given field: We use Gauss’ Law in point
form to find:
( 𝑟𝜌 )
1 𝑑
𝜌𝑣1 = ∇ ⋅ 𝐃1 = 2
𝑟2 0 = 𝜌0 C∕m3
3
𝑟 𝑑𝑟
𝜌𝑣2 = ∇ ⋅ 𝐃2 = 0
𝜌𝑣3 = ∇ ⋅ 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)𝜋𝑎3 𝜌0
𝑎𝜌
𝜌𝑠1 (𝑟 = 𝑎) = −
= − 0 C∕m2
2
3
4𝜋𝑎
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
4𝜋𝑟2 𝐷3 = 𝑄𝑒𝑛𝑐𝑙 = 4𝜋𝑏2 𝜌𝑠2 ⇒ 𝜌𝑠2 =
𝑎3 𝜌0
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 = 4𝜋𝑏2 𝜌𝑠2 =
4𝜋𝑎3 𝜌0
C
3
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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 < 𝜃 < 𝜋∕2, 0 < 𝜙 < 2𝜋.
The flux integral is
Φ1 =
∫ℎ𝑒𝑚.
𝐅1 ⋅ 𝑑𝐒 =
∫0
2𝜋
∫0
𝜋∕2
5 𝐚𝑧 ⋅ 𝐚𝑟 𝑎2 sin 𝜃 𝑑𝜃 𝑑𝜙 = −2𝜋(5)𝑎2
⏟⏟⏟
cos 𝜃
cos2 𝜃 |𝜋∕2
= 5𝜋𝑎2
|
2 |0
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 𝜋𝑎2 ) 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:
Φ2 =
∫0
2𝜋
∫0
= 10𝜋𝑎3
𝜋∕2
∫0
5𝑧 𝐚𝑧 ⋅ 𝐚𝑟 𝑎2 sin 𝜃 𝑑𝜃 𝑑𝜙 =
𝜋∕2
cos2 𝜃 sin 𝜃 𝑑𝜃 𝑑𝜙 = −
∫0
2𝜋
∫0
𝜋∕2
5(𝑎 cos 𝜃) cos 𝜃 𝑎2 sin 𝜃 𝑑𝜃 𝑑𝜙
10 3
|𝜋∕2 10 3
𝜋𝑎 cos3 𝜃 |
= 𝜋𝑎
|0
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)𝜋𝑎3 , giving the same answer as in part 𝑐.
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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∕(4𝜋𝑟2 ) 𝐚𝑟 . Using the formula for divergence in
spherical coordinates (see problem 3.21 solution), we find in this case that
)
(
1 𝑑
2 𝑄
=0
∇⋅𝐃= 2
𝑟
𝑟 𝑑𝑟
4𝜋𝑟2
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 𝜌𝑣0 for 0 < 𝑟 < 𝑎. Relate 𝜌𝑣0 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)𝜋𝑎3 𝜌𝑣0 = 𝑄, so 𝜌𝑣0 = 3𝑄∕(4𝜋𝑎3 ) C∕m3 . Gauss’ law tells us that
inside the charged sphere
𝑄𝑟3
4
4𝜋𝑟2 𝐷𝑟 = 𝜋𝑟3 𝜌𝑣0 = 3
3
𝑎
Thus
)
(
3𝑄
𝑄𝑟
𝑄𝑟3
1 𝑑
2
C∕m
and
∇
⋅
𝐃
=
=
𝐷𝑟 =
3
2
3
𝑑𝑟
4𝜋𝑎
𝑟
4𝜋𝑎
4𝜋𝑎3
as expected. Outside the charged sphere, 𝐃 = 𝑄∕(4𝜋𝑟2 ) 𝐚𝑟 as before, and the divergence is zero.
3.24. In a region in free space, electric flux density is found to be:
{
𝜌0 (𝑧 + 2𝑑) 𝐚𝑧 C∕m2
(−2𝑑 ≤ 𝑧 ≤ 0)
𝐃=
2
−𝜌0 (𝑧 − 2𝑑) 𝐚𝑧 C∕m
(0 ≤ 𝑧 ≤ 2𝑑)
Everywhere else, 𝐃 = 0.
a) Using ∇ ⋅ 𝐃 = 𝜌𝑣 , find the volume charge density as a function of position everywhere: Use
{
𝑑𝐷𝑧
𝜌0 (−2𝑑 ≤ 𝑧 ≤ 0)
𝜌𝑣 = ∇ ⋅ 𝐃 =
=
𝑑𝑧
−𝜌0 (0 ≤ 𝑧 ≤ 2𝑑)
b) determine the electric flux that passes through the surface defined by 𝑧 = 0, −𝑎 ≤ 𝑥 ≤ 𝑎,
−𝑏 ≤ 𝑦 ≤ 𝑏: In the 𝑥-𝑦 plane, 𝐃 evaluates as the constant 𝐃(0) = 2𝑑𝜌0 𝐚𝑧 . Therefore the flux
passing through the given area will be
𝑎
Φ=
𝑏
∫−𝑎 ∫−𝑏
2𝑑𝜌0 𝑑𝑥𝑑𝑦 = 8𝑎𝑏𝑑 𝜌0 C
c) determine the total charge contained within the region −𝑎 ≤ 𝑥 ≤ 𝑎, −𝑏 ≤ 𝑦 ≤ 𝑏, −𝑑 ≤ 𝑧 ≤ 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 ≤ 𝑥 ≤ 𝑎, −𝑏 ≤ 𝑦 ≤ 𝑏, 0 ≤ 𝑧 ≤ 2𝑑.
In this case,
𝑄 = −𝜌0 (2𝑎) (2𝑏) (2𝑑) = −8𝑎𝑏𝑑 𝜌0 C
This is equivalent to the net inward flux of 𝐃 into the volume, as was found in part 𝑏.
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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
5
1 𝑑
𝜌𝑣 = ∇ ⋅ 𝐃 = 2 (𝑟2 𝐷𝑟 ) = (𝑟 − 3)2 (5𝑟 − 6) C∕m3
𝑟
𝑟 𝑑𝑟
which we evaluate at 𝑟 = 4 to find 𝜌𝑣 (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 Φ =
4𝜋(4)2 (5) = 320𝜋 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 = 320𝜋 C.
3.26. If we have a perfect gas of mass density 𝜌𝑚 kg∕m3 , and assign a velocity 𝐔 m/s to each differential
element, then the mass flow rate is 𝜌𝑚 𝐔 kg∕(m2 − s). Physical reasoning then leads to the continuity
equation, ∇ ⋅ (𝜌𝑚 𝐔) = −𝜕𝜌𝑚 ∕𝜕𝑡.
a) Explain in words the physical interpretation of this equation: The quantity 𝜌𝑚 𝐔 is the flow (or
flux) density of mass. Then the divergence of 𝜌𝑚 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 ∮𝑠 𝜌𝑚 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
∮𝑠
𝜌𝑚 𝐔 ⋅ 𝑑𝐒 =
∫𝑣
∇ ⋅ (𝜌𝑚 𝐔) 𝑑𝑣 =
∫𝑣
−
𝜕𝜌𝑚
𝑑
𝑑𝑀
𝜌 𝑑𝑣 = −
𝑑𝑣 = −
𝜕𝑡
𝑑𝑡 ∫𝑣 𝑚
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
𝜌
𝐄 = 0 exp(−𝛼𝑧) 𝐚𝑧 V∕m
𝜖0 𝛼
where 𝜌0 is a positive constant (as is 𝛼). 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 𝜌𝑣 in the slab:
(
)
𝑑 𝜌0 −𝛼𝑧
𝜌𝑣 = ∇ ⋅ 𝐃 =
= −𝜌0 𝑒−𝛼𝑧
𝑒
𝑑𝑧 𝛼
so we have negative charge of decreasing magnitude with 𝑧.
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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
−𝐴𝜌0
(1 − 𝑒−𝛼𝑑 )
𝑄=𝐴
−𝜌0 𝑒−𝛼𝑧 𝑑𝑧 =
∫0
𝛼
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
∮𝑠
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:
−𝐴𝜌0
(1 − 𝑒−𝛼𝑑 )
𝐃 ⋅ 𝑑𝐒 =
𝐃(0) ⋅ (−𝐚𝑧 )𝑑𝑎 +
𝐃(𝑑) ⋅ (+𝐚𝑧 )𝑑𝑎 =
∮𝑠
∫ ∫𝐴
∫ ∫𝐴
𝛼
as before.
3.28. Repeat Problem 3.8, but use ∇ ⋅ 𝐃 = 𝜌𝑣 and take an appropriate volume integral.
We begin by finding the charge density directly through
(
)
𝐴
𝐴
1 𝑑
𝑟2
= 2
𝜌𝑣 = ∇ ⋅ 𝐃 = 2
𝑟
𝑟 𝑑𝑟
𝑟
Then, within each spherical shell of unit thickness, the contained charge is
𝑄(1) = 4𝜋
∫𝑟
𝑟+1
𝐴 ′2 ′
(𝑟 ) 𝑑𝑟 = 4𝜋𝐴(𝑟 + 1 − 𝑟) = 4𝜋𝐴
(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
𝑑𝑥𝑑𝑦𝑑𝑧
=
(9
−
4)(9
−
4)
= 3.47 C
−
∇ ⋅ 𝐃 𝑑𝑣 =
∫𝑣𝑜𝑙
∫2 ∫2 ∫2 𝑧3
4 9
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
−
∮
∫2 ∫2
∫2 ∫2
4 9
32
22
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
top
bottom
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3.30.
a) Use Maxwell’s first equation, ∇ ⋅ 𝐃 = 𝜌𝑣 , 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 𝜖(𝑥) = 𝜖1 exp(𝛼1 𝑥), where 𝜖1 and 𝛼1 are constants. Then
[
]
]
𝑑𝐸𝑥
𝑑 [ 𝛼1 𝑥
𝜖1 𝑒 𝐸𝑥 (𝑥) = 𝜖1 𝛼1 𝑒𝛼1 𝑥 𝐸𝑥 + 𝑒𝛼1 𝑥
=0
∇ ⋅ 𝐃 = ∇ ⋅ [𝜖(𝑥)𝐄(𝑥)] =
𝑑𝑥
𝑑𝑥
This reduces to
𝑑𝐸𝑥
+ 𝛼1 𝐸𝑥 = 0 ⇒ 𝐸𝑥 (𝑥) = 𝐸0 𝑒−𝛼1 𝑥
𝑑𝑥
where 𝐸0 is a constant.
b) Repeat part 𝑎, but with a radially-directed electric field (spherical coordinates), in which again
𝜌𝑣 = 0, but in which the permittivity decreases exponentially with 𝑟. In this case, the permittivity
can be written as 𝜖(𝑟) = 𝜖2 exp(−𝛼2 𝑟), where 𝜖2 and 𝛼2 are constants. Then
]
[
1 𝑑 [ 2 −𝛼2 𝑟 ] 𝜖2
2
2 𝑑𝐸𝑟
𝑟 𝜖2 𝑒
𝐸𝑟 = 2 2𝑟𝐸𝑟 − 𝛼2 𝑟 𝐸𝑟 + 𝑟
𝑒−𝛼2 𝑟 = 0
∇ ⋅ 𝐃 = ∇ ⋅ [𝜖(𝑟)𝐄(𝑟)] = 2
𝑑𝑟
𝑟 𝑑𝑟
𝑟
This reduces to
)
𝑑𝐸𝑟 ( 2
+
− 𝛼2 𝐸𝑟 = 0
𝑑𝑟
𝑟
whose solution is
[
𝐸𝑟 (𝑟) = 𝐸0 exp −
∫
(
) ]
[
] 𝐸
2
− 𝛼2 𝑑𝑟 = 𝐸0 exp −2 ln 𝑟 + 𝛼2 𝑟 = 20 𝑒𝛼2 𝑟
𝑟
𝑟
where 𝐸0 is a constant.
3.31. Given the flux density
16
cos(2𝜃) 𝐚𝜃 C∕m2 ,
𝑟
use two different methods to find the total charge within the region 1 < 𝑟 < 2 m, 1 < 𝜃 < 2 rad,
1 < 𝜙 < 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 𝜃, however,
since 𝐃 has only a 𝜃 component. On a constant-theta surface, the differential area is 𝑑𝑎 = 𝑟 sin 𝜃𝑑𝑟𝑑𝜙,
where 𝜃 is fixed at the surface location. Our flux integral becomes
𝐃=
∮
2
𝐃 ⋅ 𝑑𝐒 = −
2
2
2
16
16
cos(2) 𝑟 sin(1) 𝑑𝑟𝑑𝜙 +
cos(4) 𝑟 sin(2) 𝑑𝑟𝑑𝜙
∫1 ∫1 𝑟
∫1 ∫1 𝑟
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
𝜃=1
𝜃=2
= −16 [cos(2) sin(1) − cos(4) sin(2)] = −3.91 C
We next evaluate the volume integral side of the divergence theorem, where in this case,
[
[
]
]
1 𝑑 16
16 cos 2𝜃 cos 𝜃
1 𝑑
(sin 𝜃 𝐷𝜃 ) =
cos 2𝜃 sin 𝜃 = 2
− 2 sin 2𝜃
∇⋅𝐃=
𝑟 sin 𝜃 𝑑𝜃
𝑟 sin 𝜃 𝑑𝜃 𝑟
sin 𝜃
𝑟
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3.31 (continued) We now evaluate:
∫𝑣𝑜𝑙
2
∇ ⋅ 𝐃 𝑑𝑣 =
2
∫1 ∫1 ∫1
2
]
[
16 cos 2𝜃 cos 𝜃
−
2
sin
2𝜃
𝑟2 sin 𝜃 𝑑𝑟𝑑𝜃𝑑𝜙
sin 𝜃
𝑟2
The integral simplifies to
2
2
∫1 ∫1 ∫1
2
16[cos 2𝜃 cos 𝜃 − 2 sin 2𝜃 sin 𝜃] 𝑑𝑟𝑑𝜃𝑑𝜙 = 8
∫1
2
[3 cos 3𝜃 − cos 𝜃] 𝑑𝜃 = −3.91 C
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Engineering Electromagnetics 9th solution
Engineering Electromagnetics (Chungnam National University)
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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 𝛿:
a) along the positive 𝑥 axis... This will be 𝑑𝑊𝑎 = −𝑞𝛿 𝐄 ⋅ 𝐚𝑥 = −𝑞𝛿𝐸𝑥 J.
b) in a direction at 45 degrees
√ This will be 𝑑𝑊𝑏 = −𝑞𝛿 𝐄 ⋅ 𝐚𝑏 ,
√ from the 𝑥 axis in the first quadrant...
where 𝐚𝑏 = (𝐚𝑥 + 𝐚𝑦 )∕ 2. Finally, 𝑑𝑊𝑏 = −𝑞𝛿(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 = −𝑞𝛿 𝐄 ⋅ 𝐚𝑐 , where 𝐚𝑐 = (𝐚𝑥 + 𝐚𝑦 + 𝐚𝑧 )∕ 3, which results in
√
𝑑𝑊𝑐 = −𝑞𝛿(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, 𝜃, 𝜙), we write:
𝑑𝑊 =
−𝑞2 𝑞1
𝐚𝑟 ⋅ (−𝑑𝑥)𝐚𝑥
4𝜋𝜖0 𝑟2
where 𝑟2 = 𝑥2 + 𝑦2 + 𝑧2 , and where 𝐚𝑟 ⋅ 𝐚𝑥 = sin 𝜃 cos 𝜙. So
√
𝑞2 𝑞1
𝑞2 𝑞1 𝑥 𝑑𝑥
𝑥2 + 𝑦2
𝑥
𝑑𝑊 =
𝑑𝑥 =
√
√
2
2
2
4𝜋𝜖0 (𝑥 + 𝑦 + 𝑧 ) 𝑥2 + 𝑦2 + 𝑧2 𝑥2 + 𝑦2
4𝜋𝜖0 (𝑥2 + 𝑦2 + 𝑧2 )3∕2
⏟⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏟⏞⏞⏟
cos 𝜙
sin 𝜃
4.3. Given 𝐄 = 𝐸𝜌 𝐚𝜌 + 𝐸𝜙 𝐚𝜙 + 𝐸𝑧 𝐚𝑧 V/m, where 𝐸𝜌 , 𝐸𝜙 , and 𝐸𝑧 are constants:
a) find the incremental work done in moving charge 𝑞 through distance 𝛿 in a direction√having
equal 𝜌 and 𝜙 components... this will be 𝑑𝑊𝑎 = −𝑞𝛿 𝐄 ⋅ 𝐚1 , where 𝐚1 = (𝐚𝜌 + 𝐚𝜙 )∕ 2. So
√
𝑑𝑊𝑎 = −𝑞𝛿 (𝐸𝜌 + 𝐸𝜙 )∕ 2 J.
b) if the initial charge [location] in part 𝑎 was at radius 𝜌 = 𝑏, what change in angle 𝜙 occurred
in moving the charge? We may write 𝑑𝑊 = −𝑞𝐄 ⋅ 𝑑𝐋 where 𝑑𝐋 = 𝑑𝜌𝐚𝜌 + 𝜌𝑑𝜙𝐚𝜙 . With
√
equal distances
√ along the two directions, we would have 𝑑𝜌 = 𝜌𝑑𝜙 = 𝛿∕ 2. So, at 𝜌 = 𝑏,
𝑑𝜙 = 𝛿∕(𝑏 2) rad
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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
1𝜇C charge through this field
a) from (1,1,1) to (0,0,0): The work will be
[
𝑊 = −𝑞
∫
∫1
−6
𝐄 ⋅ 𝑑𝐋 = −10
0
𝑥𝑑𝑥 +
∫1
0
𝑦𝑑𝑦 +
∫1
]
0
𝑧𝑑𝑧 J = 1.5 𝜇J
b) from (𝜌 = 2, 𝜙 = 0) to (𝜌 = 2, 𝜙 = 90◦ ): The path involves changing 𝜙 with 𝜌 and 𝑧 fixed, and
therefore 𝑑𝐋 = 𝜌 𝑑𝜙 𝐚𝜙 . We set up the integral for the work as
𝑊 = −10−6
∫0
𝜋∕2
(𝑥𝐚𝑥 + 𝑦𝐚𝑦 + 𝑧𝐚𝑧 ) ⋅ 𝜌 𝑑𝜙 𝐚𝜙
where 𝜌 = 2, 𝐚𝑥 ⋅ 𝐚𝜙 = − sin 𝜙, 𝐚𝑦 ⋅ 𝐚𝜙 = cos 𝜙, and 𝐚𝑧 ⋅ 𝐚𝜙 = 0. Also, 𝑥 = 2 cos 𝜙 and
𝑦 = 2 sin 𝜙. Substitute all of these to get
𝑊 = −10−6
∫0
𝜋∕2 [
]
−(2)2 cos 𝜙 sin 𝜙 + (2)2 cos 𝜙 sin 𝜙 𝑑𝜙 = 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, 𝜃 = 𝜃0 ) to (𝑟 = 10, 𝜃 = 𝜃0 + 180◦ ): In this case, we are moving only in the 𝐚𝜃
direction. The work is set up as
𝑊 = −10−6
𝜃0 +𝜋
∫𝜃0
(𝑥𝐚𝑥 + 𝑦𝐚𝑦 + 𝑧𝐚𝑧 ) ⋅ 𝑟 𝑑𝜃 𝐚𝜃
Now, substitute the following relations: 𝑟 = 10, 𝑥 = 𝑟 sin 𝜃 cos 𝜙, 𝑦 = 𝑟 sin 𝜃 sin 𝜙, 𝑧 = 𝑟 cos 𝜃,
𝐚𝑥 ⋅ 𝐚𝜃 = cos 𝜃 cos 𝜙, 𝐚𝑦 ⋅ 𝐚𝜃 = cos 𝜃 sin 𝜙, and 𝐚𝑧 ⋅ 𝐚𝜃 = − sin 𝜃. Obtain
𝑊 = −10−6
𝜃0 +𝜋
∫𝜃0
[
]
(10)2 sin 𝜃 cos 𝜃 cos2 𝜙 + sin 𝜃 cos 𝜃 sin2 𝜙 − cos 𝜃 sin 𝜃 𝑑𝜃 = 0
where we use cos2 𝜙 + sin2 𝜙 = 1.
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4.5. Consider the vector field 𝐆 = (𝐴∕𝜌) 𝐚𝜙 , 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 𝜙 is 𝛼:
∫1
𝐆 ⋅ 𝑑𝐋 =
∫0
𝛼
𝐴
𝐚 ⋅ 𝐚 𝑏 𝑑𝜙 = 𝐴𝛼
𝑏 𝜙 𝜙
b) Same as part 𝑎, except the path is at radius 𝑐 > 𝑏 and the change in 𝜙 is −𝛼:
∫2
0
𝐆 ⋅ 𝑑𝐋 =
𝐴
𝐚 ⋅ 𝐚 𝑐 𝑑𝜙 = −𝐴𝛼
∫𝛼 𝑐 𝜙 𝜙
c) Evaluate ∮ 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
∮
𝐆 ⋅ 𝑑𝐋 =
4
∑
𝑖=1
∫𝑖
𝐆 ⋅ 𝑑𝐋 = 𝐴𝛼 − 𝐴𝛼 +
∫𝑏
𝑐
𝑏
𝐴
𝐴
𝐚 ⋅ 𝐚 𝑑𝜌 +
𝐚 ⋅ 𝐚 𝑑𝜌 = 0
∫𝑐 𝜌 𝜙 𝜌
𝜌 𝜙 𝜌
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 𝐚̂ 𝑥 + 4𝑧 𝐚̂ 𝑦 + 4𝑦 𝐚̂ 𝑧 . Given 𝑉 (1, 1, 1) = 10 V. Determine
𝑉 (3, 3, 3). The potential difference is expressed as
3,3,3
𝑉 (3, 3, 3) − 𝑉 (1, 1, 1) = −
=−
∫1,1,1
[
∫1
3
(𝑥 𝐚̂ 𝑥 + 4𝑧 𝐚̂ 𝑦 + 4𝑦 𝐚̂ 𝑧 ) ⋅ (𝑑𝑥 𝐚𝑥 + 𝑑𝑦 𝐚𝑦 + 𝑑𝑧 𝐚𝑧 )
]
𝑥 𝑑𝑥 +
∫1
3
4𝑧 𝑑𝑦 +
∫1
3
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:
[
]
𝑉 (3, 3, 3) − 𝑉 (1, 1, 1) = −
∫1
3
𝑥 𝑑𝑥 +
∫1
3
4(1) 𝑑𝑦 +
∫1
3
4(3) 𝑑𝑧 = −36
So
𝑉 (3, 3, 3) = −36 + 𝑉 (1, 1, 1) = −36 + 10 = −26 V
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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
( 𝜋𝑦 )
( 𝜋𝑦 )
( )
( )
𝜋
𝜋𝑥
𝜋𝑥
𝜋
sin
𝐚𝑥 − 𝐸0 sin
cos
𝐚𝑦
𝐄11 = 𝐸0 cos
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
|
|
𝐸𝑥 | 𝑑𝑥 −
𝑑𝑦
𝐸𝑦 |
|𝑦=0
|𝑥=𝑎∕2
∫𝐵
∫0
∫0
𝑎∕2
𝑏∕2
( )
( 𝜋𝑦 )
( )
( )
𝜋
𝜋
𝜋0
𝜋𝑥
𝜋𝑎
sin
𝑑𝑥 +
cos
𝑑𝑦
=−
𝐸0 sin
𝐸0 cos
∫0
∫0
𝑏
𝑎
𝑏
𝑎
2𝑎
𝑏
𝑏∕2
( 𝜋𝑦 )
𝑏
𝜋
𝑑𝑦 = 𝐸0 V
=
𝐸0 cos
∫0
𝑎
𝑏
𝑎
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 𝜙 = 0 and move to 𝜙 = 𝜋∕4. The setup is
𝑊 = −𝑞
∫
𝐄 ⋅ 𝑑𝐋 = −
∫0
𝜋∕4
𝐄 ⋅ 𝑎𝑑𝜙 𝐚𝜙 = −
∫0
𝜋∕4
(−𝑥 𝐚𝑥 ⋅ 𝐚𝜙 + 𝑦 𝐚𝑦 ⋅ 𝐚𝜙 )𝑎 𝑑𝜙
⏟⏟⏟
⏟⏟⏟
− sin 𝜙
=−
∫0
𝜋∕4
2𝑎2 sin 𝜙 cos 𝜙 𝑑𝜙 = −
∫0
𝜋∕4
𝑎2 sin(2𝜙) 𝑑𝜙 =
cos 𝜙
𝑎2
|𝜋∕4
cos(2𝜙)|
=−
|0
2
2
𝑎2
where 𝑞 = 1, 𝑥 = 𝑎 cos 𝜙, and 𝑦 = 𝑎 sin 𝜙.
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:
√
√
[
]
𝑊 =−
∫
𝐄 ⋅ 𝑑𝐋 = −
∫𝑎
𝑎∕ 2
(−𝑥) 𝑑𝑥 +
∫0
𝑎∕ 2
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 =−
∫0
2𝜋
𝑎2 sin(2𝜙) 𝑑𝜙 =
𝑎2
|2𝜋
cos(2𝜙)| = 0
|0
2
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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
𝑟
𝑉 (𝑟) = −
∫∞
𝑟
𝐄 ⋅ 𝑑𝐋 = −
𝑉0 −𝑟′ ∕𝑎
𝑒
𝑑𝑟 = 𝑉0 𝑒−𝑟∕𝑎 V
∫∞ 𝑎
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 𝜌𝑠0 C∕m2 .
a) Find the absolute potential at the sphere surface: The setup for this is
𝑎
𝑉0 = −
∫∞
𝐄 ⋅ 𝑑𝐋
where, from Gauss’ law:
𝐄=
So
𝑎
𝑉0 = −
𝑎2 𝜌𝑠0
∫ ∞ 𝜖0 𝑟 2
𝑎2 𝜌𝑠0
𝜖0 𝑟2
𝐚𝑟
𝐚𝑟 ⋅ 𝐚𝑟 𝑑𝑟 =
V∕m
𝑎2 𝜌𝑠0 |𝑎
𝑎𝜌𝑠0
V
| =
𝜖0 𝑟 |∞
𝜖0
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 = −
∫𝑏
𝑎
𝑎2 𝜌𝑠0
𝜖0 𝑟2
𝐚𝑟 ⋅ 𝐚𝑟 𝑑𝑟 =
𝑎2 𝜌𝑠0 |𝑎 𝑎2 𝜌𝑠0 [ 1 1 ]
V
−
| =
𝜖0 𝑟 |𝑏
𝜖0
𝑎 𝑏
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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 𝜃 𝐚𝜃 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 𝐚𝜃 direction, thus presenting its full length to
the transmitting antenna at the origin. The angular position of the receiving antenna is 𝜃 = 𝜃0 . As
observed from the transmitting antenna, the receiving antenna subtends angle Δ𝜃.
a) Find the voltage amplitude across the length of the receiving antenna, and express your result in
terms of 𝐴, 𝜃0 , 𝐿, and 𝑟. Using the given field, we write:
𝜃0 −Δ𝜃∕2
𝐴
sin 𝜃 𝐚𝜃 ⋅ 𝐚𝜃 𝑟𝑑𝜃
∫𝐿
∫𝜃0 +Δ𝜃∕2 𝑟
[ (
)
(
)]
Δ𝜃
Δ𝜃
= 𝐴 cos 𝜃0 −
− cos 𝜃0 +
= 2𝐴 sin 𝜃0 sin(Δ𝜃∕2) V
2
2
𝑉 =−
𝐄 ⋅ 𝑑𝐋 = −
where Δ𝜃 = tan−1 (𝐿∕𝑟).
.
b) Specialize your part 𝑎 result for the case in which 𝐿 << 𝑟 (the usual case): In this case, Δ𝜃 = 𝐿∕𝑟
.
.
and sin(Δ𝜃∕2) = Δ𝜃∕2. Therefore, 𝑉 = 𝐴(𝐿∕𝑟) sin 𝜃0 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
1
+𝐶 = 2
+𝐶
∫ (𝑟2 + 𝑎2 )2
𝑟 + 𝑎2
With a zero reference at 𝑟 → ∞, 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 ⇒ 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 ⇒ 𝐶 = 100 − 2
2
2𝑎
2𝑎
1
𝑎2 − 𝑟2
1
−
+
100
=
+ 100
𝑟2 + 𝑎2 2𝑎2
2𝑎2 (𝑟2 + 𝑎2 )
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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 × 10−12 )2 [ 1
1
× 104 = 5.76 × 10−10 J = 576 pJ
−
2𝜋𝜖0
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
2
𝑉𝑏 − 𝑉𝑎 = −
|
(𝑦 + 1)| 𝑑𝑥 −
|𝑦=0
∫0
∫0
−2
|
(𝑥 − 1)| 𝑑𝑦 −
|𝑥=2
∫0
−1
2 𝑑𝑧 = 2
b) (3,2,-1) and (-2,-3,4): Following similar reasoning,
3
𝑉𝑏 − 𝑉𝑎 = −
2
|
|
(𝑦 + 1)|
𝑑𝑥 −
(𝑥 − 1)| 𝑑𝑦 −
|𝑦=−3
|𝑥=3
∫−2
∫−3
∫4
−1
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 =
√
𝜌𝑙
𝜌
ln(𝑅1 ) − 𝑙 ln(𝑅2 ) + 𝐶
2𝜋𝜖0
2𝜋𝜖0
5, and 𝑉 = 100 V. Thus, with 𝜌𝑙 = 8 × 10−9 ,
(8 × 10−9 ) √
ln( 5) + 𝐶 ⇒ 𝐶 = 331.6 V
2𝜋𝜖0
√
√
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 × 10−9 ) [ √
ln( 10) + ln( 26) + 331.6 = −68.4 V
2𝜋𝜖0
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4.16. A spherically-symmetric charge distribution in free space (with 𝑎 < 𝑟 < ∞) 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 = −
𝑎2
𝑑𝑉
𝐚𝑟 = 2𝑉0 3 𝐚𝑟 V∕m
𝑑𝑟
𝑟
b) Find the volume charge density: Use Maxwell’s first equation:
[ (
)]
1 𝑑
𝑎2
𝑎2
2
𝜌𝑣 = ∇ ⋅ 𝐃 = ∇ ⋅ (𝜖0 𝐄) = 2
𝑟 2𝜖0 𝑉0 3
= −2𝜖0 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:
(
)
𝑎2
2
2
𝑄𝑒𝑛𝑐𝑙 =
𝐃 ⋅ 𝑑𝐒 = 4𝜋𝑎 𝐷|𝑟=𝑎 = 4𝜋𝑎 2𝜖0 𝑉0 3 = 8𝜋𝜖0 𝑎𝑉0 C
∮𝑟=𝑎
𝑎
d) Find the total energy stored in the charge (or equivalently, in its electric field) in the region
(𝑎 < 𝑟 < ∞). Use Eq. (42):
)(
)
∞(
𝜋
2𝜋
𝑎2
1
1
𝑎2
𝑊𝐸 =
𝑉0 2 𝑟2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜙
−2𝜖0 𝑉0 4
𝜌 𝑉 𝑑𝑣 =
2 ∫𝑣 𝑣
2 ∫0 ∫0 ∫𝑎
𝑟
𝑟
(1)
4
2
= − 𝜋𝜖0 𝑎𝑉0 J
3
If we integrate the energy density in the field over the region, the result is
2𝜋
𝜋
∞
𝑎4
1
2𝜖0 𝑉02 6 𝑟2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜙
𝐃 ⋅ 𝐄 𝑑𝑣 =
∫0 ∫0 ∫𝑎
∫𝑣 2
𝑟
∞
1
8
|
= −8𝜋 𝑉02 𝜖0 𝑎4 3 | = 𝜋𝜖0 𝑎𝑉02 J
3
3𝑟 |𝑎
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, 𝜌𝑠 , which covers the surface of a conducting sphere of radius 𝑎, where 𝜌𝑠 =
𝑄∕4𝜋𝑎2 , with 𝑄 as found in part 𝑐. The energy in that charge is found through
𝑊𝐸′ =
1
𝜌 𝑉 𝑑𝑎 = 4𝜋𝜖0 𝑎𝑉02
2 ∫𝑠 𝑠 0
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).
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4.17. Uniform surface charge densities of 6 and 2 nC∕m2 are present at 𝜌 = 2 and 6 cm respectively, in free
space. Assume 𝑉 = 0 at 𝜌 = 4 cm, and calculate 𝑉 at:
a) 𝜌 = 5 cm: Since 𝑉 = 0 at 4 cm, the potential at 5 cm will be the potential difference between
points 5 and 4:
𝑉5 = −
∫4
5
𝐄 ⋅ 𝑑𝐋 = −
∫4
5
𝑎𝜌𝑠𝑎
(.02)(6 × 10−9 ) ( 5 )
= −3.026 V
ln
𝑑𝜌 = −
𝜖0 𝜌
𝜖0
4
b) 𝜌 = 7 cm: Here we integrate piecewise from 𝜌 = 4 to 𝜌 = 7:
𝑉7 = −
∫4
6
7
(𝑎𝜌𝑠𝑎 + 𝑏𝜌𝑠𝑏 )
𝑎𝜌𝑠𝑎
𝑑𝜌 −
𝑑𝜌
∫6
𝜖0 𝜌
𝜖0 𝜌
With the given values, this becomes
] ( ) [
] ( )
[
(.02)(6 × 10−9 )
(.02)(6 × 10−9 ) + (.06)(2 × 10−9 )
6
7
ln
ln
𝑉7 = −
−
𝜖0
4
𝜖0
6
= −9.678 V
4.18. Find the potential at the origin produced by a line charge 𝜌𝐿 = 𝑘𝑥∕(𝑥2 + 𝑎2 ) extending along the 𝑥
axis from 𝑥 = 𝑎 to +∞, where 𝑎 > 0. Assume a zero reference at infinity.
Think of the line charge as an array of point charges, each of charge 𝑑𝑞 = 𝜌𝐿 𝑑𝑥, and each having
potential at the origin of 𝑑𝑉 = 𝜌𝐿 𝑑𝑥∕(4𝜋𝜖0 𝑥). The total potential at the origin is then the sum
of all these potentials, or
𝑉 =
∫𝑎
∞
∞
[
]
( )∞
𝜌𝐿 𝑑𝑥
𝑘
𝑘
𝑘
𝑘 𝑑𝑥
𝜋 𝜋
−1 𝑥
=
=
=
=
tan
−
2
2
4𝜋𝜖0 𝑥 ∫𝑎 4𝜋𝜖0 (𝑥 + 𝑎 ) 4𝜋𝜖0 𝑎
𝑎 𝑎
4𝜋𝜖0 𝑎 2 4
16𝜖0 𝑎
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4.19. Volume charge density is given as 𝜌𝑣 = 𝜌0 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) =
∫𝑣𝑜𝑙
𝜌𝑣 (𝐫 ′ )𝑑𝑣′
4𝜋𝜖0 |𝐫 − 𝐫 ′ |
Because we are evaluating the potential at the origin, 𝐫 = 0, 𝐫 ′ = 𝑟𝐚𝑟 , and so
𝑉 (0) =
∫0
2𝜋
𝜋
∫0 ∫0
∞
𝜌0 𝑒−𝑟
4𝜋𝜖0 𝑟2
𝑟2 sin 𝜃 𝑑𝑟𝑑𝜃𝑑𝜙 =
𝜌0 ∞ −𝑟
𝜌
𝑒 𝑑𝑟 = 0
𝜖0 ∫ 0
𝜖0
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:
∮𝑠
𝐃 ⋅ 𝑑𝐒 = 4𝜋𝑟2 𝐷(𝑟) =
Solving, find
𝐃=
∫𝑣
𝜌𝑣 𝑑𝑣 = 4𝜋
∫0
𝑟
′ |𝑟
𝑒−𝑟 ′ 2 ′
(𝑟 ) 𝑑𝑟 = −4𝜋𝜌0 (1 + 𝑟′ )𝑒−𝑟 |
′
|0
𝑟
]
𝜌0 [
𝐃
1 − (1 + 𝑟)𝑒−𝑟 𝐚𝑟 and 𝐄 =
2
𝜖0
𝑟
Now,
0
𝑉 (0) = −
′
𝜌0
∫∞
0
𝐄 ⋅ 𝑑𝐋 = −
∫∞
]
𝜌0 [
1 − (1 + 𝑟)𝑒−𝑟 𝑑𝑟
2
𝜖0 𝑟
To evaluate the integral, we use the form:
𝑒−𝑟
𝑒−𝑟
𝑒−𝑟
−
𝑑𝑟
𝑑𝑟
=
−
∫ 𝑟
∫ 𝑟2
𝑟
So the potential integral becomes
[
]
]
[
0 −𝑟
0
0 −𝑟
−𝜌0
−𝜌0
𝑒
𝑑𝑟
𝑒−𝑟 |0
𝑒
𝑒−𝑟 |0
1 |0
+
𝑑𝑟 +
𝑑𝑟 =
𝑉 (0) =
− | +
| −
|
𝜖0 ∫ ∞ 𝑟 2 ∫ ∞ 𝑟
𝑟 |∞ ∫ ∞ 𝑟
𝜖0
𝑟 |∞
𝑟 |∞
.
For the evaluation, use the small argument approximation: 𝑒−𝑟 = 1 − 𝑟. Finally,
𝜌
. −𝜌0 [ 1 1 − 𝑟 ]
𝑉 (0) =
= 0 Q.E.D.
− +
𝜖0
𝑟
𝑟 0 𝜖0
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4.20. In a certain medium, the electric potential is given by
𝑉 (𝑥) =
𝜌0
(1 − 𝑒−𝑎𝑥 )
𝑎𝜖0
where 𝜌0 and 𝑎 are constants.
a) Find the electric field intensity, 𝐄:
𝐄 = −∇𝑉 = −
]
[
𝜌
𝑑 𝜌0
−𝑎𝑥
−
𝑒
) 𝐚𝑥 = − 0 𝑒−𝑎𝑥 𝐚𝑥 V∕m
(1
𝑑𝑥 𝑎𝜖0
𝜖0
b) find the potential difference between the points 𝑥 = 𝑑 and 𝑥 = 0:
𝑉𝑑0 = 𝑉 (𝑑) − 𝑉 (0) =
)
𝜌0 (
1 − 𝑒−𝑎𝑑 V
𝑎𝜖0
c) if the medium permittivity is given by 𝜖(𝑥) = 𝜖0 𝑒𝑎𝑥 , find the electric flux density, 𝐃, and the
volume charge density, 𝜌𝑣 , in the region:
)
(
𝜌0 −𝑎𝑥
𝑎𝑥
𝐃 = 𝜖𝐄 = 𝜖0 𝑒
− 𝑒 𝐚𝑥 = −𝜌0 𝐚𝑥 C∕m2
𝜖0
Then 𝜌𝑣 = ∇ ⋅ 𝐃 = 0.
d) Find the stored energy in the region (0 < 𝑥 < 𝑑), (0 < 𝑦 < 1), (0 < 𝑧 < 1):
𝑑 𝜌2
1
1
−𝜌20 −𝑎𝑥 |𝑑
𝜌20 (
)
1
0 −𝑎𝑥
𝑊𝑒 =
𝑒 𝑑𝑥 𝑑𝑦 𝑑𝑧 =
𝐃 ⋅ 𝐄 𝑑𝑣 =
𝑒 | =
1 − 𝑒−𝑎𝑑 J
|
0
∫𝑣 2
∫0 ∫0 ∫0 2𝜖0
2𝜖0 𝑎
2𝜖0 𝑎
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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 𝜌𝑠𝑎 and 𝜌𝑠𝑏 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 𝑎1 − 1𝑏
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, 4𝜋𝑎2 𝜌𝑠𝑎 and 4𝜋𝑏2 𝜌𝑠𝑏 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 = 𝜖0 𝐸 2 =
2
2
𝑊𝐸 =
∫𝑣
𝑤𝐸 𝑑𝑣 = 4𝜋
∫𝑎
𝑏
𝜖0 𝑉02
3
(
)2 J∕m
1
1
2𝑟4 𝑎 − 𝑏
𝜖0 𝑉02
2𝜋𝜖0 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 ∫𝑠 𝜌𝑠 𝑑𝑎, use your part 𝑐 result to find the charge
density 𝜌𝑠𝑎 on the inner sphere and thereby determine the relationship between 𝜌𝑠𝑎 and 𝐄(𝑟 = 𝑎):
We have
𝑊𝐸 =
2𝜋𝜖0 𝑉02
1
𝜌𝑠𝑎 𝑉0 𝑑𝑎 = 2𝜋𝑎2 𝜌𝑠𝑎 𝑉0 = (
) (as found in part 𝑐)
1
1
2 ∫𝑖𝑛𝑛𝑒𝑟
−𝑏
𝑎
Therefore
𝜌𝑠𝑎 =
𝜖0 𝑉 0
|
(
) = 𝜖0 𝐄| ⋅ 𝐚𝑟 C∕m2
|𝑎
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.
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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 𝜌𝓁
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 (𝜌 < 𝑏) is given by
𝜌
𝑉 (𝜌) = 𝑘 − 𝓁 ln(𝜌)
2𝜋𝜖0
where 𝑘 is a constant.
a) Find 𝑘 in terms of given or known parameters: At radius 𝑏,
𝜌
𝜌
𝑉 (𝑏) = 𝑘 − 𝓁 ln(𝑏) = 0 ⇒ 𝑘 = 𝓁 ln(𝑏)
2𝜋𝜖0
2𝜋𝜖0
b) find the electric field strength, 𝐄, for 𝜌 < 𝑏:
]
[
𝜌𝓁
𝜌𝓁
𝜌𝓁
𝑑
ln(𝑏) −
ln(𝜌) 𝐚𝜌 =
𝐚 V∕m
𝐄𝑖𝑛 = −∇𝑉 = −
𝑑𝜌 2𝜋𝜖0
2𝜋𝜖0
2𝜋𝜖0 𝜌 𝜌
c) find the electric field strength, 𝐄, for 𝜌 > 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 𝜌 > 𝑎, where 𝑎 < 𝑏:
𝑏
2𝜋
1
( )
𝜌2𝓁
𝜌2𝓁
𝑏
1
ln
𝜌
𝑑𝜌
𝑑𝜙
𝑑𝑧
=
J
𝐃 ⋅ 𝐄 𝑑𝑣 =
𝑊𝑒 =
∫0 ∫0 ∫𝑎 8𝜋 2 𝜖0 𝜌2
∫𝑣 2
4𝜋𝜖0
𝑎
4.23. In free space, an electric potential is given in cylindrical coordinates by
𝑉 (𝜌) =
a) Find the electric field intensity, 𝐄:
𝑑
𝐄 = −∇𝑉 = −
𝑑𝜌
b) Find the volume charge density:
𝜌𝑣 = ∇ ⋅ 𝐃 = 𝜖0 ∇ ⋅ 𝐄 =
(
𝑎2 𝜌0 −𝜌∕𝑎
𝑒
𝜖0
)
𝑎𝜌
𝑎2 𝜌0 −𝜌∕𝑎
𝑒
𝐚𝜌 = 0 𝑒−𝜌∕𝑎 𝐚𝜌 V∕m
𝜖0
𝜖0
) 𝑎𝜌 (
𝜌 ) −𝜌∕𝑎
𝑎 𝑑 (
𝑒
C∕m3
𝜌𝜌0 𝑒−𝜌∕𝑎 = 0 1 −
𝜌 𝑑𝜌
𝜌
𝑎
c) Using Eq. (42), find the stored energy in the region (0 < 𝜌 < ∞), (0 < 𝜙 < 2𝜋), (0 < 𝑧 < 1):
Use
][
]
∞[
2𝜋
1
𝑎𝜌0 (
𝜌 ) −𝜌∕𝑎 𝑎2 𝜌0 −𝜌∕𝑎
1
1
1−
𝑒
𝑒
𝜌𝑑𝜌𝑑𝜙𝑑𝑧
𝑊𝐸 =
𝜌 𝑉 𝑑𝑣 =
2 ∫𝑣 𝑣
2 ∫0 ∫0 ∫0
𝜌
𝑎
𝜖0
[(
)
]∞
𝜋𝑎4 𝜌20
𝜋𝑎3 𝜌20 ∞ (
𝜋𝑎4 𝜌20
𝜌 ) −2𝜌∕𝑎
2𝜌 −2𝜌∕𝑎
1−
𝑒
𝑑𝜌 = −
1−
J
=
=
𝑒
𝜖0 ∫0
𝑎
4𝜖0
𝑎
4𝜖0
0
d) Repeat part 𝑐, but use Eq. (44): Here we have...
∞ 𝑎2 𝜌2
2𝜋
1
1
1
1
0 −2𝜌∕𝑎
𝑒
𝜌𝑑𝜌𝑑𝜙𝑑𝑧
𝐃 ⋅ 𝐄𝑑𝑣 =
𝜖0 𝐸 2 𝑑𝑣 =
2 ∫𝑣
2 ∫𝑣
2 ∫0 ∫0 ∫0
𝜖0
(
)
4 2
𝜋𝑎4 𝜌20
𝜋𝑎2 𝜌20 ∞ −2𝜌∕𝑎
2𝜌 −2𝜌∕𝑎 |∞ 𝜋𝑎 𝜌0
1+
J
𝜌𝑒
𝑑𝜌 = −
𝑒
=
=
|
|0
𝜖0 ∫0
4𝜖0
𝑎
4𝜖0
𝑊𝐸 =
as before.
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4.24. A certain spherically-symmetric charge configuration in free space produces an electric field given in
spherical coordinates by:
{
(𝜌0 𝑟2 )∕(100𝜖0 ) 𝐚𝑟 V∕m
(𝑟 ≤ 10)
𝐄(𝑟) =
2
(100𝜌0 )∕(𝜖0 𝑟 ) 𝐚𝑟 V∕m
(𝑟 ≥ 10)
where 𝜌0 is a constant.
a) Find the charge density as a function of position:
)
𝜌0 𝑟2
𝜌 𝑟
𝑟
= 0 C∕m3
100
25
)
(
100𝜌0
1 𝑑
𝜌𝑣 (𝑟 ≥ 10) = ∇ ⋅ (𝜖0 𝐄2 ) = 2
=0
𝑟2
𝑟 𝑑𝑟
𝑟2
1 𝑑
𝜌𝑣 (𝑟 ≤ 10) = ∇ ⋅ (𝜖0 𝐄1 ) = 2
𝑟 𝑑𝑟
(
2
b) find the absolute potential as a function of position in the two regions, 𝑟 ≤ 10 and 𝑟 ≥ 10:
𝑉 (𝑟 ≤ 10) = −
10
∫∞
100𝜌0
𝜖0 𝑟2
100𝜌0 |10 𝜌0 (𝑟′ )3 |𝑟
=
| −
|
𝜖0 𝑟 |∞
300𝜖0 |10
𝑉 (𝑟 ≥ 10) = −
𝑟
100𝜌0
∫ ∞ 𝜖0
(𝑟′ )2
𝜌0 (𝑟′ )2
𝐚 ⋅ 𝐚 𝑑𝑟′
∫10 100𝜖0 𝑟 𝑟
]
10𝜌0 [
4 − (10−3 𝑟3 ) V
=
3𝜖0
𝑟
𝐚𝑟 ⋅ 𝐚𝑟 𝑑𝑟 −
𝐚𝑟 ⋅ 𝐚𝑟 𝑑𝑟′ =
100𝜌0
100𝜌0 |𝑟
V
| =
′
𝜖0 𝑟 |∞
𝜖0 𝑟
c) check your result of part 𝑏 by using the gradient:
[
]
]
𝜌 𝑟2
10𝜌0 2
𝑑 10𝜌0 [
−3 3
(3𝑟 )(10−3 ) 𝐚𝑟 = 0 𝐚𝑟
𝐄1 = −∇𝑉 (𝑟 ≤ 10) = −
4 − (10 𝑟 ) 𝐚𝑟 =
𝑑𝑟 3𝜖0
3𝜖0
100𝜖0
[
]
100𝜌0
𝑑 100𝜌0
𝐚𝑟
𝐚𝑟 =
𝐄2 = −∇𝑉 (𝑟 ≥ 10) = −
𝑑𝑟 𝜖0 𝑟
𝜖0 𝑟2
d) find the stored energy in the charge by an integral of the form of Eq. (42):
]
[
10
𝜋
2𝜋
] 2
1
1 𝜌0 𝑟 10𝜌0 [
−3 3
4 − (10 𝑟 ) 𝑟 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜙
𝜌 𝑉 𝑑𝑣 =
𝑊𝑒 =
∫0 ∫0 ∫0 2 25
2 ∫𝑣 𝑣
3𝜖0
[
]
]10
10 [
4𝜋𝜌20
𝜌2
4𝜋𝜌20
𝑟7
𝑟6
4
3 0
3
=
10𝑟 −
= 7.18 × 10
40𝑟 −
𝑑𝑟 =
150𝜖0 ∫0
100
150𝜖0
700 0
𝜖0
e) Find the stored energy in the field by an integral of the form of Eq. (44).
𝑊𝑒 =
1
1
𝐃 ⋅ 𝐄 𝑑𝑣 +
𝐃 ⋅ 𝐄 𝑑𝑣
∫(𝑟≥10) 2 2 2
∫(𝑟≤10) 2 1 1
𝜌20 𝑟4
∞ 104 𝜌2
𝜋
2𝜋
0 2
2
𝑟 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜙
𝑟
sin
𝜃
𝑑𝑟
𝑑𝜃
𝑑𝜙
+
∫0 ∫0 ∫0 (2 × 104 )𝜖0
∫0 ∫0 ∫10 2𝜖0 𝑟4
[
]
10
∞
]
2𝜋𝜌20 [ 1 3
𝜌2
2𝜋𝜌20
𝑑𝑟
3
3 0
10−4
𝑟6 𝑑𝑟 + 104
(10
)
+
10
=
=
7.18
×
10
=
∫0
∫10 𝑟2
𝜖0
𝜖0 7
𝜖0
2𝜋
=
𝜋
10
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4.25. Consider an electric field intensity in free space that exhibits a gaussian function of radius in spherical
coordinates:
𝜌 𝑎 2 2
𝐄 = 0 𝑒−𝑟 ∕𝑎 𝐚𝑟 V∕m
𝜖0
where 𝜌0 and 𝑎 are constants, and where the field exists everywhere.
a) What charge density would produce this field?
)
) 2𝜌 𝑎 (
(
2 2
𝑟2
1 𝑑
0
2 −𝑟2 ∕𝑎2
=
𝜌0 𝑎𝑟 𝑒
1 − 2 𝑒−𝑟 ∕𝑎 C∕m3
𝜌𝑣 = ∇ ⋅ 𝐃 = 𝜖0 ∇ ⋅ 𝐄 = 2
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 =
∮𝑟→∞
𝐃 ⋅ 𝑑𝐒 = lim 4𝜋𝑟2 𝜌0 𝑎𝑒−𝑟
2 ∕𝑎2
𝑟→∞
=0
The Hard Way: Integrate the charge density found in part 𝑎 over the infinite volume:
)
(
∞
2𝜌0 𝑎
2 2
𝑟2
𝑄𝑒𝑛𝑐𝑙 = 𝜌𝑣 𝑑𝑣 = 4𝜋
1 − 2 𝑒−𝑟 ∕𝑎 𝑟2 𝑑𝑟
∫𝑣
∫0
𝑟
𝑎
⎡
⎤
⎢
⎥
∞ 3
⎢ ∞ −𝑟2 ∕𝑎2
𝑟 −𝑟2 ∕𝑎2 ⎥
= 8𝜋𝜌0 𝑎 ⎢
𝑒
𝑑𝑟⎥ = 0
𝑟𝑒
𝑑𝑟 −
∫0 𝑎 2
0
⎢∫
⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟⎥⎥
⎢⏟⏞⏞⏞⏞⏞⏞⏞⏟
⎣
⎦
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
∞
𝜌0 𝑎3 𝜋
𝜌0 𝑎 ∞ −𝑟2 ∕𝑎2
𝑉0 = −
V
𝐄 ⋅ 𝑑𝐋 = +
𝐄 ⋅ 𝑑𝐋 =
𝑒
𝑑𝑟 =
∫∞
∫0
𝜖0 ∫0
2𝜖0
d) What is the net stored energy in the field? Integrate the energy density over an infinite spherical
volume:
√
∞ 𝜌2 𝑎2
𝜋𝑎5 𝜌20 𝜋
1 2
1
0
−2𝑟2 ∕𝑎2 2
𝑒
𝑟 𝑑𝑟 =
𝐃 ⋅ 𝐄𝑑𝑣 =
𝐸 𝑑𝑣 = 4𝜋
J
𝑊𝐸 =
∫𝑣 2
∫0 2𝜖0
∫𝑣 2
4𝜖0
2
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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 = 𝜋∕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, 𝜋∕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
=−
⇒ 𝑥 = − tan 𝑦 𝑑𝑦 + 𝐶 = ln(cos 𝑦) + 𝐶
∫
𝑑𝑥
sin 𝑦
At the entry point (0, 𝜋∕3), we have 0 = ln[cos(𝜋∕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, 𝜃). A two-segment path is chosen, starting from point 𝐵(𝑟 = ∞, 𝜃 = 𝜋∕2), and integrating
along 𝑟 from ∞ to 𝑟, holding 𝜃 fixed at 𝜋∕2; then integrating along the second segment from 𝜋∕2 to
𝜃, holding radius at 𝑟. This ends at point 𝐴(𝑟, 𝜃).
[ 𝑟
]
𝜃
𝐴
2 cos 𝜃|𝜋∕2
𝑞𝑑
sin 𝜃 ′
′
′
𝐚𝑟 ⋅ 𝐚𝑟 𝑑𝑟 +
𝐚 ⋅ 𝐚 𝑟𝑑𝜃
𝑉 (𝑟, 𝜃) = −
𝐄 ⋅ 𝑑𝐋 = −
∫𝜋∕2 𝑟3 𝜃 𝜃
∫𝐵
4𝜋𝜖0 ∫∞
(𝑟′ )3
The first integral term is zero (cos(𝜋∕2)), so this leaves the second term, which evaluates as
𝑉 (𝑟, 𝜃) =
𝑞𝑑 cos 𝜃
𝑞𝑑
|𝜃
cos 𝜃 ′ |
Q.E.D.
=
2
|𝜋∕2
4𝜋𝜖0 𝑟
4𝜋𝜖0 𝑟2
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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 𝜃𝑎 and 𝜃𝑏 , each point having the same 𝑟 and 𝜙 coordinates. Under what conditions
does the answer agree with Eq. (34), for the potential at 𝜃𝑎 ?
We perform a line integral of Eq. (36) along an arc of constant 𝑟 and 𝜙:
𝜃𝑎
]
𝑞𝑑 [
𝑞𝑑
2
cos
𝜃
𝐚
+
sin
𝜃
𝐚
⋅
𝐚
𝑟
𝑑𝜃
=
−
sin 𝜃 𝑑𝜃
𝑟
𝜃
𝜃
∫𝜃𝑏 4𝜋𝜖0 𝑟3
∫𝜃𝑏 4𝜋𝜖0 𝑟2
]
𝑞𝑑 [
cos 𝜃𝑎 − cos 𝜃𝑏
=
2
4𝜋𝜖0 𝑟
𝜃𝑎
𝑉𝑎𝑏 = −
This result agrees with Eq. (34) if 𝜃𝑎 (the ending point in the path) is 90◦ (the 𝑥𝑦 plane). Under
this condition, we note that if 𝜃𝑏 > 90◦ , positive work is done when moving (against the field) to
the 𝑥𝑦 plane; if 𝜃𝑏 < 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 ′ )
4𝜋𝜖0 |𝐫 − 𝐫 ′ |3
where 𝐫 − 𝐫 ′ = 𝑃 − 𝑄 = (1, 1, 8). So
𝑉𝑃 =
(3𝐚𝑥 − 5𝐚𝑦 + 10𝐚𝑧 ) ⋅ (𝐚𝑥 + 𝐚𝑦 + 8𝐚𝑧 ) × 10−9
4𝜋𝜖0 [12 + 12 + 82 ]1.5
= 1.31 V
4.30. A dipole for which 𝐩 = 10𝜖0 𝐚𝑧 C ⋅ m is located at the origin. What is the equation of the surface on
which 𝐸𝑧 = 0 but 𝐄 ≠ 0?
First we find the 𝑧 component:
]
]
5 [
10 [
2
2
2
cos
𝜃
(𝐚
⋅
𝐚
)
+
sin
𝜃
(𝐚
⋅
𝐚
)
=
2
cos
𝜃
−
sin
𝜃
𝑟
𝑧
𝜃
𝑧
4𝜋𝑟3
2𝜋𝑟3
[
]
This will be zero when 2 cos2 𝜃 − sin2 𝜃 = 0. Using identities, we write
𝐸𝑧 = 𝐄 ⋅ 𝐚𝑧 =
1
2 cos2 𝜃 − sin2 𝜃 = [1 + 3 cos(2𝜃)]
2
The above becomes zero on the cone surfaces, 𝜃 = 54.7◦ and 𝜃 = 125.3◦ .
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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)𝜖0 𝐄 ⋅ 𝐄, and where
]
[
1
1
1
𝐚𝑧 V∕m
𝐄 = −∇𝑉 = 20 2 𝐚𝑥 + 2 𝐚𝑦 +
𝑥 𝑦𝑧
𝑥𝑦 𝑧
𝑥𝑦𝑧2
The energy is now
2
𝑊𝐸 = 200𝜖0
2
∫1 ∫1 ∫1
2[
]
1
1
1
+
+
𝑑𝑥 𝑑𝑦 𝑑𝑧
𝑥4 𝑦2 𝑧2 𝑥2 𝑦4 𝑧2 𝑥2 𝑦2 𝑧4
The integral evaluates as follows:
]2
( )
1
1
1
1
𝑑𝑦 𝑑𝑧
−
−
𝑊𝐸 = 200𝜖0
−
∫1 ∫1
3 𝑥3 𝑦2 𝑧2 𝑥𝑦4 𝑧2 𝑥𝑦2 𝑧4 1
]
2
2 [(
( )
( )
)
7
1
1
1
1
1
= 200𝜖0
+
+
𝑑𝑦 𝑑𝑧
∫1 ∫1
24 𝑦2 𝑧2
2 𝑦4 𝑧 2
2 𝑦2 𝑧 4
]2
2[ (
( )
( )
)
1
1 1
1
1
7
−
𝑑𝑧
−
−
= 200𝜖0
∫1
24 𝑦𝑧2
6 𝑦3 𝑧2
2 𝑦𝑧4 1
2 [(
( )
( ) ]
)
7 1
7 1
1 1
= 200𝜖0
𝑑𝑧
+
+
2
2
∫1
48 𝑧
48 𝑧
4 𝑧4
[ ]
7
= 200𝜖0 (3)
= 387 pJ
96
2
2[
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
]
[
1
= 2.07 × 10−10 J∕m3
𝑤𝐸 = 200𝜖0 (3)
(1.5)4 (1.5)2 (1.5)2
This, multiplied by a cube volume of 1, produces an energy value of 207 pJ.
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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, 𝜃) =
]
𝑞𝑑 [
2
cos
𝜃
𝐚
+
sin
𝜃
𝐚
𝑟
𝜃
4𝜋𝜖0 𝑟3
Then the energy will be
𝑊𝑒 =
∫𝑣𝑜𝑙
1
𝜖 𝐄 ⋅ 𝐄 𝑑𝑣 =
∫0
2 0
2𝜋
𝜋
∫0 ∫𝑎
∞
3 cos2 𝜃+1
(𝑞𝑑)2 [
]𝜋
[
]
−2𝜋(𝑞𝑑)2 1 |∞
3
2
−
cos
𝜃
−
cos
𝜃
3
cos
𝜃
+
1
sin
𝜃
𝑑𝜃
=
|
32𝜋 2 𝜖0 3𝑟3 |𝑎 ∫0
48𝜋 2 𝜖0 𝑎3 ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟0
𝜋
=
]
(𝑞𝑑)2 [
4 cos2 𝜃 + sin2 𝜃 𝑟2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜙
2
6
32𝜋 𝜖0 𝑟 ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
4
=
(𝑞𝑑)2
12𝜋𝜖0 𝑎3
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 → 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 𝜇C 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 × 10−6
𝐚
=
𝐚𝑟 C∕m2
𝑟
4𝜋𝑟2
4𝜋𝑟2
b) Calculate the total energy stored in the electrostatic field: Use
∞
𝜋
2𝜋
1 (5 × 10−6 )2 2
1
𝐃 ⋅ 𝐄 𝑑𝑣 =
𝑟 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜙
∫𝑣𝑜𝑙 2
∫0 ∫0 ∫.04 2 16𝜋 2 𝜖0 𝑟4
( ) (5 × 10−6 )2 ∞
1
𝑑𝑟 25 × 10−12 1
= 2.81 J
= (4𝜋)
=
2
2
8𝜋𝜖0 .04
16𝜋 𝜖0 ∫.04 𝑟2
𝑊𝐸 =
c) Use 𝑊𝐸 = 𝑄2 ∕(2𝐶) to calculate the capacitance of the isolated sphere: We have
𝐶=
(5 × 10−6 )2
𝑄2
=
= 4.45 × 10−12 F = 4.45 pF
2𝑊𝐸
2(2.81)
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4.34. A sphere of radius 𝑎 contains volume charge of uniform density 𝜌0 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 =
𝜌 𝑎3
𝜌0 𝑟
𝐚𝑟 𝑟 ≤ 𝑎 and 𝐄2 = 0 2 𝐚𝑟 𝑟 ≥ 𝑎
3𝜖0
3𝜖0 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:
𝑎
𝑉 (𝑟) = −
∫∞
𝐄2 ⋅ 𝐚𝑟 𝑑𝑟 −
∫𝑎
𝑟
𝑎
𝐄1 ⋅ 𝐚𝑟 𝑑𝑟′ = −
𝜌0 𝑎3
∫∞ 3𝜖0 𝑟2
𝑑𝑟 −
∫𝑎
𝑟
)
𝜌 (
𝜌0 𝑟′ ′
𝑑𝑟 = 0 3𝑎2 − 𝑟2
3𝜖0
6𝜖0
Now, using this result in (43) leads to the energy associated with the charge in the sphere:
𝑊𝑒 =
1
2 ∫0
2𝜋
𝜋
∫0 ∫0
𝑎
𝑎(
𝜌20 ( 2
4𝜋𝑎5 𝜌20
)
)
𝜋𝜌
3𝑎 − 𝑟2 𝑟2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜙 = 0
3𝑎2 𝑟2 − 𝑟4 𝑑𝑟 =
6𝜖0
3𝜖0 ∫0
15𝜖0
b) Eq. (45): Using the given fields we find the energy densities
𝜌2 𝑟2
𝜌2 𝑎6
1
1
𝑟 ≤ 𝑎 and 𝑤𝑒2 = 𝜖0 𝐄2 ⋅ 𝐄2 = 0 4 𝑟 ≥ 𝑎
𝑤𝑒1 = 𝜖0 𝐄1 ⋅ 𝐄1 = 0
2
18𝜖0
2
18𝜖0 𝑟
We now integrate these over their respective volumes to find the total energy:
𝑊𝑒 =
∫0
2𝜋
𝜋
𝑎
𝜌20 𝑟2
∫0 ∫0 18𝜖0
𝑟2 sin 𝜃 𝑑𝑟 𝑑𝜃 𝑑𝜙 +
∫0
2𝜋
𝜋
∫0 ∫𝑎
∞
𝜌20 𝑎6
4𝜋𝑎5 𝜌20
2
𝑟
sin
𝜃
𝑑𝑟
𝑑𝜃
𝑑𝜙
=
15𝜖0
18𝜖0 𝑟4
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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
1∑
𝑞 𝑉
𝑊𝐸 =
2 𝑛=1 𝑛 𝑛
4
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)
[
]
𝑞
1
1
1
+
+
𝑉1 = 𝑉21 + 𝑉31 + 𝑉41 =
√
4𝜋𝜖0 .04 .04 .04 2
Taking the summation produces a factor of 4, since the situation is the same at all four points.
Consequently,
[
]
−9 )2
(.8
×
10
1
1
2+ √
𝑊𝐸 = (4)𝑞1 𝑉1 =
= 7.79 × 10−7 J = 0.779 𝜇J
2
2𝜋𝜖0 (.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
Δ𝑊𝐸 =
4(.8 × 10−9 )2
= .813 𝜇J
√
4𝜋𝜖0 (.04 2∕2)
The total energy is now
𝑊𝐸 𝑛𝑒𝑡 = 𝑊𝐸 (part a) + Δ𝑊𝐸 = .779 + .813 = 1.59 𝜇J
4.36 Surface charge of uniform density 𝜌𝑠 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 𝜌𝑠
𝐚𝑟 V∕m
𝐄=
𝜖0 𝑟2
Then
𝑟
𝑟
𝑏2 𝜌𝑠
𝑏2 𝜌𝑠
′
𝑑𝑟
=
V
𝑉 (𝑟) = −
𝐄 ⋅ 𝑑𝐋 = −
∫∞
∫∞ 𝜖0 (𝑟′ )2
𝜖0 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
1
𝜌𝑠 𝑉 (𝑏) 𝑑𝑎 =
2 ∫𝑆
2 ∫0
2𝜋
∫0
𝜋
𝜌𝑠
2𝜋𝜌2𝑠 𝑏3
𝑏2 𝜌𝑠 2
𝑏 sin 𝜃 𝑑𝜃 𝑑𝜙 =
𝜖0 𝑏
𝜖0
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 =
∫0
∫𝑣 2
2𝜋
𝜋
∫0 ∫𝑏
∞
4 2
2𝜋𝜌2𝑠 𝑏3
1 𝑏 𝜌𝑠 2
𝑟
sin
𝜃
𝑑𝑟
𝑑𝜃
𝑑𝜙
=
2 𝜖0 𝑟4
𝜖0
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Sm ch - i hope you have a great grade
Engineering Electromagnetics 9th solution
Engineering Electromagnetics (Chungnam National University)
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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
2
1
|
|
𝐉 ⋅ 𝐧| 𝑑𝑎 =
𝐉 ⋅ 𝐚𝑦 | 𝑑𝑥 𝑑𝑧 =
−104 cos(2𝑥)𝑒−2 𝑑𝑥 𝑑𝑧
|𝑦=1
|𝑆
∫ ∫𝑆
∫0 ∫0
∫0 ∫0
1
|1
= −104 (2) sin(2𝑥)| 𝑒−2 = −1.23 MA
|0
2
𝐼=
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
3
1
1
3
3
1
|
|
|
𝐉 ⋅ (𝐚𝑥 )| 𝑑𝑦 𝑑𝑧
𝐉 ⋅ (𝐚𝑦 )| 𝑑𝑥 𝑑𝑧 +
𝐉 ⋅ (−𝐚𝑦 )| 𝑑𝑥 𝑑𝑧 +
|𝑥=1
|𝑦=1
|𝑦=0
∫2 ∫0
∫2 ∫0
∫2 ∫0
1
1[
3
3
]
sin(2)𝑒−2𝑦 𝑑𝑦 𝑑𝑧
cos(2𝑥)𝑒−0 − cos(2𝑥)𝑒−2 𝑑𝑥 𝑑𝑧 − 104
= 104
∫2 ∫0
∫2 ∫0
( )
( )
[
]
1
1
|1
|1
= 104
sin(2𝑥)| (3 − 2) 1 − 𝑒−2 + 104
sin(2)𝑒−2𝑦 | (3 − 2) = 0
|
|0
0
2
2
𝐼=
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=
∫
1
𝐉 ⋅ 𝑑𝐒 =
∫0 ∫0
2
−104 𝑥 𝐚𝑦 ⋅ (−𝐚𝑦 ) 𝑑𝑥 𝑑𝑧 = 2 × 10−4 A
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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 Ω rad/s. What total current is flowing? Begin with
𝐉 = 𝜌𝑣 𝐯 where 𝐯 is the velocity of any point in the sphere, and is given by 𝐯 = Ω𝑟 sin 𝜃 𝐚𝜙 m/s. Then
the charge density
𝜌𝑣 =
𝑄
𝑄
𝑄
=
⇒ 𝐉=
Ω𝑟 sin 𝜃 𝐚𝜙 A∕m2
3
𝑣𝑜𝑙 (4∕3)𝜋𝑏3
(4∕3)𝜋𝑏
Now the total current is the integral of 𝐉 over the half disk, extending from 0 to 𝜋 in 𝜃, and from 0 to
𝑏 in 𝑟:
𝑏
𝜋
𝑄Ω
3𝑄Ω𝑟 sin 𝜃
A
𝑟𝑑𝑟𝑑𝜃 =
𝐼=
3
∫0 ∫0
2𝜋
4𝜋𝑏
Note error in Appendix F answer.
5.4. If volume charge density is given as 𝜌𝑣 = (cos 𝜔𝑡)∕𝑟2 C∕m3 in spherical coordinates, find 𝐉. It is
reasonable to assume that 𝐉 is not a function of 𝜃 or 𝜙.
We use the continuity equation (5), along with the assumption of no angular variation to write
∇⋅𝐉=
(
)
𝜕𝜌𝑣
1 𝜕 ( 2 )
𝜕 cos 𝜔𝑡
𝜔 sin 𝜔𝑡
=
=
−
𝑟
𝐽
=
−
𝑟
2
2
𝜕𝑡
𝜕𝑡
𝑟 𝜕𝑟
𝑟
𝑟2
So we may now solve
𝜕 ( 2 )
𝑟 𝐽𝑟 = 𝜔 sin 𝜔𝑡
𝜕𝑟
by direct integration to obtain:
𝐉 = 𝐽𝑟 𝐚 𝑟 =
𝜔 sin 𝜔𝑡
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.
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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(𝜔𝑡 − 𝛽𝑧) 𝐚𝑧 A∕m2
where 𝐽0 is a constant amplitude, 𝜔 = 2𝜋𝑓 is the radian frequency, and 𝛽 is a constant to be found.
Determine 𝛽 as a function of 𝜔 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
𝜌𝑣 = −
∫
𝜕𝜌
𝑑𝐽𝑧
= 𝛽𝐽0 sin(𝜔𝑡 − 𝛽𝑧) = − 𝑣
𝑑𝑧
𝜕𝑡
𝛽𝐽0 sin(𝜔𝑡 − 𝛽𝑧)𝑑𝑡 + 𝐶 =
𝛽
𝐽 cos(𝜔𝑡 − 𝛽𝑧)
𝜔 0
where the integration constant is set to zero. Now, from Eq. (3):
𝐉 = 𝜌𝑣 𝐯 ⇒ 𝑣 =
𝜔
𝜔
⇒ 𝛽=
𝛽
𝑣
5.6. In spherical coordinates, a current density 𝐉 = −𝑘∕(𝑟 sin 𝜃) 𝐚𝜃 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:
𝑅
2𝜋
−𝑘
𝐼 = 𝐉 ⋅ 𝑑𝐒 =
𝐚𝜃 ⋅ 𝐚𝑧 𝜌 𝑑𝜌 𝑑𝜙
∫𝑠
∫0 ∫0 𝑟 sin 𝜃 ⏟⏟
⏟
− sin 𝜃
√
Then, using 𝑟 = 𝜌2 + 𝑧2 , this becomes
]
[√
√
|𝑅
𝑅2 + 𝑧 2 − 𝑧
= 2𝜋𝑘 𝜌2 + 𝑧2 | = 2𝜋𝑘
|0
∫0 ∫0
𝜌2 + 𝑧2
[√
]
At 𝑧 = 0 (part 𝑎), we have 𝐼(0) = 2𝜋𝑘𝑅 , and at 𝑧 = ℎ (part 𝑏): 𝐼(ℎ) = 2𝜋𝑘
𝑅2 + ℎ2 − ℎ .
2𝜋
𝐼=
𝑅
√
𝑘𝜌
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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 𝜌𝑣 ?
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:
𝜕𝜌𝑚
𝑑𝑣 = −
∇ ⋅ 𝐉𝑚 𝑑𝑣 = −
𝐉 ⋅ 𝐝𝐒
∫𝑣 𝜕𝑡
∫𝑣
∮𝑠 𝑚
where
∮𝑠
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
𝜕𝜌𝑚 . 0.550 × 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 × 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 𝜃𝑐 from the 𝑧 axis (in spherical coordinates). Therefore, we may write
𝓁=
2 mm
0.1mm
and tan 𝜃𝑐 =
tan 𝜃𝑐
160 + 𝓁
Solving these, we find 𝓁 = 8.4 mm, tan 𝜃𝑐 = 1.19 × 10−2 , and so 𝜃𝑐 = 0.68◦ , which gives us
a very thin cone! With this understanding, we can assume that the current density is uniform
with 𝜃 and 𝜙 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 (𝜌). This is our primary assumption.
Now, assuming constant current density at constant 𝑟, and net current, 𝐼, we may write
𝐼=
∫0
2𝜋
∫0
𝜃𝑐
𝐽 (𝑟) 𝐚𝑟 ⋅ 𝐚𝑟 𝑟2 sin 𝜃 𝑑𝜃 𝑑𝜙 = 2𝜋𝑟2 𝐽 (𝑟)(1 − cos 𝜃𝑐 )
or
𝐉(𝑟) =
(1.42 × 104 )𝐼
𝐼
𝐚
=
𝐚𝑟
𝑟
2𝜋𝑟2 (1 − cos 𝜃𝑐 )
2𝜋𝑟2
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5.8 (continued) The electric field is now
𝐄(𝑟) =
𝐉(𝑟)
(1.42 × 104 )𝐼
𝐼
𝐚𝑟 = (7.1 × 10−3 )
𝐚𝑟 V∕m
=
𝜎
2𝜋𝑟2 (2 × 106 )
2𝜋𝑟2
The voltage between the ends is now
𝑟𝑖𝑛
𝑉0 = −
∫𝑟𝑜𝑢𝑡
𝐄 ⋅ 𝐚𝑟 𝑑𝑟
.
.
where 𝑟𝑖𝑛 = 𝓁∕ cos 𝜃𝑐 = 𝓁 and where 𝑟𝑜𝑢𝑡 = (160 + 𝓁)∕ cos 𝜃𝑐 = 160 + 𝓁. The voltage is
𝓁×10−3
∫(160+𝓁)×10−3
0.40
𝐼
=
𝜋
𝑉0 = −
(7.1 × 10−3 )
[
]
1
1
𝐼
−3 𝐼
𝐚
⋅
𝐚
𝑑𝑟
=
(7.1
×
10
)
−
𝑟
𝑟
2𝜋 .0084 .1684
2𝜋𝑟2
from which we identify the resistance as 𝑅 = 0.40∕𝜋 = 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, 𝜌. Assuming each plate is of differential thickness, 𝑑𝑧, the differential
resistance of a plate will be
𝑑𝑧
𝑑𝑅 =
𝜎𝜋𝜌2
where 𝜌 = 𝑧 tan 𝜃𝑐 = 𝑧(1.19 × 10−2 ). The cone resistance will be the resistance of the stack of
plates (in series), found through
(160+𝓁)×10−3
(160+𝓁)×10−3
𝑑𝑧
=
∫𝓁×10−3
∫
∫𝓁×10−3
𝜎𝜋𝜌2
[
]
−3
3.53 × 10
1
1
=
= 0.127 ohms
−
𝜋
.0084 .1684
𝑅=
5.9.
𝑑𝑅 =
𝑑𝑧
(2 × 106 )𝜋𝑧2 (1.19 × 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=
𝑉2
𝑙
=
𝑃
𝜎(𝜋𝑎2 )
Thus the diameter will be
√
𝑑 = 2𝑎 = 2
√
𝑙𝑃
=2
𝜎𝜋𝑉 2
2(450)
= 2.8 × 10−4 m = 0.28 mm
(106 )𝜋(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
7
2
𝐽= (
)2 = 6.0 × 10 A∕m
−4
𝜋 2.8 × 10 ∕2
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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 𝜎 = 1.5 × 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∕𝜌) 𝐚𝜙 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 𝜌 inside the washer, between the two cut faces:
𝑉0 = −
∫
𝐄 ⋅ 𝑑𝐋 = −
∫𝜋
0
0.5
𝐚 ⋅ 𝐚 𝜌 𝑑𝜙 = 0.5𝜋 V
𝜌 𝜙 𝜙
b) What total current is flowing? First, the current density is 𝐉 = 𝜎𝐄, so
𝐉=
1.5 × 107 (0.5)
7.5 × 106
𝐚𝜙 =
𝐚𝜙 A∕m2
𝜌
𝜌
Current is then found by integrating 𝐉 over any transverse plane in the washer (the rectangular
cross-section):
∫𝑠
∫0
0.5×10−2
2.5×10−2
7.5 × 106
𝐚𝜙 ⋅ 𝐚𝜙 𝑑𝜌 𝑑𝑧
𝜌
∫10−2
( )
2.5
= 0.5 × 10−2 (7.5 × 106 ) ln
= 3.4 × 104 A
1
𝐼=
𝐉 ⋅ 𝑑𝐒 =
c) What is the resistance between the two faces?
𝑅=
𝑉0
0.5𝜋
=
= 4.6 × 10−5 ohms
𝐼
3.4 × 104
5.11. Two perfectly-conducting cylindrical surfaces of length 𝑙 are located at 𝜌 = 3 and 𝜌 = 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 𝜎 = 0.05 S∕m is present for 3 < 𝜌 < 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 𝜌 and length 𝑙:
𝐉=
3
𝐚 A∕m2
2𝜋𝜌𝑙 𝜌
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5.11
a) (continued)
Then the electric field is found by dividing this result by 𝜎:
𝐄=
9.55
3
𝐚𝜌 =
𝐚 V∕m
2𝜋𝜎𝜌𝑙
𝜌𝑙 𝜌
The voltage between cylinders is now:
𝑉 =−
∫5
3
𝐄 ⋅ 𝐝𝐋 =
∫3
5
Now, the resistance will be
𝑅=
( )
9.55
5
9.55
4.88
=
𝐚𝜌 ⋅ 𝐚𝜌 𝑑𝜌 =
ln
V
𝜌𝑙
𝑙
3
𝑙
𝑉
4.88 1.63
=
=
Ω
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
𝑙
𝐄 ⋅ 𝐉 𝑑𝑣 =
∫0 ∫0
2𝜋
.05
∫.03
( )
32
14.64
5
32
=
ln
W
𝜌
𝑑𝜌
𝑑𝜙
𝑑𝑧
=
2
2
2
2𝜋(.05)𝑙
3
𝑙
(2𝜋) 𝜌 (.05)𝑙
We also find the power by taking the product of voltage and current:
𝑃 =𝑉𝐼 =
14.64
4.88
(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, 𝜎(𝑧) = 𝜎0 𝑒−𝑧∕𝑑 , where 𝜎0 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
𝑑
𝑑𝑅 =
𝑒𝑧∕𝑑 𝑑𝑧
𝑑𝑧
𝑒𝑧∕𝑑 𝑑𝑧
𝑑
1.72𝑑
=
so that 𝑅 =
𝑑𝑅 =
=
(𝑒 − 1) =
Ω
∫
∫ 0 𝜎0 𝐴
𝜎𝐴
𝜎0 𝐴
𝜎0 𝐴
𝜎0 𝐴
b) the total current flowing between plates: We use
𝐼=
𝜎 𝐴𝑉
𝑉0
= 0 0
𝑅
1.72 𝑑
c) the electric field intensity 𝐄 within the material: First the current density is
𝐉=−
−𝜎0 𝑉0
−𝑉0 𝑒𝑧∕𝑑
𝐉
𝐼
𝐚𝑧 =
𝐚𝑧 so that 𝐄 =
=
𝐚 V∕m
𝐴
1.72 𝑑
𝜎(𝑧)
1.72 𝑑 𝑧
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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 𝜎 = 1.5 × 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 ×
107 )
1
[
]
−4
(2.54)(2.54∕2) × 10 − 2.54(1 − .1)(2.54∕2)(1 − .2) × 10−4
= 7.38 × 10−4 Ω
The voltage drop is now 𝑉 = 𝐼𝑅1 = 200(7.38 × 10−4 ) = 0.147 V.
b) Find the voltage drop if the interior of the tube is filled with a conducting material for which
𝜎 = 1.5 × 105 S/m: The resistance of the filling will be:
𝑅2 =
1
(1.5 ×
105 )(1∕2)(2.54)2
× 10−4 (.9)(.8)
= 2.87 × 10−2 Ω
The total resistance is now the parallel combination of 𝑅1 and 𝑅2 :
𝑅𝑇 = 𝑅1 𝑅2 ∕(𝑅1 + 𝑅2 ) = 7.19 × 10−4 Ω, 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 𝜎(𝑥) = 𝜎0 𝑒−𝑥∕𝑎 , where 𝜎0 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 = 𝜎(𝑥)𝐄 =
−𝜎0 𝑒−𝑥∕𝑎 𝑉0
𝐚𝑧
𝑑
Using this, we find
𝐼=
∫
𝑏
𝐉 ⋅ 𝑑𝐒 =
∫0 ∫0
𝑎
𝜎 𝑎𝑏𝑉0
0.63𝑎𝑏𝜎0 𝑉0
−𝜎0 𝑒−𝑥∕𝑎 𝑉0
𝐚𝑧 ⋅ (−𝐚𝑧 ) 𝑑𝑥 𝑑𝑦 = 0
(1 − 𝑒−1 ) =
A
𝑑
𝑑
𝑑
c) the resistance of the material: We use
𝑅=
𝑉0
𝑑
Ω
=
𝐼
0.63 𝑎𝑏 𝜎0
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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 𝜎(𝑟) = 𝜎0 𝑎∕𝑟 S/m, where 𝜎0 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 ∕(2𝜋𝑟) 𝐚𝑟 A∕m2
b) Find 𝐄 between surfaces in terms of 𝐼0 :
𝐼
𝐉
𝐄= = 0
𝜎
2𝜋𝑟
(
𝑟
𝑎𝜎0
)
𝐚𝑟 =
𝐼0
𝐚 V∕m
2𝜋𝑎𝜎0 𝑟 𝑟
c) Find 𝑉0 in terms of 𝐼0 :
𝑉0 = −
∫𝑏
𝑎
𝐄 ⋅ 𝑑𝐋 = −
d) Find the resistance, 𝑅:
𝑅=
∫𝑏
𝑎
𝐼0
𝐼0
ln(𝑏∕𝑎) V
𝑑𝑟 =
2𝜋𝑎𝜎0 𝑟
2𝜋𝑎𝜎0
𝑉0
ln(𝑏∕𝑎)
=
ohms
𝐼0
2𝜋𝑎𝜎0
5.16. A coaxial transmission line has inner and outer conductor radii 𝑎 and 𝑏. Between conductors (𝑎 < 𝜌 <
𝑏) lies a conductive medium whose conductivity is 𝜎(𝜌) = 𝜎0 ∕𝜌, where 𝜎0 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
2𝜋𝜌(1) 𝜌
b) Determine the electric field intensity 𝐄 in terms of 𝐼 and other parameters, given or known:
𝐄=
𝐼𝜌
𝐼
𝐉
𝐚 V∕m
=
𝐚 =
𝜎
2𝜋𝜎0 𝜌 𝜌 2𝜋𝜎0 𝜌
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
𝐚 ⋅ 𝐚 𝑑𝜌 =
V
𝑉0 = −
𝐄 ⋅ 𝑑𝐋 = −
∫𝑏
∫𝑏 2𝜋𝜎0 𝜌 𝜌
2𝜋𝜎0
d) find an expression for the conductance of the line per unit length, 𝐺:
𝐺=
2𝜋𝜎0
𝐼
=
S∕m
𝑉0
(𝑏 − 𝑎)
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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∕(𝜎𝐴) to construct the differential resistance of each layer:
𝑑𝑅 =
𝑑𝑟
𝑑𝑟
𝑑𝑟
=
=
𝜎(𝑟)(2𝜋𝑟2 ) (𝜎0 𝑎∕𝑟)(2𝜋𝑟2 ) 2𝜋𝑎𝑟𝜎0
where 𝜎(𝑟) = 𝜎0 𝑎∕𝑟 as in Problem 5.15. The net resistance will then be the series combination,
expressed through the integral:
𝑅=
∫
𝑑𝑅 =
∫𝑎
𝑏
ln(𝑏∕𝑎)
𝑑𝑟
=
ohms
2𝜋𝑎𝑟𝜎0
2𝜋𝑎𝜎0
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, 𝜎(𝜌) = 𝜎0 𝜌, where 𝜎0 is a constant.
a) Find the 𝜌-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
𝐄=−
𝑉0
𝐚 V∕m (0 < 𝜌 < 𝑎)
𝑑 𝑧
b) Find the current density, 𝐉 between plates:
𝐉 = 𝜎𝐄 = −
𝜎0 𝑉 0 𝜌
𝐚𝑧 A∕m2
𝑑
c) Find the total current, 𝐼, in the structure:
𝐼=
∫0
2𝜋
∫0
𝑎
−
2𝜋𝑎3 𝜎0 𝑉0
𝜎0 𝑉0 𝜌
𝐚𝑧 ⋅ (−𝐚𝑧 ) 𝜌 𝑑𝜌 𝑑𝜙 =
A
𝑑
3𝑑
d) Find the resistance between plates:
𝑅=
𝑉0
3𝑑
ohms
=
𝐼
2𝜋𝑎3 𝜎0
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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 𝑑𝜌. The resistance will be the parallel combination of the shell
resistances: Assuming uniform conductivity throughout the shell volume, we may use 𝑅 = 𝐿∕(𝜎𝐴)
to write the shell resistance:
𝑑
𝑑𝑅 =
𝜎0 𝜌(2𝜋𝜌𝑑𝜌)
Now
𝑎
2𝜋𝜎0 𝜌2 𝑑𝜌 2𝜋𝜎0 𝑎3
3𝑑
1
1
ohms
=
=
=
⇒ 𝑅=
𝑅 ∫ 𝑑𝑅 ∫0
𝑑
3𝑑
2𝜋𝑎3 𝜎0
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, 𝜌: Start with the field evaluated on the surface:
𝐄=
𝑞(𝐫 − 𝐫1′ )
4𝜋𝜖0 |𝐫 − 𝐫1′ |3
⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟
𝑢𝑝𝑝𝑒𝑟 𝑐ℎ𝑎𝑟𝑔𝑒
−
𝑞(𝐫 − 𝐫2′ )
4𝜋𝜖0 |𝐫 − 𝐫2′ |3
⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟
𝑖𝑚𝑎𝑔𝑒 𝑐ℎ𝑎𝑟𝑔𝑒
where 𝐫 = 𝜌 𝐚𝜌 , 𝐫1′ = 𝑑 𝐚𝑧 , and 𝐫2′ = −𝑑 𝐚𝑧 . and where |𝐫 − 𝐫1′ | = |𝐫 − 𝐫2′ | =
the above substitutions, we find
𝐄(𝜌) =
𝑞(𝜌𝐚𝜌 − 𝑑𝐚𝑧 )
4𝜋𝜖0 (𝜌2 + 𝑑 2 )3∕2
−
𝑞(𝜌𝐚𝜌 + 𝑑𝐚𝑧 )
4𝜋𝜖0 (𝜌2 + 𝑑 2 )3∕2
=
√
𝜌2 + 𝑑 2 . With
−𝑞𝑑𝐚𝑧
2𝜋𝜖0 (𝜌2 + 𝑑 2 )3∕2
b) Use your result from part 𝑎 to find the charge density, 𝜌𝑠 , and the total induced charge on the
conductor: The charge density is found through:
−𝑞𝑑𝐚𝑧
−𝑞𝑑
|
|
⋅ 𝐚𝑧 =
C∕m2
𝜌𝑠 = 𝐃 ⋅ 𝐧| = 𝜖0 𝐄 ⋅ 𝐧| =
2
2
3∕2
2
|𝑠 2𝜋(𝜌 + 𝑑 )
|𝑠
2𝜋(𝜌 + 𝑑 2 )3∕2
The total charge is now
∫𝑠
2𝜋
∞
∞
𝜌𝑑𝜌
−𝑞𝑑
𝜌𝑑𝜌𝑑𝜙
=
−𝑞𝑑
2
2
3∕2
2
∫
2𝜋(𝜌 + 𝑑 )
(𝜌 + 𝑑 2 )3∕2
0
∫0 ∫0
−1
|∞
= −𝑞𝑑 √
| = −𝑞
𝜌2 + 𝑑 2 |0
𝑄𝑖𝑛𝑑 =
𝜌𝑠 𝑑𝑎 =
No surprise.
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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:
[
]
𝜌𝑙
f inal distance f rom charge
𝑉𝑃 − 𝑉0,0,0 = −
ln
2𝜋𝜖0
initial distance f rom charge
where 𝑉0,0,0 = 0. Considering the four charges, we thus have
(√ )
[
(√ )
( √ )]
( )
𝜌𝑙
10
17
2
1
+ ln
ln
− ln
− ln
𝑉𝑃 = −
2𝜋𝜖0
2
1
1
2
)
( √ )]
[√ √ ]
[
(
(√ )
𝜌𝑙
17
10 17
1
30 × 10−9
=
+ ln
10 + ln
ln
=
ln (2) + ln √
√
2𝜋𝜖0
2
2𝜋𝜖0
2
2
= 1.20 kV
b) Find 𝐄 at 𝑃 : Use
[
𝜌𝑙
(1, 2, 0) − (0, 1, 0) (1, 2, 0) − (0, 2, 0)
+
2𝜋𝜖0
|(1, 1, 0)|2
|(1, 0, 0)|2
]
(1, 2, 0) − (0, −1, 0) (1, 2, 0) − (0, −2, 0)
−
−
|(1, 3, 0)|2
|(1, 4, 0)|2
[
]
𝜌𝑙
(1, 1, 0) (1, 0, 0) (1, 3, 0) (1, 4, 0)
=
+
−
−
= 723 𝐚𝑥 − 18.9 𝐚𝑦 V∕m
2𝜋𝜖0
2
1
10
17
𝐄𝑃 =
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5.22. The line segment 𝑥 = 0, −1 ≤ 𝑦 ≤ 1, 𝑧 = 1, carries a linear charge density 𝜌𝐿 = 𝜋|𝑦| 𝜇C/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 = 𝜌𝐿 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 =
𝜌𝐿 𝑑𝑦′ [(𝑦 − 𝑦′ ) 𝐚𝑦 − 𝐚𝑧 ]
4𝜋[(𝑦 − 𝑦′ )2 + 1]3∕2
−
𝜌𝐿 𝑑𝑦′ [(𝑦 − 𝑦′ ) 𝐚𝑦 + 𝐚𝑧 ]
4𝜋[(𝑦 − 𝑦′ )2 + 1]3∕2
=
−𝜌𝐿 𝑑𝑦′ 𝐚𝑧
2𝜋[(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
∫
1
−𝜋|𝑦′ | 𝐚𝑧 𝑑𝑦′
∫−1 2𝜋[(𝑦 − 𝑦′ )2 + 1]3∕2
]
[
𝐚𝑧 1
𝑦′
𝑦′
=−
+
𝑑𝑦′
2 ∫0 [(𝑦 − 𝑦′ )2 + 1]3∕2 [(𝑦 + 𝑦′ )2 + 1]3∕2
]1
[
𝐚𝑧
𝑦(𝑦 + 𝑦′ ) + 1
𝑦(𝑦 − 𝑦′ ) + 1
=
+
2 [(𝑦 − 𝑦′ )2 + 1]1∕2 [(𝑦 + 𝑦′ )2 + 1]1∕2 0
[
]
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
𝐃(𝑦) =
𝑑𝐃 =
Now, at the origin (part 𝑎), we find the charge density through
[
]
𝐚𝑧
1
1
|
2
=
𝜌𝑠 (0, 0, 0) = 𝐃 ⋅ 𝐚𝑧 |
√ + √ − 2 = −0.29 𝜇C∕m
|𝑦=0
2
2
2
Then, at (0,1,0) (part 𝑏), the charge density is
[
]
𝐚𝑧
3
|
=
𝜌𝑠 (0, 1, 0) = 𝐃 ⋅ 𝐚𝑧 |
1 + √ − 2 = −0.24 𝜇C∕m2
|𝑦=1
2
5
5.23. A dipole with 𝐩 = 0.1𝐚𝑧 𝜇C ⋅ 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 𝜃
𝑝
𝑧
=
4𝜋𝜖0 [𝑥2 + 𝑦2 + 𝑧2 ]1.5
4𝜋𝜖0 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:
[
]
𝑝
𝑧
𝑧
−
𝑉 =
4𝜋𝜖0 [(𝑥 − 1)2 + 𝑦2 + 𝑧2 ]1.5 [(𝑥 + 1)2 + 𝑦2 + 𝑧2 ]1.5
Substituting 𝑃 (2, 0, 1), we find
.1 × 106
𝑉 =
4𝜋𝜖0
[
]
1
1
√ − √
2 2 10 10
= 289.5 V
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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:
[
]
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 × 1019 m−3 , find the
conductivity at this temperature.
With the electron and hole charge magnitude of 1.6 × 10−19 C, the conductivity in this case can
be written:
𝜎 = |𝜌𝑒 |𝜇𝑒 + 𝜌ℎ 𝜇ℎ = (1.6 × 10−19 )(2.3 × 1019 )(0.43 + 0.21) = 2.36 S∕m
5.25. Electron and hole concentrations increase with temperature. For pure silicon, suitable expressions are
𝜌ℎ = −𝜌𝑒 = 6200𝑇 1.5 𝑒−7000∕𝑇 C∕m3 . The functional dependence of the mobilities on temperature is
given by 𝜇ℎ = 2.3 × 105 𝑇 −2.7 m2 ∕V ⋅ s and 𝜇𝑒 = 2.1 × 105 𝑇 −2.5 m2 ∕V ⋅ s, where the temperature, 𝑇 ,
is in degrees Kelvin. The conductivity will thus be
[
]
𝜎 = −𝜌𝑒 𝜇𝑒 + 𝜌ℎ 𝜇ℎ = 6200𝑇 1.5 𝑒−7000∕𝑇 2.1 × 105 𝑇 −2.5 + 2.3 × 105 𝑇 −2.7
]
1.30 × 109 −7000∕𝑇 [
=
𝑒
1 + 1.095𝑇 −.2 S∕m
𝑇
Find 𝜎 at:
a) 0◦ C: With 𝑇 = 273◦ K, the expression evaluates as 𝜎(0) = 4.7 × 10−5 S∕m.
b) 40◦ C: With 𝑇 = 273 + 40 = 313, we obtain 𝜎(40) = 1.1 × 10−3 S∕m.
c) 80◦ C: With 𝑇 = 273 + 80 = 353, we obtain 𝜎(80) = 1.2 × 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 × 1018 and 3.0 × 1015 m−3 , respectively. If 𝜇𝑒 = 0.082
m2 ∕V ⋅ s and 𝜇ℎ = 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
[
]
𝜎 = (1.6 × 10−19 ) (1.8 × 1018 )(0.082) + (3.0 × 1015 )(0.0021) = 0.0236 S∕m
The resistance is then
𝑅=
𝓁
0.011
=
= 155 kΩ
𝜎𝐴 (0.0236)(0.002)(0.0015)
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5.27. Atomic hydrogen contains 5.5×1025 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 × 10−19 m.
a) Find P: With all identical dipoles, we have
𝑃 = 𝑁𝑞𝑑 = (5.5 × 1025 )(1.602 × 10−19 )(7.1 × 10−19 ) = 6.26 × 10−12 C∕m2 = 6.26 pC∕m2
b) Find 𝜖𝑟 : We use 𝑃 = 𝜖0 𝜒𝑒 𝐸, and so
𝜒𝑒 =
6.26 × 10−12
𝑃
= 1.76 × 10−4
=
𝜖0 𝐸
(8.85 × 10−12 )(4 × 103 )
Then 𝜖𝑟 = 1 + 𝜒𝑒 = 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 = 𝜖0 𝐄 + 𝐏 = 𝜖0 𝐄 + (1∕4)𝐃. Therefore 𝐃 = (4∕3)𝜖0 𝐄, so we identify 𝜖𝑟 = 4∕3.
5.29. A coaxial conductor has radii 𝑎 = 0.8 mm and 𝑏 = 3 mm and a polystyrene dielectric for which
𝜖𝑟 = 2.56. If 𝐏 = (2∕𝜌)𝐚𝜌 nC∕m2 in the dielectric, find:
a) 𝐃 and 𝐄 as functions of 𝜌: Use
𝐄=
(2∕𝜌) × 10−9 𝐚𝜌
144.9
𝐏
=
=
𝐚 V∕m
𝜖0 (𝜖𝑟 − 1) (8.85 × 10−12 )(1.56)
𝜌 𝜌
Then
𝐃 = 𝜖0 𝐄 + 𝐏 =
] 3.28 × 10−9 𝐚𝜌
2 × 10−9 𝐚𝜌 [ 1
3.28𝐚𝜌
+1 =
C∕m2 =
nC∕m2
𝜌
1.56
𝜌
𝜌
b) Find 𝑉𝑎𝑏 and 𝜒𝑒 : Use
𝑉𝑎𝑏 = −
∫3
0.8
( )
144.9
3
𝑑𝜌 = 144.9 ln
= 192 V
𝜌
0.8
𝜒𝑒 = 𝜖𝑟 − 1 = 1.56, as found in part 𝑎.
c) If there are 4 × 1019 molecules per cubic meter in the dielectric, find 𝐩(𝜌): Use
𝐩=
(2 × 10−9 ∕𝜌)
5.0 × 10−29
𝐏
=
𝐚𝜌 C ⋅ m
𝐚
=
𝜌
𝑁
𝜌
4 × 1019
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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 𝜖𝑟 = 𝑓 𝜖𝑟1 + (1 − 𝑓 )𝜖𝑟2 , where 𝑓 is the number fraction of species 1 dipoles in the
composite, and where 𝜖𝑟1 and 𝜖𝑟2 are the dielectric constants that the unmixed species would have if
each had number density 𝑁.
We may write the total polarization vector as
(
𝐏𝑡𝑜𝑡 = 𝑁1 𝐩1 + 𝑁2 𝐩2 = 𝑁
𝑁
𝑁1
𝐩1 + 2 𝐩2
𝑁
𝑁
)
[
]
= 𝑁 𝑓 𝐩1 + (1 − 𝑓 )𝐩2 = 𝑓 𝐏1 + (1 − 𝑓 )𝐏2
[
]
In terms of the susceptibilities, this becomes 𝐏𝑡𝑜𝑡 = 𝜖0 𝑓 𝜒𝑒1 + (1 − 𝑓 )𝜒𝑒2 𝐄, where 𝜒𝑒1 and 𝜒𝑒2
are evaluated at the composite number density, 𝑁. Now
[
]
𝐃 = 𝜖𝑟 𝜖0 𝐄 = 𝜖0 𝐄 + 𝐏𝑡𝑜𝑡 = 𝜖0 1 + 𝑓 𝜒𝑒1 + (1 − 𝑓 )𝜒𝑒2 𝐄
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
𝜖𝑟
Identifying 𝜖𝑟 as shown, we may rewrite it by adding and subracting 𝑓 :
[
] [
]
𝜖𝑟 = 1 + 𝑓 − 𝑓 + 𝑓 𝜒𝑒1 + (1 − 𝑓 )𝜒𝑒2 = 𝑓 (1 + 𝜒𝑒1 ) + (1 − 𝑓 )(1 + 𝜒𝑒2 )
[
]
= 𝑓 𝜖𝑟1 + (1 − 𝑓 )𝜖𝑟2 Q.E.D.
5.31. A circular cylinder of radius 𝑎 and dielectric constant 𝜖𝑟 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 𝜙 𝐚𝜌 − 𝐸0 sin 𝜙 𝐚𝜙 = 𝐸𝜌 𝑜𝑢𝑡 𝐚𝜌 + 𝐸𝜙 𝑜𝑢𝑡 𝐚𝜙
Now, the requirement that the tangential component of 𝐄 is continuous at the boundary is stated
as 𝐸𝜙,𝑖𝑛 = 𝐸𝜙,𝑜𝑢𝑡 . Then, the requirement of normal continuity of 𝐃 at the boundary is stated as
𝐷𝜌,𝑖𝑛 = 𝐷𝜌,𝑜𝑢𝑡 ⇒ 𝜖𝑟 𝜖0 𝐸𝜌,𝑖𝑛 = 𝜖0 𝐸𝜌,𝑜𝑢𝑡
Therefore
𝐄𝑖𝑛 (𝜌, 𝜙) =
𝐸0
cos 𝜙 𝐚𝜌 − 𝐸0 sin 𝜙 𝐚𝜙
𝜖𝑟
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5.31
a) (continued). Converting to rectangular components, we find
)
𝐸 (
𝐸0
cos 𝜙 𝐚𝜌 = 0 cos2 𝜙 𝐚𝑥 + cos 𝜙 sin 𝜙 𝐚𝑦
𝜖𝑟
𝜖𝑟
and
(
)
−𝐸0 sin 𝜙 𝐚𝜙 = 𝐸0 sin2 𝜙 𝐚𝑥 − cos 𝜙 sin 𝜙 𝐚𝑦
Combining the two components, we may write
𝐄𝑖𝑛 =
)
]
𝐸0 [( 2
cos 𝜙 + 𝜖𝑟 sin2 𝜙 𝐚𝑥 + (1 − 𝜖𝑟 ) sin 𝜙 cos 𝜙 𝐚𝑦
𝜖𝑟
As a check, note that this expression reduces to 𝐄𝑖𝑛 = 𝐸0 𝐚𝑥 if 𝜖𝑟 = 1.
b) Consider a field line incident at 𝜙 = 135◦ . Find 𝜖𝑟 so that √
the interior field line will be oriented
◦
◦
at 22.5 from the 𝑥 axis: At 135 , sin 𝜙 = − cos 𝜙 = 1∕ 2, and the interior field expression
becomes:
)
) ]
𝐸 [1 (
1(
𝐄𝑖𝑛 (𝜙 = 135◦ ) = 0
𝜖𝑟 + 1 𝐚𝑥 +
𝜖𝑟 − 1 𝐚 𝑦
𝜖𝑟 2
2
The angle of the interior field from the 𝑥 axis will be
[
′
−1
𝜙 = tan
So now, we set
𝐸𝑦,𝑖𝑛
𝐸𝑥,𝑖𝑛
]
[
−1
= tan
𝜖𝑟 − 1
𝜖𝑟 + 1
]
√
𝜖𝑟 − 1
= tan(22.5◦ ) = 2 − 1
𝜖𝑟 + 1
Solve to find 𝜖𝑟 = 2.41. The 22.5◦ angle means that the interior field line will exit the cylinder
at 𝜙 = 90◦ (at the positive 𝑦 axis).
5.32. Two equal but opposite-sign point charges of 3𝜇C 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 × 10−6 )2
9 × 10−12
=
= 12(0.5 − 𝑥)
4𝜋𝜖0 𝑥2
4𝜋(8.85 × 10−12 )𝑥2
which leads to the cubic equation:
𝑥3 − 0.5𝑥2 + 6.74 × 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 × 10−6 )(0.136) = 4.08 × 10−7 C-m.
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5.33. Two perfect dielectrics have relative permittivities 𝜖𝑟1 = 2 and 𝜖𝑟2 = 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
)
(
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 = (𝜖𝑟1 ∕𝜖𝑟2 )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
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5.34. A sphere of radius 𝑏 and dielectric constant 𝜖𝑟 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, 𝜃, 𝜙) = 𝐸0 𝐚𝑥 = 𝐸0 sin 𝜃 cos 𝜙 𝐚𝑟 + cos 𝜃 cos 𝜙 𝐚𝜃 − sin 𝜙 𝐚𝜙
Because the two tangential components, 𝐸𝜃 and 𝐸𝜙 , are continuous across the sphere boundary,
and because the normal component, 𝐸𝑟 , is discontinuous by a factor of 1∕𝜖𝑟 , the interior field
will be:
)
(
1
sin 𝜃 cos 𝜙 𝐚𝑟 + cos 𝜃 cos 𝜙 𝐚𝜃 − sin 𝜙 𝐚𝜙
𝐄𝑖𝑛 (𝑟, 𝜃, 𝜙) = 𝐸0
𝜖𝑟
This field is now transformed back to rectangular components through:
)
]
[
(
1
2
2
2
2
sin 𝜃 + cos 𝜃 + sin 𝜙
𝐸𝑖𝑛,𝑥 = 𝐄𝑖𝑛 ⋅ 𝐚𝑥 = 𝐸0 cos 𝜙
𝜖𝑟
[
(
)
]
1
2
2
𝐸𝑖𝑛,𝑦 = 𝐄𝑖𝑛 ⋅ 𝐚𝑦 = 𝐸0 sin 𝜙 cos 𝜙
sin 𝜃 + cos 𝜃 − sin 𝜙 cos 𝜙
𝜖𝑟
)
(
1
−1
𝐸𝑖𝑛,𝑧 = 𝐄𝑖𝑛 ⋅ 𝐚𝑧 = 𝐸0 cos 𝜃 sin 𝜃 cos 𝜙
𝜖𝑟
Combining these and simplifying leads to the final result:
)
]
𝐸0 {[ 2 ( 2
cos 𝜙 sin 𝜃 + 𝜖𝑟 cos2 𝜃 + 𝜖𝑟 sin2 𝜙 𝐚𝑥
𝜖𝑟
[
]
}
− sin 𝜙 cos 𝜙 sin2 𝜃(𝜖𝑟 − 1) 𝐚𝑦 − cos 𝜃 sin 𝜃 cos 𝜙(𝜖𝑟 − 1)𝐚𝑧
𝐄𝑖𝑛 =
b) Specialize your result from part 𝑎 for the case in which 𝜙 = 𝜋: With this substitution, find
𝐄𝑖𝑛 (𝜙 = 𝜋) =
]
𝐸0 [ 2
(sin 𝜃 + 𝜖𝑟 cos2 𝜃)𝐚𝑥 + cos 𝜃 sin 𝜃(𝜖𝑟 − 1)𝐚𝑧
𝜖𝑟
c) Specialize your result from part 𝑎 for the case in which 𝜃 = 𝜋∕2 (sphere equator). Compare both
results (parts 𝑏 and 𝑐) to the answer to Problem 5.31. With this substitution, find
𝐄𝑖𝑛 (𝜃 = 𝜋∕2) =
]
𝐸0 [
(cos2 𝜙 + 𝜖𝑟 sin2 𝜙)𝐚𝑥 + (1 − 𝜖𝑟 ) sin 𝜙 cos 𝜙 𝐚𝑦
𝜖𝑟
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
𝜙 is in that case played by 𝜃 = 𝜙 − 90◦ .
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5.35. Let the cylindrical surfaces 𝜌 = 4 cm and 𝜌 = 9 cm enclose two wedges of perfect dielectrics, 𝜖𝑟1 = 2
for 0 < 𝜙 < 𝜋∕2, and 𝜖𝑟2 = 5 for 𝜋∕2 < 𝜙 < 2𝜋. If 𝐄1 = (2000∕𝜌)𝐚𝜌 V/m, find:
a) 𝐄2 : The interfaces between the two media will lie on planes of constant 𝜙, 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)𝜖𝑟 𝜖0 𝐸 2 . So in region 1:
1
𝑊𝐸1 =
∫0 ∫0
𝜋∕2
∫4
9
( )
(2000)2
1
𝜋
9
= 45.1 𝜇J
𝜌 𝑑𝜌 𝑑𝜙 𝑑𝑧 = 𝜖0 (2000)2 ln
(2)𝜖0
2
𝜌2
2
4
In region 2, we have
1
𝑊𝐸2 =
2𝜋
∫0 ∫𝜋∕2 ∫4
9
( )
(2000)2
9
15𝜋
1
𝜌 𝑑𝜌 𝑑𝜙 𝑑𝑧 =
(5)𝜖0
𝜖0 (2000)2 ln
= 338 𝜇J
2
𝜌2
4
4
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Engineering Electromagnetics 9th solution
Engineering Electromagnetics (Chungnam National University)
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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, 𝜖𝑟 . 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 𝜖), we
equate the two capacitances:
)
(
𝜖𝑤
𝑏
𝑑
2𝜋𝜖
=
⇒
= exp 2𝜋
ln(𝑏∕𝑎)
𝑑
𝑎
𝑤
6.2. Let 𝑆 = 100 mm2 , 𝑑 = 3 mm, and 𝜖𝑟 = 12 for a parallel-plate capacitor.
a) Calculate the capacitance:
𝐶=
𝜖𝑟 𝜖0 𝐴 12𝜖0 (100 × 10−6 )
= 0.4𝜖0 = 3.54 pf
=
𝑑
3 × 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 × 10−3 ) = 2000 V∕m, then 𝐷 = 𝜖𝑟 𝜖0 𝐸 = 2.4 × 104 𝜖0 = 0.21 𝜇C∕m2
The charge in this case is
𝑄 = 𝐃 ⋅ 𝐧|𝑠 = 𝐷𝐴 = 0.21 × (100 × 10−6 ) = 0.21 × 10−4 𝜇C = 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 × 10−3 ) = 2000 V∕m, as before. 𝐷 = 𝜖0 𝐸 = 2000𝜖0 = 17.7 nC∕m2
The charge is now 𝑄 = 𝐷𝐴 = 17.7 × (100 × 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: 𝜖𝑟 = 1, 𝐸𝐵𝐷 = 3 MV/m;
(b) barium titanate: 𝜖𝑟 = 1200, 𝐸𝐵𝐷 = 3 MV/m; (c) silicon dioxide: 𝜖𝑟 = 3.78, 𝐸𝐵𝐷 = 16 MV/m; (d)
polyethylene: 𝜖𝑟 = 2.26, 𝐸𝐵𝐷 = 4.7 MV/m? Note that 𝑉𝑚𝑎𝑥 = 𝐸𝐵𝐷𝑑, where 𝑑 is the plate separation.
Also, 𝐶 = 𝜖𝑟 𝜖0 𝐴∕𝑑, and so 𝑉𝑚𝑎𝑥 𝐶 = 𝜖𝑟 𝜖0 𝐴𝐸𝐵𝐷 , where 𝐴 is the plate area. The maximum 𝐶𝑉𝑚𝑎𝑥
product is found through the maximum 𝜖𝑟 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 = 𝜖0 𝐴∕𝑑, increasing 𝑑 by a factor of ten decreases 𝐶 by a factor of 0.1.
c) 𝐸: We require 𝐸 × 𝑑 = 𝑉0 , where 𝑉0 has not changed. Therefore, 𝐸 has decreased by a
factor of 0.1.
d) 𝐷: As 𝐷 = 𝜖0 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) 𝜌𝑆 : As 𝑄 is reduced by 0.1, 𝜌𝑆 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) 𝜌𝑆 : As 𝑄 is unchanged, 𝜌𝑆 is also unchanged, since the plate area is the same.
c) 𝐷: As 𝐷 = 𝜌𝑆 , it will remain the same also.
d) 𝐸: Since 𝐸 = 𝐷∕𝜖0 , and as 𝐷 is not changed, E will also remain the same.
e) 𝑉0 : We require 𝐸 × 𝑑 = 𝑉0 , where 𝐸 has not changed. Therefore, 𝑉0 has increased by a
factor of 10.
f) 𝐶: As 𝐶 = 𝜖0 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 𝜖𝑟 . 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, 𝜖𝑟 , and other known parameters:
We use the fact that the total capacitance is the series combination of the air- and dielectric-filled
capacitances:
𝜖 𝐴
𝜖𝜖 𝐴
1
1
1
=
+
where 𝐶𝑎𝑖𝑟 = 0
and 𝐶𝑑𝑖𝑒𝑙 = 𝑟 0
𝐶𝑇
𝐶𝑎𝑖𝑟 𝐶𝑑𝑖𝑒𝑙
𝑑−𝑏
𝑏
So
𝜖 (𝑑 − 𝑏) + 𝑏
𝑑−𝑏
𝑏
1
=
+
= 𝑟
𝐶𝑇
𝜖0 𝐴
𝜖𝑟 𝜖0 𝐴
𝜖𝑟 𝜖0 𝐴
Then, knowing that the completely air-filled capacitance will be 𝐶0 = 𝜖0 𝐴∕𝑑, the increase factor
becomes:
𝐶
𝑑
𝑓= 𝑇 =
𝐶0
(𝑑 − 𝑏) + 𝑏∕𝜖𝑟
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 𝜖(𝑧) = 𝜖0 (1 + 𝑧∕ 𝑑 2 ). Find:
a) 𝐃: With 𝑄 on the top plate of area 𝐴 = 𝜋𝑎2 , we would have a uniform surface charge density,
𝜌𝑠 = 𝑄∕𝐴 C∕m2 , and therefore 𝐃 (uniform throughout the volume between plates) is:
𝐃 = −𝜌𝑠 𝐚𝑧 = −
𝑄
𝐚𝑧 C∕m2
2
𝜋𝑎
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
V∕m
= 2
𝜖
𝜋𝑎 𝜖0 (1 + 𝑧2 ∕𝑑 2 )
c) 𝑉0 : We will assume the top plate carries voltage 𝑉0 , and the bottom plate is grounded. Therefore
𝑉0 = −
∫0
𝑑
𝑑
𝐄 ⋅ 𝑑𝐋 = −
−𝑄𝐚𝑧 ⋅ 𝐚𝑧 𝑑𝑧
∫0 𝜋𝑎2 𝜖0 (1 + 𝑧2 ∕𝑑 2 )
=
( ) 𝑑
𝑄𝑑
𝑄𝑑
−1 𝑧 |
tan
| = 2 V
2
𝑑 |0 4𝑎 𝜖0
𝜋𝑎 𝜖0
d) 𝐶: Using the result of part 𝑐, we construct:
𝐶=
4𝑎2 𝜖0
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 𝜖𝑟 (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
𝜖𝑟 𝜖0 𝐴 𝜖0 (1 + 𝑧2 ∕𝑑 2 )𝜋𝑎2
=
𝑑𝑧
𝑑𝑧
As the layers are in series, their reciprocal capacitances will sum, leading to
𝑑𝐶 =
𝑑
𝑑
1
1
𝑑𝑧
1 𝜋𝑑
=
=
=
𝐶 ∫0 𝑑𝐶 ∫0 𝜋𝑎2 𝜖0 (1 + 𝑧2 ∕𝑑 2 ) 𝜋𝑎2 𝜖0 4
Therefore 𝐶 = 4𝑎2 𝜖0 ∕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 𝜖(𝜌) = 𝜖0 (1 + 𝜌2 ∕𝑎2 ). Find:
a) 𝐄: Since 𝜖 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 = 𝜖𝐄 = −[𝜖0 (1 + 𝜌2 ∕𝑎2 )𝑉0 ∕𝑑] 𝐚𝑧 C∕m2 .
c) 𝑄: Here we find the integral of the surface charge density over the top plate:
2𝜋
𝑎
−𝜖0 (1 + 𝜌2 ∕𝑎2 )𝑉0
𝐚𝑧 ⋅ (−𝐚𝑧 ) 𝜌 𝑑𝜌 𝑑𝜙
∫𝑆
∫0 ∫0
𝑑
2𝜋𝜖0 𝑉0 𝑎
3𝜋𝜖0 𝑎2
=
𝑉0
(𝜌 + 𝜌3 ∕𝑎2 ) 𝑑𝜌 =
∫0
𝑑
2𝑑
𝑄=
𝐃 ⋅ 𝑑𝐒 =
d) 𝐶: We use 𝐶 = 𝑄∕𝑉0 and our previous result to find
𝐶=
3𝜋𝜖0 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 𝑑𝜌, and where the dielectric constant of each shell (at radius 𝜌) is 𝜖𝑟 = (1 + 𝜌2 ∕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
𝑑𝜌, radius 𝜌, and length 𝑑 will be
𝜖𝑟 𝜖0 (2𝜋𝜌)𝑑𝜌
𝜌2
where 𝜖𝑟 = 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:
(
(
)
)
𝑎
2𝜋𝜖0
2𝜋𝜖0 𝑎2
3𝜋𝜖0 𝑎2
𝜌2
𝑎4
𝐶=
𝑑𝐶 =
+ 2 =
F
1 + 2 𝜌𝑑𝜌 =
∫
∫0
𝑑
𝑑
2
2𝑑
𝑎
4𝑎
𝑑𝐶 =
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 (𝜖𝑟 = 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 =
2𝜋𝜖0
2𝜋𝜖𝑟 𝜖0
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
]
2𝜋𝜖0 [ 1.7
+ 0.15𝜖𝑟
𝐶𝑛𝑒𝑡 =
ln(𝑏∕𝑎) 2
⏟⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏟
𝑓
where the bracketed term, 𝑓 , is the capacitance increase factor that we seek. Substituting 𝜖𝑟 = 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=
𝑎𝜌𝑠
𝐚
𝜖𝜌 𝜌
where 𝜌𝑠 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
∫𝑏
𝑎
𝐄 ⋅ 𝑑𝐋 = −
∫𝑏
𝑎
𝑎𝜌
𝑎𝜌𝑠
𝑑𝜌 = 𝑠 ln(𝑏∕𝑎) = 100
𝜖𝜌
𝜖
𝑎𝜌𝑠
100
101
=
= 101 ⇒ 𝐄 =
𝐚 V∕m
𝜖
ln(2.7)
𝜌 𝜌
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 (𝜌) = 𝑉0
ln(𝑏∕𝜌)
ln(𝑏∕𝑎)
from which
𝐄 = −∇𝑉 = −
𝑉0
𝑑𝑉
100
101
𝐚 =
𝐚 =
𝐚 =
𝐚 V∕m
𝑑𝜌 𝜌 𝜌 ln(𝑏∕𝑎) 𝜌 𝜌 ln(2.7) 𝜌
𝜌 𝜌
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 𝜖𝑟1 (𝑎 < 𝜌 < 𝑐) and
𝜖𝑟2 (𝑐 < 𝜌 < 𝑏).
a) If 𝑐 = (𝑎 + 𝑏)∕2, find the ratio of 𝜖𝑟1 to 𝜖𝑟2 such that the maximum electric field intensities in the
two regions are equal: Assuming a surface charge density 𝜌𝑠 on the inner conductor and applying
Gauss’ Law in the usual manner, we would have:
𝐃=
𝑎𝜌𝑠
𝐚 C∕m2 for 𝑎 < 𝜌 < 𝑏
𝜌 𝜌
The maximum electric field intensity in region 1 is
𝐄𝑚𝑎𝑥1 =
𝜌
1
|
𝐃|
= 𝑠 𝐚𝜌 V∕m
|
𝜖𝑟1 𝜖0 𝜌=𝑎 𝜖𝑟1 𝜖0
The maximum electric field intensity in region 2 is
𝐄𝑚𝑎𝑥2 =
𝜌𝑠 𝑎
1
|
𝐃|
𝐚 V∕m
=
|
𝜌=𝑐
𝜖𝑟2 𝜖0
𝑐 𝜖𝑟2 𝜖0 𝜌
But 𝑐 = (𝑎 + 𝑏)∕2, so that
𝐄𝑚𝑎𝑥2 =
2𝜌𝑠 𝑎
𝐚
𝜖𝑟2 𝜖0 (𝑎 + 𝑏) 𝜌
So now with the requirement
𝐄𝑚𝑎𝑥1 = 𝐄𝑚𝑎𝑥2 ⇒
2𝜌𝑠 𝑎
𝜌𝑠
𝜖
𝑎+𝑏
=
⇒ 𝑟1 =
𝜖𝑟1 𝜖0
𝜖𝑟2 𝜖0 (𝑎 + 𝑏)
𝜖𝑟2
2𝑎
b) Find the capacitance per unit length under the conditions of part 𝑎: Begin by finding the voltage
between conductors:
𝑎
𝑎
𝑐
( )
( )
𝑎𝜌𝑠
𝑎𝜌𝑠
𝑎𝜌𝑠
𝑎𝜌𝑠
𝑏
𝑐
+
V
ln
ln
𝑑𝜌 −
𝑑𝜌 =
𝑉0 = −
𝐄 ⋅ 𝑑𝐋 = −
∫𝑐 𝜖𝑟1 𝜖0 𝜌
∫𝑏
∫𝑏 𝜖𝑟2 𝜖0 𝜌
𝜖𝑟2 𝜖0
𝑐
𝜖𝑟1 𝜖0
𝑎
Then with 𝑐 = (𝑎 + 𝑏)∕2, this becomes:
[
(
)
( )]
(
(
)
)] 𝑎𝜌 [
𝑎𝜌𝑠 1
𝜖
2𝑏
𝑎+𝑏
1
1
𝑏 𝜖𝑟2
1
𝑠
ln
ln
ln
ln 𝑟1
𝑉0 =
+
+
=
𝜖0 𝜖𝑟2
𝑎+𝑏
𝜖
2𝑎
𝜖
𝜖𝑟2
𝑎 𝜖𝑟1
𝜖𝑟1
𝜖𝑟2
( 𝑟1
[
) ( )] 0
)
(
𝑎𝜌
𝜖
1
1
1
𝑏
= 𝑠
+
ln
−
ln 𝑟1
𝜖0 𝜖𝑟2
𝑎
𝜖𝑟1 𝜖𝑟2
𝜖𝑟2
Finally, with the charge per unit length, 𝑄 = 2𝜋𝑎(1)𝜌𝑠 , the capacitance is:
𝐶=
2𝜋𝑎𝜌𝑠
2𝜋𝜖 𝜖
𝑄
=
= ( ) ( 𝑟2 0 ) ( ) F∕m
𝜖
𝜖
𝑉0
𝑉0
ln 𝑎𝑏 − 1 − 𝜖𝑟2 ln 𝜖𝑟1
𝑟1
𝑟2
Note that 𝜖𝑟2 should appear in the numerator as shown here, whereas the answer in Appendix F
reads 𝜖𝑟1 . Note also that 𝐶 = 2𝜋𝜖𝑟1 𝜖0 ∕ ln(𝑏∕𝑎) as expected when 𝜖𝑟2 = 𝜖𝑟1 .
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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 → ∞). If we assume charge 𝑄 on the sphere, the electric field
will be
𝑄
𝐄=
𝐚𝑟 V∕m (𝑎 < 𝑟 < ∞)
4𝜋𝜖0 𝑟2
The potential on the sphere surface will then be
𝑎
𝑉0 = −
∫∞
𝑎
𝐄 ⋅ 𝑑𝐋 = −
𝑄
𝑄 𝑑𝑟
=
∫∞ 4𝜋𝜖0 𝑟2
4𝜋𝜖0 𝑎
Capacitance is then
𝐶=
𝑄
= 4𝜋𝜖0 𝑎 F
𝑉0
b) The sphere is to be covered with a dielectric layer of thickness 𝑑 and dielectric contant 𝜖𝑟 . If
𝜖𝑟 = 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
[
]
𝑏
𝑎
𝑄
𝑄 1 1 (1 1)
𝑄
+
−
𝑉0 = −
𝑑𝑟 −
𝑑𝑟 =
∫∞ 4𝜋𝜖0 𝑟2
∫𝑏 4𝜋𝜖𝑟 𝜖0 𝑟2
4𝜋𝜖0 𝑏 𝜖𝑟 𝑎 𝑏
Then, substituting capacitance, 𝐶 = 𝑄∕𝑉0 , and solving for 𝑏, we find, after a little algebra:
𝑏=
𝑎(𝜖𝑟 − 1)
(
)
𝑎𝜖𝑟 4𝜋𝜖0 ∕𝐶 − 1
Then, substitute 𝜖𝑟 = 3 and 𝐶 = 8𝜋𝜖0 𝑎 (twice the part 𝑎 result) to obtain:
𝑏 = 4𝑎 ⇒ 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 𝜖 = 𝜖0 .
Find values for 𝐾1 , 𝜌𝐿 , 𝑎, and 𝐶: We have
[
𝐾1 =
ℎ+
√
ℎ2 + 𝑏2
𝑏
We then have
𝜌𝐿 =
Next, 𝑎 =
√
ℎ2 − 𝑏2 =
√
[
]2
=
15 +
√
(15)2 + (6)2
6
]2
= 23.0
4𝜋𝜖0 (250)
4𝜋𝜖0 𝑉0
=
= 8.87 nC∕m
ln 𝐾1
ln(23)
(15)2 − (6)2 = 13.8 m. Finally,
𝐶=
2𝜋𝜖0
2𝜋𝜖
=
= 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
2𝜋𝜖0
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,
(
)
)
(
2𝜋𝜖0
2𝜋 × 8.854
ℎ
= 1.46
= cosh
= cosh
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
𝜌𝐿 = (4𝜋𝜖0 𝑉0 )∕ ln 𝐾1 = 2.43 nC∕m. Now
[
]
𝜌𝐿 (𝑥 + 𝑎)𝐚𝑥 + 𝑦𝐚𝑦 (𝑥 − 𝑎)𝐚𝑥 + 𝑦𝐚𝑦
𝐃 = 𝜖0 𝐄 = −
−
2𝜋 (𝑥 + 𝑎)2 + 𝑦2
(𝑥 − 𝑎)2 + 𝑦2
and
𝜌𝑠, 𝑚𝑎𝑥
]
[
𝜌𝐿 ℎ − 𝑏 + 𝑎
ℎ−𝑏−𝑎
|
−
= 473 nC∕m2
= 𝐃 ⋅ (−𝐚𝑥 )|
=
|𝑥=ℎ−𝑏,𝑦=0 2𝜋 (ℎ − 𝑏 + 𝑎)2 (ℎ − 𝑏 − 𝑎)2
b) plane at a point nearest the cylinder: At 𝑥 = 𝑦 = 0,
𝜌 2
𝜌 [ 𝑎𝐚𝑥 −𝑎𝐚𝑥 ]
𝐃(0, 0) = − 𝐿
= − 𝐿 𝐚𝑥
− 2
2
2𝜋 𝑎
2𝜋 𝑎
𝑎
from which
𝜌𝑠 = 𝐃(0, 0) ⋅ 𝐚𝑥 = −
𝜌𝐿
= −15.8 nC∕m2
𝜋𝑎
c) Find the capacitance per unit length. This will be 𝐶 = 𝜌𝐿 ∕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 𝜎)
or with a perfect dielectric (permittivity 𝜖), the resulting resistance and capacitance of the structures
are related through the simple formula 𝑅𝐶 = 𝜖∕𝜎. 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
∮ 𝜖𝐄 ⋅ 𝑑𝐒
− ∫𝑏 𝐄 ⋅ 𝑑𝐋
and 𝑅 =
𝐶= 𝑠𝑎
∮𝑠 𝜎𝐄 ⋅ 𝑑𝐒
− ∫𝑏 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 =
∮𝑠 𝜖𝐄 ⋅ 𝑑𝐒
∮𝑠 𝜎𝐄 ⋅ 𝑑𝐒
=
𝜖
𝜎
The far-right result is valid provided that 𝜖 and 𝜎 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 𝜖 and 𝜎 must be constants over the capacitor (or resistor) volume; i.e., the
medium must be homogeneous. A little more subtle points are that 𝜖 and 𝜎 generally cannot vary
with field orientation (isotropic medium), and cannot vary with field intensity (linear medium),
for the simple relation 𝑅𝐶 = 𝜖∕𝜎 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 𝜖𝑅 = 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 = 𝜖0
= 𝜖0
= 59 pF∕m
𝑁𝑉
6
b) Calculate an exact value for the capacitance per unit length: This will be
2𝜋𝜖0
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 𝜖𝑅 = 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 )𝜖0 ,
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
𝜖0 =
32
𝜖 = 2𝜖0 = 17.7 pF∕m
16 0
We check this result with that using the exact formula:
𝐶=
𝜋𝜖0
−1
cosh (𝑑∕2𝑎)
=
𝜋𝜖0
−1
cosh (13∕5)
= 1.95𝜖0 = 17.3 pF∕m
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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
2𝜋𝜖
[
]
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 =
11.5𝜖0 F∕m. Use of the drawing produces:
. 22 × 2
𝜖0 = 11𝜖0 F∕m
𝐶=
4
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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 𝜖 = 𝜖0 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
𝜖0 =
.
4 × 13
𝜖0 = 10.4𝜖0 = 90 pF∕m
5
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6.21. The inner conductor of the transmission line shown in Fig. 6.14 has a square cross-section 2𝑎 ×
2𝑎, while the outer square is 5𝑎 × 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 𝜖 = 1.6𝜖0 . (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 × 2
(1.6)𝜖0 = 12.8𝜖0 = 110 pF∕m
𝐶=
4
c) If 𝑎 is changed, the result of part 𝑏 would not change, since all dimensions retain the same relative
scale.
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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 𝜖𝑟 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 = 𝜖𝑟 𝜖0 𝐴∕(𝑑∕2) and
𝐶𝑎𝑖𝑟 = 𝜖0 𝐴∕(𝑑∕2). As the two capacitors are in series, reciprocals add to give
2𝜖𝑟 𝜖0 𝐴
𝑑(1 + 𝜖𝑟 )
1
1
1
=
+
=
⇒ 𝐶𝑣𝑒𝑟𝑡 =
𝐶𝑣𝑒𝑟𝑡
𝐶𝑎𝑖𝑟 𝐶𝑑𝑖𝑒𝑙
2𝜖𝑟 𝜖0 𝐴
𝑑(1 + 𝜖𝑟 )
For the regions arranged side by side, the two resulting capacitances add, giving
𝐶ℎ𝑜𝑟𝑖𝑧 =
𝜖𝑟 𝜖0 (𝐴∕2) 𝜖0 (𝐴∕2) 𝜖0 𝐴
+
=
(1 + 𝜖𝑟 )
𝑑
𝑑
2𝑑
Now take the ratio
𝐶𝑣𝑒𝑟𝑡
4𝜖𝑟
=
𝐶ℎ𝑜𝑟𝑖𝑧
(1 + 𝜖𝑟 )2
This result is less than unity if 𝜖𝑟 > 1. Therefore, the side by side configuration yields the larger
capacitance. The increase factor over the air-filled case will be
𝑓𝑖𝑛𝑐 =
𝐶ℎ𝑜𝑟𝑖𝑧
𝜖 𝐴(1 + 𝜖𝑟 )∕(2𝑑) 1 + 𝜖𝑟
= 0
=
𝐶0
𝜖0 𝐴∕𝑑
2
As an aside, note that the increase factor achieved when the vertical arrangement is used will be
𝑓𝑖𝑛𝑐,𝑣𝑒𝑟𝑡 =
𝐶𝑣𝑒𝑟𝑡
2𝜖𝑟
=
𝐶0
(1 + 𝜖𝑟 )
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 𝜖𝑟 = 3 and 𝜎 = 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=
𝜋𝜖
𝜋 × 3 × 8.85 × 10−12
=
= 3.64 × 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 × 10−11 )(100) = 3.64 × 10−9 C∕m = 3.64 nC∕m
b) the battery current: Use
𝑅𝐶 =
𝜖
3 × 8.85 × 10−12
= 486 Ω
⇒ 𝑅=
𝜎
(1.5 × 10−3 )(3.64 × 10−11 )
Then
𝐼=
𝑉0
100
=
= 0.206 A = 206 mA
𝑅
486
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6.24. A potential field in free space is given in spherical coordinates as:
{
[𝜌0 ∕(6𝜖0 )] [3𝑎2 − 𝑟2 ] (𝑟 ≤ 𝑎)
𝑉 (𝑟) =
(𝑎3 𝜌0 )∕(3𝜖0 𝑟) (𝑟 ≥ 𝑎)
where 𝜌0 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:
(
)
)
𝜌𝑣
𝜌
𝜌 1 𝑑 ( 2
1 𝑑
2
2 𝑑𝑉1
∇ 𝑉1 = − = 2
𝑟 (−2𝑟) = − 0
𝑟
= 0 2
𝜖0
𝑑𝑟
6𝜖0 𝑟 𝑑𝑟
𝜖0
𝑟 𝑑𝑟
from which we identify 𝜌𝑣 = 𝜌0
(𝑟 ≤ 𝑎).
Outside 𝑟 = 𝑎, we use
∇ 2 𝑉2 = −
𝜌𝑣
1 𝑑
= 2
𝜖0
𝑟 𝑑𝑟
from which 𝜌𝑣 = 0 (𝑟 ≥ 𝑎).
(
𝑟2
𝑑𝑉2
𝑑𝑟
)
=
1 𝑑
𝑟2 𝑑𝑟
( ( 3 ))
−𝑎 𝜌0
=0
𝑟2
3𝜖0 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)𝜋𝑎3 𝜌0 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
(
)
𝑑𝑉
1 𝑑
𝑟2
= 0 (assuming radial variation only)
∇2 𝑉 = 2
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=
−𝐶2
𝐶2
+ 𝐶3 ⇒ 𝐶3 =
𝑏
𝑏
Applying 2:
𝑉0 =
𝑉
𝐶
𝐶
𝐶2
+ 𝐶3 = 2 − 2 ⇒ 𝐶2 = ( 0 )
1
𝑎
𝑎
𝑏
− 1𝑏
𝑎
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).
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6.26. Given the spherically-symmetric potential field in free space, 𝑉 = 𝑉0 𝑒−𝑟∕𝑎 , find:
a) 𝜌𝑣 at 𝑟 = 𝑎; Use Poisson’s equation, ∇2 𝑉 = −𝜌𝑣 ∕𝜖, which in this case becomes
−
(
) −𝑉
(
)
𝜌𝑣
𝑑 ( 2 −𝑟∕𝑎 ) −𝑉0
𝑑𝑉
𝑟 −𝑟∕𝑎
1 𝑑
𝑟2
= 20
2−
𝑒
= 2
𝑟 𝑒
=
𝜖0
𝑑𝑟
𝑎𝑟
𝑎
𝑟 𝑑𝑟
𝑎𝑟 𝑑𝑟
from which
𝜌𝑣 (𝑟) =
)
𝜖 𝑉
𝜖0 𝑉0 (
𝑟 −𝑟∕𝑎
2−
𝑒
⇒ 𝜌𝑣 (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 ⇒ 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 = 𝜖0 𝐄 = (𝜖0 𝑉0 ∕𝑎)𝑒−𝑟∕𝑎 𝐚𝑟 . Then the net outward flux of 𝐃 through a sphere of radius 𝑟 would
be
Φ(𝑟) = 𝑄𝑒𝑛𝑐𝑙 (𝑟) = 4𝜋𝑟2 𝐷 = 4𝜋𝜖0 𝑉0 𝑟2 𝑒−𝑟∕𝑎 C
As 𝑟 → ∞, 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 𝜌𝑣 = 0. It is known that both 𝐸𝑥 and
𝑉 are zero at the origin. Find 𝑓 (𝑥) and 𝑉 (𝑥, 𝑦): Since 𝜌𝑣 = 0, we know that ∇2 𝑉 = 0, and so
∇2 𝑉 =
Therefore
𝑑 2𝑓
𝜕2𝑉
𝜕2𝑉
2𝑥
+
=
16𝑒
+
−6=0
𝜕𝑥2
𝜕𝑦2
𝑑𝑥2
𝑑𝑓
𝑑 2𝑓
2𝑥
= −8𝑒2𝑥 + 6𝑥 + 𝐶1
=
−16𝑒
+
6
⇒
𝑑𝑥
𝑑𝑥2
Now
𝐸𝑥 =
𝑑𝑓
𝜕𝑉
= 8𝑒2𝑥 +
𝜕𝑥
𝑑𝑥
and at the origin, this becomes
𝐸𝑥 (0) = 8 +
𝑑𝑓 |
= 0(as given)
|
𝑑𝑥 |𝑥=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 ).
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6.28. Show that in a homogeneous medium of conductivity 𝜎, 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:
𝜕𝜌
∇⋅𝐉=− 𝑣
𝜕𝑡
where 𝐉 = 𝜎𝐄, and where, in our homogeneous medium, 𝜎 is constant with position. Now write
∇ ⋅ 𝐉 = ∇ ⋅ (𝜎𝐄) = 𝜎∇ ⋅ (−∇𝑉 ) = −𝜎∇2 𝑉 = −
𝜕𝜌𝑣
𝜕𝑡
This becomes
∇2 𝑉 = 0 (Laplace′ s equation)
when 𝜌𝑣 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 ∕𝜖0 for 𝑟 ≤ 1? Use Poisson’s equation with radial variation
only:
))
( (
(
)
𝜌
𝑑𝑉
1 𝑑
180𝑟3
1 𝑑
−30𝑟4
=−
= − 𝑣 ⇒ 𝜌𝑣 = 180𝑟3 C∕m3
𝑟2
𝑟2
= 2
∇2 𝑉 = 2
𝑑𝑟
𝜖0
𝜖0
𝜖0
𝑟 𝑑𝑟
𝑟 𝑑𝑟
The total charge in the sphere is then:
𝑄=
∫𝑣
𝜌𝑣 𝑑𝑣 = 4𝜋
∫0
1
180𝑟3 𝑟𝑑 𝑟 = 120𝜋 C
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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 𝜌0 C∕m3 , and which has permittivity 𝜖.
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 =
−𝜌0 𝑧2
𝜌0
𝑑 2𝑉
=
−
⇒
𝑉
=
+ 𝐶1 𝑧 + 𝐶2
𝜖
2𝜖
𝑑𝑧2
At 𝑧 = 0, 𝑉 = 0, and so 𝐶2 = 0. Then, at 𝑧 = 𝑑, 𝑉 = 0 as well, so we find 𝐶1 = 𝜌0 𝑑∕2𝜖.
Finally, 𝑉 (𝑧) = (𝜌0 𝑧∕2𝜖)[𝑑 − 𝑧].
b) Determine the electric field intensity, 𝐄 between plates: Taking the answer to part 𝑎, we find 𝐄
through
[
] 𝜌
𝑑 𝜌0 𝑧
𝑑𝑉
𝐚𝑧 = −
(𝑑 − 𝑧) = 0 (2𝑧 − 𝑑) 𝐚𝑧 V∕m
𝐄 = −∇𝑉 = −
𝑑𝑧
𝑑𝑧 2𝜖
2𝜖
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
−𝜌0 𝑧2
𝑉 (𝑧) =
+ 𝐶1 𝑧 + 𝐶2
2𝜖
with 𝐶2 = 0 as before, since 𝑉 (𝑧 = 0) = 0. Then
𝑉 (𝑧 = 𝑑) = 𝑉0 =
−𝜌0 𝑑 2
𝑉
𝜌 𝑑
+ 𝐶1 𝑑 ⇒ 𝐶1 = 0 + 0
2𝜖
𝑑
2𝜖
So that
𝜌 𝑧
𝑉0
𝑧 + 0 (𝑑 − 𝑧)
𝑑
2𝜖
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
𝜌
𝑑𝑉
𝐚𝑧 =
𝐚𝑧 + 0 (2𝑧 − 𝑑) 𝐚𝑧 V∕m
𝑑𝑧
𝑑
2𝜖
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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 𝜖𝑟 = 𝜖𝑟1 for 0 < 𝑧 < 𝑏 (region 1), and 𝜖𝑟 = 𝜖𝑟2 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
3. 𝐷1 = 𝐷2 at 𝑧 = 𝑏 ⇒ 𝜖𝑟1
𝑑𝑉1 |
|
𝑑𝑧 |𝑏
= 𝜖𝑟2
𝑑𝑉2 |
|
𝑑𝑧 |𝑏
4. 𝑉1 = 𝑉2 at 𝑧 = 𝑏
The boundary conditions are now applied in order. Beginning with the first condition:
0 = 𝐴(0) + 𝐵 ⇒ 𝐵 = 0
Second condition:
𝑉0 = 𝐶𝑑 + 𝐷 ⇒ 𝐷 = 𝑉0 − 𝐶𝑑
Third condtion:
𝜖𝑟1 𝐴 = 𝜖𝑟2 𝐶 ⇒ 𝐶 =
𝜖𝑟1
𝐴
𝜖𝑟2
Fourth condition:
𝐴𝑏 + 𝐵 = 𝐶𝑏 + 𝐷 ⇒ 𝐴𝑏 + 0 =
𝜖𝑟1
𝜖
𝐴𝑏 + 𝑉0 − 𝑟1 𝐴𝑑
𝜖𝑟2
𝜖𝑟2
Solving the above for 𝐴 gives:
𝐴=
𝑉0
𝜖𝑟1
(𝑑
𝜖𝑟2
𝑏+
− 𝑏)
Making all substitutions back into the general solutions, we finally get:
𝑉1 =
and
𝑏+
⎡
𝑉2 (𝑧) = 𝑉0 ⎢1 −
⎢
⎣
𝑉0 𝑧
𝜖𝑟1
(𝑑
𝜖𝑟2
− 𝑏)
(0 < 𝑧 < 𝑏)
⎤
(𝑑 − 𝑧) ⎥
(𝑏 < 𝑧 < 𝑑)
𝜖𝑟2
𝑏 + (𝑑 − 𝑏) ⎥
𝜖𝑟1
⎦
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6.32. A uniform volume charge has constant density 𝜌𝑣 = 𝜌0 C∕m3 , and fills the region 𝑟 < 𝑎, in which
permittivity 𝜖 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:
(
) −𝜌
−𝜌0 𝑟2 𝐶1
𝑑𝑉
1 𝑑
0
𝑟2
=
+
⇒ 𝑉 (𝑟) =
+ 𝐶2
∇2 𝑉 = 2
𝑑𝑟
𝜖
6𝜖0
𝑟
𝑟 𝑑𝑟
We require that 𝑉 is finite at the orgin (or as 𝑟 → 0), and so therefore 𝐶1 = 0. Next, 𝑉 = 0 at
𝑟 = 𝑎, which gives 𝐶2 = 𝜌0 𝑎2 ∕6𝜖. 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 ⇒ 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
{𝜌
0
(𝑎2 − 𝑟2 ) 𝑟 < 𝑎
6𝜖
𝑉 (𝑟) =
0 𝑟>𝑎
b) the electric field intensity, 𝐄, everywhere: Use
𝐄 = −∇𝑉 =
𝜌 𝑟
−𝑑𝑉
𝐚𝑟 = 0 𝐚𝑟 𝑟 < 𝑎
𝑑𝑟
3𝜖
Outside (𝑟 > 𝑎), the potential is zero, and so 𝐄 = 0 there as well.
6.33. The functions 𝑉1 (𝜌, 𝜙, 𝑧) and 𝑉2 (𝜌, 𝜙, 𝑧) both satisfy Laplace’s equation in the region 𝑎 < 𝜌 < 𝑏,
0 ≤ 𝜙 < 2𝜋, −𝐿 < 𝑧 < 𝐿; each is zero on the surfaces 𝜌 = 𝑏 for −𝐿 < 𝑧 < 𝐿; 𝑧 = −𝐿 for 𝑎 < 𝜌 < 𝑏;
and 𝑧 = 𝐿 for 𝑎 < 𝜌 < 𝑏; and each is 100 V on the surface 𝜌 = 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 (𝜌 = 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 .
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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:
−𝜌0 𝑧
−𝜌0
𝑑𝑉1
𝑑 2 𝑉1
⇒
=
+ 𝐶1 (𝑧 < 𝑏)
=
2
𝜖
𝑑𝑧
𝜖
𝑑𝑧
In Region 2 (𝑧 > 𝑏), we solve Laplace’s equation, assuming 𝑧 variation only:
𝑑 2 𝑉2
𝑑𝑉2
= 𝐶1′ (𝑧 > 𝑏)
=0 ⇒
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
𝜖
−𝜌0 𝑏
𝑑𝑉1 |
𝑑𝑉 |
𝜖
+ 𝐶1
= 𝜖0 2 |
⇒ −𝜌0 𝑏 + 𝜖𝐶1 = 𝜖0 𝐶1′ ⇒ 𝐶1′ =
|
𝑑𝑧 |𝑧=𝑏
𝑑𝑧 |𝑧=𝑏
𝜖0
𝜖0
Substituting the above expression for 𝐶1′ , and performing a second integration on the Poisson
and Laplace equations, we find
𝑉1 (𝑧) = −
and
𝑉2 (𝑧) = −
𝜌0 𝑧2
+ 𝐶1 𝑧 + 𝐶2 (𝑧 < 𝑏)
2𝜖
𝜌0 𝑏𝑧
𝜖
+ 𝐶1 𝑧 + 𝐶2′ (𝑧 > 𝑏)
2𝜖0
𝜖0
Next, requiring 𝑉1 = 0 at 𝑧 = 0 leads to 𝐶2 = 0. Then, the requirement that 𝑉2 = 0 at 𝑧 = 𝑑
leads to
𝜌 𝑏𝑑
𝜌 𝑏𝑑
𝜖
𝜖
0 = − 0 + 𝐶1 𝑑 + 𝐶2′ ⇒ 𝐶2′ = 0 − 𝐶1 𝑑
𝜖0
𝜖0
𝜖0
𝜖0
With 𝐶2 and 𝐶2′ known, the voltages now become
𝑉1 (𝑧) = −
𝜌0 𝑧2
𝜌 𝑏
𝜖
+ 𝐶1 𝑧 and 𝑉2 (𝑧) = 0 (𝑑 − 𝑧) − 𝐶1 (𝑑 − 𝑧)
2𝜖
𝜖0
𝜖0
Finally, to evaluate 𝐶1 , we equate the two voltage expressions at 𝑧 = 𝑏:
[
]
𝜌0 𝑏 𝑏 + 2𝜖𝑟 (𝑑 − 𝑏)
𝑉1 |𝑧=𝑏 = 𝑉2 |𝑧=𝑏 ⇒ 𝐶1 =
2𝜖 𝑏 + 𝜖𝑟 (𝑑 − 𝑏)
where 𝜖𝑟 = 𝜖∕𝜖0 . Substituting 𝐶1 as found above into 𝑉1 and 𝑉2 leads to the final expressions for
the voltages:
[(
)
]
𝜌0 𝑏𝑧
𝑏 + 2𝜖𝑟 (𝑑 − 𝑏)
𝑧
−
(𝑧 < 𝑏)
𝑉1 (𝑧) =
2𝜖
𝑏 + 𝜖𝑟 (𝑑 − 𝑏)
𝑏
[
]
𝜌0 𝑏2
𝑑−𝑧
𝑉2 (𝑧) =
(𝑧 > 𝑏)
2𝜖0 𝑏 + 𝜖𝑟 (𝑑 − 𝑏)
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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
[
(
)]
𝜌0
𝑑𝑉1
𝑏 𝑏 + 2𝜖𝑟 (𝑑 − 𝑏)
𝐚 =
𝑧−
𝐚𝑧 V∕m (𝑧 < 𝑏)
𝐄1 = −
𝑑𝑧 𝑧
𝜖
2 𝑏 + 𝜖𝑟 (𝑑 − 𝑏)
[
]
𝜌0 𝑏2
𝑑𝑉2
1
𝐚 =
𝐚 V∕m (𝑧 > 𝑏)
𝐄2 = −
𝑑𝑧 𝑧
2𝜖0 𝑏 + 𝜖𝑟 (𝑑 − 𝑏) 𝑧
6.35. In spherical coordinates, a potential is known to be a function of 𝜃 only.
a) Find the function 𝑉 (𝜃) if 𝑣 = 10 V at 𝜃 = 90◦ and 𝐄 = −500 𝐚𝜃 V/m at 𝜃 = 30◦ , 𝑟 = 04 m:
Begin with the general solution of Laplace’s equation for theta variation (Eq. (41)):
( )
𝜃
+𝐵
𝑉 (𝜃) = 𝐴 ln tan
2
where the constants, 𝐴 and 𝐵, are to be found. Using the first boundary condition, 𝑉 = 10 at
𝜃 = 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 𝑑𝜃 𝜃
𝑟 sin 𝜃 𝜃
Next apply the second boundary condition:
𝐄(𝑟 = 0.4, 𝜃 = 30◦ ) = −
𝐴
𝐚 = −5𝐴𝐚𝜃 = −500𝐚𝜃 ⇒ 𝐴 = 100 V
0.4 sin 30◦ 𝜃
With the constants found, we construct the final answer:
( )
𝜃
+ 10 V
𝑉 (𝜃) = 100 ln tan
2
b) Find the electric field intensity in rectangular coordinates at 𝜃 = 90◦ , 𝑟 = 1 m: Use the previous
result
𝐴
𝐚 so that 𝐄(𝑟 = 1, 𝜃 = 90◦ ) = −100𝐚𝜃
𝐄=−
𝑟 sin 𝜃 𝜃
As 𝜃 = 90◦ , the field is evaluated in the 𝑥𝑦 plane, on which 𝐚𝜃 = −𝐚𝑦 .
Therefore 𝐄(𝑟 = 1, 𝜃 = 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,
∇ ⋅ (𝜓𝐆) = 𝐆 ⋅ ∇𝜓 + 𝜓∇ ⋅ 𝐆, where 𝜓 and 𝐆 are scalar and vector functions, respectively. Determine
a general rule on the allowed directions in which 𝜖 may vary with respect to the electric field.
In the original derivation of Poisson’s equation, we started with ∇ ⋅ 𝐃 = 𝜌𝑣 , where 𝐃 = 𝜖𝐄.
Therefore
∇ ⋅ 𝐃 = ∇ ⋅ (𝜖𝐄) = −∇ ⋅ (𝜖∇𝑉 ) = −∇𝑉 ⋅ ∇𝜖 − 𝜖∇2 𝑉 = 𝜌𝑣
We see from this that Poisson’s equation, ∇2 𝑉 = −𝜌𝑣 ∕𝜖, results when ∇𝑉 ⋅ ∇𝜖 = 0. In words, 𝜖
is allowed to vary, provided it does so in directions that are normal to the local electric field.
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6.37. Coaxial conducting cylinders are located at 𝜌 = 0.5 cm and 𝜌 = 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 (𝜌) = 100
ln(.012∕𝜌)
V
ln(.012∕.005)
We seek 𝜌 at which 𝑉 = 20 V, and thus we need to solve:
20 = 100
ln(.012∕𝜌)
.012
= 1.01 cm
⇒ 𝜌=
ln(2.4)
(2.4)0.2
b) 𝐸𝜌 𝑚𝑎𝑥 : We have
𝐸𝜌 = −
𝑑𝑉
100
𝜕𝑉
=−
=
𝜕𝜌
𝑑𝜌
𝜌 ln(2.4)
whose maximum value will occur at the inner cylinder, or at 𝜌 = .5 cm:
𝐸𝜌 𝑚𝑎𝑥 =
100
= 2.28 × 104 V∕m = 22.8 kV∕m
.005 ln(2.4)
c) 𝜖𝑟 if the charge per meter length on the inner cylinder is 20 nC/m: The capacitance per meter
length is
2𝜋𝜖0 𝜖𝑟
𝑄
=
𝐶=
ln(2.4) 𝑉0
We solve for 𝜖𝑟 :
𝜖𝑟 =
(20 × 10−9 ) ln(2.4)
= 3.15
2𝜋𝜖0 (100)
6.38. Repeat Problem 6.37, but with the dielectric only partially filling the volume, within 0 < 𝜙 < 𝜋, and
with free space in the remaining volume.
We note that the dielectric changes with 𝜙, and not with 𝜌. 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 = 𝜖𝑟 𝜖0 𝐸𝜌,𝑚𝑎𝑥 (𝜋𝑎) + 𝜖0 𝐸𝜌,𝑚𝑎𝑥 (𝜋𝑎) = 𝜖0 𝐸𝜌,𝑚𝑎𝑥 (𝜋𝑎)(1 + 𝜖𝑟 ) C∕m
Using the numbers given or found in Problem 6.37, we obtain
1 + 𝜖𝑟 =
20 × 10−9 C∕m
= 6.31 ⇒ 𝜖𝑟 = 5.31
(8.852 × 10−12 )(22.8 × 103 V∕m)(0.5 × 10−2 m)𝜋
We may also note that the average dielectric constant in this problem, (𝜖𝑟 + 1)∕2, is the same as
that of the uniform dielectric constant found in Problem 6.37.
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6.39. The two conducting planes illustrated in Fig. 6.14 are defined by 0.001 < 𝜌 < 0.120 m, 0 < 𝑧 < 0.1 m,
𝜙 = 0.179 and 0.188 rad. The medium surrounding the planes is air. For region 1, 0.179 < 𝜙 < 0.188,
neglect fringing and find:
a) 𝑉 (𝜙): The general solution to Laplace’s equation will be 𝑉 = 𝐶1 𝜙 + 𝐶2 , and so
20 = 𝐶1 (.188) + 𝐶2 and 200 = 𝐶1 (.179) + 𝐶2
Subtracting one equation from the other, we find
−180 = 𝐶1 (.188 − .179) ⇒ 𝐶1 = −2.00 × 104
Then
20 = −2.00 × 104 (.188) + 𝐶2 ⇒ 𝐶2 = 3.78 × 103
Finally, 𝑉 (𝜙) = (−2.00 × 104 )𝜙 + 3.78 × 103 V.
b) 𝐄(𝜌): Use
𝐄(𝜌) = −∇𝑉 = −
1 𝑑𝑉
2.00 × 104
=
𝐚𝜙 V∕m
𝜌 𝑑𝜙
𝜌
c) 𝐃(𝜌) = 𝜖0 𝐄(𝜌) = (2.00 × 104 𝜖0 ∕𝜌) 𝐚𝜙 C∕m2 .
d) 𝜌𝑠 on the upper surface of the lower plane: We use
2.00 × 104
2.00 × 104
|
=
𝜌𝑠 = 𝐃 ⋅ 𝐧|
𝐚𝜙 ⋅ 𝐚𝜙 =
C∕m2
|𝑠𝑢𝑟𝑓 𝑎𝑐𝑒
𝜌
𝜌
e) 𝑄 on the upper surface of the lower plane: This will be
.1
𝑄𝑡 =
.120
∫0 ∫.001
2.00 × 104 𝜖0
𝑑𝜌 𝑑𝑧 = 2.00 × 104 𝜖0 (.1) ln(120) = 8.47 × 10−8 C = 84.7 nC
𝜌
f) Repeat 𝑎) to 𝑐) for region 2 by letting the location of the upper plane be 𝜙 = .188 − 2𝜋, and then
find 𝜌𝑠 and 𝑄 on the lower surface of the lower plane. Back to the beginning, we use
20 = 𝐶1′ (.188 − 2𝜋) + 𝐶2′ and 200 = 𝐶1′ (.179) + 𝐶2′
Subtracting one from the other, we find
−180 = 𝐶1′ (.009 − 2𝜋) ⇒ 𝐶1′ = 28.7
Then 200 = 28.7(.179) + 𝐶2′ ⇒ 𝐶2′ = 194.9. Thus 𝑉 (𝜙) = 28.7𝜙 + 194.9 in region 2. Then
𝐄=−
28.7𝜖0
28.7
𝐚𝜙 V∕m and 𝐃 = −
𝐚𝜙 C∕m2
𝜌
𝜌
𝜌𝑠 on the lower surface of the lower plane will now be
𝜌𝑠 = −
28.7𝜖0
28.7𝜖0
𝐚𝜙 ⋅ (−𝐚𝜙 ) =
C∕m2
𝜌
𝜌
The charge on that surface will then be 𝑄𝑏 = 28.7𝜖0 (.1) ln(120) = 122 pC.
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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
Δ𝑉
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 𝜖(𝜌) = 𝜖0 (1 + 𝜌2 ∕𝑎2 ). Find:
a) 𝑉 (𝑧): Since 𝜖 varies in the direction normal to 𝐄, Laplace’s equation applies, and we write
∇2 𝑉 =
𝑑 2𝑉
= 0 ⇒ 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 = 𝜖𝐄 = −𝜖0 (1 + 𝜌2 ∕𝑎2 )(𝑉0 ∕𝑑) 𝐚𝑧 C∕m2 . The charge
density on the top plate is then 𝜌𝑠 = 𝐃 ⋅ −𝐚𝑧 = 𝜖0 (1 + 𝜌2 ∕𝑎2 )(𝑉0 ∕𝑑) C∕m2 . From this we find
the charge on the top plate:
𝑄=
∫0
2𝜋
∫0
𝑎
𝜖0 (1 + 𝜌2 ∕𝑎2 )(𝑉0 ∕𝑑) 𝜌 𝑑𝜌 𝑑𝜙 =
3𝜋𝑎2 𝜖0 𝑉0
C
2𝑑
d) 𝐶. The capacitance is 𝐶 = 𝑄∕𝑉0 = 3𝜋𝑎2 𝜖0 ∕(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
−
−
1
𝑏
1
𝑏
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 = −
𝐸𝑟,𝑚𝑎𝑥 = 𝐸(𝑟 = 𝑎) =
𝑉 𝐚
𝑑𝑉
𝐚𝑟 = ( 0 𝑟 )
𝑑𝑟
𝑟2 𝑎1 − 1𝑏
𝑉0
100
=
= 26.7 V∕mm = 26.7 kV∕m
𝑎(1 − (𝑎∕𝑏)) 5(1 − (5∕20))
c) Find 𝜖𝑟 if the surface charge density on the inner sphere is 1.0 𝜇C∕m2 : 𝜌𝑠 will be equal in magnitude to the electric flux density at 𝑟 = 𝑎. So 𝜌𝑠 = (2.67 × 104 V∕m)𝜖𝑟 𝜖0 = 10−6 C∕m2 . Thus
𝜖𝑟 = 4.23. Note, in the first printing, the given charge density was 100 𝜇C∕m2 , leading to a
ridiculous answer of 𝜖𝑟 = 423.
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6.42. The hemisphere 0 < 𝑟 < 𝑎, 0 < 𝜃 < 𝜋∕2, is composed of homogeneous conducting material of
conductivity 𝜎. The flat side of the hemisphere rests on a perfectly-conducting plane. Now, the material
within the conical region 0 < 𝜃 < 𝛼, 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 𝜃-variation only in the potential. Laplace’s equation is therefore:
(
)
𝑑
1
𝑑𝑉
∇2 𝑉 = 2
sin 𝜃
=0
𝑑𝜃
𝑟 sin 𝜃 𝑑𝜃
This reduces to
𝐶
𝑑𝑉
= 1 ⇒ 𝑉 (𝜃) = 𝐶1 ln tan (𝜃∕2) + 𝐶2
𝑑𝜃
sin 𝜃
We assume zero potential on the plane (at 𝜃 = 𝜋∕2), which means that 𝐶2 = 0. On the cone (at
𝜃 = 𝛼), we assume potential 𝑉0 , and so 𝑉0 = 𝐶1 ln tan(𝛼∕2)
⇒ 𝐶1 = 𝑉0 ∕ ln tan(𝛼∕2) The potential function is now
𝑉 (𝜃) = 𝑉0
ln tan(𝜃∕2)
𝛼 < 𝜃 < 𝜋∕2
ln tan(𝛼∕2)
The electric field is then
𝐄 = −∇𝑉 = −
𝑉0
1 𝑑𝑉
𝐚𝜃 = −
𝐚 V∕m
𝑟 𝑑𝜃
𝑟 sin 𝜃 ln tan(𝛼∕2) 𝜃
The total current can now be found by integrating the current density, 𝐉 = 𝜎𝐄, over any crosssection. Choosing the lower plane at 𝜃 = 𝜋∕2, this becomes
𝐼=
∫0
2𝜋
∫0
𝑎
−
𝜎𝑉0
2𝜋𝑎𝜎𝑉0
𝐚𝜃 ⋅ 𝐚𝜃 𝑟 𝑑𝑟 𝑑𝜙 = −
A
𝑟 sin(𝜋∕2) ln tan(𝛼∕2)
ln tan(𝛼∕2)
The resistance is finally
𝑉0
ln tan(𝛼∕2)
=−
ohms
𝐼
2𝜋𝑎𝜎
Note that 𝑅 is in fact positive (despite the minus sign) since ln tan(𝛼∕2) is negative when 𝛼 < 𝜋∕2
(which it must be).
𝑅=
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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 𝜖𝑟1 (𝑎 < 𝑟 < 𝑐) and 𝜖𝑟2 (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 𝜖𝑟1 = 𝜖𝑟2 : In spherical coordinates, Laplace’s equation with radial variation
only is:
(
)
𝑑𝑉
1 𝑑
𝑟2
=0
∇2 𝑉 = 2
𝑑𝑟
𝑟 𝑑𝑟
which we work with to find:
(
)
(
)
𝐶
𝐶
𝑑
𝑑𝑉
𝑑𝑉
𝑑𝑉
𝑟2
= 0 ⇒ 𝑟2
= 𝐶𝑎 ⇒
= 2𝑎 ⇒ 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=
𝐶
𝐶3
+ 𝐶4 ⇒ 𝐶4 = − 3
𝑏
𝑏
The second one gives:
𝐶1
𝐶
+ 𝐶2 ⇒ 𝐶2 = 𝑉0 − 1
𝑎
𝑎
As the electric field is the negative gradient of the potential field, the third condition can be written as:
(
)
)
(
−𝐶3
𝑑𝑉2 |
−𝐶1
𝑑𝑉1 |
= 𝜖2
⇒ 𝜖1 𝐶1 = 𝜖2 𝐶3
𝜖1
| = 𝜖2
| ⇒ 𝜖1
𝑑𝑟 |𝑐
𝑑𝑟 |𝑐
𝑐2
𝑐2
𝑉0 =
Finally, the fourth condition gives:
𝐶
𝐶
𝐶
𝐶1
𝐶1
𝐶
+ 𝐶2 = 3 + 𝐶4 ⇒
+ 𝑉0 − 1 = 3 − 3
𝑐
𝑐
𝑐
𝑎
𝑐
𝑏
Using the result of the third condition, the above becomes:
(
𝐶1
)
(
)
𝜖
1 1
1 1
+ 𝑉0 = 1 𝐶1
−
−
𝑐 𝑎
𝜖2
𝑐 𝑏
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6.43 (continued) The constants can now be tabulated:
𝐶1 =
𝜖1
𝜖2
⎡
⎢
𝐶2 = 𝑉0 ⎢1 −
⎢
⎣
𝐶3 = (
𝐶4 = (
1
𝑐
𝑉0
) (
)
−
+ 𝑎1 − 1𝑐
𝜖1
𝜖2
⎤
⎥
1∕𝑎
(
) (
)⎥
1
1
1
1
−𝑏 + 𝑎−𝑐 ⎥
𝑐
⎦
(
1
𝑐
1
𝑐
−
1
𝑏
1
𝑏
)
𝑉0
+
𝜖2
𝜖1
(
1
𝑎
−
1
𝑐
)
−𝑉0 ∕𝑏
)
(
)
𝜖
+ 𝜖2 𝑎1 − 1𝑐
−
1
𝑏
1
Substituting the four constants into the earlier voltage expressions gives the final results:
(
)
⎤
− 1𝑟
⎥
) (
) ⎥ (𝑎 ≤ 𝑟 ≤ 𝑐)
(
𝜖1 1
− 1𝑏 + 𝑎1 − 1𝑐 ⎥
𝜖2 𝑐
⎦
)
(
𝑉0 1𝑟 − 1𝑏
𝑉2 (𝑟) = (
(
)
) (𝑐 ≤ 𝑟 ≤ 𝑏)
𝜖2 1
1
1
1
+
−
−
𝑐
𝑏
𝜖
𝑎
𝑐
⎡
⎢
𝑉1 (𝑟) = 𝑉0 ⎢1 −
⎢
⎣
1
𝑎
1
6.44. A potential field in free space is given as 𝑉 = 100 ln tan(𝜃∕2) + 50 V.
a) Find the maximum value of |𝐄𝜃 | on the surface 𝜃 = 40◦ for 0.1 < 𝑟 < 0.8 m, 60◦ < 𝜙 < 90◦ .
First
𝐄=−
100
100
100
1 𝑑𝑉
𝐚𝜃 = −
𝐚𝜃 = −
𝐚
𝐚𝜃 = −
2
𝑟 𝑑𝜃
2𝑟 sin(𝜃∕2) cos(𝜃∕2)
𝑟 sin 𝜃 𝜃
2𝑟 tan(𝜃∕2) cos (𝜃∕2)
This will maximize at the smallest value of 𝑟, or 0.1:
𝐄𝑚𝑎𝑥 (𝜃 = 40◦ ) = 𝐄(𝑟 = 0.1, 𝜃 = 40◦ ) = −
100
𝐚 = 1.56 𝐚𝜃 kV∕m
0.1 sin(40) 𝜃
b) Describe the surface 𝑉 = 80 V: Set 100 ln tan 𝜃∕2+50 = 80 and solve for 𝜃: Obtain ln tan 𝜃∕2 =
0.3 ⇒ tan 𝜃∕2 = 𝑒.3 = 1.35 ⇒ 𝜃 = 107◦ (the cone surface at 𝜃 = 107 degrees).
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6.45. In free space, let 𝜌𝑣 = 200𝜖0 ∕𝑟2.4 .
a) Use Poisson’s equation to find 𝑉 (𝑟) if it is assumed that 𝑟2 𝐸𝑟 → 0 when 𝑟 → 0, and also that
𝑉 → 0 as 𝑟 → ∞: With 𝑟 variation only, we have
(
)
𝜌
𝑑𝑉
1 𝑑
∇2 𝑉 = 2
𝑟2
= − 𝑣 = −200𝑟−2.4
𝑑𝑟
𝜖
𝑟 𝑑𝑟
(
)
𝑑𝑉
𝑑
𝑟2
= −200𝑟−.4
𝑑𝑟
𝑑𝑟
or
Integrate once:
or
(
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 → 0 when 𝑟 → 0 Therefore 𝐶1 = 0. Integrate again
to find:
333.3 −.4
𝑉 (𝑟) =
𝑟 + 𝐶2
.4
From our second boundary condition, 𝑉 → 0 as 𝑟 → ∞, 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
200𝜖0 ′ 2
𝑟.6
(𝑟
)
𝑑𝑟
=
800𝜋𝜖
4𝜋𝑟2 𝐷𝑟 = 4𝜋
0
∫0 (𝑟′ )2.4
.6
Thus
𝐸𝑟 =
800𝜋𝜖0 𝑟.6
𝐷𝑟
=
= 333.3𝑟−1.4 V∕m
𝜖0
.6(4𝜋)𝜖0 𝑟2
Now
𝑟
𝑉 (𝑟) = −
∫∞
333.3(𝑟′ )−1.4 𝑑𝑟′ = 833.3𝑟−.4 V
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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 𝜌0 . Assume permittivity 𝜖0
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 ≤ 𝑎) becomes
(
)
𝜌 𝑟2 𝐶
𝜌
1 𝑑
2 𝑑𝑉1
2
∇ 𝑉1 = 2
𝑟
= − 0 ⇒ 𝑉1 (𝑟) = − 0 + 1 + 𝐶2 (𝑟 ≤ 𝑎)
𝑑𝑟
𝜖0
6𝜖0
𝑟
𝑟 𝑑𝑟
For region 2 (𝑟 ≥ 𝑎) there is no charge and so Laplace’s equation becomes
(
)
𝐶
𝑑𝑉
1 𝑑
𝑟2 2 = 0 ⇒ 𝑉2 (𝑟) = 3 + 𝐶4 (𝑟 ≥ 𝑎)
∇2 𝑉2 = 2
𝑑𝑟
𝑟
𝑟 𝑑𝑟
Now, as 𝑟 → ∞, 𝑉2 → 0, so therefore 𝐶4 = 0. Also, as 𝑟 → 0, 𝑉1 must be finite, so therefore
𝐶1 = 0. Then, 𝑉 must be continuous across the boundary, 𝑟 = 𝑎:
𝐶
𝐶
𝜌 𝑎2
𝜌 𝑎2
|
|
⇒ − 0 + 𝐶2 = 3 ⇒ 𝐶2 = 3 + 0
𝑉1 | = 𝑉2 |
|𝑟=𝑎
|𝑟=𝑎
6𝜖0
𝑎
𝑎
6𝜖0
So now
𝑉1 (𝑟) =
𝜌0 2
𝐶
𝐶
(𝑎 − 𝑟2 ) + 3 and 𝑉2 (𝑟) = 3
6𝜖0
𝑎
𝑟
Finally, since the permittivity is 𝜖0 everywhere, the electric field will be continuous at 𝑟 = 𝑎.
This is equivalent to the continuity of the voltage derivatives:
𝐶
𝜌 𝑎3
𝜌 𝑎
𝑑𝑉2 |
𝑑𝑉1 |
⇒ − 0 = − 23 ⇒ 𝐶3 = 0
| =
|
𝑑𝑟 |𝑟=𝑎
𝑑𝑟 |𝑟=𝑎
3𝜖0
3𝜖0
𝑎
So the potentials in their final forms are
𝑉1 (𝑟) =
𝜌0
𝜌 𝑎3
(3𝑎2 − 𝑟2 ) and 𝑉2 (𝑟) = 0
6𝜖0
3𝜖0 𝑟
The requested absolute potential at the origin is now 𝑉1 (𝑟 = 0) = 𝜌0 𝑎2 ∕(2𝜖0 ) V.
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Engineering Electromagnetics 9th solution
Engineering Electromagnetics (Chungnam National University)
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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 =
∫−∞
∞ 𝐼𝑑𝑧 𝐚 × [2𝐚 + 3𝐚 + (4 − 𝑧)𝐚 ]
∞
𝐼𝑑𝑧[2𝐚𝑦 − 3𝐚𝑥 ]
𝐼𝑑𝐋 × 𝐚𝑅
𝑧
𝑥
𝑦
𝑧
=
=
∫−∞
∫−∞ 4𝜋(𝑧2 − 8𝑧 + 29)3∕2
4𝜋𝑅2
4𝜋(𝑧2 − 8𝑧 + 29)3∕2
Using integral tables, this evaluates as
]∞
[
𝐼 2(2𝑧 − 8)(2𝐚𝑦 − 3𝐚𝑥 )
𝐼
𝐇𝑎 =
=
(2𝐚 − 3𝐚𝑥 )
2
1∕2
4𝜋 52(𝑧 − 8𝑧 + 29)
26𝜋 𝑦
−∞
Then with 𝐼 = 8 mA, we finally obtain 𝐇𝑎 = −294𝐚𝑥 + 196𝐚𝑦 𝜇A∕m
b) Repeat if the filament is located at 𝑥 = −1, 𝑦 = 2: In this case the Biot-Savart integral becomes
∞
𝐇𝑏 =
∫−∞
𝐼𝑑𝑧 𝐚𝑧 × [(2 + 1)𝐚𝑥 + (3 − 2)𝐚𝑦 + (4 − 𝑧)𝐚𝑧 ]
4𝜋(𝑧2
− 8𝑧 +
26)3∕2
∞
=
𝐼𝑑𝑧[3𝐚𝑦 − 𝐚𝑥 ]
∫−∞ 4𝜋(𝑧2 − 8𝑧 + 26)3∕2
Evaluating as before, we obtain with 𝐼 = 8 mA:
]∞
[
𝐼
𝐼 2(2𝑧 − 8)(3𝐚𝑦 − 𝐚𝑥 )
=
𝐇𝑏 =
(3𝐚 − 𝐚𝑥 ) = −127𝐚𝑥 + 382𝐚𝑦 𝜇A∕m
2
1∕2
4𝜋 40(𝑧 − 8𝑧 + 26)
20𝜋 𝑦
−∞
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 𝜇A∕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 √
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 × (𝑥0 𝐚𝑥 − 𝑧 𝐚𝑧 )
𝐼𝑑𝐋 × 𝐚𝑅
=
=
4𝜋𝑅2
4𝜋(𝑥20 + 𝑧2 )3∕2
4𝜋(𝑥20 + 𝑧2 )3∕2
where 𝐚𝑦 would be normal to the plane of the triangle. The magnetic field at 𝑥0 is then
𝓁∕2
𝐇=
𝐼 𝑑𝑧 𝑥0 𝐚𝑦
∫−𝓁∕2 4𝜋(𝑥2 + 𝑧2 )3∕2
0
=
𝐼 𝑧 𝐚𝑦
𝐼𝓁 𝐚𝑦
|𝓁∕2
=
|
√
√
|−𝓁∕2
4𝜋𝑥0 𝑥20 + 𝑧2
2𝜋𝑥0 𝓁 2 + 4𝑥20
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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∕(2𝜋𝓁) 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∕(2𝜋𝓁) 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 𝐚𝜙 direction, and is of radius 𝑎. Find 𝐇 everywhere on the 𝑧 axis: Use the Biot-Savart
Law:
𝐼𝑑𝐋 × 𝐚𝑅
𝐇=
∫𝑙𝑜𝑜𝑝 4𝜋𝑅2
√
√
where in this case, 𝑑𝐋 = 𝑎𝑑𝜙𝐚𝜙 , 𝑅 = 𝑎2 + 𝑧2 , and 𝐚𝑅 = (−𝑎𝐚𝜌 + 𝑧𝐚𝑧 )∕ 𝑎2 + 𝑧2 . With these
substitutions, we get
𝐇=
∫0
2𝜋
𝐼𝑎𝑑𝜙𝐚𝜙 × (−𝑎𝐚𝜌 + 𝑧𝐚𝑧 )
4𝜋(𝑎2 + 𝑧2 )3∕2
=
∫0
2𝜋
𝐼𝑎2 𝐚𝑧 + 𝐼𝑎𝑧𝐚𝜌
4𝜋(𝑎2 + 𝑧2 )3∕2
𝑑𝜙
Note that when substituting 𝐚𝜌 = 𝐚𝑥 cos 𝜙 + 𝐚𝑦 sin 𝜙, the second term in the last integral will integrate
to zero, leaving
𝜋𝑎2 𝐼𝐚𝑧
𝐦
𝐇=
=
2
2
3∕2
2
2𝜋(𝑎 + 𝑧 )
2𝜋(𝑎 + 𝑧2 )3∕2
where the magnetic moment is defined by 𝐦 = 𝜋𝑎2 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 𝐚𝜙 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 𝐚𝜙 direction. Using the Biot-Savart law, we have 𝐼𝑑𝐋 = 𝐼𝑎𝑑𝜙 𝐚𝜙 ,
√
√
𝑅 = 𝑎2 + 𝑧2 , and 𝐚𝑅 = (𝑧𝐚𝑧 − 𝑎𝐚𝜌 )∕ 𝑎2 + 𝑧2 . The field on the 𝑧 axis is then
𝐇=
∫0
2𝜋
𝐼𝑎𝑑𝜙 𝐚𝜙 × (𝑧𝐚𝑧 − 𝑎𝐚𝜌 )
4𝜋(𝑎2
+
𝑧2 )3∕2
=
∫0
2𝜋
𝐼𝑎2 𝑑𝜙 𝐚𝑧
4𝜋(𝑎2
+
𝑧2 )3∕2
=
2(𝑎2
𝑎2 𝐼
𝐚𝑧 A∕m
+ 𝑧2 )3∕2
In obtaining this result, the term involving 𝐚𝜙 × 𝑧𝐚𝑧 = 𝑧𝐚𝜌 has integrated to zero, when taken
over the range 0 < 𝜙 < 2𝜋. Substitute 𝐚𝜌 = 𝐚𝑥 cos 𝜙 + 𝐚𝑦 sin 𝜙 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:
[
]
𝐇=
𝑎2 𝐼
2
1
1
]3∕2 + [
]3∕2
[
2
2
2
(𝑧 − ℎ) + 𝑎
(𝑧 + ℎ) + 𝑎2
𝐚𝑧 A∕m
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7.4 (continued) We can rewrite this in terms of normalized distances, 𝑧∕𝑎, and ℎ∕𝑎:
[(
)−3∕2 ((
)−3∕2 ]
)
)
(
𝑧 ℎ 2
𝑧 ℎ 2
𝐼
𝑎𝐇 =
+
𝐚𝑧 A
+1
+1
−
+
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.
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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∕(2𝜋𝜌) 𝐚𝜙 , which we transform
to cartesian to obtain:
−𝐼𝑦
𝐼𝑥
𝐚𝑥 +
𝐚𝑦
𝐇=
2𝜋(𝑥2 + 𝑦2 )
2𝜋(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
2𝜋(𝑦 + 𝑧 )
2𝜋(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
]
[
]
𝐼(𝑦 − 1)
𝐼(𝑦 + 1)
𝐼𝑧
𝐼𝑧
−
−
𝐚𝑦 +
𝐚𝑧
𝐇=
2𝜋[(𝑦 + 1)2 + 𝑧2 ] 2𝜋[(𝑦 − 1)2 + 𝑧2 ]
2𝜋[(𝑦 − 1)2 + 𝑧2 ] 2𝜋[(𝑦 + 1)2 + 𝑧2 ]
[
√
We now evaluate this at 𝑧 = 2, and find the magnitude ( 𝐇 ⋅ 𝐇), resulting in
𝐼
|𝐇| =
2𝜋
[(
2
2
− 2
2
𝑦 + 2𝑦 + 5 𝑦 − 2𝑦 + 5
)2
(
+
(𝑦 + 1)
(𝑦 − 1)
− 2
2
𝑦 − 2𝑦 + 5 𝑦 + 2𝑦 + 5
)2 ]1∕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 𝜌𝑠 lies on the disk, which rotates about the 𝑧 axis at angular velocity Ω 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 = 𝜌𝑠 𝐯 =
√
2
2
𝜌𝑠 𝜌Ω 𝐚𝜙 , 𝑅 = 𝑧 + 𝜌 , and 𝐚𝑅 = (𝑧 𝐚𝑧 − 𝜌 𝐚𝜌 )∕𝑅. The differential field at point 𝑧 is
𝑑𝐇 =
𝜌𝑠 𝜌 Ω 𝐚𝜙 × (𝑧 𝐚𝑧 − 𝜌 𝐚𝜌 )
𝜌𝑠 𝜌 Ω (𝑧 𝐚𝜌 + 𝜌 𝐚𝑧 )
𝐊𝑑𝑎 × 𝐚𝑅
=
𝜌 𝑑𝜌 𝑑𝜙 =
𝜌 𝑑𝜌 𝑑𝜙
2
2
2
3∕2
4𝜋𝑅
4𝜋(𝑧 + 𝜌 )
4𝜋(𝑧2 + 𝜌2 )3∕2
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7.6. (continued). On integrating the above over 𝜙 around a complete circle, the 𝐚𝜌 components cancel from
symmetry, leaving us with
2𝜋
𝑎
𝜌𝑠 𝜌 Ω 𝜌 𝐚𝑧
𝑎
𝜌𝑠 Ω 𝜌3 𝐚𝑧
𝑑𝜌
∫0 4𝜋(𝑧2 + 𝜌2 )3∕2
∫0 2(𝑧2 + 𝜌2 )3∕2
)
(
√
[
]𝑎
⎡ 2
2 1−
2 ∕𝑧2 ⎤
1
+
𝑎
𝑎
+
2𝑧
𝜌 Ω⎢
𝜌Ω √ 2
⎥
𝑧2
𝑧 + 𝜌2 + √
𝐚𝑧 = 𝑠 ⎢
= 𝑠
√
⎥ 𝐚𝑧 A∕m
2
2𝑧 ⎢
1 + 𝑎2 ∕𝑧2
𝑧2 + 𝜌2 0
⎥
⎦
⎣
𝐇(𝑧) =
∫0
𝜌 𝑑𝜌 𝑑𝜙 =
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 𝐚𝜙 in terms of 𝐚𝑥
and 𝐚𝑦 and angle 𝜙 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 × 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 𝐚𝜌 direction and is uniform at a given radius.
The surface current density can therefore be written as 𝐊(𝜌) = 2𝐼∕(𝜋𝜌) 𝐚𝜌 A∕m2 over the region
(0 < 𝜙 < 𝜋∕2). The Biot-Savart law applicable to surface current is written as
𝐊 × 𝐚𝑅
𝑑𝐴
∫𝑠 4𝜋𝑅2
√
√
where 𝑅 = 𝑧2 + 𝜌2 and 𝐚𝑅 = (𝑧𝐚𝑧 − 𝜌𝐚𝜌 )∕ 𝑧2 + 𝜌2 Substituting these and integrating over
the first quadrant yields the setup:
𝐇=
𝐇=
∫0
𝜋∕2
∫0
∞
2𝐼 𝐚𝜌 × (𝑧𝐚𝑧 − 𝜌𝐚𝜌 )
4𝜋 2 𝜌(𝑧2 + 𝜌2 )3∕2
𝜌 𝑑𝜌 𝑑𝜙 =
∫0
𝜋∕2
∫0
∞
−𝐼𝑧 𝐚𝜙
2𝜋 2 (𝑧2 + 𝜌2 )3∕2
𝑑𝜌 𝑑𝜙
Following the hint, we now substitute 𝐚𝜙 = 𝐚𝑦 cos 𝜙 − 𝐚𝑥 sin 𝜙, and write:
∞ (𝐚 cos 𝜙 − 𝐚 sin 𝜙)
∞
𝜋∕2
𝑑𝜌
𝑦
𝑥
𝐼𝑧
−𝐼𝑧
𝑑𝜌
𝑑𝜙
=
(𝐚
−
𝐚
)
𝑥
𝑦 ∫
2
2
2
3∕2
2
2
∫
∫
2𝜋 0
(𝑧 + 𝜌 )
2𝜋
(𝑧 + 𝜌2 )3∕2
0
0
∞
𝜌
𝐼𝑧
𝐼
|
= 2 (𝐚𝑥 − 𝐚𝑦 ) √
| = 2 (𝐚𝑥 − 𝐚𝑦 ) A∕m
|
0
2𝜋
2𝜋 𝑧
𝑧2 𝑧2 + 𝜌 2
𝐇=
b) repeat part 𝑎, but with the copper sheet occupying the entire 𝑥𝑦 plane. In this case, the 𝜙 limits
are (0 < 𝜙 < 2𝜋). The cos 𝜙 and sin 𝜙 terms would then integrate to zero, so the answer is just
that: 𝐇 = 0.
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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:
𝑧2
𝐇=
∫𝑧1
𝜌 tan 𝛼2 𝐼𝑑𝑧𝐚 × (𝜌𝐚 − 𝑧𝐚 )
𝜌 tan 𝛼2
𝐼𝜌𝐚𝜙 𝑑𝑧
𝐼𝑑𝐋 × 𝐚𝑅
𝑧
𝜌
𝑧
=
=
∫𝜌 tan 𝛼1
∫𝜌 tan 𝛼1 4𝜋(𝜌2 + 𝑧2 )3∕2
4𝜋𝑅2
4𝜋(𝜌2 + 𝑧2 )3∕2
The integral is evaluated (using tables) and gives the desired result:
]
[
𝐼𝑧𝐚𝜙
tan 𝛼1
tan 𝛼2
𝐼
|𝜌 tan 𝛼2
𝐇=
−√
𝐚𝜙
=
|
√
√
1 + tan2 𝛼
1 + tan2 𝛼
4𝜋𝜌 𝜌2 + 𝑧2 |𝜌 tan 𝛼1 4𝜋𝜌
2
1
𝐼
=
(sin 𝛼2 − sin 𝛼1 )𝐚𝜙
4𝜋𝜌
7.9. A uniform surface charge density 𝜌𝑠 = 𝜌0 C∕m2 is in the shape of a sphere of radius 𝑎. The sphere
rotates about the 𝑧 axis at angular velocity Ω rad/s. Find
a) the total current flowing: The surface current density will be 𝐊 = 𝜌𝑠 𝐯 = 𝜌0 Ω𝑎 sin 𝜃 𝐚𝜙 A/m. The
total current will be
𝐼=
∫
𝐊 ⋅ 𝐧𝑑𝐿 =
∫0
𝜋
𝜌0 𝑎Ω sin 𝜃 𝐚𝜙 ⋅ 𝐚𝜙 𝑎𝑑𝜃 = 2𝑎2 Ω𝜌0 A
b) 𝐇 at the origin. Make appropriate use of Eq. (6): The setup is
𝐇=
𝐊 × 𝐚𝑅
𝑑𝑎
∫ ∫𝑠 4𝜋𝑅2
where 𝑅 = 𝑎 and 𝐚𝑅 = −𝐚𝑟 . Substituting these, and the given current density, we have
𝐇=
∫0
2𝜋
∫0
𝜋
𝜌0 Ω𝑎 sin 𝜃 𝐚𝜙 × (−𝐚𝑟 )
4𝜋𝑎2
𝑎2 sin 𝜃𝑑𝜃𝑑𝜙 =
2𝜋𝑎3 𝜌0 Ω
4𝜋𝑎2
∫0
𝜋
− sin2 𝜃 𝐚𝜃 𝑑𝜃
Now, the 𝜃 dependence in 𝐚𝜃 must be included – done by converting 𝐚𝜃 to cylindrical components:
𝐚𝜃 = 𝐚𝜌 cos 𝜃 − 𝐚𝑧 sin 𝜃. With this substitution, we have
]𝜋
]
𝑎𝜌0 Ω 𝜋 [ 3
−𝑎𝜌0 Ω [ 1
sin 𝜃 𝐚𝑧 − cos 𝜃 sin2 𝜃 𝐚𝜌 𝑑𝜃 =
cos 𝜃(sin2 𝜃 + 2) 𝐚𝑧
2 ∫0
2
3
0
2𝑎𝜌0 Ω
𝐚𝑧 A∕m
=
3
𝐇=
Note that in terms of the total current (from part 𝑎), this is 𝐇 = 𝐼∕(3𝑎) 𝐚𝑧 A/m.
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7.10. A uniform volume charge density 𝜌𝑣 = 𝜌0 C∕m3 is in the shape of a sphere of radius 𝑎. The sphere
rotates about the 𝑧 axis at angular velocity Ω rad/s. Find
a) the total current flowing: This will be found through
𝐼=
∫ ∫
𝐉 ⋅ 𝑑𝐒 where 𝐉 = 𝜌𝑣 𝐯 = 𝜌0 Ω𝑟 sin 𝜃 𝐚𝜙 A∕m2
from which:
𝜋
𝐼=
∫0 ∫0
𝑎
𝜌0 Ω𝑟 sin 𝜃 𝐚𝜙 ⋅ 𝐚𝜙 𝑟𝑑𝑟𝑑𝜃 = 2𝜌0 Ω
∫0
𝑎
𝑟2 𝑑𝑟 =
2𝜌0 Ω𝑎3
A
3
b) the magnetic field, 𝐇, at the origin: Use the volume current version of the Biot-Savart Law:
2𝜋
𝜋
𝑎 𝜌 Ω𝑟 sin 𝜃 𝐚 × (−𝐚 )
𝐉 × 𝐚𝑅 𝑑𝑣
0
𝜙
𝑟 2
=
𝑟 sin 𝜃 𝑑𝑟𝑑𝜃𝑑𝜙
∫0 ∫0 ∫0
∫𝑣 4𝜋𝑅2
4𝜋𝑟2
𝑎
𝜋
−2𝜋𝜌0 Ω𝑟 sin2 𝜃 𝐚𝜃
=
𝑑𝑟𝑑𝜃𝑑𝜙
∫0 ∫0
4𝜋
𝐇=
where we used 𝑅 = 𝑟 and 𝐚𝑅 = −𝐚𝑟 as observed from the origin. Next, include the 𝜃 dependence
in 𝐚𝜃 by converting it to cylindrical components: 𝐚𝜃 = 𝐚𝜌 cos 𝜃 − 𝐚𝑧 sin 𝜃. The field expression
now becomes:
]
𝜌0 Ω 𝜋 𝑎 [
𝜌 Ω𝑎2 𝜋 3
𝑟 sin3 𝜃 𝐚𝑧 − 𝑟 cos 𝜃 sin2 𝜃 𝐚𝜌 𝑑𝑟𝑑𝜃 = − 0
sin 𝜃 𝐚𝑧 𝑑𝜃
2 ∫0 ∫0
4 ∫0
] 𝜋 𝑎2 𝜌 Ω
𝑎2 𝜌0 Ω [ 1
0
=−
cos 𝜃(sin2 𝜃 + 2) 𝐚𝑧 =
𝐚𝑧 A∕m
4
3
3
0
𝐇=−
Note that in terms of the total current (from part 𝑎), this is 𝐇 = 𝐼∕(2𝑎) 𝐚𝑧 A/m.
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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:
𝜋𝑎2 𝑛𝐼𝑑𝑧 𝐚𝑧
𝑑𝐇 =
2𝜋(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=
∫
𝐿∕2
𝑑𝐇 =
𝜋𝑎2 𝑛𝐼𝑑𝑧 𝐚𝑧
∫−𝐿∕2 2𝜋(𝑎2 + 𝑧2 )3∕2
=
𝑛𝐼𝐚𝑧
𝜋𝑎2 𝑛𝐼𝐚𝑧
𝑧
|𝐿∕2
A∕m
=√
|
√
2𝜋 𝑎2 𝑎2 + 𝑧2 |−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 → ∞ clearly yields 𝐇 → 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
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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
∮
𝐇 ⋅ 𝑑𝐋 = 2𝐻𝑖𝑛 × 𝑤 = 𝐼𝑒𝑛𝑐𝑙 = 𝐽 × 𝑤 × 𝑑 ⇒ 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 × 𝑤 = 𝐽 × 𝑤 × 𝐷 ⇒ 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
⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟ ⏟⏞⏞⏟⏞⏞⏟
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(0.3)
−10(0.2)
𝐚𝑥 −
𝐚𝑥 = −2.5𝐚𝑥 A∕m
𝐇=
2
2
⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟ ⏟⏞⏞⏟⏞⏞⏟
upper slab
lower slab
e) 𝑧 = 1.2m: This point lies above both slabs, where again fields cancel completely: Thus 𝐇 = 0.
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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 𝐚𝜙 .
a) Show that 𝐻 is not a function of 𝜙 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
𝜌 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 𝜙 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 𝐻𝜙 and 𝐻𝜌 are everywhere zero. First, if 𝐻𝜙 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 𝜙
direction. This we cannot do, and so 𝐻𝜙 must be zero. Another argument is that when applying
the Biot-Savart law, there is no current element that would produce a 𝜙 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 𝜌 > 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 𝜌 < 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
∮
𝐇 ⋅ 𝑑𝐋 = 𝐻𝑧 𝑑 = 𝐼𝑒𝑛𝑐𝑙 = 𝐾𝑎 𝑑 ⇒ 𝐻𝑧 = 𝐾𝑎
where 𝑑 is the length of the loop in the 𝑧 direction.
e) A second shell, 𝜌 = 𝑏, carries a current 𝐾𝑏 𝐚𝜙 . Find 𝐇 everywhere: For 𝜌 < 𝑎 we would have
both cylinders contributing, or 𝐻𝑧 (𝜌 < 𝑎) = 𝐾𝑎 + 𝐾𝑏 . Between the cylinders, we are outside the
inner one, so its field will not contribute. Thus 𝐻𝑧 (𝑎 < 𝜌 < 𝑏) = 𝐾𝑏 . Outside (𝜌 > 𝑏) the field
will be zero.
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7.14. A toroid having a cross section of rectangular shape is defined by the following surfaces: the cylinders
𝜌 = 2 and 𝜌 = 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 𝜌 = 3 cm. Find 𝐇 at the point 𝑃 (𝜌, 𝜙, 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
2𝜋𝜌 𝜙
(2 < 𝜌 < 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(2𝜋 × 3 × 10−2 ) = −3𝜋 A
This current is directed along negative 𝑧, which means that the current on the inner radius (𝜌 = 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 𝐚𝜙 within the toroid that is used in Ampere’s law.
So 𝐼𝑒𝑛𝑐𝑙 = +3𝜋 A. We can now proceed with what is requested:
a) 𝑃𝐴 (1.5cm, 0, 2cm): The radius, 𝜌 = 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 𝜙 and 𝑧 values do not
matter. We find
3𝐚𝜙
𝐼
= 71.4 𝐚𝜙 A∕m
𝐇𝐵 = 𝑒𝑛𝑐𝑙 𝐚𝜙 =
2𝜋𝜌
2(2.1 × 10−2 )
c) 𝑃𝐶 (2.7cm, 𝜋∕2, 2cm): again, 𝜙 and 𝑧 values make no difference, so
𝐇𝐶 =
3𝐚𝜙
2(2.7 × 10−2 )
= 55.6 𝐚𝜙 A∕m
d) 𝑃𝐷 (3.5cm, 𝜋∕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,
𝜃 = 0) and bottom (𝑟 = 𝑎, 𝜃 = 𝜋). 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
𝐚𝜙 = −
𝐚 A∕m (𝑟 > 𝑎)
𝐇=−
2𝜋𝜌
2𝜋𝑟 sin 𝜃 𝜙
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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 𝜌 will be the total current, 𝐼, divided by
the circumference at radius 𝜌:
𝐼
𝐊=
𝐚 A∕m
2𝜋𝜌 𝜌
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 −𝐚𝜙 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:
∮
𝐇 ⋅ 𝑑𝐋 = 2𝜋𝜌𝐻𝜙 = 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
𝐻𝜙 = −
𝐼
2𝜋𝜌
⇒
𝐇=−
𝐼
𝐚 A∕m
2𝜋𝜌 𝜙
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 = 2𝜋𝜌[𝐼∕(2𝜋𝜌)] = 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).
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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 𝜎(𝜌) = 𝜎0 𝜌 S/m, where 𝜎0 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 , 𝜎0 , and 𝜌:
This will be 𝐉 = 𝜎𝐄 = 𝜎0 𝐸0 𝜌 𝐚𝑧 A∕m2
b) Find 𝐸0 in terms of 𝐼:
∫ ∫
𝐼=
∫0
𝐉 ⋅ 𝐧 𝑑𝑎 =
2𝜋
∫0
𝑎
𝜎0 𝐸0 𝜌 𝐚𝑧 ⋅ 𝐚𝑧 𝜌𝑑𝜌𝑑𝜙 = 2𝜋𝜎0 𝐸0
𝑎3
3𝐼
V∕m
⇒ 𝐸0 =
3
2𝜋𝜎0 𝑎3
c) Express 𝐉 in terms of 𝐼 and other known parameters, excluding 𝐸0 : Using the results of part 𝑏,
we have again:
3𝐼𝜌
𝐉 = 𝜎𝐄 =
𝐚𝑧 A∕m2
2𝜋𝑎3
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 𝜌 < 𝑎:
∮
𝐇𝑖𝑛 ⋅ 𝑑𝐋 = 2𝜋𝜌𝐻𝜙 =
∫ ∫
𝐉 ⋅ 𝐧 𝑑𝑎 =
So that
𝐇𝑖𝑛 = 𝐻𝜙 𝑖𝑛 𝐚𝜙 =
Outside (𝜌 ≥ 𝑎), we have
∫0
2𝜋
∫0
𝜌
3𝐼𝜌′
𝜌3
′
′
𝐚
⋅
𝐚
𝜌
𝑑𝜌
𝑑𝜙
=
𝐼
𝑧
𝑧
2𝜋𝑎3
𝑎3
𝐼𝜌2
𝐚𝜙 A∕m2 (𝜌 ≤ 𝑎)
2𝜋𝑎3
2𝜋𝜌𝐻𝜙 𝑜𝑢𝑡 = 𝐼 ⇒ 𝐇𝑜𝑢𝑡 =
𝐼
𝐚 A∕m (𝜌 ≥ 𝑎)
2𝜋𝜌 𝜙
e) Verify your result of part 𝑑 by applying ∇ × 𝐇 = 𝐉: The curl in cylindrical coordinates will in
this case (with 𝐇 𝜙-directed and varying only with 𝜌) be the single term:
∇×𝐇=
1 𝑑(𝜌𝐻𝜙 )
𝐚𝑧
𝜌 𝑑𝜌
Inside the wire, we have
1 𝑑
∇ × 𝐇𝑖𝑛 =
𝜌 𝑑𝜌
(
𝐼𝜌3
2𝜋𝑎3
)
𝐚𝑧 =
3𝐼𝜌
𝐚𝑧 = 𝐉 (as found in part c)
2𝜋𝑎3
(
)
Outside the wire, we have
∇ × 𝐇𝑜𝑢𝑡
1 𝑑
=
𝜌 𝑑𝜌
𝐼𝜌
2𝜋𝜌
𝐚𝑧 = 0 (as expected)
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7.18. A wire of 3-mm radius is made up of an inner material (0 < 𝜌 < 2 mm) for which 𝜎 = 107 S/m, and
an outer material (2mm < 𝜌 < 3mm) for which 𝜎 = 4 × 107 S/m. If the wire carries a total current of
100 mA dc, determine 𝐇 everywhere as a function of 𝜌.
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 = 𝜋(.002)2 𝐽1 + 𝜋[(.003)2 − (.002)2 ]𝐽2 = 𝜋 (.002)2 𝜎1 + [(.003)2 − (.002)2 ]𝜎2 𝐸
Solve for 𝐸 to obtain
𝐸=
0.1
= 1.33 × 10−4 V∕m
𝜋[(4 × 10−6 )(107 ) + (9 × 10−6 − 4 × 10−6 )(4 × 107 )]
We next apply Ampere’s circuital law to a circular path of radius 𝜌, where 𝜌 < 2mm:
2𝜋𝜌𝐻𝜙1 = 𝜋𝜌2 𝐽1 = 𝜋𝜌2 𝜎1 𝐸 ⇒ 𝐻𝜙1 =
𝜎1 𝐸𝜌
= 663 A∕m
2
Next, for the region 2mm < 𝜌 < 3mm, Ampere’s law becomes
2𝜋𝜌𝐻𝜙2 = 𝜋[(4 × 10−6 )(107 ) + (𝜌2 − 4 × 10−6 )(4 × 107 )]𝐸
⇒ 𝐻𝜙2 = 2.7 × 103 𝜌 −
8.0 × 10−3
A∕m
𝜌
Finally, for 𝜌 > 3mm, the field outside is that for a long wire:
𝐻𝜙3 =
0.1
1.6 × 10−2
𝐼
=
=
A∕m
2𝜋𝜌 2𝜋𝜌
𝜌
7.19. In spherical coordinates, the surface of a solid conducting cone is described by 𝜃 = 𝜋∕4 and a conducting plane by 𝜃 = 𝜋∕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
2𝜋𝑟 𝑟
(𝜃 = 𝜋∕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 𝜃 = 𝜋∕4:
]−1
[
𝜋∕4
2𝜋
𝐼 𝐚𝑟
2
𝐚𝑟 =
𝑟2 sin 𝜃 ′ 𝑑𝜃 ′ 𝑑𝜙
𝐉(𝑟) = 𝐼
√ A∕m (0 < 𝜃 < 𝜋∕4)
∫0 ∫0
2
2𝜋𝑟 (1 − 1∕ 2)
c) Determine 𝐇 as a function of 𝑟 and 𝜃 in the region between the cone and the plane: From symmetry, we expect 𝐇 to be 𝜙-directed and uniform at constant 𝑟 and 𝜃. Ampere’s circuital law can
therefore be stated as:
∮
𝐇 ⋅ 𝑑𝐋 = 2𝜋𝑟 sin 𝜃 𝐻𝜙 = 𝐼 ⇒ 𝐇 =
𝐼
𝐚 A∕m (𝜋∕4 < 𝜃 < 𝜋∕2)
2𝜋𝑟 sin 𝜃 𝜙
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7.19
d) Determine 𝐇 as a function of 𝑟 and 𝜃 inside the cone: Again, 𝜙-directed 𝐇 is anticipated, so we
apply Ampere’s law in the following way:
∮
𝐇 ⋅ 𝑑𝐋 = 2𝜋𝑟 sin 𝜃 𝐻𝜙 =
∫𝑠
𝐉 ⋅ 𝑑𝐒 =
∫0
This becomes
2𝜋𝑟 sin 𝜃 𝐻𝜙 = −2𝜋
2𝜋
𝜃
𝐼 𝐚𝑟
2
′
′
√ ⋅ 𝐚 𝑟 sin 𝜃 𝑑𝜃 𝑑𝜙
∫0 2𝜋𝑟2 (1 − 1∕ 2) 𝑟
𝜃
′|
√ cos 𝜃 ||0
2𝜋(1 − 1∕ 2)
𝐼
]
(1 − cos 𝜃)
𝐚𝜙 A∕m (0 < 𝜃 < 𝜋∕4)
𝐇=
√
sin 𝜃
2𝜋𝑟(1 − 1∕ 2)
or
𝐼
[
As a test of this, note that the inside and outside fields (results of parts 𝑐 and 𝑑) are equal at the
cone surface (𝜃 = 𝜋∕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 𝜌2 𝐚𝜙 A/m.
a) Find 𝜎 as a function of 𝜌: Start by finding 𝐉 from 𝐇 by taking the curl. With 𝐇 𝜙-directed, and
varying with radius only, the curl becomes:
𝐉=∇×𝐇=
)
1 𝑑 (
1 𝑑 ( 5 3)
𝜌𝐻𝜙 𝐚𝑧 =
10 𝜌 𝐚𝑧 = 3 × 105 𝜌 𝐚𝑧 A∕m2
𝜌 𝑑𝜌
𝜌 𝑑𝜌
Then 𝐄 = 0.1∕20 = 0.005 𝐚𝑧 V/m, which we then use with 𝐉 = 𝜎𝐄 to find
𝜎=
3 × 105 𝜌
𝐽
=
= 6 × 107 𝜌 S∕m
𝐸
0.005
b) What is the resistance between the two ends? The current in the wire is
𝐼=
∫𝑠
𝐉 ⋅ 𝑑𝐒 = 2𝜋
∫0
𝑎
(3 × 105 𝜌) 𝜌 𝑑𝜌 = 6𝜋 × 105
(
)
1 3
𝑎 = 2𝜋 × 105 (0.005)3 = 0.079 A
3
Finally, 𝑅 = 𝑉0 ∕𝐼 = 0.1∕0.079 = 1.3 Ω
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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 = 𝑏𝜌 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
𝐉 ⋅ 𝑑𝐒 =
∫0
2𝜋
∫0
𝑎
𝑏𝜌 𝐚𝑧 ⋅ 𝐚𝑧 𝜌 𝑑𝜌 𝑑𝜙 =
2𝜋𝑏𝑎3
A
3
b) find 𝐇𝑖𝑛 (0 < 𝜌 < 𝑎), as a function of 𝜌: From the symmetry, 𝜙-directed 𝐇 (= 𝐻𝜙 𝐚𝜙 ) is expected
in the interior; this will be constant at a fixed radius, 𝜌. Apply Ampere’s circuital law to a circular
path of radius 𝜌 inside:
∮
𝐇𝑖𝑛 ⋅ 𝑑𝐋 = 2𝜋𝜌𝐻𝜙,𝑖𝑛 =
∫𝑠
So that
𝐇𝑖𝑛 =
𝐉 ⋅ 𝑑𝐒 =
∫0
2𝜋
∫0
𝜌
𝑏𝜌′ 𝐚𝑧 ⋅ 𝐚𝑧 𝜌′ 𝑑𝜌′ 𝑑𝜙 =
2𝜋𝑏𝜌3
3
𝑏𝜌2
𝐚 A∕m (0 < 𝜌 < 𝑎)
3 𝜙
c) find 𝐇𝑜𝑢𝑡 (𝜌 > 𝑎), as a function of 𝜌 Same as part 𝑏, except the path integral is taken at a radius
outside the wire:
∮
𝐇𝑜𝑢𝑡 ⋅ 𝑑𝐋 = 2𝜋𝜌𝐻𝜙,𝑜𝑢𝑡 =
∫𝑠
So that
𝐇𝑜𝑢𝑡 =
𝐉 ⋅ 𝑑𝐒 =
∫0
2𝜋
∫0
𝑎
𝑏𝜌 𝐚𝑧 ⋅ 𝐚𝑧 𝜌 𝑑𝜌 𝑑𝜙 =
2𝜋𝑏𝑎3
3
𝑏𝑎3
𝐚 A∕m (𝜌 > 𝑎)
3𝜌 𝜙
d) verify your results of parts 𝑏 and 𝑐 by using ∇ × 𝐇 = 𝐉: With a 𝜙 component of 𝐇 only, varying
only with 𝜌, the curl in cylindrical coordinates reduces to
𝐉=∇×𝐇=
1 𝑑
(𝜌𝐻𝜙 ) 𝐚𝑧
𝜌 𝑑𝜌
Apply this to the inside field to get
1 𝑑
𝐉𝑖𝑛 =
𝜌 𝑑𝜌
(
𝑏𝜌3
3
)
𝐚𝑧 = 𝑏𝜌 𝐚𝑧
For the outside field, we find
𝐉𝑜𝑢𝑡
1 𝑑
=
𝜌 𝑑𝜌
(
𝜌𝑏𝑎3
3𝜌
)
𝐚𝑧 = 0
as expected.
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7.22. A solid cylinder of radius 𝑎 and length 𝐿, where 𝐿 >> 𝑎, contains volume charge of uniform density
𝜌0 C∕m3 . The cylinder rotates about its axis (the 𝑧 axis) at angular velocity Ω rad/s.
a) Determine the current density 𝐉, as a function of position within the rotating cylinder: Use 𝐉 =
𝜌0 𝐯 = 𝜌0 𝜌Ω 𝐚𝜙 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 → ∞ (since 𝐿 >> 𝑎) thereby specializing our evaluation to
positions near the half-length. The on-axis field is therefore:
(
)
√
⎡ 2
2 1−
2 ∕𝑧2 ⎤
𝑎
+
2𝑧
1
+
𝑎
∞
𝜌0 Ω ⎢
⎥
𝐻𝑧 (𝜌 = 0) =
√
⎥ 𝑑𝑧
⎢
∫−∞ 2𝑧 ⎢
1 + 𝑎2 ∕𝑧2
⎥
⎣
⎦
[
]
∞
𝜌0 Ω
𝑎2
2𝑧2
+√
− 2𝑧 𝑑𝑧
=2
√
∫0
2
2
2
2
2
𝑧 +𝑎
𝑧 +𝑎
[ 2
]∞
√
√
𝑧√ 2
𝑎2
𝑧2
𝑎
2
2
2
2
2
= 2𝜌0 Ω
𝑧 +𝑎 −
ln(𝑧 + 𝑧 + 𝑎 ) +
ln(𝑧 + 𝑧 + 𝑎 ) −
2
2
2
2 0
]∞
]
[ √
[ √
= 𝜌0 Ω 𝑧 𝑧2 + 𝑎2 − 𝑧2
= 𝜌0 Ω 𝑧 𝑧2 + 𝑎2 − 𝑧2
0
𝑧→∞
Using the large 𝑧 approximation in the radical, we obtain
)
]
[ (
𝜌 Ω𝑎2
𝑎2
2
2
𝐻𝑧 (𝜌 = 0) = 𝜌0 Ω 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:
∮
𝐇 ⋅ 𝑑𝐋 = 𝐻𝑧 (𝜌 = 0)𝓁 + 𝐻𝑧 (𝜌 > 𝑎)𝓁 = 𝐼𝑒𝑛𝑐𝑙 =
∫𝑠
𝓁
𝐉 ⋅ 𝑑𝐒 =
∫0 ∫0
𝑎
𝜌0 𝜌Ω 𝐚𝜙 ⋅ 𝐚𝜙 𝑑𝜌 𝑑𝑧
𝜌0 Ω𝑎2
2
But we found earlier that 𝐻𝑧 (𝜌 = 0) = 𝜌0 Ω𝑎2 ∕2. Therefore, we identify the outside field,
𝐻𝑧 (𝜌 > 𝑎) = 0. Next, change the rectangular path only by displacing the central path component
off-axis by distance 𝜌, but still lying within the cylinder. The enclosed current is now somewhat
less, and Ampere’s law becomes
=𝓁
∮
𝐇 ⋅ 𝑑𝐋 = 𝐻𝑧 (𝜌)𝓁 + 𝐻𝑧 (𝜌 > 𝑎)𝓁 = 𝐼𝑒𝑛𝑐𝑙 =
=𝓁
∫𝑠
𝓁
𝐉 ⋅ 𝑑𝐒 =
∫0 ∫𝜌
𝑎
𝜌0 𝜌′ Ω 𝐚𝜙 ⋅ 𝐚𝜙 𝑑𝜌 𝑑𝑧
𝜌0 Ω 2
𝜌 Ω
(𝑎 − 𝜌2 ) ⇒ 𝐇(𝜌) = 0 (𝑎2 − 𝜌2 ) 𝐚𝑧 A∕m
2
2
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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 𝜌,
the curl in cylindrical coordinates becomes
∇×𝐇=−
𝑑𝐻𝑧
𝐚 = 𝜌0 Ω𝜌 𝐚𝜙 A∕m2 = 𝐉
𝑑𝜌 𝜙
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 𝐚𝜙 ,
where 𝐸0 and 𝐻0 are constants.
a) Using the cylindrical coordinate form of ∇ × 𝐇 = 𝐉, determine the required functional form of
the medium conductivity, 𝜎, in order for the preceding condition to be met: For a constant 𝐇 that
is 𝜙-directed, the curl of 𝐇 in cylindrical coordinates reduces to
∇×𝐇=
1 𝑑(𝜌𝐻𝜙 )
1
𝐚𝑧 = 𝐻0 𝐚𝑧 = 𝐉 = 𝜎𝐸0 𝐚𝑧
𝜌 𝑑𝜌
𝜌
We therefore identify:
𝜎=
𝐻0 1
(inverse radial dependence)
𝐸0 𝜌
b) Repeat part 𝑎, but use the spherical coordinate form of ∇ × 𝐇 = 𝐉: Again, for a constant 𝜙directed magnetic field, the curl in spherical coordinates reduces to:
∇×𝐇=
𝐻𝜙
𝐻
1 𝑑(𝐻𝜙 sin 𝜃)
1 𝑑(𝑟𝐻𝜙 )
𝐚𝑟 −
𝐚𝜃 =
cot 𝜃 𝐚𝑟 − 0 𝐚𝜃
𝑟 sin 𝜃
𝑑𝜃
𝑟 𝑑𝑟
𝑟
𝑟
Now use 𝐚𝜃 = 𝐚𝜌 cos 𝜃 − 𝐚𝑧 sin 𝜃 and 𝐚𝑟 = 𝐚𝜌 sin 𝜃 + 𝐚𝑧 cos 𝜃 to obtain
𝐻0
𝑟
𝐻0
=
𝑟
∇×𝐇=
[
]
cot 𝜃(𝐚𝜌 sin 𝜃 + 𝐚𝑧 cos 𝜃) − 𝐚𝜌 cos 𝜃 + 𝐚𝑧 sin 𝜃
]
[ 2
𝐻0
𝐻
cos 𝜃
+ sin 𝜃 𝐚𝑧 =
𝐚𝑧 = 0 𝐚𝑧
sin 𝜃
𝑟 sin 𝜃
𝜌
....thus leading to the same result as in part 𝑎.
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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,
∞
∑
𝑛=1
𝑦2
𝑦
1
𝜋
𝜋
+ 2𝜋𝑦
= −
2
2 2𝑦 𝑒 − 1
+𝑛
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:
𝐇=
𝐼𝑑𝐋 × 𝐚𝑅
∫
4𝜋𝑅2
(
)2
where 𝑅 = 𝑛2 + 𝑦2 + 𝑧2 for both wires, and where, for the wire at 𝑥 = +𝑛
−𝑛𝐚𝑥 + 𝑦𝐚𝑦 + 𝑧𝐚𝑧
𝐚+
=
(
)1∕2
𝑅
𝑛 2 + 𝑦2 + 𝑧 2
𝑛𝐚𝑥 + 𝑦𝐚𝑦 + 𝑧𝐚𝑧
𝐚−
=
(
)1∕2
𝑅
𝑛2 + 𝑦2 + 𝑧2
and
for the wire located at 𝑥 = −𝑛. Then with 𝐼𝑑𝐋 = 𝑑𝑧 𝐚𝑧 (𝐼 = 1) for both wires, the Biot-Savart
construction becomes
∞
𝐇=
∫−∞
[
]
𝑑𝑧 𝐚𝑧 × (−𝑛𝐚𝑥 + 𝑦𝐚𝑦 + 𝑧𝐚𝑧 ) + (𝑛𝐚𝑥 + 𝑦𝐚𝑦 + 𝑧𝐚𝑧 )
−𝑦𝐚𝑥 ∞
𝑑𝑧
=
(
)3∕2
(
)3∕2
∫
2𝜋
−∞ 𝑛2 + 𝑦2 + 𝑧2
4𝜋 𝑛2 + 𝑦2 + 𝑧2
This evaluates as
1
𝐇=−
𝜋
(
𝑦
2
𝑛 + 𝑦2
)
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,
𝐇𝑛𝑒𝑡
∞
1∑
=−
𝜋 𝑛=1
(
𝑦
2
𝑛 + 𝑦2
)
𝐚𝑥 A∕m
Using the given closed form of the series expansion, our final answer is
]
[
1
1
1
𝐚 A∕m
−
+
𝐇𝑛𝑒𝑡 = −
2 2𝜋𝑦 𝑒2𝜋𝑦 − 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 × 𝐚𝑁 = 𝐚𝑧 × 𝐚𝑦 = − 𝐚𝑥
2
2
2
Our answer of part 𝑎 approaches this value as 𝑦 → ∞, demonstrating that at large distances, the
parallel wires act like a current sheet. Interestingly, at close-in locations, such that 2𝜋𝑦 << 1, we
.
may expand 𝑒2𝜋𝑦 = 1 + 2𝜋𝑦, leading to the cancellation of the last two terms in the part 𝑎 result,
.
and again, 𝐇𝑛𝑒𝑡 = −1∕2 𝐚𝑥 .
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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 ≤ 𝑦 ≤ 4, 3 ≤ 𝑧 ≤ 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
)
(
)
(
𝜕𝐻𝑧 𝜕𝐻𝑦
𝑥𝑧2
2
+ 6𝑥 𝑧 𝐚𝑥
𝐽𝑥 = (∇ × 𝐇)𝑥 =
−
=−
𝜕𝑦
𝜕𝑧
(𝑦 + 1)2
The current through the strip is then
)
)4
4(
−2𝑧2
2𝑧2
+ 24𝑧 𝑑𝑦 𝑑𝑧 = −
𝐼 = 𝐉 ⋅ 𝐚𝑥 𝑑𝑎 = −
+ 24𝑧𝑦 𝑑𝑧
∫3
∫𝑠
∫3 ∫1
(𝑦 + 1)
(𝑦 + 1)2
1
4(
)4
)
(
1
3 2
𝑧 + 72𝑧 𝑑𝑧 = − 𝑧3 + 36𝑧2 = −259
=−
∫3 5
5
3
4
4(
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
3
1
2(1)𝑧2
2(4)𝑧2
−
3(2)2 (4)2 𝑑𝑦 +
𝑑𝑧 +
𝑑𝑧
∫4
∫4
∫3
∮
∫1
(4 + 1)
(1 + 1)
1
8
= 108(3) − (43 − 33 ) + 192(1 − 4) − (33 − 43 ) = −259
15
3
4
4
𝐼=
𝐇 ⋅ 𝑑𝐋 =
3(2)2 (3)2 𝑑𝑦 +
−
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 ∫𝑆 (∇ × 𝐇) ⋅ 𝑑𝐒 if 𝐇 = 3𝜌 𝐚𝜙 in cylindrical coordinates: First, the
1
√
√
intersection of the sphere with the 𝑥-𝑦 plane is a disk in the plane of radius 𝜌𝑑 = 42 − 32 = 7.
The curl of the given field (having a 𝜙 component that varies only with 𝜌) is
∇×𝐇=
)
1 𝑑 ( 2)
1 𝑑 (
𝜌𝐻𝜙 𝐚𝑧 =
3𝜌 𝐚𝑧 = 6 𝐚𝑧
𝜌 𝑑𝜌
𝜌 𝑑𝜌
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:
√
Φ = 6 × 𝜋( 7)2 = 42𝜋
Another way to solve the problem is to use Stokes’ theorem and write the flux as
Φ=
∮
𝐇 ⋅ 𝑑𝐋
where the path integral is taken around the disk perimeter. Doing this gives:
√
√
Φ = 2𝜋𝜌𝑑 𝐻𝜙 |𝜌𝑑 = 2𝜋 7 × 3 7 = 42𝜋
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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 ∇ × 𝐇: For this field, the general curl expression in rectangular coordinates simplifies to
∇×𝐇=−
𝜕𝐻𝑦
𝜕𝑧
𝐚𝑥 +
𝜕𝐻𝑦
𝜕𝑥
𝐚𝑧 =
2(𝑥 + 2𝑦)
1
𝐚𝑥 + 2 𝐚𝑧 A∕m
𝑧3
𝑧
b) Find 𝐉: This will be the answer of part 𝑎, since ∇ × 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
5
𝐼=
2
1
|
𝐉| ⋅ 𝐚𝑧 𝑑𝑥 𝑑𝑦 =
𝑑𝑥 𝑑𝑦 = 1∕8 A
∫ ∫ |𝑧=4
∫3 ∫1 42
d) Show that the same result is obtained using the other side of Stokes’ theorem: We take ∮ 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=
3
5
1 + 2𝑦
2 + 2𝑦
1
1
|
𝑑𝑦 +
𝑑𝑦 = (2) − (2) = 1∕8 A
𝐇| ⋅ 𝑑𝐋 =
∫5
∫3
∮ |𝑧=4
16
16
8
16
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7.28. Given 𝐇 = (3𝑟2 ∕ sin 𝜃)𝐚𝜃 + 54𝑟 cos 𝜃𝐚𝜙 A/m in free space:
a) find the total current in the 𝐚𝜃 direction through the conical surface 𝜃 = 20◦ , 0 ≤ 𝜙 ≤ 2𝜋,
0 ≤ 𝑟 ≤ 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 𝜙 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 𝐚𝜃 direction (right hand rule).
We find
∮
𝐇 ⋅ 𝑑𝐋 =
∫0
2𝜋
[ 2
]
(3𝑟 ∕ sin 𝜃)𝐚𝜃 + 54𝑟 cos 𝜃𝐚𝜙 𝑟=5,𝜃=20 ⋅ 5 sin(20◦ ) 𝑑𝜙 (−𝐚𝜙 )
= −2𝜋(54)(25) cos(20◦ ) sin(20◦ ) = −2.73 × 103 A
This result means that there is a component of current that enters the cone surface in the −𝐚𝜃
direction, to which is associated a component of 𝐇 in the positive 𝐚𝜙 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:
( 3 )
)
1 𝜕 ( 2
1 𝜕
3𝑟
1 𝜕
𝐚𝜙 = 𝐉
54𝑟 cos 𝜃 𝐚𝜃 +
∇×𝐇=
(54𝑟 cos 𝜃 sin 𝜃)) 𝐚𝑟 −
𝑟 sin 𝜃 𝜕𝜃
𝑟 𝜕𝑟
𝑟 𝜕𝑟 sin 𝜃
Thus
9𝑟
𝐚
sin 𝜃 𝜙
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 𝐚𝜃 , as defined in part 𝑎:
𝐉 = 54 cot 𝜃 𝐚𝑟 − 108 cos 𝜃 𝐚𝜃 +
∫𝑆
(∇ × 𝐇) ⋅ 𝑑𝐒 =
∫0
=−
2𝜋
∫0
∫0
2𝜋
5[
∫0
54 cot 𝜃 𝐚𝑟 − 108 cos 𝜃 𝐚𝜃 +
9𝑟
𝐚
sin 𝜃 𝜙
]
20◦
⋅ 𝐚𝜃 𝑟 sin(20◦ ) 𝑑𝑟 𝑑𝜙
5
108 cos(20◦ ) sin(20◦ )𝑟𝑑𝑟𝑑𝜙 = −2𝜋(54)(25) cos(20◦ ) sin(20◦ )
= −2.73 × 103 A
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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
𝐚𝑧 = 1.59 × 107 𝐚𝑧 A∕m2
𝐉=
𝜋(0.2 × 10−3 )2
b) Use Ampere’s circuital law to find 𝐇 and 𝐁 within the conductor: Inside, at radius 𝜌, we have
𝜌𝐽
𝐚 = 7.96 × 106 𝜌 𝐚𝜙 A∕m
2𝜋𝜌𝐻𝜙 = 𝜋𝜌2 𝐽 ⇒ 𝐇 =
2 𝜙
Then 𝐁 = 𝜇0 𝐇 = (4𝜋 × 10−7 )(7.96 × 106 )𝜌𝐚𝜙 = 10𝜌 𝐚𝜙 Wb∕m2 .
c) Show that ∇ × 𝐇 = 𝐉 within the conductor: Using the result of part 𝑏, we find,
(
)
1 𝑑
1 𝑑 1.59 × 107 𝜌2
∇×𝐇=
(𝜌𝐻𝜙 ) 𝐚𝑧 =
𝐚𝑧 = 1.59 × 107 𝐚𝑧 A∕m2 = 𝐉
𝜌 𝑑𝜌
𝜌 𝑑𝜌
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 𝜌, and so
1
𝐼
𝐚 =
𝐚 A∕m
𝐇=
2𝜋𝜌 𝜙 𝜋𝜌 𝜙
Now 𝐁 = 𝜇0 𝐇 = 𝜇0 ∕(𝜋𝜌) 𝐚𝜙 Wb∕m2 .
e) Show that ∇ × 𝐇 = 𝐉 outside the conductor: Here we use 𝐇 outside the conductor and write:
(
)
1 𝑑
1 𝑑
1
(𝜌𝐻𝜙 ) 𝐚𝑧 =
∇×𝐇=
𝜌
𝐚𝑧 = 0 (as expected)
𝜌 𝑑𝜌
𝜌 𝑑𝜌
𝜋𝜌
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 𝜎 = 106 (1 + 106 𝜌2 ) 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 𝜙 symmetry, 𝐇 will be 𝜙directed only. We find 𝐄 = (𝑉0 ∕𝑑)𝐚𝑧 = 10−3 𝐚𝑧 V∕m. Then 𝐉 = 𝜎𝐄 = 103 (1 + 106 𝜌2 )𝐚𝑧 A∕m2 .
Next we apply Ampere’s circuital law to a circular path of radius 𝜌, centered on the 𝑧 axis and
normal to the axis:
∮
𝐇 ⋅ 𝑑𝐋 = 2𝜋𝜌𝐻𝜙 =
∫ ∫𝑆
𝐉 ⋅ 𝑑𝐒 =
∫0
2𝜋
∫0
𝜌
103 (1 + 106 (𝜌′ )2 )𝐚𝑧 ⋅ 𝐚𝑧 𝜌′ 𝑑𝜌′ 𝑑𝜙
[
]
𝜌
103
103 𝜌2 106 4
′
6 ′ 3
′
𝐻𝜙 =
+
𝜌
𝜌 + 10 (𝜌 ) 𝑑𝜌 =
𝜌 ∫0
𝜌
2
4
Thus
Finally, 𝐇 = 500𝜌(1 + 5 × 105 𝜌3 )𝐚𝜙 A∕m (0 < 𝜌 < 2mm).
b) the total magnetic flux inside the conductor: With field in the 𝜙 direction, a plane normal to 𝐁
will be that in the region 0 < 𝜌 < 2 mm, 0 < 𝑧 < 1 m. The flux will be
Φ=
∫ ∫𝑆
1
𝐁 ⋅ 𝑑𝐒 = 𝜇0
∫0 ∫0
2×10−3
(
)
500𝜌 + 2.5 × 108 𝜌3 𝑑𝜌𝑑𝑧 = 8𝜋 × 10−10 Wb = 2.5 nWb
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7.31. The cylindrical shell defined by 1 cm < 𝜌 < 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
𝜙 = 0, 0 < 𝑧 < 1:
a) 0 < 𝜌 < 1.2 cm: We first need to find 𝐉, 𝐇, and 𝐁: The current density will be:
𝐉=
50
𝐚𝑧 = 1.66 × 105 𝐚𝑧 A∕m2
−2
2
− (1.0 × 10 ) ]
10−2 )2
𝜋[(1.4 ×
Next we find 𝐻𝜙 at radius 𝜌 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:
2𝜋𝜌𝐻𝜙 = 𝐼𝑒𝑛𝑐𝑙 =
∫0
2𝜋
𝜌
∫10−2
1.66 × 105 𝜌′ 𝑑𝜌′ 𝑑𝜙
⇒ 𝐻𝜙 = 8.30 × 104
(𝜌2 − 10−4 )
A∕m (10−2 m < 𝜌 < 1.4 × 10−2 m)
𝜌
Then 𝐁 = 𝜇0 𝐇, or
𝐁 = 0.104
(𝜌2 − 10−4 )
𝐚𝜙 Wb∕m2
𝜌
Now,
1
𝐁 ⋅ 𝑑𝐒 =
∫0 ∫10−2
[
(1.2 × 10−2 )2 − 10−4
= 0.104
2
Φ𝑎 =
∫ ∫
]
[
10−4
𝑑𝜌 𝑑𝑧
0.104 𝜌 −
𝜌
]
( )
1.2
= 3.92 × 10−7 Wb = 0.392 𝜇Wb
− 10−4 ln
1.0
1.2×10−2
b) 1.0 cm < 𝜌 < 1.4 cm (note typo in book): This is part 𝑎 over again, except we change the upper
limit of the radial integration:
1
𝐁 ⋅ 𝑑𝐒 =
∫0 ∫10−2
[
(1.4 × 10−2 )2 − 10−4
= 0.104
2
Φ𝑏 =
∫ ∫
]
[
10−4
𝑑𝜌 𝑑𝑧
0.104 𝜌 −
𝜌
( )]
1.4
−4
= 1.49 × 10−6 Wb = 1.49 𝜇Wb
− 10 ln
1.0
1.4×10−2
c) 1.4 cm < 𝜌 < 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 × 10−2 )2 − 10−4 ]
10−5
𝐚𝜙 =
𝐚 Wb∕m2
𝜌
𝜌 𝜙
We now find
1
Φ𝑐 =
20×10−2
∫0 ∫1.4×10−2
( )
10−5
20
= 2.7 × 10−5 Wb = 27 𝜇Wb
𝑑𝜌 𝑑𝑧 = 10−5 ln
𝜌
1.4
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7.32. The free space region defined by 1 < 𝑧 < 4 cm and 2 < 𝜌 < 3 cm is a toroid of rectangular crosssection. Let the surface at 𝜌 = 3 cm carry a surface current 𝐊 = 2𝐚𝑧 kA/m.
a) Specify the current densities on the surfaces at 𝜌 = 2 cm, 𝑧 = 1cm, and 𝑧 = 4cm. All surfaces
must carry equal currents. With this requirement, we find: 𝐊(𝜌 = 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 𝜌 = 2 and 𝜌 = 3. Knowing the the radial current density will vary as 1∕𝜌, we find
𝐊(𝑧 = 1) = (60∕𝜌)𝐚𝜌 A∕m with 𝜌 in meters. Similarly, 𝐊(𝑧 = 4) = −(60∕𝜌)𝐚𝜌 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:
∮
2𝜋
𝐊(𝜌 = 2) ⋅ 𝐚𝑧 (2 × 10−2 ) 𝑑𝜙
∫0
2𝜋(3000)(.02)
𝐚𝜙 = −60∕𝜌 𝐚𝜙 A∕m (inside)
⇒ 𝐇=−
𝜌
𝐇 ⋅ 𝑑𝐋 = 2𝜋𝜌𝐻𝜙 =
c) Calculate the total flux within the toriod: We have 𝐁 = −(60𝜇0 ∕𝜌)𝐚𝜙 Wb∕m2 . Then
.03
.04
Φ=
∫.01 ∫.02
( )
−60𝜇0
3
= 0.92 𝜇Wb
𝐚𝜙 ⋅ (−𝐚𝜙 ) 𝑑𝜌 𝑑𝑧 = (.03)(60)𝜇0 ln
𝜌
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
( )]
( )
( )]
[ ( )
𝜕 𝜕𝐺
𝜕 𝜕𝐺
𝜕 𝜕𝐺
𝜕 𝜕𝐺
−
−
𝐚𝑦
𝐚𝑥 +
∇ × ∇𝐺 =
𝜕𝑦 𝜕𝑧
𝜕𝑧 𝜕𝑦
𝜕𝑧 𝜕𝑥
𝜕𝑥 𝜕𝑧
[ ( )
( )]
𝜕 𝜕𝐺
𝜕 𝜕𝐺
+
𝐚𝑧 = 0 for any 𝐺
−
𝜕𝑥 𝜕𝑦
𝜕𝑦 𝜕𝑥
[
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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 𝜌 = 6 carries a total current of 12A in the −𝐚𝑧 direction, and another shell at 𝜌 = 10 carries a total
current of 4A in the −𝐚𝑧 direction.
a) Find 𝐇 for 0 < 𝜌 < 12: Ampere’s circuital law states that ∮ 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 𝐚𝜙 direction. We proceed as follows:
0 < 𝜌 < 6 ∶ 2𝜋𝜌𝐻𝜙 = 16 ⇒ 𝐇 = 16∕(2𝜋𝜌)𝐚𝜙
6 < 𝜌 < 10 ∶ 2𝜋𝜌𝐻𝜙 = 16 − 12 ⇒ 𝐇 = 4∕(2𝜋𝜌)𝐚𝜙
𝜌 > 10 ∶ 2𝜋𝜌𝐻𝜙 = 16 − 12 − 4 = 0 ⇒ 𝐇 = 0
b) Plot 𝐻𝜙 vs. 𝜌:
c) Find the total flux Φ crossing the surface 1 < 𝜌 < 7, 0 < 𝑧 < 1: This will be
1
Φ=
∫0 ∫1
6
7
1
]
4𝜇0
2𝜇 [
16𝜇0
𝑑𝜌 𝑑𝑧 +
𝑑𝜌 𝑑𝑧 = 0 4 ln 6 + ln(7∕6) = 5.9 𝜇Wb
∫0 ∫6 2𝜋𝜌
2𝜋𝜌
𝜋
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7.35. A current sheet, 𝐊 = 20 𝐚𝑧 A/m, is located at 𝜌 = 2, and a second sheet, 𝐊 = −10 𝐚𝑧 A/m is located
at 𝜌 = 4.
a.) Let 𝑉𝑚 = 0 at 𝑃 (𝜌 = 3, 𝜙 = 0, 𝑧 = 5) and place a barrier at 𝜙 = 𝜋. Find 𝑉𝑚 (𝜌, 𝜙, 𝑧) for
−𝜋 < 𝜙 < 𝜋: Since the current is cylindrically-symmetric, we know that 𝐇 = 𝐼∕(2𝜋𝜌) 𝐚𝜙 ,
where 𝐼 is the current enclosed, equal in this case to 2𝜋(2)𝐾 = 80𝜋 A. Thus, using the result of
Section 8.6, we find
𝐼
80𝜋
𝑉𝑚 = − 𝜙 = −
𝜙 = −40𝜙 A
2𝜋
2𝜋
which is valid over the region 2 < 𝜌 < 4, −𝜋 < 𝜙 < 𝜋, and −∞ < 𝑧 < ∞. For 𝜌 > 4, the outer
current contributes, leading to a total enclosed current of
𝐼𝑛𝑒𝑡 = 2𝜋(2)(20) − 2𝜋(4)(10) = 0
With zero enclosed current, 𝐻𝜙 = 0, and the magnetic potential is zero as well.
b) Let 𝐀 = 0 at 𝑃 and find 𝐀(𝜌, 𝜙, 𝑧) for 2 < 𝜌 < 4: Again, we know that 𝐇 = 𝐻𝜙 (𝜌), 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 𝜌. We can then write:
∇×𝐀=−
𝑑𝐴𝑧
𝜇 𝐼
𝐚 = 𝐁 = 0 𝐚𝜙
𝑑𝜌 𝜙
2𝜋𝜌
Thus
𝑑𝐴𝑧
𝜇 𝐼
𝜇 𝐼
=− 0
⇒ 𝐴𝑧 = − 0 ln(𝜌) + 𝐶
𝑑𝜌
2𝜋𝜌
2𝜋
We require that 𝐴𝑧 = 0 at 𝜌 = 3. Therefore 𝐶 = [(𝜇0 𝐼)∕(2𝜋)] ln(3), Then, with 𝐼 = 80𝜋, we
finally obtain
𝜇 (80𝜋)
𝐀=− 0
[ln(𝜌) − ln(3)] 𝐚𝑧 = 40𝜇0 ln
2𝜋
( )
3
𝐚𝑧 Wb∕m
𝜌
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 = ∇ × 𝐀 = −2𝑥𝐚𝑥 − 𝐚𝑦 + (2𝑧 − 3)𝐚𝑧 ⇒ 𝐁𝑃 = −4𝐚𝑥 − 𝐚𝑦 + 3𝐚𝑧 Wb∕m2
Now
𝐇𝑃 = (1∕𝜇0 )𝐁𝑃 = −3.2 × 106 𝐚𝑥 − 8.0 × 105 𝐚𝑦 + 2.4 × 106 𝐚𝑧 A∕m
Then 𝐉 = ∇ × 𝐇 = (1∕𝜇0 )∇ × 𝐁 = 0, as the curl formula in cartesian coordinates shows.
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7.37. Let 𝑁 = 1000, 𝐼 = 0.8 A, 𝜌0 = 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 𝜌 = 2.5 cm, 𝜙 = 0.3𝜋. Keep 𝜙 within the range 0 < 𝜙 < 2𝜋:
Well-within the toroid, we have
𝐇=
1 𝑑𝑉𝑚
𝑁𝐼
𝐚𝜙 = −∇𝑉𝑚 = −
𝐚
2𝜋𝜌
𝜌 𝑑𝜙 𝜙
Thus
𝑉𝑚 = −
𝑁𝐼𝜙
+𝐶
2𝜋
Then,
0=−
or C = 120. Finally
1000(0.8)(0.3𝜋)
+𝐶
2𝜋
[
]
400
𝑉𝑚 = 120 −
𝜙 A (0 < 𝜙 < 2𝜋)
𝜋
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 𝐚𝜙 direction.
a) Assuming that 𝑟 >> 𝑑𝐿, and following a method similar to that in Sec. 4.7, show that
𝑑𝐀 =
𝜇0 𝐼(𝑑𝐿)2 sin 𝜃
4𝜋𝑟2
𝐚𝜙
We begin with the expression for the differential vector potential, Eq. (48):
𝑑𝐀 =
𝜇0 𝐼𝑑𝐋
4𝜋𝑅
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:
)
)]
[ (
(
𝜇0 𝐼𝑑𝐿
1
1
1
1
𝑑𝐀 =
−
+ 𝐚𝑦
−
𝐚𝑥
4𝜋
𝑅2 𝑅1
𝑅3 𝑅4
The challenge is to determine the four distances, 𝑅1 through 𝑅4 , in terms of spherical coordinates,
𝑟, 𝜃, and 𝜙, 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.
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b) ℎ = 𝑎∕4,
7.38
c) ℎ = 𝑎∕2:
a) (continued). We now may write:
]
[
.
𝑑𝐿
𝐚𝑦 ⋅ 𝐚𝑟
𝑅1 = 𝑟 −
2
and
[
]
.
𝑑𝐿
𝑅2 = 𝑟 +
𝐚𝑦 ⋅ 𝐚𝑟
2
Referring to the right-hand figure and applying similar reasoning there leads to
[
]
.
𝑑𝐿
𝑅3 = 𝑟 −
𝐚𝑥 ⋅ 𝐚𝑟
2
and
[
]
.
𝑑𝐿
𝑅4 = 𝑟 +
𝐚𝑥 ⋅ 𝐚𝑟
2
where we know that 𝐚𝑥 ⋅ 𝐚𝑟 = sin 𝜃 cos 𝜙 and 𝐚𝑦 ⋅ 𝐚𝑟 = sin 𝜃 sin 𝜙 We now substitute all these
relations into the original expression for 𝑑𝐀:
[((
)−1 (
)−1 )
𝜇0 𝐼𝑑𝐿
𝑑𝐿
𝑑𝐿
𝑑𝐀 =
𝑟+
sin 𝜃 sin 𝜙
sin 𝜃 sin 𝜙
− 𝑟−
𝐚𝑥
4𝜋
2
2
((
)
]
)−1 (
)−1
𝑑𝐿
𝑑𝐿
𝑟−
+
sin 𝜃 cos 𝜙
sin 𝜃 cos 𝜙
− 𝑟+
𝐚𝑦
2
2
[((
)−1 (
)−1 )
𝜇0 𝐼𝑑𝐿
𝑑𝐿
𝑑𝐿
1+
sin 𝜃 sin 𝜙
sin 𝜃 sin 𝜙
− 1−
𝐚𝑥
=
4𝜋𝑟
2𝑟
2𝑟
((
)−1 (
)−1 ) ]
𝑑𝐿
𝑑𝐿
sin 𝜃 cos 𝜙
sin 𝜃 cos 𝜙
1−
− 1+
𝐚𝑦
+
2𝑟
2𝑟
Now, since 𝑑𝐿∕𝑟 << 1, this simplifies to
) (
))
. 𝜇 𝐼𝑑𝐿 [((
𝑑𝐿
𝑑𝐿
𝑑𝐀 = 0
1−
sin 𝜃 sin 𝜙 − 1 +
sin 𝜃 sin 𝜙
𝐚𝑥
4𝜋𝑟
2𝑟
2𝑟
) (
)) ]
((
𝑑𝐿
𝑑𝐿
sin 𝜃 cos 𝜙 − 1 −
sin 𝜃 cos 𝜙
𝐚𝑦
+
1+
2𝑟
2𝑟
]
𝜇 𝐼(𝑑𝐿)2 sin 𝜃 [
− sin 𝜙 𝐚𝑥 + cos 𝜙 𝐚𝑦
= 0
2
4𝜋𝑟
𝜇0 𝐼(𝑑𝐿)2 sin 𝜃
𝐚𝜙
=
4𝜋𝑟2
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7.38
b) Show that
)
𝐼(𝑑𝐿)2 (
2 cos 𝜃 𝐚𝑟 + sin 𝜃 𝐚𝜃
3
4𝜋𝑟
Using the part 𝑎 expression, we construct 𝑑𝐇 = (1∕𝜇0 )∇ × 𝑑𝐀, where in this case, we have a
𝜙-directed 𝑑𝐀 that varies with 𝑟 and 𝜃. The curl expression in spherical coordinates reduces to:
1 𝜕(𝑟𝑑𝐴)
1 𝜕(𝑑𝐴 sin 𝜃)
∇ × 𝑑𝐀 =
𝐚𝑟 −
𝐚𝜃
𝑟 sin 𝜃
𝜕𝜃
𝑟 𝜕𝑟
or
(
(
)
)
𝜕 𝐼(𝑑𝐿)2 sin2 𝜃
1
1 𝜕 𝐼(𝑑𝐿)2 sin 𝜃
𝐚𝜃
∇ × 𝑑𝐀 =
𝐚𝑟 −
𝑟 sin 𝜃 𝜕𝜃
𝑟 𝜕𝑟
4𝜋𝑟
4𝜋𝑟2
)
𝐼(𝑑𝐿)2 (
=
2
cos
𝜃
𝐚
+
sin
𝜃
𝐚
𝑟
𝜃
4𝜋𝑟3
𝑑𝐇 =
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 × 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 × −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 ⇒ 𝑉𝑚 = 30𝑦 + 𝐶1
𝑑𝑦
Then
0 = 30(0.2) + 𝐶1 ⇒ 𝐶1 = −6 ⇒ 𝑉𝑚 = 30𝑦 − 6 A
c) Find B: 𝐁 = 𝜇0 𝐇 = −30𝜇0 𝐚𝑦 Wb∕m2 .
d) Obtain an expression for 𝐀 if 𝐀 = 0 at 𝑃 : We expect 𝐀 to be 𝑧-directed (with the current), and
so from ∇ × 𝐀 = 𝐁, where 𝐁 is 𝑦-directed, we set up
𝑑𝐴
− 𝑧 = −30𝜇0 ⇒ 𝐴𝑧 = 30𝜇0 𝑥 + 𝐶2
𝑑𝑥
Then 0 = 30𝜇0 (0.1) + 𝐶2 ⇒ 𝐶2 = −3𝜇0 . So finally 𝐀 = 𝜇0 (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 ∮ 𝐀 ⋅ 𝑑𝐋 = ∫ 𝐁 ⋅ 𝑑𝐒.
We use the fact that 𝐁 = ∇ × 𝐀, and substitute this into the desired relation to find
𝐀 ⋅ 𝑑𝐋 =
∇ × 𝐀 ⋅ 𝑑𝐒
∮
∫
This is just a statement of Stokes’ theorem (already proved), so we are done.
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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
∮
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 𝜙), 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 𝜙-directed 𝐁 field is given as
𝐁=
𝜇0 𝐼
𝐚
2𝜋𝜌 𝜙
In addition, 𝐀 was derived in the development that lead to Eq. (66), and is:
( )
𝜇0 𝐼
𝑏
ln
𝐚𝑧
𝐀=
2𝜋
𝜌
in which the boundary condition of 𝐀 = 0 at 𝜌 = 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 Δ𝑧 + 𝐴2 (𝑏 − 𝜌) + 𝐴3 Δ𝑧 − 𝐴4 (𝑏 − 𝜌)
∮
The only surviving term is 𝐴1 Δ𝑧 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
( )
𝜇0 𝐼
𝑏
ln
Δ𝑧
𝐀 ⋅ 𝑑𝐋 = 𝐴1 Δ𝑧 =
∮
2𝜋
𝜌
Then
∫ ∫
𝐁 ⋅ 𝑑𝐒 =
∫0
Δ𝑧
∫𝜌
𝑏
𝜇0 𝐼
𝜇 𝐼
𝑑𝜌′ 𝑑𝑧 = 0 ln
′
2𝜋𝜌
2𝜋
( )
𝑏
Δ𝑧
𝜌
thus completing the demonstration.
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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 𝜇0 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 𝜌 only. So
∮
𝐀 ⋅ 𝑑𝐋 = 2𝜋𝜌𝐴 =
or
𝐀=
As a check,
∇×𝐀=
∫ ∫𝑠
𝐁 ⋅ 𝑑𝐒 = 𝜇0 𝑛𝐼𝜋𝜌2
𝜇0 𝑛𝐼𝜋𝜌
𝐚𝜙 A
2𝜋
1 𝑑(𝜌𝐴𝜙 )
𝐚𝑧 = 𝜇0 𝑛𝐼 𝐚𝑧
𝜌 𝑑𝜌
7.42. Show that ∇2 (1∕𝑅12 ) = −∇1 (1∕𝑅12 ) = 𝐑21 ∕𝑅312 . First
(
∇2
1
𝑅12
)
]−1∕2
[
= ∇2 (𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 + (𝑧2 − 𝑧1 )2
]
[
𝐑21
−𝐑12
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
𝑅312
𝑅312
Also note that ∇1 (1∕𝑅12 ) would give the same result, but of opposite sign.
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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
𝐚𝑧 = −
𝐚𝑧
𝐉𝑜𝑢𝑡 = −
2
2
𝜋(7𝑎) − 𝜋(5𝑎)
24𝜋𝑎2
Since current 𝐼 flows in both conductors, but in opposite directions, Ampere’s circuital law inside the
outer conductor gives:
2𝜋𝜌𝐻𝜙 = 𝐼 −
∫0
2𝜋
𝜌
∫5𝑎
[
]
49𝑎2 − 𝜌2
𝐼
𝐼
′
′
𝜌 𝑑𝜌 𝑑𝜙 ⇒ 𝐻𝜙 =
2𝜋𝜌
24𝜋𝑎2
24𝑎2
Now, with 𝐁 = 𝜇0 𝐇, we note that ∇ × 𝐀 will have a 𝜙 component only, and from the direction and
symmetry of the current, we expect 𝐀 to be 𝑧-directed, and to vary only with 𝜌. Therefore
∇×𝐀=−
𝑑𝐴𝑧
𝐚 = 𝜇0 𝐇
𝑑𝜌 𝜙
[
]
𝑑𝐴𝑧
𝜇0 𝐼 49𝑎2 − 𝜌2
=−
𝑑𝜌
2𝜋𝜌
24𝑎2
and so
Then by direct integration,
[
]
𝜇0 𝐼𝜌
𝜇0 𝐼 𝜌2
−𝜇0 𝐼(49)
𝑑𝜌 +
𝑑𝜌 + 𝐶 =
− 98 ln 𝜌 + 𝐶
𝐴𝑧 =
∫ 48𝜋𝑎2
∫
48𝜋𝜌
96𝜋 𝑎2
As per Fig. 8.20, we establish a zero reference at 𝜌 = 5𝑎, enabling the evaluation of the integration
constant:
𝜇 𝐼
𝐶 = − 0 [25 − 98 ln(5𝑎)]
96𝜋
Finally,
[( 2
)
( )]
𝜇0 𝐼
𝜌
5𝑎
Wb∕m
− 25 + 98 ln
𝐴𝑧 =
2
96𝜋
𝜌
𝑎
A plot of this continues the plot of Fig. 8.20, in which the curve goes negative at 𝜌 = 5𝑎, and then
approaches a minimum of −.09𝜇0 𝐼∕𝜋 at 𝜌 = 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) − ∇ × ∇ × 𝐀
We begin with
∇⋅𝐀=
𝜕𝐴𝑥 𝜕𝐴𝑦 𝜕𝐴𝑧
+
+
𝜕𝑥
𝜕𝑦
𝜕𝑧
Then the 𝑥 component of ∇(∇ ⋅ 𝐀) is
[∇(∇ ⋅ 𝐀)]𝑥 =
(
Now
∇×𝐀=
𝜕𝐴𝑧 𝜕𝐴𝑦
−
𝜕𝑦
𝜕𝑧
)
𝜕 2 𝐴𝑥
𝜕𝑥2
(
𝐚𝑥 +
+
𝜕 2 𝐴𝑦
𝜕𝑥𝜕𝑦
𝜕𝐴𝑥 𝜕𝐴𝑧
−
𝜕𝑧
𝜕𝑥
+
𝜕 2 𝐴𝑧
𝜕𝑥𝜕𝑧
)
(
𝐚𝑦 +
𝜕𝐴𝑦
𝜕𝐴𝑥
−
𝜕𝑥
𝜕𝑦
)
𝐚𝑧
and the 𝑥 component of ∇ × ∇ × 𝐀 is
[∇ × ∇ × 𝐀]𝑥 =
𝜕 2 𝐴𝑦
𝜕𝑥𝜕𝑦
−
𝜕 2 𝐴𝑥
𝜕𝑦2
−
𝜕 2 𝐴𝑥
𝜕𝑧2
+
𝜕 2 𝐴𝑧
𝜕𝑧𝜕𝑦
Then, using the underlined results
[∇(∇ ⋅ 𝐀) − ∇ × ∇ × 𝐀]𝑥 =
𝜕 2 𝐴𝑥
𝜕𝑥2
+
𝜕 2 𝐴𝑥
𝜕𝑦2
+
𝜕 2 𝐴𝑥
𝜕𝑧2
= ∇2 𝐴𝑥
Similar results will be found for the other two components, leading to
∇(∇ ⋅ 𝐀) − ∇ × ∇ × 𝐀 = ∇2 𝐴𝑥 𝐚𝑥 + ∇2 𝐴𝑦 𝐚𝑦 + ∇2 𝐴𝑧 𝐚𝑧 ≡ ∇2 𝐀
QED
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Engineering Electromagnetics 9th solution
Engineering Electromagnetics (Chungnam National University)
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CHAPTER 8 – 9th Edition
8.1. A point charge, 𝑄 = −0.3 𝜇C and 𝑚 = 3 × 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 × 105 𝐚𝑥 m/s at the origin. At 𝑡 = 3 𝜇s, 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 × 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 × 10−6 )(30) 2
𝑡 = −1.5 × 1010 𝑡2 m
−16
2 × 3 × 10
At 𝑡 = 3 𝜇s, 𝑧 = −(1.5 × 1010 )(3 × 10−6 )2 = −.135 cm. Now, considering the initial constant
velocity in 𝑥, the charge in 3 𝜇s attains an 𝑥 coordinate of 𝑥 = 𝑣𝑡 = (3 × 105 )(3 × 10−6 ) = .90 m.
In summary, at 𝑡 = 3 𝜇s 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 × 1010 )(3 × 10−6 ) = −9 × 104 m∕s
𝑚
Including the intial 𝑥-directed velocity, we finally obtain 𝐯 = 3 × 105 𝐚𝑥 − 9 × 104 𝐚𝑧 m∕s.
c) the kinetic energy of the charge: Have
1
1
K.E. = 𝑚|𝑣|2 = (3 × 10−16 )(1.13 × 105 )2 = 1.5 × 10−5 J
2
2
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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 × 𝐁), where |𝐁| = 0.5 G = 5.0 × 10−5 T. We find
|𝐅𝑒 | = (1.6 × 10−19 C)(105 V∕m) = 1.6 × 10−14 N
|𝐅𝑚 | = (1.6 × 10−19 C)(107 m∕s)(5.0 × 10−5 T) = 8.0 × 10−17 N = 0.005|𝐅𝑒 |
8.3. A point charge for which 𝑄 = 2 × 10−16 C and 𝑚 = 5 × 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 ) × 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) × 𝐁)], where
𝐯(0) × 𝐁 = (2𝐚𝑥 − 3𝐚𝑦 − 4𝐚𝑧 )105 × (−3𝐚𝑥 + 2𝐚𝑦 − 𝐚𝑧 )10−3 = 1100𝐚𝑥 + 1400𝐚𝑦 − 500𝐚𝑧
So the force in newtons becomes
𝐅(0) = (2×10−16 )[(100+1100)𝐚𝑥 +(1400− 200)𝐚𝑦 +(300− 500)𝐚𝑧 ] = 4×10−14 [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 × 10−26 kg)(5.39 × 105 m∕s)2 = 7.25 × 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 Ω, the velocity
is 𝐯 = 𝑅Ω 𝐚𝜙 . The magnetic force is then 𝐅𝑚 = 𝑞𝐯 × 𝐁 = 𝑞𝑅Ω 𝐚𝜙 × 𝐵0 𝐚𝑧 = 𝑞𝑅Ω𝐵0 𝐚𝜌 . This
force will be negative (pulling the particle toward the center of the path) if the charge is positive
and motion is in the −𝐚𝜙 direction, or if the charge is negative, and motion is in positive 𝐚𝜙 . In
either case, the centripital force must counteract the magnetic force. Assuming particle mass 𝑚,
the force balance equation is 𝑞𝑅Ω𝐵0 = 𝑚Ω2 𝑅, from which Ω = 𝑞𝐵0 ∕𝑚. The revolution period
is 𝑇 = 2𝜋∕Ω = 2𝜋𝑚∕(𝑞𝐵0 ), which is independent of 𝑅. For an electron, we have 𝑞 = 1.6 × 10−9
C, and 𝑚 = 9.1 × 1031 kg. The cyclotron frequency is therefore
Ω𝑐 =
𝑞
𝐵 = 1.76 × 1011 𝐵0 s−1
𝑚 0
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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 𝐚𝜌 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 × 𝑑𝐅, where 𝐑 = 𝑎𝐚𝜌 , and
𝑑𝐅 = 𝐊𝑑𝑎 × 𝐁 = 𝐾0 𝐚𝑧 𝑎𝑑𝜙𝑑𝑧 × −𝐵0 𝐚𝜌 = −𝐾0 𝐵0 𝑎𝑑𝜙𝑑𝑧𝐚𝜙
So
𝑑𝐓 = 𝑎𝐚𝜌 × −𝐾0 𝐵0 𝑎𝑑𝜙𝑑𝑧𝐚𝜙 = −𝑎2 𝐾0 𝐵0 𝑑𝜙𝑑𝑧𝐚𝑧
Summing up all contributions:
𝐓=
∫
𝑑
𝑑𝐓 =
∫0 ∫0
2𝜋
−𝑎2 𝐾0 𝐵0 𝑑𝜙𝑑𝑧 = −2𝜋𝑎2 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 × 𝐁) ⋅ 𝑑𝐥 and
𝑑𝑊 ′ = (𝐼𝑑𝐥 × 𝐁) ⋅ 𝑑𝐋
We now apply the vector identity, Eq.(A.6), Appendix A: (𝐀 × 𝐁) ⋅ 𝐂 = (𝐁 × 𝐂) ⋅ 𝐀, and write:
(𝐼𝑑𝐋 × 𝐁) ⋅ 𝑑𝐥 = (𝐁 × 𝑑𝐥) ⋅ 𝐼𝑑𝐋 = −(𝐼𝑑𝐥 × 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) = −𝜇0 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 × 𝐁 = 𝐼2 𝑑𝑥 𝐚𝑥 ×
−𝜇0 𝐼1
−𝜇0 𝐼1 𝐼2
𝐚𝑦 =
𝐚𝑧 𝑑𝑥
2𝑏
2𝑏
The force per unit length is then
𝐅=
∫0
1
−
−𝜇0 𝐼1 𝐼2
𝜇0 𝐼1 𝐼2
𝐚𝑧 𝑑𝑥 =
𝐚𝑧 N∕m
2𝑏
2𝑏
where again, 𝑑 << 𝑏.
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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:
𝐇(𝑥0 ) =
𝐊 × 𝐚𝑅
𝑑𝑎
∫𝑠 4𝜋𝑅2
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 ) =
− 𝑑2
∫−∞ ∫−( 𝑑 +𝑏)
2
[
]
−𝐾0 𝐚𝑧 × −(𝑥 − 𝑥0 ) 𝐚𝑥 − 𝑧 𝐚𝑧
𝑑𝑥 𝑑𝑧
[
]3∕2
4𝜋 (𝑥 − 𝑥0 )2 + 𝑧2
Taking the cross product leaves only a 𝑦 component:
∞
𝐇(𝑥0 ) =
− 𝑑2
𝐾0 𝐚𝑦 (𝑥 − 𝑥0 )
∫−∞ ∫−( 𝑑 +𝑏) 4𝜋 [(𝑥 − 𝑥 )2 + 𝑧2 ]3∕2
2
0
𝑑𝑥 𝑑𝑧
It is easiest to evaluate the 𝑧 integral first, leading to
𝑑
𝑑
−2
−2
𝐾
𝐾0
𝑧
𝑑𝑥
|∞
𝐇0 = 0 𝐚𝑦
𝐚
𝑑𝑥
=
|
√
4𝜋 ∫−( 𝑑 +𝑏) (𝑥 − 𝑥 ) (𝑥 − 𝑥 )2 + 𝑧2 |−∞
2𝜋 𝑦 ∫−( 𝑑 +𝑏) (𝑥 − 𝑥0 )
0
0
2
2
Evaluate the 𝑥 integral to find:
𝑑
𝐾
𝐾
|−
𝐇0 = 0 𝐚𝑦 ln(𝑥 − 𝑥0 )| 2𝑑
= − 0 ln
|
−(
+𝑏)
2𝜋
2𝜋
2
[𝑑
2
+ 𝑏 + 𝑥0
𝑑
2
+ 𝑥0
]
𝐚𝑦 A∕m
Now the force acting on the right-hand strip per unit length is
𝐅=
∫𝑠
𝐊 × 𝐁 𝑑𝑎
where 𝐊 is the surface current density in the right-hand strip, and 𝐁 is the magnetic flux density
(𝜇0 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:
[𝑑
]
(𝑏+ 𝑑2 )
1
+ 𝑏 + 𝑥0
−𝜇0 𝐾0
2
ln
𝑘0 𝐚𝑧 ×
𝐚𝑦 𝑑𝑥0 𝑑𝑧
𝐅=
𝑑
∫0 ∫ 𝑑
2𝜋
+ 𝑥0
2
2
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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=
=
𝜇0 𝐾02
2𝜋
𝐚𝑥
∫𝑑
(𝑏+ 𝑑2 ) [
(
ln
2
𝜇0 𝐾02
2𝜋
[(
𝐚𝑥
)
(
)]
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=
𝜇0 𝑑𝐾02
2𝜋
[
𝐚𝑥
)
(
2𝑏
ln
1+
𝑑
(
2𝑏 )
𝑑
+ 𝑑𝑏
1+
1
(
𝑏
− ln 1 +
𝑑
]
)
b) let 𝑏 approach zero while maintaining constant current, 𝐼 = 𝐾0 𝑏, and show that the force per
unit length approaches 𝜇0 𝐼 2 ∕(2𝜋𝑑) N/m.
As 𝑏 gets small, so does the ratio 𝑏∕𝑑. We may then write:
[
( )2 ]
𝑏
1 .
𝑏
= 1− +
𝑏
𝑑
𝑑
1+
𝑑
The force expression now becomes:
𝐅=
𝜇0 𝑑𝐾02
2𝜋
[(
𝐚𝑥
[
) [(
)(
]]
( )2 )]
𝑏
2𝑏
𝑏
𝑏
2𝑏
1− +
1+
− ln 1 +
ln 1 +
𝑑
𝑑
𝑑
𝑑
𝑑
The product in the natural log function is expanded:
𝐅=
𝜇0 𝑑𝐾02
2𝜋
]
[(
[
) [
]]
2𝑏
𝑏
𝑏 𝑏2 2𝑏 2𝑏2 2𝑏2
ln 1 − + 2 +
𝐚𝑥 1 +
− 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:
)
[(
]
2
)(
. 𝜇0 𝑑𝐾0
2𝑏
𝑏
𝑏 𝑏2 2𝑏3
+ 3 −
𝐚𝑥 1 +
−
𝐅=
2𝜋
𝑑
𝑑 𝑑2
𝑑
𝑑
Expanding and simplifying, the final result is
. 𝜇 (𝑏𝐾0 )2
𝐚𝑥 N∕m
𝐅= 0
2𝜋𝑑
(
𝑏
<< 1
𝑑
)
where the term, 4𝑏4 ∕𝑑 4 , has been neglected.
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8.9. A current of −100𝐚𝑧 A/m flows on the conducting cylinder 𝜌 = 5 mm and +500𝐚𝑧 A/m is present
on the conducting cylinder 𝜌 = 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 × 𝐁, where 𝐁 is the field from the inner cylinder, evaluated at the
outer cylinder location:
2𝜋(1)(500)𝜇0
𝐚𝜙 = 100𝜇0 𝐚𝜙 T
𝐁=
2𝜋(5)
Thus 𝑑𝐅 = −100𝐚𝑧 × 100𝜇0 𝐚𝜙 = 104 𝜇0 𝐚𝜌 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 < 𝜙 < 𝜋), and integrate the 𝑦 component of 𝑑𝐅 over this
range, and over a unit length in the 𝑧 direction:
1
𝐹𝑦 =
∫0 ∫0
𝜋
104 𝜇0 𝐚𝜌 ⋅ 𝐚𝑦 (5 × 10−3 ) 𝑑𝜙 𝑑𝑧 =
∫0
𝜋
50𝜇0 sin 𝜙 𝑑𝜙 = 100𝜇0 = 4𝜋 × 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
1
𝐅=
∫0 ∫0
𝑏
𝐊𝑏 × 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
1
𝐅=
∫0 ∫0
𝑏
𝜇 𝐼
𝜇 𝐼2
𝐼
− 𝐚𝑧 × 0 𝐚𝑥 𝑑𝑆 = − 0 𝐚𝑦 N∕m
𝑏
2𝑏
2𝑏
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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
𝜇0 𝐼1 𝐼2 ∕(2𝜋𝑑): The force on 𝐼2 is given by
]
[
𝐼1 𝐼2
𝐚𝑅12 × 𝑑𝐋1
𝐅2 = 𝜇0
× 𝑑𝐋2
4𝜋 ∮ ∮
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:
∞
[(𝑧2 − 𝑧1 )𝐚𝑧 − 𝑑𝐚𝑦 ] × 𝑑𝑧1 𝐚𝑧
−𝑑 𝑑𝑧1 𝐚𝑥
𝐚𝑅12 × 𝑑𝐋1
=
=
2
2
1.5
2
∮
∫−∞ [(𝑧2 − 𝑧1 )2 + 𝑑 2 ]1.5
∮
[(𝑧2 − 𝑧1 ) + 𝑑 ]
𝑅
12
The force expression now becomes
[ ∞
]
𝑑 𝑑𝑧1 𝑑𝑧2 𝐚𝑦
−𝑑 𝑑𝑧1 𝐚𝑥
𝐼1 𝐼2 1 ∞
𝐼1 𝐼2
=
𝜇
×
𝑑𝑧
𝐚
𝐅2 = 𝜇0
0
2
𝑧
2
2
1.5
4𝜋 ∮ ∫−∞ [(𝑧2 − 𝑧1 ) + 𝑑 ]
4𝜋 ∫0 ∫−∞ [(𝑧2 − 𝑧1 )2 + 𝑑 2 ]1.5
Note that the “outside” integral is taken over a unit length of current 𝐼2 . Evaluating, obtain,
𝐅2 = 𝜇0
𝐼1 𝐼2 𝑑 𝐚𝑦
4𝜋𝑑 2
(2)
∫0
1
𝑑𝑧2 =
𝜇0 𝐼1 𝐼2
𝐚 N∕m
2𝜋𝑑 𝑦
as expected.
b) Show how a simpler method can be used to check your result: We use 𝑑𝐅2 = 𝐼2 𝑑𝐋2 × 𝐁12 , where
the field from current 1 at the location of current 2 is
𝜇 𝐼
𝐁12 = 0 1 𝐚𝑥 T
2𝜋𝑑
so over a unit length of 𝐼2 , we obtain
𝜇 𝐼
𝐼 𝐼
𝐅2 = 𝐼2 𝐚𝑧 × 0 1 𝐚𝑥 = 𝜇0 1 2 𝐚𝑦 N∕m
2𝜋𝑑
2𝜋𝑑
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 = 2𝜋𝑎. The magnetic induction at points very
.
close to either current loop is approximately that of an infinite straight wire, or 𝐵 = 𝜇0 𝐼∕(2𝜋𝑑).
The force is therefore found using Eq. (11) to be
𝐹 = 𝐼𝐿𝐵 = 𝐼(2𝜋𝑎)
𝜇 𝑎𝐼 2
𝜇0 𝐼
= 0
2𝜋𝑑
𝑑
If the force is attractive, then the currents must be in the same direction.
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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
𝜇0 𝐼1
2𝜋(𝑦2
+
𝑧20 )1∕2
(𝑦2
+
𝑧20 )1∕2
=
𝜇0 𝐼1 (𝑥𝐚𝑧 − 𝑧0 𝐚𝑦 )
2𝜋(𝑥2
+
𝑧20 )
Wb∕m2
where the substitution 𝑦 = 𝑥 was used, as this is true along the second current. Now 𝑑𝐅 =
𝐼2 𝑑𝐋 × 𝐁, where 𝑑𝐋 = 𝑑𝑥𝐚𝑥 + 𝑑𝑦𝐚𝑦 , so that
𝑑𝐅 = 𝐼2 (𝑑𝑥𝐚𝑥 + 𝑑𝑦𝐚𝑦 ) ×
𝜇0 𝐼1 (𝑥𝐚𝑧 − 𝑧0 𝐚𝑦 )
2𝜋(𝑥2 + 𝑧20 )
=
𝜇0 𝐼1 𝐼2
2𝜋(𝑥2 + 𝑧20 )
[
]
𝑥(𝐚𝑥 − 𝐚𝑦 ) − 𝑧0 𝐚𝑧 𝑑𝑥 N
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 =
∫−∞
−𝜇0 𝐼1 𝐼2 𝑧0 1
=
tan−1
2
2
2𝜋
𝑧
2𝜋(𝑥 + 𝑧0 )
0
−𝜇0 𝐼1 𝐼2 𝑧0 𝑑𝑥
(
𝑥
𝑧0
)
𝜇 𝐼 𝐼
|∞
=− 0 1 2 N
|
|−∞
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.
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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 × 𝑑𝐅
where 𝐑 is the vector directed from the origin to the current element, and is given by 𝐑 =
𝑧0 𝐚𝑧 + 𝑎𝐚𝜌 , and where
𝑑𝐅 = 𝐼𝑑𝐋 × 𝐁 = 𝐼𝑎 𝑑𝜙 𝐚𝜙 × 𝐵0 𝐚𝑥 = −𝐼𝑎𝐵0 cos 𝜙 𝑑𝜙 𝐚𝑧
So now
(
) (
)
𝑑𝐓 = 𝑧0 𝐚𝑧 + 𝑎𝐚𝜌 × −𝐼𝑎𝐵0 cos 𝜙 𝑑𝜙 𝐚𝑧 = 𝑎2 𝐼𝐵0 cos 𝜙 𝑑𝜙 𝐚𝜙
Using 𝐚𝜙 = cos 𝜙 𝐚𝑦 − sin 𝜙 𝐚𝑥 , the differential torque becomes
(
)
𝑑𝐓 = 𝑎2 𝐼𝐵0 cos 𝜙 cos 𝜙 𝐚𝑦 − sin 𝜙 𝐚𝑥 𝑑𝜙
Note that there is no dependence on 𝑧0 . The net torque on the loop is now
𝐓=
∫
𝑑𝐓 =
∫0
2𝜋
(
)
𝑎2 𝐼𝐵0 cos2 𝜙 𝐚𝑦 − cos 𝜙 sin 𝜙 𝐚𝑥 𝑑𝜙 = 𝜋𝑎2 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 = 400𝜋𝑎2 𝐼𝐵0 𝐚𝑦 = 400𝜋(1.5 × 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 ×
15𝜇0 𝑥 𝑑𝑥
5𝜇0
15𝜇0 cos 𝜙 𝑑𝑥
𝐚𝑧
𝐚𝑧 = −
𝐚𝜙 = −
√
2𝜋𝜌
2𝜋(𝑥2 + 4)
2𝜋 𝑥2 + 4
So now the differential torque about the (0, 2, 0) origin is
𝑑𝐓 = 𝐑𝑇 × 𝑑𝐅 = 𝑥 𝐚𝑥 × −
15𝜇0 𝑥 𝑑𝑥
2𝜋(𝑥2 + 4)
𝐚𝑧 =
15𝜇0 𝑥2 𝑑𝑥
2𝜋(𝑥2 + 4)
𝐚𝑦
The torque is then
( )]𝑏
15𝜇0 [
−1 𝑥
𝑥
−
2
tan
𝐚
𝑦
∫−𝑏 2𝜋(𝑥2 + 4)
2𝜋
2 −𝑏
( )]
[
𝑏
= (6 × 10−6 ) 𝑏 − 2 tan−1
𝐚𝑦 N ⋅ m
2
𝑏
𝐓=
15𝜇0 𝑥2 𝑑𝑥
𝐚𝑦 =
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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 𝜔∕2, where 𝜔 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 = 2𝜋∕𝜔, and so 𝐼 = 𝑒𝜔∕(2𝜋).
Now the dipole moment magnitude will be 𝑚 = 𝐼𝐴, where 𝐴 is the loop area. Thus
𝑚=
𝑒𝜔 2 1 2
𝜋𝑎 = 𝑒𝑎 𝜔 ∕∕
2𝜋
2
b) Show that the torque produced by a magnetic field parallel to the plane of the orbit is 𝑒𝑎2 𝜔𝐵∕2:
With 𝐵 assumed constant over the loop area, we would have 𝐓 = 𝐦 × 𝐁. With 𝐁 parallel to the
loop plane, 𝐦 and 𝐁 are orthogonal, and so 𝑇 = 𝑚𝐵. So, using part 𝑎, 𝑇 = 𝑒𝑎2 𝜔𝐵∕2.
c) by equating the Coulomb and centrifugal forces, show that 𝜔 is (4𝜋𝜖0 𝑚𝑒 𝑎3 ∕𝑒2 )−1∕2 , where 𝑚𝑒 is
the electron mass: The force balance is written as
𝑒2
= 𝑚𝑒 𝜔2 𝑎 ⇒ 𝜔 =
4𝜋𝜖0 𝑎2
(
4𝜋𝜖0 𝑚𝑒 𝑎3
)−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 × 10−11 m; let 𝐵 = 0.5 T: First
[
(1.60 × 10−19 )2
𝜔=
4𝜋(8.85 × 10−12 )(9.1 × 10−31 )(6 × 10−11 )3
]1∕2
= 3.42 × 1016 rad∕s
1
𝑇 = (3.42 × 1016 )(1.60 × 10−19 )(0.5)(6 × 10−11 )2 = 4.93 × 10−24 N ⋅ m
2
Finally,
𝑚=
𝑇
= 9.86 × 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 ⇒
𝑒2
= 𝑚𝑒 𝜔2 𝑎 + 𝑒𝜔𝑎𝐵
2
⏟⏟⏟
4𝜋𝜖0 𝑎
𝑄𝑣𝐵
With 𝐵 = 0, we solve for 𝜔 to find:
√
𝜔 = 𝜔0 =
𝑒2
4𝜋𝜖0 𝑚𝑒 𝑎3
Then with 𝐵 present, we find
𝜔2 =
Therefore
𝑒𝜔𝐵
𝑒𝜔𝐵
𝑒2
−
= 𝜔20 −
3
𝑚𝑒
𝑚𝑒
4𝜋𝜖0 𝑚𝑒 𝑎
𝜔 = 𝜔0
.
But 𝜔 = 𝜔0 , and so
𝑒𝜔𝐵 .
= 𝜔0
1− 2
𝜔0 𝑚𝑒
(
.
𝜔 = 𝜔0 1 −
)
(
√
𝑒𝐵
2𝜔0 𝑚𝑒
𝑒𝜔𝐵
1−
2𝜔20 𝑚𝑒
)
= 𝜔0 −
𝑒𝐵
∕∕
2𝑚𝑒
As for the magnetic moment, we have
)
(
1
1 𝑒2 𝑎2 𝐵
𝑒𝐵
𝑒𝜔 2 1
2 . 1 2
= 𝜔0 𝑒𝑎2 −
∕∕
𝜋𝑎 = 𝜔𝑒𝑎 = 𝑒𝑎 𝜔0 −
𝑚 = 𝐼𝑆 =
2𝜋
2
2
2𝑚𝑒
2
4 𝑚𝑒
Finally, for 𝑎 = 6 × 10−11 m, 𝐵 = 0.5 T, we have
Δ𝜔
1.60 × 10−19 × 0.5
𝑒𝐵 1 . 𝑒𝐵 1
= 1.3 × 10−6
=
=
=
𝜔
2𝑚𝑒 𝜔 2𝑚𝑒 𝜔0
2 × 9.1 × 10−31 × 3.4 × 1016
where 𝜔0 = 3.4 × 1016 sec−1 is found from Problem 16. Finally,
Δ𝑚 𝑒2 𝑎2 𝐵
2 . 𝑒𝐵
×
= 1.3 × 10−6
=
=
𝑚
4𝑚𝑒
2𝑚𝑒 𝜔0
𝜔𝑒𝑎2
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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 × 𝐁 about any origin, where 𝐼 = 0.6 A and 𝐒 = 16𝐚𝑧 m2 .
We find 𝐓 = 0.6(16)𝐚𝑧 × 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 × (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 ≥ 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) × 𝑑𝐅, where
𝑑𝐅 = 𝐼𝑑𝐋 × 𝐁 = 0.6 𝑑𝑦 𝐚𝑦 × [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) × 𝑑𝐅 = 𝐚𝑥 + (𝑦 − 2)𝐚𝑦 − 3𝐚𝑧 × [−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 𝜒𝑚 = 3.1 and within which 𝐁 = 0.4𝑦𝐚𝑧 T, find:
a) 𝐇: We use 𝐁 = 𝜇0 (1 + 𝜒𝑚 )𝐇, or
𝐇=
0.4𝑦𝐚𝑦
(1 + 3.1)𝜇0
= 77.6𝑦𝐚𝑧 kA∕m
b) 𝜇 = (1 + 3.1)𝜇0 = 5.15 × 10−6 H∕m.
c) 𝜇𝑟 = (1 + 3.1) = 4.1.
d) 𝐌 = 𝜒𝑚 𝐇 = (3.1)(77.6𝑦𝐚𝑦 ) = 241𝑦𝐚𝑧 kA∕m
e) 𝐉 = ∇ × 𝐇 = (𝑑𝐻𝑧 )∕(𝑑𝑦) 𝐚𝑥 = 77.6 𝐚𝑥 kA∕m2 .
f) 𝐉𝑏 = ∇ × 𝐌 = (𝑑𝑀𝑧 )∕(𝑑𝑦) 𝐚𝑥 = 241 𝐚𝑥 kA∕m2 .
g) 𝐉𝑇 = ∇ × 𝐁∕𝜇0 = 318𝐚𝑥 kA∕m2 .
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8.20. Find 𝐇 in a material where:
a) 𝜇𝑟 = 4.2, there are 2.7 × 1029 atoms/m3 , and each atom has a dipole moment of 2.6 × 10−30 𝐚𝑦
A ⋅ m2 . Since all dipoles are identical, we may write 𝐌 = 𝑁𝐦 = (2.7 × 1029 )(2.6 × 10−30 𝐚𝑦 ) =
0.70𝐚𝑦 A/m. Then
0.70 𝐚𝑦
𝐌
=
= 0.22 𝐚𝑦 A∕m
𝐇=
𝜇𝑟 − 1 4.2 − 1
b) 𝐌 = 270 𝐚𝑧 A/m and 𝜇 = 2 𝜇H∕m: Have 𝜇𝑟 = 𝜇∕𝜇0 = (2 × 10−6 )∕(4𝜋 × 10−7 ) = 1.59. Then
𝐇 = 270𝐚𝑧 ∕(1.59 − 1) = 456 𝐚𝑧 A∕m.
c) 𝜒𝑚 = 0.7 and 𝐁 = 2𝐚𝑧 T: Use
𝐇=
2𝐚𝑧
𝐁
=
= 936 𝐚𝑧 kA∕m
𝜇0 (1 + 𝜒𝑚 ) (4𝜋 × 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
𝜌 = 0.3 m and 𝜌 = 0.4 m, respectively: We use ∮ 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 𝜙-directed and will vary only with
radius. Note first that for 𝜌 < 0.3 m, no bound current will be enclosed by a path integral, so we
conclude that 𝐌 = 0 for 𝜌 < 0.3m. At radii between the currents the path integral will enclose
only the inner current so,
∮
𝐌 ⋅ 𝑑𝐋 = 2𝜋𝜌𝑀𝜙 = 2𝜋(0.3)12 ⇒ 𝐌 =
3.6
𝐚 A∕m (0.3 < 𝜌 < 0.4m)
𝜌 𝜙
Finally, for 𝜌 > 0.4 m, the total enclosed bound current is 𝐼𝑏,𝑡𝑜𝑡 = 2𝜋(0.3)(12) − 2𝜋(0.4)(9) = 0,
so therefore 𝐌 = 0 (𝜌 > 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 = 𝜇0 (𝐇 + 𝐌) and from
𝐌 = 𝜒𝑚 𝐇, we write
(
)−1
1
0.02
𝐵
𝐵
+1
=
=
𝑀=
= 47.7 A∕m
𝜇0 𝜒𝑚
𝜇0 (334) (4𝜋 × 10−7 )(334)
b) the magnetic field intensity is 1200 A/m and the relative permeability is 1.005: From 𝐁 = 𝜇0
(𝐇 + 𝐌) = 𝜇0 𝜇𝑟 𝐇, we write
𝑀 = (𝜇𝑟 − 1)𝐻 = (.005)(1200) = 6.0 A∕m
c) there are 7.2 × 1028 atoms per cubic meter, each having a dipole moment of 4 × 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 × 1028 )(4 × 10−30 ) = 0.288 A∕m
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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 𝜇𝑟 = 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 𝜌 around the wire axis, and where
𝜌 < 𝑎:
1000𝜇0 𝐼𝜌
𝐼𝜌
(103 )4𝜋 × 10−7 (1)𝜌
𝜋𝜌2
𝐼
⇒
𝐻
=
⇒
𝐁
=
𝐚
=
𝐚𝜙
𝜙
𝜋𝑎2
2𝜋𝑎2
2𝜋𝑎2
2𝜋 × 10−6
= 200𝜌 𝐚𝜙 Wb∕m2
2𝜋𝜌𝐻 =
b) 𝐇: Using part 𝑎, 𝐇 = 𝐁∕𝜇𝑟 𝜇0 = 𝜌∕(2𝜋) × 106 𝐚𝜙 A∕m.
c) 𝐌:
𝐌 = 𝐁∕𝜇0 − 𝐇 =
(2000 − 2)𝜌
× 106 𝐚𝜙 = 1.59 × 108 𝜌 𝐚𝜙 A∕m
4𝜋
d) 𝐉:
𝐉=∇×𝐇=
1 𝑑(𝜌𝐻𝜙 )
𝐚𝑧 = 3.18 × 105 𝐚𝑧 A∕m
𝜌 𝑑𝜌
𝐉𝑏 = ∇ × 𝐌 =
1 𝑑(𝜌𝑀𝜙 )
𝐚𝑧 = 3.18 × 108 𝐚𝑧 A∕m2
𝜌 𝑑𝜌
e) 𝐉𝑏 within the wire:
8.23. Calculate values for 𝐻𝜙 , 𝐵𝜙 , and 𝑀𝜙 at 𝜌 = 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 𝜇 = 3 𝜇H∕m for 2.5 < 𝜌 < 3.5 mm,
𝜇 = 5 𝜇H∕m for 3.5 < 𝜌 < 4.5 mm, and 𝜇 = 10 𝜇H∕m for 4.5 < 𝜌 < 6 mm. Compute for:
a) 𝑐 = 3 mm: Have
𝐻𝜙 =
12
𝐼
= 637 A∕m
=
2𝜋𝜌 2𝜋(3 × 10−3 )
Then 𝐵𝜙 = 𝜇𝐻𝜙 = (3 × 10−6 )(637) = 1.91 × 10−3 Wb∕m2 .
Finally, 𝑀𝜙 = (1∕𝜇0 )𝐵𝜙 − 𝐻𝜙 = 884 A∕m.
b) 𝑐 = 4 mm: Have
𝐻𝜙 =
12
𝐼
=
= 478 A∕m
2𝜋𝜌 2𝜋(4 × 10−3 )
Then 𝐵𝜙 = 𝜇𝐻𝜙 = (5 × 10−6 )(478) = 2.39 × 10−3 Wb∕m2 .
Finally, 𝑀𝜙 = (1∕𝜇0 )𝐵𝜙 − 𝐻𝜙 = 1.42 × 103 A∕m.
c) 𝑐 = 5 mm: Have
𝐻𝜙 =
𝐼
12
= 382 A∕m
=
2𝜋𝜌 2𝜋(5 × 10−3 )
Then 𝐵𝜙 = 𝜇𝐻𝜙 = (10 × 10−6 )(382) = 3.82 × 10−3 Wb∕m2 .
Finally, 𝑀𝜙 = (1∕𝜇0 )𝐵𝜙 − 𝐻𝜙 = 2.66 × 103 A∕m.
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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 𝜇𝑟 = 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 × 𝐚𝑛 = 𝐾0 𝐚𝑦 × 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
𝑧2
1∕(𝑧2 −𝑧1 )
𝑧2
𝜇0 𝐾0
(𝑎𝑧 + 1) 𝑑𝑧
∫𝑠
∫𝑧1 ∫0
(𝑧2 − 𝑧1 ) ∫𝑧1
[
]
]
𝜇0 𝐾0 [ 𝑎 2
𝑎
(𝑧2 − 𝑧21 ) + (𝑧2 − 𝑧1 ) = 𝜇0 𝐾0 (𝑧2 + 𝑧1 ) + 1 Wb∕m2
=
(𝑧2 − 𝑧1 ) 2
2
Φ𝑚 =
𝜇𝐇 ⋅ 𝑑𝐒 =
𝜇0 𝐾0 (𝑎𝑧 + 1) 𝐚𝑥 ⋅ 𝐚𝑥 𝑑𝑦 𝑑𝑧 =
8.25. A conducting filament at 𝑧 = 0 carries 12 A in the 𝐚𝑧 direction. Let 𝜇𝑟 = 1 for 𝜌 < 1 cm, 𝜇𝑟 = 6 for
1 < 𝜌 < 2 cm, and 𝜇𝑟 = 1 for 𝜌 > 2 cm. Find
a) 𝐇 everywhere: This result will depend on the current and not the materials, and is:
𝐼
1.91
𝐇=
𝐚 =
A∕m (0 < 𝜌 < ∞)
2𝜋𝜌 𝜙
𝜌
b) 𝐁 everywhere: We use 𝐁 = 𝜇𝑟 𝜇0 𝐇 to find:
𝐁(𝜌 < 1 cm) = (1)𝜇0 (1.91∕𝜌) = (2.4 × 10−6 ∕𝜌)𝐚𝜙 T
𝐁(1 < 𝜌 < 2 cm) = (6)𝜇0 (1.91∕𝜌) = (1.4 × 10−5 ∕𝜌)𝐚𝜙 T
𝐁(𝜌 > 2 cm) = (1)𝜇0 (1.91∕𝜌) = (2.4 × 10−6 ∕𝜌)𝐚𝜙 T where 𝜌 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 < 𝜌 < 𝑎 within the solenoid has 𝜇𝑟 = 5, while 𝜇𝑟 = 1 for 𝑎 < 𝜌 < 3 cm. Determine 𝑎 so that
a) a total flux of 10 𝜇Wb is present: First, the magnetic flux density in the coil is written in general
as 𝐁 = 𝜇𝑛𝐼 𝐚𝑧 Wb∕m2 . Using 𝑏 = 0.03m as the outer radius, the total flux in the coil becomes
2𝜋
𝑎
2𝜋
𝑏
5𝜇0 𝑛𝐼𝜌 𝑑𝜌 𝑑𝜙 +
𝜇 𝑛𝐼 𝜌 𝑑𝜌 𝑑𝜙
𝐁 ⋅ 𝑑𝐒 =
∫𝑠
∫0 ∫0
∫0 ∫𝑎 0
[
]
[
]
= 2𝜋𝜇0 𝑛𝐼 5𝑎2 + (𝑏2 − 𝑎2 ) = 2𝜋𝜇0 𝑛𝐼 4𝑎2 + 𝑏2
Φ𝑚 =
Substituting the given numbers, we have
[
]
Φ𝑚 = 2𝜋(4𝜋 × 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 < 𝜌 < 𝑎 and 𝑎 < 𝜌 < 3 cm:
Using the expression for the flux in part 𝑎, we set
3
𝑏
5𝑎2 = 𝑏2 − 𝑎2 ⇒ 𝑎 = √ = √ = 1.22 cm
6
6
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8.27. Let 𝜇𝑟1 = 2 in region 1, defined by 2𝑥 + 3𝑦 − 4𝑧 > 1, while 𝜇𝑟2 = 5 in region 2 where 2𝑥 + 3𝑦 − 4𝑧 < 1.
In region 1, 𝐇1 = 50𝐚𝑥 − 30𝐚𝑦 + 20𝐚𝑧 A/m. Find:
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 𝜇1 𝐇𝑁1 = 𝜇2 𝐇𝑁2 or
𝐇𝑁2 =
𝜇𝑟1
2
𝐇 = (−4.83𝐚𝑥 − 7.24𝐚𝑦 + 9.66𝐚𝑧 ) = −1.93𝐚𝑥 − 2.90𝐚𝑦 + 3.86𝐚𝑧 A∕m
𝜇𝑅 2 𝑁1 5
e) 𝜃1 , the angle between 𝐇1 and 𝐚𝑁21 : This will be
]
[
50𝐚𝑥 − 30𝐚𝑦 + 20𝐚𝑧
𝐇1
cos 𝜃1 =
⋅ (.37𝐚𝑥 + .56𝐚𝑦 − .74𝐚𝑧 ) = −0.21
⋅𝐚
=
|𝐇1 | 𝑁21
(502 + 302 + 202 )1∕2
Therefore 𝜃1 = cos−1 (−.21) = 102◦ .
f) 𝜃2 , 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
[
]
52.90𝐚𝑥 − 25.66𝐚𝑦 + 14.20𝐚𝑧
𝐇2
cos 𝜃2 =
⋅𝐚
=
⋅ (.37𝐚𝑥 + .56𝐚𝑦 − .74𝐚𝑧 ) = −0.09
|𝐇2 | 𝑁21
60.49
Therefore 𝜃2 = cos−1 (−.09) = 95◦ .
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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 𝜇 = 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 = Φ𝑅, where, in the central leg,
𝑅𝑐 =
𝐿𝑖𝑛
3 × 10−2
= 2.4 × 104 H
=
−3
−4
𝜇𝐴𝑖𝑛
(5 × 10 )(2.5 × 10 )
In each outer leg, the reluctance is
𝑅𝑜 =
𝐿𝑜𝑢𝑡
10 × 10−2
=
= 1.25 × 105 H
−3
−4
𝜇𝐴𝑜𝑢𝑡
(5 × 10 )(1.6 × 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 × 104
H. We now have
𝑚𝑚𝑓
14.4
=
Φ=
= 1.66 × 10−4 Wb
𝑅𝑇
8.65 × 104
The flux density in the center leg is now
𝐵=
Φ 1.66 × 10−4
=
= 0.666 T
𝐴
2.5 × 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 × 10−3
= 9.55 × 105 H
−7
−4
(4𝜋 × 10 )(2.5 × 10 )
Now the total reluctance is 𝑅𝑛𝑒𝑡 = 𝑅𝑇 + 𝑅𝑎𝑖𝑟 = 8.56 × 104 + 9.55 × 105 = 1.04 × 106 . The flux
in the center leg is now
14.4
Φ=
= 1.38 × 10−5 Wb
1.04 × 106
and
1.38 × 10−5
𝐵=
= 55.3 mT
2.5 × 10−4
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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 Φ = 0.666(2.5 × 10−4 ) = 1.66 × 10−4 Wb. This divides equally in
the two outer legs, so that the flux density in each outer leg is
( )
1 1.66 × 10−4
= 0.52 Wb∕m2
2 1.6 × 10−4
.
Using Fig. 8.11 with this result, we find 𝐻𝑜𝑢𝑡 = 90 A ⋅ t∕m We now use
𝐵𝑜𝑢𝑡 =
∮
𝐇 ⋅ 𝑑𝐋 = 𝑁𝐼
to find
𝐼=
) (120)(3 × 10−2 ) + (90)(10 × 10−2 )
1 (
𝐻𝑖𝑛 𝐿𝑖𝑛 + 𝐻𝑜𝑢𝑡 𝐿𝑜𝑢𝑡 =
= 10.5 mA
𝑁
1200
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8.30. A rectangular core has fixed permeability 𝜇𝑟 >> 1, a square cross-section of dimensions 𝑎 × 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
∮
𝐇 ⋅ 𝑑𝐋 = (2𝑏 + 2𝑑)𝐻
At all other points in the core interior, but off the midline, the path integral becomes
∮
𝐇 ⋅ 𝑑𝐋 = [2(𝑑 + 𝑎 − 2𝑥) + 2(𝑏 + 𝑎 − 2𝑥)] 𝐻 = 𝐼𝑒𝑛𝑐𝑙 = 𝑁1 𝐼1
The flux density magnitudes are therefore
𝐵11 = 𝐵12 = 𝜇𝐻 =
𝜇𝑟 𝜇0 𝑁1 𝐼1
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:
𝑎
𝑎
𝜇𝑟 𝜇0 𝑁1 𝐼1
1
|𝑎
𝑑𝑥 𝑑𝑦 = − 𝜇𝑟 𝜇0 𝑁1 𝐼1 𝑎 ln [𝑑 + 𝑏 + 2𝑎 − 4𝑥] |
|0
∫𝑠
∫0 ∫0 2(𝑑 + 𝑏 + 2𝑎 − 4𝑥)
8
]
[
𝑑 + 𝑏 + 2𝑎
1
Wb
= 𝜇𝑟 𝜇0 𝑁1 𝐼1 𝑎 ln
8
𝑑 + 𝑏 − 2𝑎
Φ𝑚 =
𝐁 ⋅ 𝑑𝐒 =
c) Find the self-inductance of coil 1:
𝐿11 =
]
[
𝑁1 𝐵11
1
𝑑 + 𝑏 + 2𝑎
= 𝜇𝑟 𝜇0 𝑁12 𝑎 ln
H
𝐼1
8
𝑑 + 𝑏 − 2𝑎
d) find the mutual inductance between coils 1 and 2.
𝑀12 = 𝑀 =
]
[
𝑁2 𝐵12
1
𝑑 + 𝑏 + 2𝑎
H
= 𝜇𝑟 𝜇0 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∕(𝜇𝐴), is zero, leaving only the gap reluctance. This is
𝑅𝑔 =
0.25 × 10−3
𝑑
= 7.1 × 105 H
=
𝜇 0 𝐴𝑔
(4𝜋 × 10−7 )(2.5 × 10−4 )
Now
Φ=
𝑚𝑚𝑓
200
= 2.8 × 10−4 Wb
=
𝑔
7.1 × 105
b) the magnetic material is assumed to be linear with 𝜇𝑟 = 1000: Now the core reluctance is no
longer zero, but
8 × 10−2
= 2.6 × 105 H
𝑅𝑐 =
−7
−4
(1000)(4𝜋 × 10 )(2.5 × 10 )
The flux is then
Φ=
𝑚𝑚𝑓
200
= 2.1 × 10−4 Wb
=
𝑅𝑐 + 𝑅𝑔
9.7 × 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 Φ found
in part 𝑎, assuming infinite permeability: Φ(1) = 2.8 × 10−4 Wb. The flux density in the core
is then 𝐵𝑐(1) = (2.8 × 10−4 )∕(2.5 × 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:
∮
𝐇 ⋅ 𝑑𝐋 = 𝐻𝑐(1) 𝐿𝑐 + 𝐻𝑔(1) 𝑑
where 𝐻𝑐(1) 𝐿𝑐 = (270)(8 × 10−2 ) = 22, and where 𝐻𝑔(1) 𝑑 = Φ(1) 𝑔 = (2.8 × 10−4 )(7.1 × 105 ) =
199. But we require that
∮
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 Φ(2) = 2.5 × 10−4 Wb. Now
∮
𝐇 ⋅ 𝑑𝐋 = 𝐻𝑐(2) 𝐿𝑐 + Φ(2) 𝑅𝑔 = 200(8 × 10−2 ) + (2.5 × 10−4 )(7.1 × 105 ) = 194
This is less than 200, meaning that the actual flux is slightly higher than 2.5 × 10−4 Wb.
I will leave the answer at that, considering the lack of fine resolution in Fig. 8.11.
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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 𝜇𝑟 fills the region between conductors. Assume current 𝐼 flows in both
conductors, in opposite directions.
Within the coax, the magnetic field is 𝐇 = 𝐼∕(2𝜋𝜌) 𝐚𝜙 . The energy density is then
𝜇 𝜇 𝐼2
1
𝑤𝑚 = 𝐁 ⋅ 𝐇 = 𝑟 20 2 J∕m3
2
8𝜋 𝜌
The energy per unit length in 𝑧 is therefore
𝑊𝑚 =
∫𝑣
1
𝑤𝑚 𝑑𝑣 =
∫0 ∫0
2𝜋
∫𝑎
𝑏
𝜇𝑟 𝜇0 𝐼 2
8𝜋 2 𝜌2
𝜌 𝑑𝜌 𝑑𝜙 𝑑𝑧 =
𝜇𝑟 𝜇0 𝐼 2 ( 𝑏 )
J∕m
ln
4𝜋
𝑎
b) Obtain the inductance, 𝐿, per unit length of line by equating the energy to (1∕2)𝐿𝐼 2 .
𝐿=
2𝑊𝑚
𝐼2
=
𝜇𝑟 𝜇0 ( 𝑏 )
H∕m
ln
2𝜋
𝑎
8.33. A toroidal core has a square cross section, 2.5 cm < 𝜌 < 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 𝜇𝑟 = 10, while the
lower half, −0.5 cm < 𝑧 < 0, has 𝜇𝑟 = 20. An mmf of 150 A ⋅ t establishes a flux in the 𝐚𝜙 direction.
For 𝑧 > 0, find:
a) 𝐻𝜙 (𝜌): Ampere’s circuital law gives:
2𝜋𝜌𝐻𝜙 = 𝑁𝐼 = 150 ⇒ 𝐻𝜙 =
150
= 23.9∕𝜌 A∕m
2𝜋𝜌
b) 𝐵𝜙 (𝜌): We use 𝐵𝜙 = 𝜇𝑟 𝜇0 𝐻𝜙 = (10)(4𝜋 × 10−7 )(23.9∕𝜌) = 3.0 × 10−4 ∕𝜌 Wb∕m2 .
c) Φ𝑧>0 : This will be
Φ𝑧>0 =
∫ ∫
𝐁 ⋅ 𝑑𝐒 =
= 5.0 × 10
−7
∫0
.005
.035
∫.025
)
(
3.0 × 10−4
.035
𝑑𝜌𝑑𝑧 = (.005)(3.0 × 10−4 ) ln
𝜌
.025
Wb
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 𝐻𝜙 = 23.9∕𝜌 A∕m. Next, 𝐵𝜙 is modified only by the
new permeability, which is twice the value used in part 𝑎: Thus 𝐵𝜙 = 6.0 × 10−4 ∕𝜌 Wb∕m2 .
Finally, since 𝐵𝜙 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 Φ𝑧<0 = 1.0 × 10−6 Wb.
e) Find Φtotal : This will be the sum of the values found for 𝑧 < 0 and 𝑧 > 0, or Φtotal =
1.5 × 10−6 Wb.
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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 = 𝐼𝜌∕(2𝜋𝑎2 ) 𝐚𝜙 , and find the integral of the energy density over the unit length
in 𝑧:
1
2𝜋
𝑎
𝜇0 𝐼 2
𝜇0 𝜌2 𝐼 2
1
𝜌
𝑑𝜌
𝑑𝜙
𝑑𝑧
=
𝜇0 𝐻 2 𝑑𝑣 =
J∕m
𝑊𝑒 =
∫0 ∫0 ∫0 8𝜋 2 𝑎4
∫𝑣𝑜𝑙 2
16𝜋
8.35. The cones 𝜃 = 21◦ and 𝜃 = 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◦ < 𝜃 < 159◦ , 0 < 𝜙 < 2𝜋: We can apply Ampere’s
circuital law and take advantage of symmetry. We expect to see 𝐇 in the 𝐚𝜙 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:
∮
𝐇 ⋅ 𝑑𝐋 =
∫0
2𝜋
𝐻𝜙 𝐚𝜙 ⋅ 𝑟 sin 𝜃𝐚𝜙 𝑑𝜙 = 𝐼𝑒𝑛𝑐𝑙. = 40 A
Assuming that 𝐻𝜙 is constant over the integration path, we take it outside the integral and solve:
𝐻𝜙 =
40
20
⇒ 𝐇=
𝐚 A∕m
2𝜋𝑟 sin 𝜃
𝜋𝑟 sin 𝜃 𝜙
b) How much energy is stored in this region? This will be
159
2𝜋
.25
100𝜇0 159 𝑑𝜃
200𝜇0
1
2
𝑟
sin
𝜃
𝑑𝑟
𝑑𝜃
𝑑𝜙
=
𝜇0 𝐻𝜙2 =
∫0 ∫21◦ ∫0 𝜋 2 𝑟2 sin2 𝜃
∫𝑣 2
𝜋 ∫21◦ sin 𝜃
[
]
100𝜇0
tan(159∕2)
ln
= 1.35 × 10−4 J
=
𝜋
tan(21∕2)
◦
◦
𝑊𝐻 =
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 𝐵𝜙 = 20𝜇0 ∕(𝜋𝑟 sin 𝜃). We
integrate this over the cross-sectional area defined by 0 < 𝑟 < 0.25 and 21◦ < 𝜃 < 159◦ , to find
the total flux:
[
]
159◦
0.25
20𝜇0
5𝜇0
5𝜇
tan(159∕2)
Φ=
𝑟 𝑑𝑟 𝑑𝜃 =
ln
= 0 (3.37) = 6.74 × 10−6 Wb
∫21◦ ∫0
𝜋𝑟 sin 𝜃
𝜋
tan(21∕2)
𝜋
Now 𝐿 = Φ∕𝐼 = 6.74 × 10−6 ∕40 = 0.17 𝜇H.
Second method: Use the energy computation in part 𝑏, and write
𝐿=
2𝑊𝐻
2(1.35 × 10−4 )
=
= 0.17 𝜇H
𝐼2
(40)2
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8.36. The dimensions of the outer conductor of a coaxial cable are 𝑏 and 𝑐, where 𝑐 > 𝑏. Assuming 𝜇 = 𝜇0 ,
find the magnetic energy stored per unit length in the region 𝑏 < 𝜌 < 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 𝜌, where 𝑏 < 𝜌 < 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 𝜌. We obtain:
]
[ 2
]
[ 2
𝑐 − 𝜌2
𝜌 − 𝑏2
=𝐼 2
2𝜋𝜌𝐻 = 𝐼 − 𝐼 2
𝑐 − 𝑏2
𝑐 − 𝑏2
So that
]
[ 2
𝑐 − 𝜌2
𝐼
𝐇=
𝐚𝜙 A∕m (𝑏 < 𝜌 < 𝑐)
2𝜋𝜌 𝑐 2 − 𝑏2
The energy within the outer conductor is now
]
[ 2
𝑐
2𝜋
1
𝜇0 𝐼 2
1
𝑐
2
2
2
𝜇 𝐻 𝑑𝑣 =
− 2𝑐 + 𝜌 𝜌 𝑑𝜌 𝑑𝜙, 𝑑𝑧
𝑊𝑚 =
∫0 ∫0 ∫𝑏 8𝜋 2 (𝑐 2 − 𝑏2 )2 𝜌2
∫𝑣𝑜𝑙 2 0
[
]
𝜇0 𝐼 2
1
2 2
4 4
ln(𝑐∕𝑏)
−
(1
−
𝑏
∕𝑐
)
+
=
(1
−
𝑏
∕𝑐
)
J
4
4𝜋(1 − 𝑏2 ∕𝑐 2 )2
8.37. A toroid has known core reluctance . Two windings, having 𝑁1 and 𝑁2 turns are present. Find
a) the self inductances of coils 1 and 2: In general, we have 𝑁𝐼 = Φ𝑚 , and 𝐿 = 𝑁Φ𝑚 ∕𝐼. So
𝐿1 =
2
𝑁2
𝑁1 (𝑁1 𝐼1 ∕) 𝑁1
and likewise 𝐿2 = 2
=
𝐼1


b) the mutual inductance between the two coils:
𝑀=
𝑁2 (𝑁1 𝐼1 ∕) 𝑁1 (𝑁2 𝐼2 ∕) 𝑁1 𝑁2
=
=
𝐼1
𝐼2

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8.38. A toroidal core has a rectangular cross section defined by the surfaces 𝜌 = 2 cm, 𝜌 = 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 𝜌 in the interior of the toroid, and in the 𝐚𝜙 direction. Doing this for coil 1 (carrying
current 𝐼1 ), we find:
∮
𝐇 ⋅ 𝑑𝐋 = 2𝜋𝜌𝐻𝜙 = 𝑁1 𝐼1 ⇒ 𝐻𝜙 =
𝑁1 𝐼1
2𝜋𝜌
The flux in the toroid is then the integral over the cross section of 𝐁:
Φ11 =
∫ ∫
.045
𝐁 ⋅ 𝑑𝐋 =
∫.04
.03
∫.02
( )
𝜇𝜇 𝑁 𝐼
𝜇𝑟 𝜇0 𝑁1 𝐼1
.03
𝑑𝜌 𝑑𝑧 = (.005) 𝑟 0 1 1 ln
2𝜋𝜌
2𝜋
.02
The self linkage is then given by Λ11 = 𝑁1 Φ, and the inductance is
𝐿11 =
𝑁1 Φ11
(.005)(80)(4𝜋 × 10−7 )(1000)2
=
ln(1.5) = 32 mH
𝐼1
2𝜋
Performing the same procedure for coil 2, we find:
𝐿22 =
𝑁2 Φ22
(.005)(80)(4𝜋 × 10−7 )(2500)2
=
ln(1.5) = 203 mH
𝐼2
2𝜋
b) Find the mutual inductance between the two coils: Because the coils share the same core, we
have Φ12 = Φ11 and Φ21 = Φ22 . Then
Λ12 = 𝑁2 Φ12 and Λ21 = 𝑁1 Φ21
and thus
𝐿12 =
𝑁 Φ
𝑁2 Φ12
(.005)(80)(4𝜋 × 10−7 )(1000)(2500)
= 𝐿21 = 1 21 =
ln(1.5) = 80 mH
𝐼1
𝐼2
2𝜋
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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
𝑊𝐻 =
𝑤
1
1
1
1
𝜇0 𝐻 2 𝑑𝑣 =
𝜇0 𝐾02 𝑑𝑥 𝑑𝑦 𝑑𝑧 = 𝑤𝑑𝜇0 𝐾02 J∕m
∫𝑣 2
∫0 ∫0 ∫0 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=
𝜇0 𝑑
2 𝑑𝑤
2 𝑤𝑑
2
2
𝜇
𝐾
=
𝜇
𝐾
=
H∕m
0
0
0
0
𝑤
𝐼2 2
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:
𝑑
Φ=
∫0 ∫0
1
𝑑
𝜇0 𝐻𝐚𝑦 ⋅ 𝐚𝑦 𝑑𝑥 𝑑𝑧 =
Then
𝐿=
∫0 ∫0
1
𝜇0 𝐾0 𝑑𝑥 𝑑𝑦 = 𝜇0 𝑑𝐾0
𝜇 𝑑
Φ 𝜇0 𝑑𝐾0
= 0 H∕m
=
𝐼
𝑤𝐾0
𝑤
8.40. A coaxial cable has conductor radii 𝑎 and 𝑏, where 𝑎 < 𝑏. Material of permeability 𝜇𝑟 ≠ 1 exists in
the region 𝑎 < 𝜌 < 𝑐, while the region 𝑐 < 𝜌 < 𝑏 is air-filled. Find an expression for the inductance
per unit length.
In both regions, the magnetic field will be 𝐇 = 𝐼∕(2𝜋𝜌) 𝐚𝜙 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 𝜙, unit length in 𝑧, and between radii 𝑎 and 𝑏:
Φ𝑚 =
∫𝑠
𝑏
1
𝜇0 𝐼
𝜇𝑟 𝜇0 𝐼
𝑑𝜌 𝑑𝑧 +
𝑑𝜌 𝑑𝑧
∫0 ∫𝑐 2𝜋𝜌
∫0 ∫𝑎 2𝜋𝜌
( )
( )]
𝜇 𝐼[
𝑐
𝑏
= 0 𝜇𝑟 ln
+ ln
Wb∕m
2𝜋
𝑎
𝑐
1
𝑐
𝐁 ⋅ 𝑑𝐒 =
The inductance per unit length is then 𝐿 = Φ𝑚 ∕𝐼 (with one turn), or
( )
( )]
𝜇 [
𝑐
𝑏
+ ln
H∕m
𝐿 = 0 𝜇𝑟 ln
2𝜋
𝑎
𝑐
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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
Φ=
∫0 ∫1
3
𝜇 𝐼
−𝜇0 𝐼
𝐚𝑥 ⋅ (−𝐚𝑥 ) 𝑑𝑦 𝑑𝑧 = 0 ln 3
2𝜋𝑦
2𝜋
The mutual inductance is then
𝑀=
𝑁Φ 150𝜇0
=
ln 3 = 33 𝜇H
𝐼
2𝜋
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=
𝜇0 𝐼𝐚𝜙
√
2𝜋 𝑦2 + 1
The flux through the coil is now
Φ=
=
1
∫0 ∫1
3
1
3
∫0 ∫1
3
1
𝜇0 𝐼𝐚𝜙
𝜇0 𝐼 sin 𝜙
⋅ (−𝐚𝑥 ) 𝑑𝑦 𝑑𝑧 =
𝑑𝑦 𝑑𝑧
√
√
∫0 ∫1 2𝜋 𝑦2 + 1
2𝜋 𝑦2 + 1
𝜇 𝐼
𝜇0 𝐼𝑦
|3
𝑑𝑦 𝑑𝑧 = 0 ln(𝑦2 + 1)| = (1.6 × 10−7 )𝐼
2
|1
2𝜋
2𝜋(𝑦 + 1)
The mutual inductance is then
𝑀=
𝑁Φ
= (150)(1.6 × 10−7 ) = 24 𝜇H
𝐼
8.42. Find the mutual inductance between two filaments forming circular rings of radii 𝑎 and Δ𝑎, where
Δ𝑎 << 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 =
∫0
2𝜋
𝐼𝑎𝑑𝜙 𝐚𝜙 × (−𝐚𝜌 )
4𝜋𝑎2
=
∫0
2𝜋
𝐼𝑑𝜙 𝐚𝑧
𝐼
=
𝐚 A∕m
4𝜋𝑎
2𝑎 𝑧
We now approximate that field as constant over a circular area of radius Δ𝑎, and write the flux
linkage (for the single turn) as
Φ
𝜇 𝜋(Δ𝑎)2
𝜇 𝐼𝜋(Δ𝑎)2
.
⇒ 𝑀= 𝑚 = 0
Φ𝑚 = 𝜋(Δ𝑎)2 𝐵𝑜𝑢𝑡𝑒𝑟 = 0
2𝑎
𝐼
2𝑎
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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 𝜇0 ∕(8𝜋) H∕m. We first find the magnetic
field inside the conductor, then calculate the energy stored there. From Ampere’s circuital law:
2𝜋𝜌𝐻𝜙 =
Now
𝜋𝜌2
𝐼𝜌
𝐼 ⇒ 𝐻𝜙 =
A∕m
2
𝜋𝑎
2𝜋𝑎2
1
𝑊𝐻 =
1
𝜇 𝐻 2 𝑑𝑣 =
∫0 ∫0
∫𝑣 2 0 𝜙
2𝜋
∫0
𝑎
𝜇 0 𝐼 2 𝜌2
8𝜋 2 𝑎4
𝜌 𝑑𝜌 𝑑𝜙 𝑑𝑧 =
𝜇0 𝐼 2
J∕m
16𝜋
Now, with 𝑊𝐻 = (1∕2)𝐿𝐼 2 , we find 𝐿𝑖𝑛𝑡 = 𝜇0 ∕(8𝜋) as expected.
b) Find the internal inductance if the portion of the conductor for which 𝜌 < 𝑐 < 𝑎 is removed: The
hollowed-out conductor still carries current 𝐼, so Ampere’s circuital law now reads:
]
[ 2
𝜌 − 𝑐2
𝜋(𝜌2 − 𝑐 2 )
𝐼
A∕m
2𝜋𝜌𝐻𝜙 =
⇒ 𝐻𝜙 =
2𝜋𝜌 𝑎2 − 𝑐 2
𝜋(𝑎2 − 𝑐 2 )
and the energy is now
1
∫0 ∫0
2𝜋
𝜇0 𝐼 2
𝜇0 𝐼 2 (𝜌2 − 𝑐 2 )2
𝑎
𝜌 𝑑𝜌 𝑑𝜙 𝑑𝑧 =
∫𝑐 8𝜋 2 𝜌2 (𝑎2 − 𝑐 2 )2
4𝜋(𝑎2 − 𝑐 2 )2 ∫𝑐
( )]
[
𝜇0 𝐼 2
1 4
𝑎
4
2 2
2
4
J∕m
(𝑎
−
𝑐
)
−
𝑐
(𝑎
−
𝑐
)
+
𝑐
ln
=
2
2
2
𝑐
4𝜋(𝑎 − 𝑐 ) 4
𝑊𝐻 =
𝑎[
]
𝐶4
𝜌 − 2𝑐 𝜌 +
𝑑𝜌
𝜌
3
2
The internal inductance is then
𝐿𝑖𝑛𝑡
]
[
𝜇0 𝑎4 − 4𝑎2 𝑐 2 + 3𝑐 4 + 4𝑐 4 ln(𝑎∕𝑐)
2𝑊𝐻
=
=
H∕m
8𝜋
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 (𝜇∕𝜋) 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) =
𝜇𝐼
𝐚 Wb∕m2 (𝑎 < 𝑥 < 𝑑 − 𝑎)
𝜋𝑥 𝑦
The flux per unit length in 𝑧 between conductors is now:
Φ𝑚 =
∫𝑠
1
𝐁 ⋅ 𝑑𝐒 =
∫0 ∫𝑎
(𝑑−𝑎)
𝜇𝐼 ( 𝑑 − 𝑎 )
𝜇𝐼
𝐚𝑦 ⋅ 𝐚𝑦 𝑑𝑥 𝑑𝑧 =
ln
𝜋𝑥
𝜋
𝑎
Now, the external inductance will be 𝐿 = Φ𝑚 ∕𝐼, which becomes
. 𝜇 (𝑑 )
H∕m
𝐿 = ln
𝜋
𝑎
under the assumption that 𝑎 << 𝑑.
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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 =
−𝜌𝑚
𝜇0
where 𝜌𝑚 = −𝜇0 ∇ ⋅ 𝐌 is an equivalent magnetic charge density. What happens when 𝐌 is
uniform? Eq. (25) can be written as
𝐇=
𝐁
− 𝐌 = −∇𝑉𝑚
𝜇0
Taking the divergence, we find
−∇ ⋅ ∇𝑉𝑚 = −∇2 𝑉𝑚 =
1
∇⋅𝐁−∇⋅𝐌
𝜇0
Noting that ∇ ⋅ 𝐁 = 0, we obtain the desired result if we define 𝜌𝑚 = −𝜇0 ∇ ⋅ 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 = ∇ × 𝐀, and Eq. (25), show that when
a permanent magnetization exists, and with zero current:
∇ × ∇ × 𝐀 = 𝜇0 𝐉𝑒𝑞
where 𝐉𝑒𝑞 = ∇ × 𝐌, in analogy to Eq. (57) in Chapter 7: Take
∇ × 𝐁 = ∇ × ∇ × 𝐀 = ∇ × 𝜇0 (𝐇 + 𝐌) = 𝜇0 ∇ × 𝐌 = 𝜇0 𝐉𝑒𝑞
where 𝐇 = 0 because we have zero current.
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Engineering Electromagnetics 9th solution
Engineering Electromagnetics (Chungnam National University)
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CHAPTER 9 – 9th Edition
9.1. In Fig. 9.4, let 𝐵 = 0.2 cos 120𝜋𝑡 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 Φ = 𝜋(0.15)2 𝐵 = 1.41 × 10−2 cos 120𝜋𝑡
Wb. Now, 𝑒𝑚𝑓 = 𝑉𝑏𝑎 (𝑡) = −𝑑Φ∕𝑑𝑡 = (120𝜋)(1.41 × 10−2 ) sin 120𝜋𝑡. Then 𝑉𝑎𝑏 (𝑡) = −𝑉𝑏𝑎 (𝑡) =
−5.33 sin 120𝜋𝑡 V.
b) 𝐼(𝑡) = 𝑉𝑏𝑎 (𝑡)∕𝑅 = 5.33 sin(120𝜋𝑡)∕250 = 21.3 sin(120𝜋𝑡) 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 𝜔𝑡 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
Φ𝑚 =
∫𝑠
𝑣𝑡
𝐁 ⋅ 𝑑𝐒 =
∫0 ∫0
𝑑
𝐵0 sin 𝜔𝑡 (𝐚𝑧 ⋅ 𝐚𝑧 ) 𝑑𝑥 𝑑𝑦 = 𝐵0 𝑣 𝑡 𝑑 sin 𝜔𝑡
Then the emf is
𝑒𝑚𝑓 =
∮
𝐄 ⋅ 𝑑𝐋 = −
𝑑Φ𝑚
= −𝐵0 𝑑 𝑣 [sin 𝜔𝑡 + 𝜔𝑡 cos 𝜔𝑡] V
𝑑𝑡
9.3. Given 𝐇 = 300 𝐚𝑧 cos(3 × 108 𝑡 − 𝑦) A/m in free space, find the emf developed in the general 𝐚𝜙
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
300𝜇0 cos(3 × 108 𝑡 − 𝑦) 𝑑𝑥 𝑑𝑦 = 300𝜇0 sin(3 × 108 𝑡 − 𝑦)|10
∫0 ∫0
[
]
= 300𝜇0 sin(3 × 108 𝑡 − 1) − sin(3 × 108 𝑡) Wb
Φ=
Then
[
]
𝑑Φ
= −300(3 × 108 )(4𝜋 × 10−7 ) cos(3 × 108 𝑡 − 1) − cos(3 × 108 𝑡)
𝑑𝑡
[
]
= −1.13 × 105 cos(3 × 108 𝑡 − 1) − cos(3 × 108 𝑡) V
emf = −
b) corners at (0,0,0), (2𝜋,0,0), (2𝜋,2𝜋,0), (0,2𝜋,0): In this case, the flux is
Φ = 2𝜋 × 300𝜇0 sin(3 × 108 𝑡 − 𝑦)|2𝜋
=0
0
The emf is therefore 0.
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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 𝜔, 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
Φ𝑚 =
∫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 𝜔𝑡 𝐚𝑧 − sin 𝜔𝑡 𝐚𝑦
Therefore,
𝑏∕2
Φ𝑚 =
𝑎∕2
(
)
𝐵0 𝐚𝑧 ⋅ cos 𝜔𝑡 𝐚𝑧 − sin 𝜔𝑡 𝐚𝑦 𝑑𝑥 𝑑𝑦 = 𝑎𝑏𝐵0 cos 𝜔𝑡
∫−𝑏∕2 ∫−𝑎∕2
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 =
∮
𝐄 ⋅ 𝑑𝐋 = −
𝑑Φ𝑚
= 𝑎𝑏 𝜔𝐵0 sin 𝜔𝑡 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
Φ=
∫0
0.2
∫0
𝑥
0.8(𝑥′ )2 𝑑𝑥′ 𝑑𝑦 =
0.16 3 0.16
𝑥 =
(5𝑡 + 2𝑡3 )3 Wb
3
3
Then
emf = −
𝑑Φ 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
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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(𝜔𝑡 − 𝛽𝑦) 𝐚𝑧 , where 𝜔 and 𝛽 are constants.
We begin by finding the net magnetic flux through the loop:
Φ𝑚 =
=
∫𝑠
𝑏∕2
𝐁 ⋅ 𝑑𝐒 =
𝑎∕2
∫−𝑏∕2 ∫−𝑎∕2
𝐵0 cos(𝜔𝑡 − 𝛽𝑦) 𝐚𝑧 ⋅ 𝐚𝑧 𝑑𝑥 𝑑𝑦
]
𝐵0 𝑎 [
sin(𝜔𝑡 + 𝛽𝑏∕2) − sin(𝜔𝑡 − 𝛽𝑏∕2)
𝛽
Now the emf is
𝑒𝑚𝑓 =
∮
𝐄 ⋅ 𝑑𝐋 = −
]
𝑑Φ𝑚
𝐵 𝑎𝜔 [
=− 0
cos(𝜔𝑡 + 𝛽𝑏∕2) − cos(𝜔𝑡 − 𝛽𝑏∕2)
𝑑𝑡
𝛽
Using the trig identity, cos(𝑎 ± 𝑏) = cos 𝑎 cos 𝑏 ∓ sin 𝑎 sin 𝑏, we may write the above result as
𝜔
𝑒𝑚𝑓 = +2𝐵0 𝑎 sin(𝜔𝑡) sin(𝛽𝑏∕2) V
𝛽
9.7. The rails in Fig. 9.7 each have a resistance of 2.2 Ω∕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 Ω resistor is present across the left end with the right end open-circuited: The flux in the
left-hand closed loop is
Φ𝑙 = 𝐵 × area = (0.8)(0.2)(2 + 9𝑡)
Then, emf 𝑙 = −𝑑Φ𝑙 ∕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 Ω 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
Φ𝑟 = (0.8)(0.2)[(16 − 2) − 9𝑡]
and emf 𝑟 = −𝑑Φ𝑟 ∕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
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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
∮
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 𝐚𝜙
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, 𝐚𝜙 . Since the wire is perfectly-conducting, the only voltage appears
across the resistor, and is given as 𝑉𝑅 . Faraday’s law becomes
∮
𝐄 ⋅ 𝑑𝐋 = 𝑉𝑅 = −
𝑑Φ𝑚
𝑑
=−
𝐵 𝑡 𝐚 ⋅ 𝐚 𝑑𝑎 = −𝜋𝑎2 𝐵0
𝑑𝑡
𝑑𝑡 ∫𝑠 0 𝑧 𝑧
This indicates that the resistor voltage, 𝑉𝑅 = 𝜋𝑎2 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 = 𝜋𝑎2 𝐵0 ∕𝑅.
9.9. A square filamentary loop of wire is 25 cm on a side and has a resistance of 125 Ω 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 × 108 𝑡 − 0.5𝑥) 𝜇T.
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:
∫0
.25
∫0
.25
8 × 10−6 cos(1.5 × 108 𝑡 − 0.5𝑥) 𝑑𝑥 𝑑𝑦
[
]
= −(4 × 10−6 ) sin(1.5 × 108 𝑡 − 0.13) − sin(1.5 × 108 𝑡) Wb
Φ(𝑡) =
[
]
Now, 𝑒𝑚𝑓 = 𝑉 (𝑡) = −𝑑Φ∕𝑑𝑡 = 6.0 × 102 cos(1.5 × 108 𝑡 − 0.13) − cos(1.5 × 108 𝑡) , The total loop
resistance is 𝑅 = 125(0.25 + 0.25 + 0.25 + 0.25) = 125 Ω. Then the ohmic power is
𝑃 (𝑡) =
[
]2
𝑉 2 (𝑡)
= 2.9 × 103 cos(1.5 × 108 𝑡 − 0.13) − cos(1.5 × 108 𝑡) Watts
𝑅
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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 𝜎∕𝜔𝜖 for the applied field 𝐸 = 𝐸𝑚 cos 𝜔𝑡. Assume 𝜇 = 𝜇0 . First, 𝐷 =
𝜖𝐸 = 𝜖𝐸𝑚 cos 𝜔𝑡. Then the displacement current density is 𝜕𝐷∕𝜕𝑡 = −𝜔𝜖𝐸𝑚 sin 𝜔𝑡. Second,
𝐽𝑐 = 𝜎𝐸 = 𝜎𝐸𝑚 cos 𝜔𝑡. Using these results we find |𝐽𝑐 |∕|𝐽𝑑 | = 𝜎∕𝜔𝜖.
b) What is the amplitude ratio if the applied field is 𝐸 = 𝐸𝑚 𝑒−𝑡∕𝜏 , where 𝜏 is real? As before, find
𝐷 = 𝜖𝐸 = 𝜖𝐸𝑚 𝑒−𝑡∕𝜏 , and so 𝐽𝑑 = 𝜕𝐷∕𝜕𝑡 = −(𝜖∕𝜏)𝐸𝑚 𝑒−𝑡∕𝜏 . Also, 𝐽𝑐 = 𝜎𝐸𝑚 𝑒−𝑡∕𝜏 . Finally,
|𝐽𝑐 |∕|𝐽𝑑 | = 𝜎𝜏∕𝜖.
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 𝜖 = 10−11 F/m, 𝜇 = 10−5 H/m, and
𝜎 = 10−5 S/m. If the electric field intensity is 𝐄 = (106 ∕𝜌) cos(105 𝑡)𝐚𝜌 V/m, find:
a) 𝐉: Use
𝐉 = 𝜎𝐄 =
(
10
𝜌
)
cos(105 𝑡)𝐚𝜌 A∕m2
b) the total conduction current, 𝐼𝑐 , through the capacitor: Have
𝐼𝑐 =
∫ ∫
𝐉 ⋅ 𝑑𝐒 = 2𝜋𝜌𝑙𝐽 = 20𝜋𝑙 cos(105 𝑡) = 8𝜋 cos(105 𝑡) A
c) the total displacement current, 𝐼𝑑 , through the capacitor: First find
𝐉𝑑 =
(105 )(10−11 )(106 )
𝜕
1
𝜕𝐃
= (𝜖𝐄) = −
sin(105 𝑡)𝐚𝜌 = − sin(105 𝑡) A∕m
𝜕𝑡
𝜕𝑡
𝜌
𝜌
Now
𝐼𝑑 = 2𝜋𝜌𝑙𝐽𝑑 = −2𝜋𝑙 sin(105 𝑡) = −0.8𝜋 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
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9.12. The magnetic flux density 𝐁 = 𝐵0 cos(𝜔𝑡) 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
∇×𝐇=
= 𝜖0
𝜕𝑡
𝜕𝑡
The given field field is 𝑦-directed and varies spatially only with 𝑧. So the above becomes
∇×𝐇=∇×
1 𝜕𝐵𝑦
𝜕𝐃 𝑘0 𝐵0
𝐁
=−
cos(𝜔𝑡) sin(𝑘0 𝑧) 𝐚𝑥 A∕m2
𝐚𝑥 =
=
𝜇0
𝜇0 𝜕𝑧
𝜕𝑡
𝜇0
b) the electric field intensity: We write
𝐄=
𝑘 𝐵
𝑘 𝐵
1
𝜕𝐃
cos(𝜔𝑡)𝑑𝑡 = 0 0 sin(𝑘0 𝑧) sin(𝜔𝑡) 𝐚𝑥
𝑑𝑡 + 𝐶 = 0 0 sin(𝑘0 𝑧) 𝐚𝑥
∫
𝜖0 ∫ 𝜕𝑡
𝜇0 𝜖0
𝜔𝜇0 𝜖0
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
∇×𝐄=−
𝜕𝐁
⇒ 𝐁 = − ∇ × 𝐄 𝑑𝑡
∫
𝜕𝑡
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=−
=
∫
∇ × 𝐄 𝑑𝑡 = −
𝑘20 𝐵0
𝜔2 𝜇0 𝜖0
−𝑘20 𝐵0
𝜕𝐸𝑥
cos(𝑘0 𝑧) sin(𝜔𝑡)𝐚𝑦 𝑑𝑡
𝐚𝑦 𝑑𝑡 =
∫ 𝜔𝜇0 𝜖0
∫ 𝜕𝑧
cos(𝑘0 𝑧) cos(𝜔𝑡) 𝐚𝑦
Requiring this result to be the same as the given field leads to the identification:
√
𝑘0 = 𝜔 𝜇0 𝜖0 m−1 .
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9.13. In free space it is known that 𝐄 = 𝐸0 ∕𝑟 sin 𝜃 cos(𝜔𝑡 − 𝑘0 𝑟) 𝐚𝜃 = 𝐸𝜃 𝐚𝜃 . Show that a component of
𝐇 in the 𝐚𝜙 direction arises, where 𝐻𝜙 = 𝐸𝜃 (𝜖0 ∕𝜇0 )1∕2 . Do this by applying Eqs. (20) and (21) and
requiring consistency:
Start with Eq. (20) in free space:
∇ × 𝐄 = −𝜇0
𝜕𝐇
𝜕𝑡
where
∇ × 𝐸𝜃 (𝑟, 𝜃) 𝐚𝜃 =
𝜕𝐻𝜙
1
1 𝜕(𝑟𝐸𝜃 )
𝐚𝜙 = − 𝐸0 𝑘0 sin 𝜃 sin(𝜔𝑡 − 𝑘0 𝑟)𝐚𝜙 = −𝜇0
𝐚
𝑟 𝜕𝑟
𝑟
𝜕𝑡 𝜙
After integrating the above result over time, we find
𝐇 = 𝐻𝜙 𝐚𝜙 =
𝐸0 𝑘0
sin 𝜃 cos(𝜔𝑡 − 𝑘0 𝑟)𝐚𝜙
𝜔𝜇0 𝑟
Now, use Eq. (21) in free space:
∇ × 𝐇 = 𝜖0
𝜕𝐄
𝜕𝑡
where
1 𝜕(𝐻𝜙 sin 𝜃)
1 𝜕(𝑟𝐻𝜙 )
𝐚𝑟 −
𝐚𝜃
𝑟 sin 𝜃
𝜕𝜃
𝑟 𝜕𝑟
𝐸0 𝑘2
2𝐸 𝑘
= 2 0 0 2 cos 𝜃 sin(𝜔𝑡 − 𝑘0 𝑟)𝐚𝑟 + 2 0 2 sin 𝜃 cos(𝜔𝑡 − 𝑘0 𝑟)𝐚𝜃
𝜔 𝜇 0 𝜖0 𝑟
𝜔 𝜇 0 𝜖0 𝑟
√
The second term in this field will be equal to the given field when 𝑘0 = 𝜔 𝜇0 𝜖0 . With this
substitution, the 𝜙 component of 𝐇 becomes:
√
𝜖0 𝐸0
sin 𝜃 cos(𝜔𝑡 − 𝑘0 𝑟) Q.E.D
𝐻𝜙 =
𝜇0 𝑟
∇ × 𝐻𝜙 (𝑟, 𝜃)𝐚𝜙 =
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9.14. A voltage source, 𝑉0 sin 𝜔𝑡, is connected between two concentric conducting spheres, 𝑟 = 𝑎 and 𝑟 = 𝑏,
𝑏 > 𝑎, where the region between them is a material for which 𝜖 = 𝜖𝑟 𝜖0 , 𝜇 = 𝜇0 , and 𝜎 = 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 =
(1∕𝑟) − (1∕𝑏)
𝑉 sin 𝜔𝑡
(1∕𝑎) − (1∕𝑏) 0
𝑉0 sin 𝜔𝑡
𝐚𝑟
2
𝑟 (1∕𝑎 − 1∕𝑏)
Now
𝐉𝐝 =
⇒ 𝐃=
𝜖𝑟 𝜖0 𝑉0 sin 𝜔𝑡
𝑟2 (1∕𝑎 − 1∕𝑏)
𝐚𝑟
𝜕𝐃 𝜖𝑟 𝜖0 𝜔𝑉0 cos 𝜔𝑡
𝐚𝑟
= 2
𝜕𝑡
𝑟 (1∕𝑎 − 1∕𝑏)
The displacement current is then
𝐼𝑑 = 4𝜋𝑟2 𝐽𝑑 =
4𝜋𝜖𝑟 𝜖0 𝜔𝑉0 cos 𝜔𝑡
𝑑𝑉
=𝐶
(1∕𝑎 − 1∕𝑏)
𝑑𝑡
where, from Eq. 6, Chapter 6,
𝐶=
4𝜋𝜖𝑟 𝜖0
(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
∇ × 𝐇 = 𝐉 (the source of 𝐇 is conduction current only)
𝜕𝐁
= 0 ⇒ 𝐁 is constant
𝜕𝑡
𝜌𝑣 = 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
∇ × 𝐄 = 0 (𝐄 is constant in time and conservative)
𝜕𝐃
𝜕𝐃
=0 ⇒ 𝐉=
=0
𝜕𝑡
𝜕𝑡
∇ ⋅ 𝐃 = 𝜌𝑣 (𝐄 arises f rom f ree charge only)
∇×𝐇=𝐉+
∇ ⋅ 𝐁 = 0 (irrelevant as 𝐁 = 0 anyway)
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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:
𝜕𝜌
𝜕
∇⋅∇×𝐇=∇⋅𝐉+ ∇⋅𝐃=∇⋅𝐉+ 𝑣 =0
⏟⏞⏞⏟⏞⏞⏟
𝜕𝑡
𝜕𝑡
0
where we have used ∇ ⋅ 𝐃 = 𝜌𝑣 , another Maxwell equation.
9.17. The electric field intensity in the region 0 < 𝑥 < 5, 0 < 𝑦 < 𝜋∕12, 0 < 𝑧 < 0.06 m in free space is
given by 𝐄 = 𝐶 sin(12𝑦) sin(𝑎𝑧) cos(2 × 1010 𝑡) 𝐚𝑥 V/m. Beginning with the ∇ × 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
𝜕𝐸𝑧
𝜕𝐸𝑥
𝐚𝑦 −
𝐚
𝜕𝑧
𝜕𝑦 𝑧
[
]
𝜕𝐁
= 𝐶 𝑎 sin(12𝑦) cos(𝑎𝑧)𝐚𝑦 − 12 cos(12𝑦) sin(𝑎𝑧)𝐚𝑧 cos(2 × 1010 𝑡) = −
𝜕𝑡
∇×𝐄=
Then
1
∇ × 𝐄 𝑑𝑡 + 𝐶1
𝜇0 ∫
[
]
𝐶
=−
𝑎 sin(12𝑦) cos(𝑎𝑧)𝐚𝑦 − 12 cos(12𝑦) sin(𝑎𝑧)𝐚𝑧 sin(2 × 1010 𝑡) A∕m
10
𝜇0 (2 × 10
𝐇=−
where the integration constant, 𝐶1 = 0, since there are no initial conditions. Using this result, we now
find
]
[
𝜕𝐻𝑧 𝜕𝐻𝑦
𝐶(144 + 𝑎2 )
𝜕𝐃
−
𝐚𝑥 = −
sin(12𝑦) sin(𝑎𝑧) sin(2 × 1010 𝑡) 𝐚𝑥 =
∇×𝐇=
𝜕𝑦
𝜕𝑧
𝜕𝑡
𝜇0 (2 × 1010 )
Now
𝐄=
𝐶(144 + 𝑎2 )
1
𝐃
=
∇ × 𝐇 𝑑𝑡 + 𝐶2 =
sin(12𝑦) sin(𝑎𝑧) cos(2 × 1010 𝑡) 𝐚𝑥
𝜖0 ∫ 𝜖0
𝜇0 𝜖0 (2 × 1010 )2
where 𝐶2 = 0. This field must be the same as the original field as stated, and so we require that
𝐶(144 + 𝑎2 )
=1
𝜇0 𝜖0 (2 × 1010 )2
Using 𝜇0 𝜖0 = (3 × 108 )−2 , we find
[
(2 × 1010 )2
𝑎=
− 144
(3 × 108 )2
]1∕2
= 66 m−1
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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 𝜇𝑟 = 1, 𝜖𝑟 = 20, and 𝜎 = 0. Neglect fields outside the
dielectric. Given the field 𝐇 = 5 cos(109 𝑡 − 𝛽𝑧)𝐚𝑦 A/m, use Maxwell’s equations to help find:
a) 𝛽, if 𝛽 > 0: Take
∇×𝐇=−
So
𝐄=
𝜕𝐻𝑦
𝜕𝑧
𝐚𝑥 = −5𝛽 sin(109 𝑡 − 𝛽𝑧)𝐚𝑥 = 20𝜖0
𝜕𝐄
𝜕𝑡
−5𝛽
𝛽
cos(109 𝑡 − 𝛽𝑧)𝐚𝑥
sin(109 𝑡 − 𝛽𝑧)𝐚𝑥 𝑑𝑡 =
∫ 20𝜖0
(4 × 109 )𝜖0
Then
∇×𝐄=
𝜕𝐸𝑥
𝛽2
𝜕𝐇
𝐚𝑦 =
sin(109 𝑡 − 𝛽𝑧)𝐚𝑦 = −𝜇0
𝜕𝑧
𝜕𝑡
(4 × 109 )𝜖0
So that
𝐇=
𝛽2
−𝛽 2
9
sin(10
𝑡
−
𝛽𝑧)𝐚
𝑑𝑡
=
cos(109 𝑡 − 𝛽𝑧)
𝑥
18
∫ (4 × 109 )𝜇0 𝜖0
(4 × 10 )𝜇0 𝜖0
= 5 cos(109 𝑡 − 𝛽𝑧)𝐚𝑦
where the last equality is required to maintain consistency. Therefore
𝛽2
= 5 ⇒ 𝛽 = 14.9 m−1
(4 × 1018 )𝜇0 𝜖0
b) the displacement current density at 𝑧 = 0: Since 𝜎 = 0, we have
∇ × 𝐇 = 𝐉𝑑 = −5𝛽 sin(109 𝑡 − 𝛽𝑧) = −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
0.1
[
]
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 𝜌𝑒𝑘𝑡 𝐚𝜙
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 ∇×𝐇 = 𝜕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
𝜇s. Use the ∇ × 𝐇 equation to show that the rate at which 𝐵𝑧 should (but does not) change with
𝜌 is only about 5 × 10−6 T/m in free space at 𝑡 = 0: Assuming that 𝐁 varies with 𝜌, we write
𝜕𝐻
1 𝑑𝐵0 𝑘𝑡
𝜕𝐄
1
𝑒 = 𝜖0
= − 𝜖0 𝑘2 𝐵0 𝜌𝑒𝑘𝑡
∇ × 𝐇 = − 𝑧 𝐚𝜙 = −
𝜕𝜌
𝜇0 𝑑𝜌
𝜕𝑡
2
Thus
𝑑𝐵0
1012 (1)𝜌
1
= 𝜇0 𝜖0 𝑘2 𝜌𝐵0 =
= 5.6 × 10−6 𝜌
𝑑𝜌
2
2(3 × 108 )2
which is near the stated value if 𝜌 is on the order of 1m.
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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:
∇ × 𝐄0 (𝐫) 𝑒𝑠𝑡 = −
∇ × 𝐇0 (𝐫) 𝑒𝑠𝑡 = 𝐉0 (𝐫)𝑒𝑠𝑡 +
)
𝜕 (
𝐁0 (𝐫) 𝑒𝑠𝑡 = −𝑠𝐁0 (𝐫) 𝑒𝑠𝑡
𝜕𝑡
)
𝜕 (
𝐃0 (𝐫) 𝑒𝑠𝑡 = 𝐉0 (𝐫)𝑒𝑠𝑡 + 𝑠𝐃0 (𝐫) 𝑒𝑠𝑡
𝜕𝑡
∇ ⋅ 𝐃0 (𝐫) 𝑒𝑠𝑡 = 𝜌0 (𝐫) 𝑒𝑠𝑡
∇ ⋅ 𝐁0 (𝐫) 𝑒𝑠𝑡 = 0
In all cases, the 𝑒𝑠𝑡 terms divide out, leaving:
∇ × 𝐄0 (𝐫) = −𝑠𝐁0 (𝐫)
∇ × 𝐇0 (𝐫) = 𝐉0 (𝐫) + 𝑠𝐃0 (𝐫)
∇ ⋅ 𝐃0 (𝐫) = 𝜌0 (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 = −∇ × ∇ × 𝐀 + ∇(∇ ⋅ 𝐀) = −𝜇𝐉
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 = ∇ × 𝐀, (55) becomes
−∇ × 𝐁 = −𝜇𝐉 ⇒ ∇ × 𝐇 = 𝐉
b) Show that Eq. (51) becomes Faraday’s law when taking the curl: Doing this gives
∇ × 𝐄 = −∇ × ∇𝑉 −
𝜕
∇×𝐀
𝜕𝑡
The curl of the gradient is identially zero, and ∇ × 𝐀 = 𝐁. We are left with
∇ × 𝐄 = −𝜕𝐁∕𝜕𝑡
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9.22. In a sourceless medium, in which 𝐉 = 0 and 𝜌𝑣 = 0, assume a rectangular coordinate system in which
𝐄 and 𝐇 are functions only of 𝑧 and 𝑡. The medium has permittivity 𝜖 and permeability 𝜇.
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
⇒
𝐚𝑦 = −𝜇
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
∇×𝐄=−
∇×𝐇=
𝜕𝐻𝑦
𝜕𝐸
𝜕𝐃
⇒ −
𝐚𝑥 = 𝜖 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
= −𝜇
𝜕 2 𝐻𝑦
𝜕𝑡𝜕𝑧
and
𝜕 2 𝐻𝑦
𝜕𝑧𝜕𝑡
= −𝜖
𝜕 2 𝐸𝑥
𝜕𝑡2
Combining these two, we find
𝜕 2 𝐸𝑥
𝜕𝑧2
= 𝜇𝜖
𝜕 2 𝐸𝑥
𝜕𝑡2
b) Show that 𝐸𝑥 = 𝐸0 cos(𝜔𝑡 − 𝛽𝑧) is a solution of that equation for a particular value of 𝛽: Substituting, we find
𝜕 2 𝐸𝑥
𝜕𝑧2
= −𝛽 2 𝐸0 cos(𝜔𝑡 − 𝛽𝑧) and 𝜇𝜖
𝜕 2 𝐸𝑥
𝜕𝑡2
= −𝜔2 𝜇𝜖𝐸0 cos(𝜔𝑡 − 𝛽𝑧)
These two will be equal provided the constant multipliers of cos(𝜔𝑡 − 𝛽𝑧) are equal.
√
c) Find 𝛽 as a function of given parameters. Equating the two constants in part 𝑏, we find 𝛽 = 𝜔 𝜇𝜖.
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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(𝜔𝑡 − 𝛽𝑧)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
∇×𝐄=−
from which
𝐇=
𝜕𝐸𝑦
𝜕𝑧
𝐚𝑥 = +𝛽𝐸0 sin(𝜔𝑡 − 𝛽𝑧)𝐚𝑥 = −𝜇0
𝜕𝐇
𝜕𝑡
𝛽𝐸
−𝛽𝐸0
sin(𝜔𝑡 − 𝛽𝑧)𝐚𝑥 𝑑𝑡 + 𝐶 = + 0 cos(𝜔𝑡 − 𝛽𝑧)𝐚𝑥
𝜇0 ∫
𝜔𝜇0
where the integration constant is zero. Next we have
𝛽𝐸0
∇ × 𝐀 = 𝐁 = 𝜇0 𝐇 =
cos(𝜔𝑡 − 𝛽𝑧)𝐚𝑥 =
𝜔
(
𝜕𝐴𝑧 𝜕𝐴𝑦
−
𝜕𝑦
𝜕𝑧
)
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
𝛽𝐸0
𝛽𝐸0
=
cos(𝜔𝑡 − 𝛽𝑧) ⇒ 𝐀 =
𝑦 cos(𝜔𝑡 − 𝛽𝑧)𝐚𝑧
𝜕𝑦
𝜔
𝜔
which is seen to satisfy the boundary condition at 𝑦 = 0.
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9.24. A vector potential is given as 𝐀 = 𝐴0 cos(𝜔𝑡 − 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=
𝑘𝐴
1 𝜕𝐴𝑦
1
∇×𝐀=−
𝐚𝑥 = − 0 sin(𝜔𝑡 − 𝑘𝑧) 𝐚𝑥 A∕m
𝜇
𝜇 𝜕𝑧
𝜇
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:
∇×𝐇=
𝜕𝐃
𝜕𝐄
=𝜖
𝜕𝑡
𝜕𝑡
Using the magnetic field as found above, we find
∇×𝐇=
Now,
𝜕𝐻𝑥
𝑘 2 𝐴0
𝑘 2 𝐴0
𝜕𝐄
𝐚𝑦 =
cos(𝜔𝑡 − 𝑘𝑧) 𝐚𝑦 = 𝜖
⇒ 𝐄=
sin(𝜔𝑡 − 𝑘𝑧) 𝐚𝑦 V∕m
𝜕𝑧
𝜇
𝜕𝑡
𝜔𝜇𝜖
[
]
𝜕𝐀
𝜕𝐀
⇒ ∇𝑉 = −
+𝐄
𝜕𝑡
𝜕𝑡
]
[
𝜕𝑉
𝑘2
sin(𝜔𝑡 − 𝑘𝑧) 𝐚𝑦 =
∇𝑉 = 𝐴0 𝜔 1 − 2
𝐚
𝜕𝑦 𝑦
𝜔 𝜇𝜖
𝐄 = −∇𝑉 −
or
Integrating over 𝑦 we find
[
]
𝑘2
sin(𝜔𝑡 − 𝑘𝑧) + 𝐶
𝑉 = 𝐴0 𝜔 𝑦 1 − 2
𝜔 𝜇𝜖
√
where 𝐶, the integration constant, can be taken as zero. In part 𝑏, it will be shown that 𝑘 = 𝜔 𝜇𝜖,
which means that 𝑉 = 0.
b) Specify 𝑘 in terms of 𝐴0 , 𝜔, and the constants of the lossless medium, 𝜖 and 𝜇. Use the other
Maxwell curl equation:
𝜕𝐁
𝜕𝐇
∇×𝐄=−
= −𝜇
𝜕𝑡
𝜕𝑡
so that
𝑘3 𝐴
𝜕𝐇
1
1 𝜕𝐸𝑦
=− ∇×𝐄=
𝐚𝑥 = − 2 0 cos(𝜔𝑡 − 𝑘𝑧) 𝐚𝑥
𝜕𝑡
𝜇
𝜇 𝜕𝑧
𝜔𝜇 𝜖
Integrate over 𝑡 (and set the integration constant to zero) and require the result to be consistant
with part 𝑎:
𝑘𝐴
𝑘3 𝐴
𝐇 = − 2 20 sin(𝜔𝑡 − 𝑘𝑧) 𝐚𝑥 = − 0 sin(𝜔𝑡 − 𝑘𝑧) 𝐚𝑥
𝜇
𝜔 𝜇 𝜖
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
f rom part a
We identify
√
𝑘 = 𝜔 𝜇𝜖
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9.25. In a region where 𝜇𝑟 = 𝜖𝑟 = 1 and 𝜎 = 0, the retarded potentials are given by 𝑉 = 𝑥(𝑧 − 𝑐𝑡) V and
√
𝐀 = 𝑥[(𝑧∕𝑐) − 𝑡]𝐚𝑧 Wb/m, where 𝑐 = 1∕ 𝜇0 𝜖0 .
a) Show that ∇ ⋅ 𝐀 = −𝜇𝜖(𝜕𝑉 ∕𝜕𝑡):
First,
∇⋅𝐀=
Second,
√
𝜕𝐴𝑧
𝑥
= = 𝑥 𝜇0 𝜖0
𝜕𝑧
𝑐
𝜕𝑉
𝑥
= −𝑐𝑥 = − √
𝜕𝑡
𝜇0 𝜖0
so we observe that ∇ ⋅ 𝐀 = −𝜇0 𝜖0 (𝜕𝑉 ∕𝜕𝑡) in free space, implying that the given statement would
hold true in general media.
b) Find 𝐁, 𝐇, 𝐄, and 𝐃:
Use
𝐁=∇×𝐀=−
Then
𝐇=
)
(
𝜕𝐴𝑥
𝑧
𝐚 T
𝐚𝑦 = 𝑡 −
𝜕𝑥
𝑐 𝑦
)
(
1
𝑧
𝐁
=
𝐚 A∕m
𝑡−
𝜇0
𝜇0
𝑐 𝑦
Now,
𝐄 = −∇𝑉 −
𝜕𝐀
= −(𝑧 − 𝑐𝑡)𝐚𝑥 − 𝑥𝐚𝑧 + 𝑥𝐚𝑧 = (𝑐𝑡 − 𝑧)𝐚𝑥 V∕m
𝜕𝑡
Then
𝐃 = 𝜖0 𝐄 = 𝜖0 (𝑐𝑡 − 𝑧)𝐚𝑥 C∕m2
c) Show that these results satisfy Maxwell’s equations if 𝐉 and 𝜌𝑣 are zero:
i. ∇ ⋅ 𝐃 = ∇ ⋅ 𝜖0 (𝑐𝑡 − 𝑧)𝐚𝑥 = 0
ii. ∇ ⋅ 𝐁 = ∇ ⋅ (𝑡 − 𝑧∕𝑐)𝐚𝑦 = 0
iii.
∇×𝐇=−
𝜕𝐻𝑦
𝜕𝑧
𝐚𝑥 =
1
𝐚 =
𝜇0 𝑐 𝑥
√
𝜖0
𝐚
𝜇0 𝑥
which we require to equal 𝜕𝐃∕𝜕𝑡:
𝜕𝐃
= 𝜖0 𝑐𝐚𝑥 =
𝜕𝑡
iv.
∇×𝐄=
√
𝜖0
𝐚
𝜇0 𝑥
𝜕𝐸𝑥
𝐚 = −𝐚𝑦
𝜕𝑧 𝑦
which we require to equal −𝜕𝐁∕𝜕𝑡:
𝜕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 𝜌𝑣 are both zero. Replace 𝜖 by 𝜇, 𝜇 by 𝜖, 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:
∇ × 𝐄 = −𝜇
𝜕𝐇
𝜕𝑡
(1)
𝜕𝐄
𝜕𝑡
∇ ⋅ 𝜖𝐄 = 0
(3)
∇ ⋅ 𝜇𝐇 = 0
(4)
∇×𝐇=𝜖
(2)
In making the above substitutions, we find that (1) converts to (2), (2) converts to (1), and (3)
and (4) convert to each other.
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Engineering Electromagnetics 9th solution
Engineering Electromagnetics (Chungnam National University)
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CHAPTER 10 – 9th Edition
10.1. The parameters of a certain transmission line, operating at 𝜔 = 6.0 × 108 rad/s are 𝐿 = 0.350 𝜇H∕m,
𝐶 = 40.0 pF/m, 𝐺 = 0, and 𝑅 = 15.0 Ω/m. Evaluate 𝛼, 𝛽, 𝜆, and 𝑍0 .
We use
𝛾=
=
√
√
𝑍𝑌 =
√
(𝑅 + 𝑗𝜔𝐿)(𝐺 + 𝑗𝜔𝐶)
[15 + 𝑗(6 × 108 )(0.350 × 10−6 )][0 + 𝑗(6 × 108 )(40 × 10−12 )]
= 0.080 + 𝑗2.25 m−1 = 𝛼 + 𝑗𝛽
Therefore, 𝛼 = 0.080 Np∕m, 𝛽 = 2.25 rad∕m, and 𝜆 = 2𝜋∕𝛽 = 2.80 m. Finally,
√
𝑍0 =
√
√
𝑍
=
𝑌
15 + 𝑗2.1 × 102
= 93.6 − 𝑗3.34 Ω = 93.67∠ − 0.036 rad
0 + 𝑗2.4 × 10−2
𝑅 + 𝑗𝜔𝐿
=
𝐺 + 𝑗𝜔𝐶
10.2. A sinusoidal wave on a transmission line is specified by voltage and current in phasor form:
𝑉𝑠 (𝑧) = 𝑉0 𝑒𝛼𝑧 𝑒𝑗𝛽𝑧
and 𝐼𝑠 (𝑧) = 𝐼0 𝑒𝛼𝑧 𝑒𝑗𝛽𝑧 𝑒𝑗𝜙
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+𝑗𝛽𝑧 .
b) It is found that 𝛼 = 0, 𝑍0 = 50 Ω, and the wave velocity is 𝑣𝑝 = 2.5×108 m/s, with 𝜔 = 108 s−1 .
Evaluate 𝑅, 𝐺, 𝐿, 𝐶, 𝜆, and 𝜙: First, the fact that 𝛼 = 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=
1
1
= 8.0 × 10−11 F = 80 pF
=
𝑍0 𝑣𝑝
50(2.5 × 108 )
Then
𝐿 = 𝐶𝑍02 = (8.0 × 10−11 )(50)2 = 2.0 × 10−7 = 0.20 𝜇H
Now,
𝜆=
𝑣𝑝
𝑓
=
2𝜋𝑣𝑝
𝜔
=
2𝜋(2.5 × 108 )
= 15.7 m
108
Finally, the current phase is found through
𝐼0 𝑒𝑗𝜙 =
𝑉0
𝑍0
Since 𝑉0 , 𝐼0 , and 𝑍0 are all real, it follows that 𝜙 = 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
𝑉 (𝑡) = 𝑉0 𝑒−𝑡
2 ∕(2𝑇 2 )
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 − |Γ|2 ), where Γ =
(100 − 50)∕(100 + 50) = 1∕3. We find:
𝑃𝐿 (𝑡) =
)
(10)2 −𝑡2 ∕𝑇 2 (
16 2 2
1
= 𝑒−𝑡 ∕𝑇 W
1−
𝑒
50
9
9
The energy dissipated by the load is the time integral of the above, or
∞
𝑊 =
∫−∞
√
16 −𝑡2 ∕𝑇 2
16
𝑒
𝑑𝑡 = (20 × 10−9 ) 𝜋 = 63 nJ
9
9
10.4. A sinusoidal voltage wave of amplitude 𝑉0 , frequency 𝜔, and phase constant, 𝛽, 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−𝑗𝛽𝑧 + 𝑉0 𝑒𝑗𝛽𝑧
The current is found from the voltage by dividing by 𝑍0 (while incorporating the proper sign
for forward and backward waves):
𝐼𝑠𝑇 (𝑧) =
)
𝑉0 −𝑗𝛽𝑧
𝑉
𝑉 (
2𝑉
𝑒
− 0 𝑒𝑗𝛽𝑧 = − 0 𝑒𝑗𝛽𝑧 − 𝑒−𝑗𝛽𝑧 = −𝑗 0 sin(𝛽𝑧)
𝑍0
𝑍0
𝑍0
𝑍0
The real instantaneous current is now
⎫
⎧
{
}
⎪
⎪ 2𝑉0
𝑗𝜔𝑡
sin(𝛽𝑧)[cos(𝜔𝑡) + 𝑗 sin(𝜔𝑡)]⎬
(𝑧, 𝑡) = 𝑒 𝐼𝑠𝑇 (𝑧)𝑒
= 𝑒 ⎨−𝑗
𝑍
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⎪
0
⎪
⎭
⎩
𝑒𝑗𝜔𝑡
2𝑉
= 0 sin(𝛽𝑧) sin(𝜔𝑡)
𝑍0
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10.5. Two voltage waves of equal amplitude, 𝑉0 , and frequency, 𝜔, propagate in the forward 𝑧 direction in a
lossy transmission line having attenuation coefficient 𝛼, and characteristic impedance, 𝑍0 = |𝑍0 |𝑒𝑗𝛿
One wave is phase-shifted from the other by 𝜙 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−𝛼𝑧 𝑒−𝑗𝛽𝑧 + 𝑉0 𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝑒𝑗𝜙
(
)
= 𝑉0 𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 1 + 𝑒𝑗𝜙
(
)
= 𝑉0 𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝑒𝑗𝜙∕2 𝑒−𝑗𝜙∕2 + 𝑒𝑗𝜙∕2
= 2𝑉0 cos(𝜙∕2)𝑒𝑗𝜙∕2 𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧
In real instantaneous form, this becomes:
{
}
(𝑧, 𝑡) = 𝑒 𝑉𝑠𝑇 (𝑧)𝑒𝑗𝜔𝑡 = 2𝑉0 cos(𝜙∕2)𝑒−𝛼𝑧 cos(𝜔𝑡 − 𝛽𝑧 + 𝜙∕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 |𝑒𝑗𝛿 :
𝐼𝑠𝑇 (𝑧) =
2𝑉0
cos(𝜙∕2)𝑒𝑗(𝜙∕2−𝛿) 𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧
|𝑍0 |
So that
{
}
2𝑉0 −𝛼𝑧
𝑒 cos(𝜙∕2) cos(𝜔𝑡 − 𝛽𝑧 + 𝜙∕2 − 𝛿) A
(𝑧, 𝑡) = 𝑒 𝐼𝑠𝑇 (𝑧)𝑒𝑗𝜔𝑡 =
|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:
{
} 2𝑉02
1
∗
cos2 (𝜙∕2) cos 𝛿 𝑒−2𝛼𝑧 W
=
𝑃 = 𝑒 𝑉𝑠𝑇 𝐼𝑠𝑇
2
|𝑍0 |
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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 𝛼 = 0.080 Np/m. Then
Loss[dB∕m] = 8.69𝛼 = 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
Γ𝐿 =
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 × 𝑒−2𝛼𝐿 × (1 − |Γ𝐿 |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
[ ]
[
]
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 × 𝑒−2𝛼(2𝐿) × |Γ𝐿 |2 = 100 𝑒−200(0.080) (0.304)2 = 1.04 × 10−6 W = 1.04 mW
As a decibel reduction from the original input power, this becomes
]
[
]
[
𝑃𝑖𝑛
100
= 10 log10
= 79.8 dB
𝑃𝑜𝑢𝑡 [dB] = 10 log10
𝑃𝑜𝑢𝑡
1.04 × 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 × 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
)
(
1
Loss𝑙𝑜𝑎𝑑 [dB] = 10 log10
1 − |Γ𝐿 |2
where Γ𝐿 = (75 − 50)∕(75 + 50) = 0.20. So
(
Loss𝑙𝑜𝑎𝑑 = 10 log10
1
1 − (0.20)2
)
= 0.18 dB
The transit loss will be
(
Loss𝑡𝑟𝑎𝑛𝑠 = 10 log10
1
𝑒−2𝛼𝐿
)
= (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 × 10−2 mW = 83 𝜇W
Method 2 – using loss factors: The 0.09 dB/m line loss corresponds to an exponential voltage
attenuation coefficient of 𝛼 = 0.09∕8.69 = 1.04 × 10−2 Np/m. Now, the power dropped at the
load will be
]
[
][
𝑃𝑙𝑜𝑎𝑑 = 𝑃𝑖𝑛 𝑒−2𝛼𝐿 (1 − |Γ𝐿 |2 ) = 𝑃𝑖𝑛 exp −2(1.04 × 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 × 10−2 mW = 83 𝜇W as before
0.12
)
(
In dBm this is 𝑃𝑖𝑛 = 10 log10 8.3 × 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)
=
=
Γ= 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:
(
)(
)
−(25 + 𝑗10)
−(25 − 𝑗10)
2
|Γ| =
= 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 − |Γ|2 =
1 − 0.046 = 0.954. The loss this represents in dB is then
(
)
(
)
1
1
10 log10
= 0.205 dB
=
10
log
10
0.954
1 − |Γ|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,
√
√
𝐿2
2
𝑍02 =
⇒ 𝐿2 = 𝐶2 𝑍02
Then 𝛽2 = 𝜔 𝐿2 𝐶2 = 2𝜋𝑓 𝐶2 𝑍02
𝐶2
The line length at 𝜆∕4 (the shortest length that will work) is then
( )
𝜆2
1
1
1 2𝜋
= 0.05 m
𝓁2 =
=
=
=
9
4
4 𝛽2
4𝑓 𝐶2 𝑍02
(4 × 10 )(10−10 )(50)
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
Γ𝑖𝑛 = (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 |Γ|2 = (3∕5)2 = 0.36, or 36%.
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10.11. Two voltage waves of equal amplitude, 𝑉0 , and which have differing frequencies, 𝜔 and 3𝜔 (with corresponding phase constants, 𝛽 and 3𝛽) 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 𝛽𝑧 at 𝑡 = 0.
The total complex instantaneous voltage will be:
𝑉𝑐𝑇 (𝑧, 𝑡) = 𝑉0 𝑒𝑗𝜔𝑡 𝑒−𝑗𝛽𝑧 + 𝑉0 𝑒𝑗3𝜔𝑡 𝑒−𝑗3𝛽𝑧
(
)
= 𝑉0 𝑒𝑗2𝜔𝑡 𝑒−𝑗2𝛽𝑧 𝑒−𝑗𝜔𝑡 𝑒+𝑗𝛽𝑧 + 𝑒+𝑗𝜔𝑡 𝑒−𝑗𝛽𝑧
= 2𝑉0 𝑒𝑗2𝜔𝑡 𝑒−2𝑗𝛽𝑧 cos (𝜔𝑡 − 𝛽𝑧)
In real instantaneous form, this becomes:
{
}
(𝑧, 𝑡) = 𝑒 𝑉𝑐𝑇 (𝑧, 𝑡) = 2𝑉0 cos(𝜔𝑡 − 𝛽𝑧) cos(2𝜔𝑡 − 2𝛽𝑧) V
This function (with normalized amplitude) is plotted below as a function of 𝛽𝑧.
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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
]
[
(𝑅𝐿 + 𝑗𝑋𝐿 ) cos(𝛽𝓁) + 𝑗𝑍0 sin(𝛽𝓁)
= 𝑅𝑔 − 𝑗𝑋𝑔
𝑍𝑖𝑛 = 𝑍0
𝑍0 cos(𝛽𝓁) + 𝑗(𝑅𝐿 + 𝑗𝑋𝐿 ) sin(𝛽𝓁)
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(𝛽𝓁) + 𝑗𝑍02 sin(𝛽𝓁) = (𝑅𝑔 − 𝑗𝑋𝑔 )[𝑍0 cos(𝛽𝓁) + 𝑗(𝑅𝐿 + 𝑗𝑋𝐿 ) sin(𝛽𝓁)]
We next separate the equation by equating the real parts of both sides and the imaginary parts
of both sides, giving
(𝑅𝐿 − 𝑅𝑔 ) cos(𝛽𝓁) = −
and
(𝑋𝐿 + 𝑋𝑔 ) cos(𝛽𝓁) =
(𝑅𝑔 𝑋𝐿 + 𝑅𝐿 𝑋𝑔 )
𝑍0
𝑅𝑔 𝑅𝐿 − 𝑋𝑔 𝑋𝐿 − 𝑍02
𝑍0
sin(𝛽𝓁) (real parts)
sin(𝛽𝓁) (imaginary parts)
Using the two equations, we find two conditions on the tangent of 𝛽𝓁:
tan(𝛽𝓁) =
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
√
√
.
. √
. 1
𝐶
𝐿
𝑍0 =
and 𝛽 = 𝜔 𝐿𝐶
𝛼= 𝑅
2
𝐿
𝐶
Let 𝓁 be the line length. We then may write:
√
1 √
8.69 √ 8 100
𝐶
10 dB = 8.69𝛼𝓁 = 8.69 𝐴0 𝑓
= 8.69 × 104 𝐴0
𝓁=
𝐴 10
2
𝐿
2 0
𝑍0
So 𝐴0 = (10∕8.69) × 10−4 = 1.15 × 10−4
√
and therefore at 100 MHz, 𝑅 = 1.15 × 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 𝛼 scales as 𝑓 , changing 𝑓 from 100 to 400 MHz doubles 𝛼 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
Γ=
=
100 + 50 3
and so the fraction of power incident on the load that is transferred to it is 1 − |Γ|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 × 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
[
]
𝑍𝑖𝑛
25 + 𝑗25
= 10
= 4 + 𝑗2
𝑉𝑖𝑛 = 𝑉𝑠0
𝑍𝑔 + 𝑍𝑖𝑛
50 + 25 + 𝑗25
The power will be
1
𝑃𝑖𝑛 = 𝑃𝐿 = 𝑒
2
{
𝑉𝑖𝑛 𝑉𝑖𝑛∗
}
∗
𝑍𝑖𝑛
1
= 𝑒
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−𝑗𝛽𝑧 + 𝑉0− 𝑒+𝑗𝛽𝑧
where 𝑉0− = Γ𝐿 𝑉0+ , and where
Γ𝐿 =
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 = −𝜆∕4. Thus 𝛽𝓁 = 𝜋∕2,
and
[
]
𝑉𝑖𝑛 = 𝑉𝑠 (−𝓁) = 𝑉0+ 𝑒𝑗𝛽𝓁 + Γ𝐿 𝑒−𝑗𝛽𝓁 = 𝑉0+ [𝑗 + (0.2 − 𝑗0.4)(−𝑗)] = 𝑉0+ (−0.4 + 𝑗0.8)
Using 𝑉𝑖𝑛 from part 𝑏, we have
𝑉0+ =
(4 + 𝑗2)
(−0.4 + 𝑗0.8)
Now, the voltage at the load will be
𝑉𝐿 = 𝑉0+ (1 + Γ𝐿 ) =
(4 + 𝑗2)
(1 + 0.2 − 𝑗0.4) = −2 − 𝑗6 V
(−0.4 + 𝑗0.8)
As a check,
1
𝑃𝐿 = 𝑒
2
{
𝑉𝐿 𝑉𝐿∗
𝑍𝐿∗
}
1
= 𝑒
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,
𝛽=
𝜔
2𝜋 × 60
= 1.9 × 10−6 rad∕m So 𝛽𝑙 = (1.9 × 10−6 )(80) = 1.5 × 10−4 << 1
=
𝑣𝑝
(2∕3)(3 × 108 )
.
The line is thus essentially a lumped circuit, where 𝑍𝑖𝑛 = 𝑍𝐿 = 80 Ω. Therefore
[
𝑉𝑠,𝑜𝑢𝑡 = 120
]
80
= 104 V
12 + 80
b) 500 kHz: In this case
𝛽=
Now
2𝜋 × 5 × 105
= 1.57 × 10−2 rad∕s So 𝛽𝑙 = 1.57 × 10−2 (80) = 1.26 rad
2 × 108
[
]
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
]
[
]
[
𝑍𝑖𝑛
33.17 − 𝑗9.57
= 120
= 89.5 − 𝑗6.46 = 89.7∠ − .071
𝑉𝑖𝑛 = 120
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−𝑗𝛽𝑧 + 𝑉0− 𝑒𝑗𝛽𝑧 , where
𝑉0− = Γ𝐿 𝑉0+ =
80 − 50 +
3 +
𝑉 =
𝑉
80 + 50 0
13 0
At 𝑧 = −80 m we thus have
[
]
89.5 − 𝑗6.46
3
𝑉𝑖𝑛 = 𝑉0+ 𝑒𝑗1.26 + 𝑒−𝑗1.26 ⇒ 𝑉0+ = 𝑗1.26
= 42.7 − 𝑗100 V
13
𝑒
+ (3∕13)𝑒−𝑗1.26
Now
𝑉𝑠,𝑜𝑢𝑡 = 𝑉0+ (1 + Γ𝐿 ) = (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:
1 (134)2
1 |𝑉𝑠,𝑜𝑢𝑡 |
=
= 112 W
=
2 𝑅𝐿
2 80
2
𝑃𝑎𝑣𝑔,𝐿
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.
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10.16. A 100-Ω lossless transmission line is connected to a second line of 40-Ω impedance, whose length is
𝜆∕4. The other end of the short line is terminated by a 25-Ω resistor. A sinusoidal wave (of frequency
𝑓 ) having 50 W average power is incident from the 100-Ω 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
Γ𝑖𝑛 =
𝑍𝑖𝑛 − 𝑍01
64 − 100
9
=
=−
𝑍𝑖𝑛 + 𝑍01
64 + 100
41
The dissipated power there is then
(
( )2 )
(
)
9
2
= 47.6 W
𝑃𝑖𝑛 = 𝑃𝐿 = 50 1 − |Γ𝑖𝑛 | = 50 1 −
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 = 𝜆∕8. 𝛽𝓁 is now 𝜋∕4, and the input impedance, from
Eq. (98) is:
]
[
25 cos(𝜋∕4) + 𝑗40 sin(𝜋∕4)
′
= 36.0 + 𝑗17.5 ohms
𝑍𝑖𝑛 = 40
40 cos(𝜋∕4) + 𝑗25 sin(𝜋∕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
Γ′𝑖𝑛 =
′ −𝑍
𝑍𝑖𝑛
01
′
𝑍𝑖𝑛
+ 𝑍01
=
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|Γ′𝑖𝑛 |2 = 50(0.234) = 11.7 W
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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 2.6𝜆 length line, terminated by the 25 ohm resistor: We use 𝛽𝑙 = (2𝜋∕𝜆)(2.6𝜆) =
16.33 rad. This value, modulo 2𝜋 is (by subtracting 2𝜋 twice) 3.77 rad. Now
]
[
25 cos(3.77) + 𝑗50 sin(3.77)
= 33.7 + 𝑗24.0
𝑍𝑖𝑛 = 50
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
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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 |Γ𝐿 | = (1∕2) (1 − 𝑗)(1 + 𝑗) = 1∕ 2. Finally
Γ𝐿 =
√
1 + |Γ𝐿 | 1 + 1∕ 2
=
𝑠2 =
√ = 5.83
1 − |Γ𝐿 | 1 − 1∕ 2
For section 1, we need the reflection coefficient at the junction (location of the 100 Ω resistor) seen
by waves incident from section 1: We first need the input impedance of the .2𝜆 length of section 2:
[
]
[
]
(50 − 𝑗100) cos(𝛽2 𝑙) + 𝑗50 sin(𝛽2 𝑙)
(1 − 𝑗2)(0.309) + 𝑗0.951
𝑍𝑖𝑛2 = 50
= 50
50 cos(𝛽2 𝑙) + 𝑗(50 − 𝑗100) sin(𝛽2 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 100Ω resistor, leading to a net junction impedance
found by
1
1
1
=
+
⇒ 𝑍𝑖𝑛𝑇 = 8.06 + 𝑗3.23 = 8.69∠0.38 rad
𝑍𝑖𝑛𝑇
100 8.63 + 𝑗3.82
The reflection coefficient will be
Γ𝑗 =
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.
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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 𝜇H∕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
√
√
𝐿
2 × 10−7
=
= 50 Ω
𝑍0 =
𝐶
8 × 10−11
and
1
1
=√
= 2.5 × 108 m∕s
𝑣𝑝 = √
(2 × 10−7 )(9 × 10−11 )
𝐿𝐶
Now 𝛽 = 𝜔∕𝑣𝑝 = (2𝜋 × 108 )∕(2.5 × 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 𝛽𝑙 = (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
𝛽𝑙 = (2.5)(.3) = 0.75):
]
[
(7.93 + 𝑗19.9) cos(.75) + 𝑗50 sin(.75)
= 35.9 + 𝑗98.0 = 104.3∠1.22
𝑍𝑖𝑛2 = 50
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
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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
Γ𝐿 =
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 2.7𝜆, we have 𝛽𝑙 = (2𝜋)(2.7) = 16.96
rad. The input impedance is then
[
]
[
]
(40 + 𝑗30) cos(16.96) + 𝑗50 sin(16.96)
−1.236 − 𝑗5.682
𝑍𝑖𝑛 = 50
= 50
= 61.8 − 𝑗37.5 Ω
50 cos(16.96) + 𝑗(40 + 𝑗30) sin(16.96)
1.308 − 𝑗3.804
c) If 𝜔𝐿 = 10 Ω, find 𝐼𝑠 : The source drives a total impedance given by 𝑍𝑛𝑒𝑡 = 20 + 𝑗𝜔𝐿 + 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 𝜔 = 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 𝜔𝐿 = 37.5, so that 𝐿 = 37.5∕𝜔 = 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 Ω: 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 Ω. The line is operating
at 200 MHz and 𝑍𝑖𝑛 = 200 − 𝑗200 Ω. 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
[
] [
]
𝑍𝑖𝑛
𝑧𝐿 cos(𝛽𝐿) + 𝑗 sin(𝛽𝐿)
𝑧𝐿 + 𝑗 tan(𝛽𝐿)
𝑧𝑖𝑛 =
=
=
𝑍0
cos(𝛽𝐿) + 𝑗𝑧𝐿 sin(𝛽𝐿)
1 + 𝑗𝑧𝐿 tan(𝛽𝐿)
Solve for 𝑧𝐿 :
[
𝑧𝑖𝑛 − 𝑗 tan(𝛽𝐿)
𝑧𝐿 =
1 − 𝑗𝑧𝑖𝑛 tan(𝛽𝐿)
]
where, with 𝜆 = 𝑐∕𝑓 = 3 × 108 ∕2 × 108 = 1.50 m, we find 𝛽𝐿 = (2𝜋)(1)∕(1.50) = 4.19, and so
tan(𝛽𝐿) = 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 × 103 + 𝑗69.8 Ω. Next
Γ=
𝑍𝐿 − 𝑍0
6.42 × 102 + 𝑗69.8
= .446 + 𝑗2.68 × 10−2 = .447∠6.0 × 10−2 rad
=
𝑍𝐿 + 𝑍0
1.44 × 103 + 𝑗69.8
212
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10.21. (continued) Now
𝑠=
1 + |Γ| 1 + .447
=
= 2.62
1 − |Γ| 1 − .447
Finally
𝑧𝑚𝑎𝑥 = −
𝜆𝜙
𝜙
(6.0 × 10−2 )(1.50)
=−
=−
= −7.2 × 10−3 m = −7.2 mm
2𝛽
4𝜋
4𝜋
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 𝜆 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 0.6667𝜆, or equivalently, 0.1667𝜆 (since 0.5𝜆 is once around the chart),
obtaining (0.0881 + 0.1667)𝜆) = 0.2548𝜆, which is the position of the load. A straight line is now
drawn from the origin though the 0.2548𝜆 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 0.2548𝜆
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 0.0048𝜆 (marked as .005𝜆 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 0.10𝜆 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 Γ𝐿 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
|Γ𝐿 | = 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 𝜙 = −108◦ . In summary, Γ𝐿 = 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 0.10𝜆.
(+
jX
/Z
45
1.2
1.6
1.8
2.0
6
0.0
0.4
4
70
0
5
14
20
N
75
T
0.4
5
0.0
0.4
3.0
PO
NE
0.6
EC
0.8
NC
80
>
TA
TO
R
OM
4
0.0
6
15
0
0.4
0.3
1.0
EA
C
C TI
85
5.0
10
IN D
U
0.8
0.6
90
15
170
10
0.1
0.49
0.4
20
0.2
50
20
10
55.0
44.0
33.0
22.0
11.4
11.8
11.6
11.2
00.9
00.8
00.7
00.6
00.5
00.4
00.3
00.2
0.1
1.0
AN
PT
CE
US
ES
IV
CT
DU
IN
R
-75
,O
o)
2.0
0.38
0.11
-10
1000
CT
AN
-1110
00..1
0.0
6
-70
-1
40
(-j
0.0
T
4
EN
0
-65 .5
1.8
0.37
0.12
RE
A
N
0.6
1.6
-90
0.13
0.7
1.4
1.0
1.2
0.36
0.9
0.14
-80
-4
0
5
-4
0.15
0.35
0
-70
-5
6
4
0.88
0.1
0.3
ITI
VE
-55
-35
AC
-60
7
0.3
3
CA P
-60
PO
5
0.0
Z
X/
0.1
OM
AD <
A RD LO
S TO W
-170
)
0.22
-30
2
CE
C
0.4
0.3
1
0.3
-90
o
jB/ Y
0.4
0.1
9
8
0.1
0
-5
-25
-85
(CE
0.6
-20
0
0.10 lambda WTL
0.3
-4
0.4
0.2
5
0.4
0.8
9
3.0
4
0.0
0
-15 -80
4.0
<
1.0
-15
0.47
5.0
j0.68
0.2
zL
0.28
0.3
0.2
1
-30
6
0.4
0.8
0.22
EN
VEL
WA
-160
0.6
-20
0.2
-10
0.48
VSWR = 5
0.4
10
G TH
00.2
0.1
20
± 180
11.0
50
0.0
50
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE
ONDUCTANCE COMPONENT (G/Yo)
0.49
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
A RD
4.0
1.0
VE
R
0.47
160
0.2
ER A
25
0.4
20
0.48
0.1
8
0.3
2
50
0.28
G EN
0.2
0.22
TO W
3
1
0.2
9
0.2
30
G TH S
CI
7
0.3
30
0.2
> W A VELEN
PA
0.1
60
0.3
0.0
CA
R
Yo )
jB/
E (+
40
,O
NC
4
1
0.3
o)
TA
EP
SC
SU
VE
TI
6
0.3
35
9
0.1
0.5
65
3
0.4
0
13
1.0
0.8
55
0.6 60
120
0.1
70
40
1.4
2
0.4
0.15
0.35
80
0.7
0
7
0.0
0.9
110
1
0.4
0.36
90
50
0
.08
0.14
0.37
0.38
0.39
100
0.4
0.13
0.12
0.11
0.1
.09
0.0
7
-1
30
0.4
3
-1
20
0.0
8
0.4
2
9
1
0.4
00..4
0.39
214
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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 0.35𝜆 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 Γ𝐿 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
|Γ𝐿 | = 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 𝜙 = +72◦ . In summary, Γ𝐿 = 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 0.10𝜆.
(+
jX
/Z
45
11.4
1.2
1.0
50
1.6
2.0
6
0.0
0.4
4
70
0
5
14
N
75
T
0.4
5
0.0
0.4
NE
PO
OM
4
0.0
6
15
0
TA
TO
NC
80
>
EC
0.4
0.3
R
0.88
1.0
EA
C
C TI
85
VE
R
0.47
0.2
160
5.0
10
IN D
U
0.8
0.6
.6
170
10
0.1
0.49
00.4
20
0.2
0.22
50
20
10
55.0
44.0
33.0
22.0
11.8
11.4
11.6
11.2
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
D LO A D <
OWAR
-170
)
C
CE
US
ES
IV
CT
DU
IN
-75
2.0
1.8
1.0
-90
0.13
0.37
-110
0
0.14
-80
-4
0
0.36
0.9
1.2
1.4
0.15
0.35
0.12
0.38
ITI
VE
0.11
-100
0.1
RE
A
CT
AN
40
-70
(-j
-1
T
6
EN
0.6
1.6
4
5
-4
0.3
-70
-5
6
0.8
0.1
AC
-55
-35
0.3
3
CA P
-60
-60
7
0.7
0.1
N
5
0.0
,O
o)
R
0.2
-30
2
PO
5
Z
X/
0.3
OM
0
-65 .5
0.3
1
CE
C
0.0
9
0.1
4
0.4
0.4
<
1.0
AN
PT
0
8
0.1
0
-5
-25
0.4
0.6
-85
o
jB/ Y
E (-
0.8
9
0.3
-4
0.4
0.2
-20
3.0
4
0.0
0
-15 -80
4.0
0.47
1.0
-15
0.2
0.3
0.28
0.2
1
-30
6
0.4
0.8
0.22
E
VEL
WA
-160
0.6
-20
0.2
5.0
-90
10
0.48
VSWR = 5
0.4
-10
HS T
N GT
0.2
0.1
20
± 180
11.0
50
0.0
50
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
0.49
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
0.23
90
15
20
A RD
4.0
1.0
0.28
ER A
20
3.0
0.6
0.22
G EN
zL = 0.54 +j1.23
0.4
9
0.2
0.48
0.1
8
0.3
2
50
25
1
0.2
TO W
0.2
30
G TH S
C
3
0.2
> W A VELEN
PA
7
0.3
30
0.3
0.0
CA
R
C
0.1
60
40
,O
AN
4
1
0.3
o)
PT
CE
US
ES
IV
IT
6
0.3
35
9
0.1
0.5
65
3
0.4
0
13
0.1
70
40
Yo )
jB/
E (+
1.8
7
0.0
0.8
55
0.6 60
120
0.3355
80
0.7
2
0.4
0.1155
0.36
90
0.9
110
1
0.4
8
0.0
0.35 lambda WTL
0.14
0.37
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
-12
0
0.0
8
0.4
2
0.0
9
1
0.4
0.4
0.39
215
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10.23. The normalized load on a lossless transmission line is 𝑧𝐿 = 2 + 𝑗1. Let 𝜆 = 20 m Make use of the
Smith chart to find:
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 0.213𝜆. Moving from here toward the generator, we cross
the positive Γ𝑅 axis (at which the impedance is purely real and greater than 1) at 0.250𝜆. The
distance is then (0.250 − 0.213)𝜆 = 0.037𝜆 from the load. With 𝜆 = 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 𝜆 − 0.037 𝜆 = 0.463 𝜆.
With 𝜆 = 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∕𝜆 < 0.25. The required input impedance is that at the actual line input
(to the left of the two 20Ω resistors. The input to the line section occurs just to the right of the 20Ω
resistors, and the input impedance there we first find with the Smith chart. This impedance is in series
with the two 20Ω resistors, so we add 40Ω to the calculated impedance from the Smith chart to find
the net line input impedance. To begin, the 20Ω 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 < 𝜆∕4.
Problem 10.24
On the chart, radial lines are drawn at positions corresponding to .025𝜆 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 40Ω resistors and obtain the normalized impedance
at the line input. The magnitudes of these values are then found, and the results are multiplied by
50Ω. The table below summarizes the results.
𝑙∕𝜆
𝑧𝑖𝑛𝑙 (to right of 40Ω)
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
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10.24. (continued) As a check, the line input input impedance can be found analytically through
[
]
[
]
20 cos(2𝜋𝑙∕𝜆) + 𝑗50 sin(2𝜋𝑙∕𝜆)
60 cos(2𝜋𝑙∕𝜆) + 𝑗66 sin(2𝜋𝑙∕𝜆)
𝑍𝑖𝑛 = 40 + 50
= 50
50 cos(2𝜋𝑙∕𝜆) + 𝑗20 sin(2𝜋𝑙∕𝜆)
50 cos(2𝜋𝑙∕𝜆) + 𝑗20 sin(2𝜋𝑙∕𝜆)
from which
[
|𝑍𝑖𝑛 | = 50
36 cos2 (2𝜋𝑙∕𝜆) + 43.6 sin2 (2𝜋𝑙∕𝜆)
]1∕2
25 cos2 (2𝜋𝑙∕𝜆) + 4 sin2 (2𝜋𝑙∕𝜆)
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 𝜆∕4. Since this distance is given as 25 cm, we see that 𝜆 = 100 cm = 1 m.
Thus the maximum voltage location is 5∕100 = 0.05𝜆 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 𝜆 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 𝜆
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 Ω.
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 → ∞ point on the chart. The distance of 20 cm
corresponds to 20∕100 = 0.20 𝜆, placing the load position at 0.45 𝜆 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 𝜆 WTL position. Thus
.
𝑍𝐿 = 𝑗300(0.327) = 𝑗98 Ω.
Problem 10.25
219
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10.26. A 75-ohm lossless line is of length 1.2𝜆. 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
0.15𝜆 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 Γ 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 1.2𝜆 (equivalent to 0.2𝜆) 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
(+
jX
/Z
45
1.4
1.2
1.6
T
N
75
NE
PO
EC
NC
80
>
TA
TO
R
OM
4
0.0
6
15
0
0.4
0.3
1.0
EA
C
85
VE
R
0.47
160
0.2
C TI
10
8
0.
IN D
U
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
0.6
90
5.0
20
A RD
15
0.28
0.1
0.49
20
0.2
170
10
1st Vmin location
r = 0.25
0.4
0.48
4.0
1.0
0.22
ER A
2.0
06
0.
0.
44
0
5
14
0.4
5
0.0
0.
4
zL2 = 1.23 +
j1.60
0.8
9
0.2
G EN
20
3.0
0.6
1
0.2
TO W
25
0.4
30
G TH S
C
0.3
2
50
0.2
50
20
10
5.0
44.0
33.0
22.0
11.8
11.6
11.4
11.0
11.2
00.9
00.8
00.7
00.6
00.5
00.4
00.3
0.2
0.1
± 180
0.0
50
50
> W A VELEN
PA
0.11
8
30
0.2
0.3
0.0
CA
R
C
0.1
40
,O
AN
Yo )
jB/
E (+
31
0.
o)
PT
CE
US
ES
IV
IT
4
7
0.314 lambda
0.3 WTL
60
3
19
0.
0.5
65
3
0.4
0
13
6
0.3
35
1.8
7
0.0
0.8
55
0.6 60
120
0.1
70
40
0.7
2
0.4
0.15
0.35
80
1.0
1
0.4
8
0.0
0.9
110
0.36
90
50
0
0.14
0.37
0.38
0.39
100
0.4
0.13
0.12
0.11
0.1
.09
.22
00.2
VSWR = 4
10
o
jB/ Y
E (-
8
C
1.0
N
TA
CT
DU
IN
Z
X/
2.0
1.8
1.6
14
1.
1.2
0.36
10
1.0
0.14
-80
-4
0
5
-4
0...11155
35
0..335
00.99
-770
--90
0
0.13
0.37
-110
0
-5
6
4
0.8
0.1
0.3
-35
0.12
0.38
0.7
7
0.3
RE
A
CT
AN
-55
0.1
0.15 lambda WTL
3
CA
-60
P AC
ITI
VE
CE
EN
40
(-j
-1
T
-70
N
06
0.
PO
0.6
0.2
-30
2
-60
8
0.1
0
-5
0.3
CO
M
0
-65 .5
0.
31
44
0.
0.4
0.
19
-25
5
0.0
,O
o)
R
-75
0.6
3.0
-20
<
P
CE
US
ES
IV
9
0
0.0
7
-1
30
0.4
3
-1 2
0
0.0
8
0.4
2
9
0.114 lambda WTL
0.3
-4
4
0.
0.2
5
0.4
1
4
0.0
0
-15 -80
0..88
0.47
1.0
4.0
j1.16
0.2
zL1
0.2
j0.777
zin2
0.28
0.22
0.3
-15
0.2
5.0
-85
0.
-10
)
0.66
-30
6
0.4
20
0.44
0.1
-20
0.49
0.48
AD <
A RD LO
S TO W
G TH
0
N
-17
E
VEL
WA
-90
-160
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT
ONENT (G/Yo)
00.1
.111
-10
100
0.1
0.0
1
0.4
0.4
0.39
220
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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 𝜆: 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 𝜆.
To this we add 0.15 𝜆, obtaining about 0.437 𝜆, which we then mark on the chart (0.287 𝜆 is not
the precise value, but I have added 0.15 𝜆 to that mark to obtain the point shown on the chart that
is near to 0.437 𝜆. This “eyeballing” method increases the accuracy a little). A line drawn from
the 0.437 𝜆 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 Γ𝑟 axis. This is at the zero position on the WTL scale. The
load is at the approximate 0.213 𝜆 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 0.312𝜆
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 0.2𝜆, finally arriving at (again) slightly more than 0.012𝜆 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 Ω 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 𝜆 = 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 0.36𝜆 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 Ω. As a check, I will do the problem analytically. First, we use
[
]
4(18)
1
𝑧𝑚𝑖𝑛 = −18 cm = − (𝜙 + 𝜋) ⇒ 𝜙 =
− 1 𝜋 = 1.382 rad = 79.2◦
2𝛽
50
Now
𝑠 − 1 2.5 − 1
=
= 0.4286
𝑠 + 1 2.5 + 1
and so Γ𝐿 = 0.4286∠1.382. Using this, we find
|Γ𝐿 | =
𝑧𝐿 =
1 + Γ𝐿
= 0.798 + 𝑗0.823
1 − Γ𝐿
and thus 𝑍𝐿 = 𝑧𝐿 (60) = 47.8 + 𝑗49.3 Ω.
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, 𝜆 = 2(16) = 32cm, and (assuming an
air-filled line), 𝑓 = 𝑐∕𝜆 = 3 × 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)𝜆 = 0.1563𝜆 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 0.1563𝜆 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 𝜆 = 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−𝑗𝛽𝑧 + Γ𝑒𝑗𝛽𝑧 )
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
Γ = |Γ| =
𝑠−1 4−1
=
= 0.6
𝑠+1 4+1
The voltage magnitude is then
√
[
]1∕2
𝑉 (𝑧)𝑉 ∗ (𝑧) = 𝑉 + (𝑒−𝑗𝛽𝑧 + Γ𝑒𝑗𝛽𝑧 )(𝑒𝑗𝛽𝑧 + Γ𝑒−𝑗𝛽𝑧 )
[
]1∕2
= 𝑉 + 1 + 2Γ cos(2𝛽𝑧) + Γ2
|𝑉 (𝑧)| =
Note that with cos(2𝛽𝑧) = ±1, we obtain |𝑉 | = 𝑉 + (1 ± Γ) as expected. With 𝑠 = 4 and
with |𝑉 |𝑚𝑖𝑛 = 1, we find |𝑉 |𝑚𝑎𝑥 = 4. Then with Γ = 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(2𝛽𝑧)
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(2𝛽𝑧) ⇒ 2𝛽𝑧 = cos−1 (−0.9726)
Thus 2𝛽𝑧 = 2.904. At this stage, we note the the |𝑉 |𝑚𝑖𝑛 point will occur at 2𝛽𝑧 = 𝜋. We
therefore compute the range, Δ𝑧, over which |𝑉 | < 1.1 through the equation:
2𝛽(Δ𝑧) = 2(𝜋 − 2.904) ⇒ Δ𝑧 =
𝜋 − 2.904
= 0.0378 m = 3.8 cm
2𝜋∕1
where 𝜆 = 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(2𝛽𝑧) ⇒ cos(2𝛽𝑧) = 0.6298
Since the maximum corresponds to 2𝛽𝑧 = 0, we find the range through
2𝛽Δ𝑧 = 2 cos−1 (0.6298) ⇒ Δ𝑧 =
0.8896
= 0.142 m = 14.2 cm
2𝜋∕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 = 0.183𝜆. The distance in wavelenghs between 𝑦𝐿 and 𝑦𝑖𝑛2 is again noted on the WTG scale,
or 𝑑𝑏 = 0.317𝜆. These are the two possible attachment points for the shorted stub. The shortest of
these is 𝑑𝑎 = 0.183𝜆. 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 0.331𝜆 (WTG). The (shortest) stub
length is thus 𝑑1𝑎 = (0.331 − 0.250)𝜆 = 0.81𝜆.
226
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10.33. In Fig. 10.17, let 𝑍𝐿 = 100 + 𝑗150 Ω, and 𝑍0 = 100 Ω.
(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 = 0.25𝜆), 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 0.344𝜆 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)𝜆 = 0.094𝜆.
0
65
.43
0
6
0.0
0.4
4
70
0
(+
jX
/Z
5
14
5
0.0
0.4
45
1.4
1.6
NE
PO
OM
4
0.0
6
15
0
0.8
TA
TO
NC
80
>
EC
0.4
0.3
R
1.2
N
75
T
0.4
20
5.0
1.0
EA
C
VE
R
0.47
C TI
85
0.2
160
15
10
0.8
IN D
U
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
A RD
4.0
1.0
20
0.6
170
10
0.1
0.49
0.4
0.48
3.0
.6
0
zL = yB = 1.00
+j1.50
0.28
ER A
25
0.4
0.22
90
0.3
2
50
9
0.2
G EN
0.2
1
0.2
TO W
0.1
8
30
30
G TH S
C
3
0.2
20
0.2
50
20
10
5.0
4.0
3.0
2.0
1.8
1.6
1.4
1.0
1.2
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
± 180
0.0
50
50
> W A VELEN
PA
7
0.3
0.3
0.0
CA
R
0.1
60
40
,O
Yo )
jB/
E (+
1
0.3
o)
P
CE
US
ES
IV
IT
C
4
9
0.1
0.5
13
N
TA
6
0.3
35
2.0
0
0.1
70
40
1.8
.07
1.0
55
0.6 60
120
0.15
0.35
80
0.7
2
0.4
0.8
8
0.0
0.9
110
1
0.4
0.14
0.36
90
50
0
0.37
0.38
0.39
100
0.4
0.13
0.12
0.11
0.1
.09
0.22
0.2
20
10
jB/
E (-
8
0.8
1.0
C
AN
PT
CT
DU
IN
2.0
1.8
0.37
0.12
0.38
CT
AN
0.11
-100
0.1
0.0
40
(-j
-70
T
6
EN
0.6
1.6
1.4
1.0
0.9
1.2
0.36
-90
0.13
-110
0
0.14
-80
-4
0
5
-4
0.15
0.35
-5
-70
0.8
4
0.3
6
0.11
RE
A
-55
-35
ITI
VE
0.7
7
0.3
AC
0.0
7
-1
30
0.4
3
-60
0.1
0.344 lambda WTG
(stub input location)
3
CA P
-60
N
0
-65 .5
0.2
-30
2
PO
-1
Z
X/
0.3
OM
4
0.3
1
CE
C
0.0
0.4
9
0.1
0.4
0
8
0.1
0
-5
-25
5
0.0
,O
o)
R
-75
0.6
3.0
5
0.4
9
0.3
-4
0.4
0.2
-20
<
CE
US
ES
IV
0.8
4
0.0
0
-15 -80
4.0
0.47
1.0
-15
0.2
0.3
0.28
j0.462
0.22
yL
5.0
-85
Yo )
0.66
-10
0.2
0.2
1
-30
6
0.4
Psc (0.25 lambda WTG)
0.44
0.1
-20
0.49
0.48
D<
RD LO A
TO W A
TH S
-170
EN G
VEL
WA
-90
-160
RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo)
-12
0
0.0
8
0.4
2
9
1
0.4
0.4
0.39
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)𝜆 =
0.344𝜆 (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 𝜆 = 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 0.1𝜆 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 Ω.
Problem 10.34
228
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10.35. A load, 𝑍𝐿 = 25 + 𝑗75 Ω, is located at 𝑧 = 0 on a lossless two-wire line for which 𝑍0 = 50 Ω 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 𝜆 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 𝜆 on the WTG scale.
The total distance between load and input is then 𝑑 = (0.088 + 0.308)𝜆 = 0.396𝜆. At 300 MHz,
and with 𝑣 = 𝑐, the wavelength is 𝜆 = 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
𝜔𝐶 =
2.23
2.23
= 2.4 × 10−11 F = 24 pF
⇒ 𝐶=
𝑍0
(50)(2𝜋 × 3 × 108 )
Problem 10.35
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10.36. The two-wire lines shown in Fig. 10.36 are all lossless and have 𝑍0 = 200 Ω. Find 𝑑 and the
shortest possible value for 𝑑1 to provide a matched load if 𝜆 = 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 = .098𝜆 = 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 = 0.347𝜆 = 37.7 cm.
Problem 10.36
230
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10.37. In the transmission line of Fig. 10.20, 𝑅𝑔 = 𝑍0 = 50 Ω, and 𝑅𝐿 = 25 Ω. 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+ =
𝑉0 𝑍 0
50
1
=
𝑉0 = 𝑉0
𝑅 𝑔 + 𝑍0
100
2
Now, Γ𝐿 = (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 + Γ𝐿 ) = 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 = 50Ω, and 𝑅𝐿 = 𝑅𝑔 = 25Ω. Carry out the analysis for the time period
0 < 𝑡 < 8𝑙∕𝑣. At the generator end, we have Γ𝑔 = −1∕3. At the load end, we have Γ𝐿 = −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:
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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 → ∞.
10.39. In the transmission line of Fig. 10.20, 𝑍0 = 50 Ω and 𝑅𝐿 = 𝑅𝑔 = 25 Ω. 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
Γ𝐿 = (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:
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10.39. (continued)
𝑙
5𝑙
<𝑡<
𝑣
4𝑣
13𝑙
3𝑙
<𝑡<
𝑣
4𝑣
21𝑙
5𝑙
<𝑡<
𝑣
4𝑣
7𝑙
29𝑙
<𝑡<
𝑣
4𝑣
)
(
1
𝑉 + = 0.44 𝑉0
∶ 𝑉1 = 1 −
3
(
)
1
1
𝑉 + = −0.15 𝑉0
∶ 𝑉2 = − 1 −
3
3
( )2 (
)
1
1
∶ 𝑉3 =
1−
𝑉 + = 0.049 𝑉0
3
3
)
( )3 (
1
1
1−
𝑉 + = −0.017 𝑉0
∶ 𝑉4 = −
3
3
233
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10.40. In the charged line of Fig. 10.25, the characteristic impedance is 𝑍0 = 100Ω, and 𝑅𝑔 = 300Ω. 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 Γ𝑔 = 1∕2 and Γ𝐿 = 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 Γ𝑔 = −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
3𝑙
𝑙
<𝑡<
∶ 𝑉𝐿 = 0
2𝑣
2𝑣
2
3𝑙
∶ 𝑉𝐿 = 𝑉0
𝑡>
2𝑣
0<𝑡<
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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.
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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 Γ𝐺 = (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.
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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.
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Engineering Electromagnetics 9th solution
Engineering Electromagnetics (Chungnam National University)
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CHAPTER 11 – 9th Edition
11.1. Show that 𝐸𝑥𝑠 = 𝐴𝑒𝑗𝑘0 𝑧+𝜙 is a solution to the vector Helmholtz equation, Sec. 11.1, Eq. (30), for
√
𝑘0 = 𝜔 𝜇0 𝜖0 and any 𝜙 and 𝐴: We take
𝑑2
𝐴𝑒𝑗𝑘0 𝑧+𝜙 = (𝑗𝑘0 )2 𝐴𝑒𝑗𝑘0 𝑧+𝜙 = −𝑘20 𝐸𝑥𝑠
𝑑𝑧2
11.2. A 20-GHz uniform plane wave propagates in the forward 𝑧 direction in a lossless medium for which
𝜖𝑟 = 𝜇𝑟 = 3. Assume a real electric field amplitude, 𝐸0 , and take the wave as polarized in the 𝑥
direction. Find:
√
√
a) 𝑣𝑝 : 𝑣𝑝 = 1∕ 𝜇𝜖 = 𝑐∕ 𝜇𝑟 𝜖𝑟 = 𝑐∕3 = 1.0 × 108 m∕s.
b) 𝛽: In this lossless medium, 𝑘 = 𝛽 = 𝜔∕𝑣𝑝 = (4𝜋 × 1010 )∕(1.0 × 108 ) = 1.26 × 103 m−1 .
c) 𝜆: 𝜆 = 2𝜋∕𝛽 = 5.0 × 10−3 m = 5.0 mm.
d) 𝐄𝑠 : With real amplitude 𝐸0 , forward 𝑧 travel, and 𝑥 polarization, we write
𝐄𝑠 = 𝐸0 exp(−𝑗𝛽𝑧)𝐚𝑥 = 𝐸0 exp(−𝑗1.26 × 103 𝑧) 𝐚𝑥 V∕m.
√
√
e) 𝐇𝑠 : First, the intrinsic impedance of the medium is 𝜂 = 𝜇∕𝜖 = 𝜂0 𝜇𝑟 ∕𝜖𝑟 = 𝜂0 = 377 ohms.
Then 𝐇𝑠 = (𝐸0 ∕𝜂0 ) exp(−𝑗𝛽𝑧) 𝐚𝑦 = (𝐸0 ∕377) exp(−𝑗1.26 × 103 𝑧) 𝐚𝑦 A∕m.
{
}
f) < 𝐒 >= (1∕2)𝑒 𝐄𝑠 × 𝐇∗𝑠 = (𝐸02 ∕754) 𝐚𝑧 W∕m2
11.3. An 𝐇 field in free space is given as (𝑥, 𝑡) = 10 cos(108 𝑡 − 𝛽𝑥)𝐚𝑦 A/m. Find
a) 𝛽: Since we have a uniform plane wave, 𝛽 = 𝜔∕𝑐, where we identify 𝜔 = 108 sec−1 . Thus
𝛽 = 108 ∕(3 × 108 ) = 0.33 rad∕m.
b) 𝜆: We know 𝜆 = 2𝜋∕𝛽 = 18.9 m.
c) (𝑥, 𝑡) at 𝑃 (0.1, 0.2, 0.3) at 𝑡 = 1 ns: Use 𝐸(𝑥, 𝑡) = −𝜂0 𝐻(𝑥, 𝑡) = −(377)(10) cos(108 𝑡 −
𝛽𝑥) = −3.77 × 103 cos(108 𝑡 − 𝛽𝑥). The vector direction of 𝐄 will be −𝐚𝑧 , since we require that
𝐒 = 𝐄 × 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 × 103 cos[(108 )(10−9 ) − (0.33)(0.1)]𝐚𝑧 = −3.77 × 103 cos(6.7 × 10−2 )𝐚𝑧
= −3.76 × 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 𝜆∕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
𝜆𝑎𝑖𝑟 =
3.0 × 108 m∕s
𝑐
0.12
= 300 m, so that the f raction =
=
= 4.0 × 10−4
𝑓
300
106 s−1
b) The same antenna is embedded in a ferrite material for which 𝜖𝑟 = 20 and 𝜇𝑟 = 2, 000. What
fraction of a wavelength is it now?
𝜆
0.12
300
𝜆𝑓 𝑒𝑟𝑟𝑖𝑡𝑒 = √ 𝑎𝑖𝑟 = √
= 0.08
= 1.5m ⇒ f raction =
1.5
𝜇𝑟 𝜖𝑟
(20)(2000)
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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 𝜔2 ; the backward 𝑧-directed wave is of frequency 𝜔1 , where
𝜔1 < 𝜔2 . Both frequencies are very slightly detuned on either side of their mean frequency 𝜔0 , such
that 𝜔0 − 𝜔1 = 𝜔2 − 𝜔0 << 𝜔0 . 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 = 𝜔2 ∕𝑐 and 𝑘1 = 𝜔1 ∕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(𝑗𝜔2 𝑡) + exp(+𝑗𝑘1 𝑧) exp(𝑗𝜔1 𝑡) 𝐚𝑥
Now define the mean frequency and wavenumber:
𝜔0 =
𝜔2 + 𝜔1
2
and
𝑘𝑚 =
𝑘2 + 𝑘1
2
then define Δ𝑘 = 𝑘2 − 𝑘1 and Δ𝜔 = 𝜔2 − 𝜔1 , from which
𝑘2 = 𝑘𝑚 +
Δ𝑘
,
2
𝑘1 = 𝑘𝑚 −
Δ𝑘
,
2
𝜔2 = 𝜔0 +
Δ𝜔
,
2
and
𝜔1 = 𝜔0 −
Δ𝜔
2
The total complex field can now be expressed as
[
]
𝐄𝑐𝑇 (𝑧) = 𝐸0 𝑒−𝑗Δ𝑘𝑧∕2 𝑒𝑗𝜔0 𝑡 𝑒−𝑗𝑘𝑚 𝑧 𝑒𝑗Δ𝜔𝑡∕2 + 𝑒+𝑗𝑘𝑚 𝑧 𝑒−𝑗Δ𝜔𝑡∕2 𝐚𝑥
)
(
Δ𝜔
𝑡 − 𝑘𝑚 𝑧 𝐚𝑥
= 2𝐸0 𝑒−𝑗Δ𝑘𝑧∕2 𝑒𝑗𝜔0 𝑡 cos
2
The power density (light intensity) is now proportional to
(
)
Δ𝜔
∗
𝐸𝑐𝑇 𝐸𝑐𝑇
= 4𝐸02 cos2
𝑡 − 𝑘𝑚 𝑧
2
The intensity pattern moves in the forward 𝑧 direction at velocity 𝑣 = Δ𝜔∕2𝑘𝑚 = 𝑐Δ𝜔∕2𝜔0 .
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11.6. A uniform plane wave has electric field 𝐄𝑠 = (𝐸𝑦0 𝐚𝑦 − 𝐸𝑧0 𝐚𝑧 ) 𝑒−𝛼𝑥 𝑒−𝑗𝛽𝑥 V∕m. The intrinsic
impedance of the medium is given as 𝜂 = |𝜂| 𝑒𝑗𝜙 , where 𝜙 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−𝑗𝛽𝑥 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
] [
]
[
𝐸𝑦
𝐸𝑦0
𝐸𝑧
𝐸𝑧0
𝐇𝑠 =
𝐚 +
𝐚 =
𝐚 +
𝐚 𝑒−𝛼𝑥 𝑒−𝑗𝜙 𝑒−𝑗𝛽𝑥 A∕m
𝜂 𝑧
𝜂 𝑦
|𝜂| 𝑧 |𝜂| 𝑦
} [
{
]
c) Find (𝑥, 𝑡) and (𝑥, 𝑡): (𝑥, 𝑡) = 𝑒 𝐄𝑠 𝑒𝑗𝜔𝑡 = 𝐸𝑦0 𝐚𝑦 − 𝐸𝑧0 𝐚𝑧 𝑒−𝛼𝑥 cos(𝜔𝑡 − 𝛽𝑥)
]
{
}
1 [
𝐸𝑦0 𝐚𝑧 + 𝐸𝑧0 𝐚𝑦 𝑒−𝛼𝑥 cos(𝜔𝑡 − 𝛽𝑥 − 𝜙)
(𝑥, 𝑡) = 𝑒 𝐇𝑠 𝑒𝑗𝜔𝑡 =
|𝜂|
where all amplitudes are assumed real.
d) Find < 𝐒 > in W∕m2 :
)
(
{
}
1
1
2
2
< 𝐒 >= 𝑒 𝐄𝑠 × 𝐇∗𝑠 =
𝑒−2𝛼𝑥 cos 𝜙 𝐚𝑥 W∕m2
𝐸𝑦0
+ 𝐸𝑧0
2
2|𝜂|
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
(𝑤ℎ) 𝐸𝑦0
+ 𝐸𝑧0
𝑒−2𝛼𝑑 cos 𝜙 W
𝑃 =
< 𝐒 > ⋅ 𝑑𝐒 = | < 𝐒 > |𝑥=𝑑 × 𝑎𝑟𝑒𝑎 =
∫ ∫𝑝𝑙𝑎𝑡𝑒
2|𝜂|
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11.7. Express in terms of the penetration depth, 𝛿, 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) = 8.69𝛼𝐿 =
𝛿
8.69𝐿
⇒ 𝐿1 =
= 0.12𝛿
𝛿
8.69
b) 3 dB: This would be 𝐿3 = 3𝐿1 = 0.35𝛿
c) 30 dB: 𝐿30 = 30𝐿1 = 3.45𝛿.
11.8. An electric field in free space is given in spherical coordinates as 𝐄𝑠 (𝑟) = 𝐸0 (𝑟)𝑒−𝑗𝑘𝑟 𝐚𝜃 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 𝐚𝜙 A∕m
𝜂0
b) Find < 𝐒 >: This will be
{
} 𝐸 2 (𝑟)
1
< 𝐒 >= 𝑒 𝐄𝑠 × 𝐇∗𝑠 = 0 𝐚𝑟 W∕m2
2
2𝜂0
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)
W
𝑃 = 4𝜋𝑟2 0
2𝜂0
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
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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−𝛼𝑥 𝑒−𝑗𝛽𝑧 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 ∇ × 𝐄𝑠 = −𝑗𝜔𝜇0 𝐇𝑠 , which leads to
(
)
𝜕𝐸𝑦
𝜕𝐸𝑦
)
𝑗
𝑗 (
𝑗
𝐇𝑠 =
−𝛼𝐸𝑦 𝐚𝑧 + 𝑗𝛽𝐸𝑦 𝐚𝑥
∇ × 𝐸𝑦 𝐚𝑦 =
𝐚𝑧 −
𝐚𝑥 =
𝜔𝜇0
𝜔𝜇0 𝜕𝑥
𝜕𝑧
𝜔𝜇0
where the subscript 𝑠 has been dropped everywhere, with the understanding that that all fields
are in phasor form. Finally,
𝐇(𝑥, 𝑧) = −
)
𝐸0 (
𝛽𝐚𝑥 + 𝑗𝛼𝐚𝑧 𝑒−𝛼𝑥 𝑒−𝑗𝛽𝑧
𝜔𝜇0
b) Find 𝛼 as a function of 𝛽, 𝜔, 𝜖0 , and 𝜇0 , 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 ∇ × 𝐄 = −𝑗𝜔𝜇0 𝐇 equation
is of course automatically satisfied. This leaves ∇ × 𝐇 = 𝑗𝜔𝜖0 𝐄. Substituting the known fields,
we find
)
(
𝜕𝐻𝑥 𝜕𝐻𝑧
∇×𝐇=
−
𝐚𝑦 = 𝑗𝜔𝜖0 𝐸𝑦 𝐚𝑦
𝜕𝑧
𝜕𝑥
or:
)
𝑗𝐸0 ( 2
𝛽 − 𝛼 2 𝑒−𝛼𝑥 𝑒−𝑗𝛽𝑧 = 𝑗𝜔𝜖0 𝐸0 𝑒−𝛼𝑥 𝑒−𝑗𝛽𝑧
𝜔𝜇0
which simplifies to
𝛽 2 − 𝛼 2 = 𝜔2 𝜇0 𝜖0 ⇒ 𝛼 =
√
𝛽 2 − 𝜔2 𝜇0 𝜖0
11.10. In a medium characterized by intrinsic impedance
𝜂 = |𝜂|𝑒)𝑗𝜙 , a linearly-polarized plane wave prop(
agates, with magnetic field given as 𝐇𝑠 = 𝐻0𝑦 𝐚𝑦 + 𝐻0𝑧 𝐚𝑧 𝑒−𝛼𝑥 𝑒−𝑗𝛽𝑥 . Find:
a) 𝐄𝑠 : Requiring orthogonal components of 𝐄𝑠 for each component of 𝐇𝑠 , we find
[
]
𝐄𝑠 = |𝜂| 𝐻0𝑧 𝐚𝑦 − 𝐻0𝑦 𝐚𝑧 𝑒−𝛼𝑥 𝑒−𝑗𝛽𝑥 𝑒𝑗𝜙
[
]
b) (𝑥, 𝑡) = 𝑒 {𝐄𝑠 𝑒𝑗𝜔𝑡 } = |𝜂| 𝐻0𝑧 𝐚𝑦 − 𝐻0𝑦 𝐚𝑧 𝑒−𝛼𝑥 cos(𝜔𝑡 − 𝛽𝑥 + 𝜙).
[
]
c) (𝑥, 𝑡) = 𝑒 {𝐇𝑠 𝑒𝑗𝜔𝑡 } = 𝐻0𝑦 𝐚𝑦 + 𝐻0𝑧 𝐚𝑧 𝑒−𝛼𝑥 cos(𝜔𝑡 − 𝛽𝑥).
]
[
2 + 𝐻 2 𝑒−2𝛼𝑥 cos 𝜙 𝐚 W∕m2
d) < 𝐒 >= 21 𝑒{𝐄𝑠 × 𝐇∗𝑠 } = 12 |𝜂| 𝐻0𝑦
𝑥
0𝑧
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11.11. A uniform plane wave at frequency 𝑓 = 100 MHz propagates in a material having conductivity
𝜎 = 3.0 S/m and dielectric constant 𝜖𝑟′ = 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
𝜎
=
= 67.5
′
8
𝜔𝜖
(2𝜋 × 10 )(8.00)(8.85 × 10−12 )
This falls within the good conductor approximation.
b) Calculate 𝛼, 𝛽, and 𝜂. Assuming a good conductor, we have:
√
.
. √
𝛼 = 𝛽 = 𝜋𝑓 𝜇0 𝜎 = 𝜋(108 )(4𝜋 × 10−7 )(3) = 34.4 m−1
Then
. 1+𝑗
(1 + 𝑗)𝛼 (1 + 𝑗)(34.4)
(Eq. (85)) =
=
= 11.5(1 + 𝑗) ohms
𝜂=
𝜎𝛿
𝜎
3.0
c) Assuming forward 𝑧 propagation and 𝑥 polarization, write the electric field in phasor form:
𝐄𝑠 = 100𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝐚𝑥
where 𝛼 and 𝛽 are as found in part 𝑏.
d) Write the magnetic field in phasor form:
𝐇𝑠 =
100
𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝐚𝑦 = 6.15 𝑒−𝑗0.79 𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝐚𝑦
11.5(1 + 𝑗)
e) Write the time-average Poynting vector, < 𝐒 > in W∕m2 .
{
} 1
1
< 𝐒 >= 𝑒 𝐄𝑠 × 𝐇∗𝑠 = (100)(6.15)𝑒−2𝛼𝑧 cos(0.79) 𝐚𝑧 = 216 𝑒−2𝛼𝑧 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 × 10−2 m = 2.0 cm
8.69𝛼 (8.69)(34.4)
g) What dB power drop corresponds to one penetration depth, 𝛿? In general, the decibel power
loss is given by 8.69𝛼𝑧. So, when 𝑧 = 𝛿 = 1∕𝛼, 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 𝜎 =
10−3 S/m, dielectric constant 𝜖𝑟′ = 80.0, and permeability 𝜇0 :
a) Calculate the loss tangent and determine whether the medium would qualify as a good dielectric
or a good conductor:
10−3
𝜎
=
= 0.002
𝜔𝜖 ′
(2𝜋 × 108 )(80.0)(8.85 × 10−12 )
This falls within the good dielectric approximation.
b) Calculate 𝛼, 𝛽, and 𝜂. Assuming a good dielectric, we have:
√
𝜂0 𝜎
. 𝜎 𝜇0
377(10−3 )
𝛼=
= 0.0211 Np∕m
=
= √
√
2 𝜖0
2 𝜖𝑟′
2 80
√
. √
𝜔
2𝜋 × 108 √
𝛽 = 𝜔 𝜇0 𝜖0 𝜖𝑟′ =
𝜖𝑟′ =
80 = 18.7 rad∕m
𝑐
3 × 108
√
𝜇0
.
377
𝜂=
= √ = 42.2 ohms
′
𝜖𝑟 𝜖0
80
c) Assuming forward 𝑧 propagation and 𝑥 polarization, write the electric field in phasor form:
𝐄𝑠 = 100𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝐚𝑥
where 𝛼 and 𝛽 are as found in part 𝑏.
d) Write the magnetic field in phasor form:
𝐇𝑠 =
100 −𝛼𝑧 −𝑗𝛽𝑧
𝑒 𝑒
𝐚𝑦 = 2.37𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝐚𝑦
42.2
e) Write the time-average Poynting vector, < 𝐒 > in W∕m2 .
{
} 1
1
< 𝐒 >= 𝑒 𝐄𝑠 × 𝐇∗𝑠 = (100)(2.37)𝑒−2𝛼𝑧 𝐚𝑧 = 118 𝑒−2𝛼𝑧 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
8.69𝛼 (8.69)(0.0211)
g) What dB power drop corresponds to one penetration depth, 𝛿? In general, the decibel power
loss is given by 8.69𝛼𝑧. So, when 𝑧 = 𝛿 = 1∕𝛼, we have 8.69 dB.
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11.13. Let 𝑗𝑘 = 0.2+𝑗1.5 m−1 and 𝜂 = 450+𝑗60 Ω for a uniform plane wave propagating in the 𝐚𝑧 direction.
If 𝜔 = 300 Mrad/s, find 𝜇, 𝜖 ′ , and 𝜖 ′′ : We begin with
√
𝜇
1
= 450 + 𝑗60
𝜂=
√
′
𝜖
1 − 𝑗(𝜖 ′′ ∕𝜖 ′ )
and
√
√
𝑗𝑘 = 𝑗𝜔 𝜇𝜖 ′ 1 − 𝑗(𝜖 ′′ ∕𝜖 ′ ) = 0.2 + 𝑗1.5
Then
𝜂𝜂 ∗ =
𝜇
1
= (450 + 𝑗60)(450 − 𝑗60) = 2.06 × 105
√
′
𝜖
′′
′
2
1 + (𝜖 ∕𝜖 )
and
(𝑗𝑘)(𝑗𝑘)∗ = 𝜔2 𝜇𝜖 ′
√
1 + (𝜖 ′′ ∕𝜖 ′ )2 = (0.2 + 𝑗1.5)(0.2 − 𝑗1.5) = 2.29
(1)
(2)
Taking the ratio of (2) to (1),
(
)
(𝑗𝑘)(𝑗𝑘)∗
2.29
′′ ′ 2
2 ′ 2
=
𝜔
(𝜖
)
1
+
(𝜖
∕𝜖
)
=
= 1.11 × 10−5
∗
𝜂𝜂
2.06 × 105
Then with 𝜔 = 3 × 108 ,
(𝜖 ′ )2 =
1.23 × 10−22
1.11 × 10−5
(
)=(
)
(3 × 108 )2 1 + (𝜖 ′′ ∕𝜖 ′ )2
1 + (𝜖 ′′ ∕𝜖 ′ )2
(3)
Now, we use Eqs. (35) and (36). Squaring these and taking their ratio gives
√
1 + (𝜖 ′′ ∕𝜖 ′ )2
(0.2)2
𝛼2
=
=
√
𝛽2
(1.5)2
1 + (𝜖 ′′ ∕𝜖 ′ )2
We solve this to find 𝜖 ′′ ∕𝜖 ′ = 0.271. Substituting this result into (3) gives 𝜖 ′ = 1.07 × 10−11 F/m.
Since 𝜖 ′′ ∕𝜖 ′ = 0.271, we then find 𝜖 ′′ = 2.90 × 10−12 F/m. Finally, using these results in either (1)
or (2) we find 𝜇 = 2.28 × 10−6 H/m. Summary: 𝜇 = 2.28 × 10−6 H∕m,
𝜖 ′ = 1.07 × 10−11 F∕m, and 𝜖 ′′ = 2.90 × 10−12 F∕m.
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11.14. A certain nonmagnetic material has the material constants 𝜖𝑟′ = 2 and 𝜖 ′′ ∕𝜖 ′ = 4 × 10−4 at 𝜔 = 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, 𝜖 ′′ = (4 × 104 )(2)(8.854 × 10−12 ) = 7.1 × 10−15 F/m. Then, since
𝜖 ′′ ∕𝜖 ′ << 1, we use the approximate form for 𝛼, given by Eq. (51) (written in terms of 𝜖 ′′ ):
√
. 𝜔𝜖 ′′ 𝜇
(1.5 × 109 )(7.1 × 10−15 ) 377
−3
=
𝛼=
√ = 1.42 × 10 Np∕m
2
𝜖′
2
2
The required distance is now 𝑧1 = (1.42 × 10−3 )−1 = 706 m
b) the power level is reduced by one-half: The governing relation is 𝑒−2𝛼𝑧1∕2 = 1∕2, or 𝑧1∕2 =
ln 2∕2𝛼 = ln 2∕2(1.42 × 10− 3) = 244 m.
c) the phase shifts 360◦ : This distance is defined as
√ one wavelength, where 𝜆 = 2𝜋∕𝛽
√
= (2𝜋𝑐)∕(𝜔 𝜖𝑟′ ) = [2𝜋(3 × 108 )]∕[(1.5 × 109 ) 2] = 0.89 m.
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) 𝜖𝑟′ = 1 and 𝜖𝑟′′ = 0: In a non-magnetic material, we would have:
√
1∕2
√
( ′′ )2
⎤
𝜇0 𝜖0 𝜖𝑟′ ⎡
𝜖𝑟
⎢ 1+
− 1⎥
𝛼=𝜔
⎥
2 ⎢
𝜖𝑟′
⎣
⎦
and
√
1∕2
√
( ′′ )2
⎤
𝜖𝑟
𝜇0 𝜖0 𝜖𝑟′ ⎡
⎢ 1+
𝛽=𝜔
+ 1⎥
⎥
2 ⎢
𝜖𝑟′
⎣
⎦
√
′
′′
With the given values of 𝜖𝑟 and 𝜖𝑟 , it is clear that 𝛽 = 𝜔 𝜇0 𝜖0 = 𝜔∕𝑐, and so
𝜆 = 2𝜋∕𝛽 = 2𝜋𝑐∕𝜔 = 3 × 1010 ∕1010 = 3 cm. It is also clear that 𝛼 = 0.
. √
b) 𝜖𝑟′ = 1.04 and 𝜖𝑟′′ = 9.00 × 10−4 : In this case 𝜖𝑟′′ ∕𝜖𝑟′ << 1, and so 𝛽 = 𝜔 𝜖𝑟′ ∕𝑐 = 2.13 cm−1 .
Thus 𝜆 = 2𝜋∕𝛽 = 2.95 cm. Then
√
√
′′
𝜔𝜖𝑟′′ 𝜇0 𝜖0
. 𝜔𝜖 ′′ 𝜇
𝜔 𝜖𝑟
2𝜋 × 1010 (9.00 × 10−4 )
=
=
=
𝛼=
√
√
√
2
𝜖′
2
2𝑐 𝜖 ′
2 × 3 × 108
𝜖𝑟′
1.04
𝑟
−2
= 9.24 × 10 Np∕m
c) 𝜖𝑟′ = 2.5 and 𝜖𝑟′′ = 7.2: Using the above formulas, we obtain
√
1∕2
√
⎤
⎡
( )2
2𝜋 × 1010 2.5 ⎢
7.2
+ 1⎥ = 4.71 cm−1
1+
𝛽=
√
⎥
⎢
2.5
10
(3 × 10 ) 2 ⎣
⎦
and so 𝜆 = 2𝜋∕𝛽 = 1.33 cm. Then
√
1∕2
√
⎤
⎡
( )2
2𝜋 ×
2.5 ⎢
7.2
− 1⎥ = 335 Np∕m
1+
𝛼=
√
⎥
⎢
2.5
8
(3 × 10 ) 2 ⎣
⎦
1010
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11.16. Consider the power dissipation term, ∫ 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
in Eq.(77),
{ that resulted
}
the time-average power dissipation per volume will be < 𝑝𝑑 >= (1∕2)𝑒 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 >= (𝜎∕2)|𝐸0 |2 𝑒−2𝛼𝑧 : Begin with the phasor expression
for the electric field, assuming complex amplitude 𝐸0 , and 𝑥-polarization:
𝐄𝑠 = 𝐸0 𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝐚𝑥 V∕m2
Then
𝐉𝑠 = 𝜎𝐄𝑠 = 𝜎𝐸0 𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝐚𝑥 A∕m2
So that
{
}
< 𝑝𝑑 >= (1∕2)𝑒 𝐸0 𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝐚𝑥 ⋅ 𝜎𝐸0∗ 𝑒−𝛼𝑧 𝑒+𝑗𝛽𝑧 𝐚𝑥 = (𝜎∕2)|𝐸0 |2 𝑒−2𝛼𝑧
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), 𝜂𝑐 = (1 + 𝑗)∕(𝜎𝛿), where the skin depth, 𝛿 = 1∕𝛼. The magnetic field phasor
is then
𝐸
𝜎
𝐇𝑠 = 𝑠 𝐚𝑦 =
𝐸 𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 𝐚𝑦 A∕m
𝜂𝑐
(1 + 𝑗)𝛼 0
The time-average Poynting vector is then
{
}
𝜎
1
|𝐸 |2 𝑒−2𝛼𝑧 𝐚𝑧 W∕m2
< 𝐒 >= 𝑒 𝐄𝑠 × 𝐇∗𝑠 =
2
4𝛼 0
Now, consider a rectangular volume of side lengths, Δ𝑥, Δ𝑦, and Δ𝑧, 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 = Δ𝑧). With small 𝑧, we may approximate 𝑒−2𝛼𝑧 = 1 − 2𝛼𝑧, and the dissipated power in
the volume becomes
]
[
]
[
𝜎
𝜎
𝜎
𝑃𝑑 = 𝑃𝑖𝑛 − 𝑃𝑜𝑢𝑡 =
|𝐸0 |2 Δ𝑥Δ𝑦 −
|𝐸0 |2 (1 − 2𝛼Δ𝑧) Δ𝑥Δ𝑦 = |𝐸0 |2 (Δ𝑥Δ𝑦Δ𝑧)
4𝛼
4𝛼
2
This is just the result of part 𝑎, evaluated at 𝑧 = 0 and multiplied by the volume. The relation
becomes exact as Δ𝑧 → 0, in which case < 𝑝𝑑 >→ (𝜎∕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 𝜎. 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 𝜌.
∮
𝐇 ⋅ 𝑑𝐋 =
∫ ∫
𝐉 ⋅ 𝐧𝑑𝑎 ⇒ 2𝜋𝜌𝐻𝜙 = 𝜋𝜌2 𝐽 = 𝜋𝜌2 𝜎𝑉0 ∕𝐿
where 𝐉 = 𝜎𝐄 = (𝜎𝑉0 ∕𝐿) 𝐚𝑧 . So
𝜎𝜌𝑉0
𝐚 A∕m (0 < 𝜌 < 𝑎)
2𝐿 𝜙
𝐇=
c. Find the Poynting vector 𝐒 = 𝐄 × 𝐇:
𝐒=
−𝜎𝜌𝑉02
𝑉0
𝜎𝜌𝑉0
𝐚𝜌 W∕m2
𝐚𝑧 ×
𝐚𝜙 =
𝐿
2𝐿
2𝐿2
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:
−
∮
𝐒 ⋅ 𝐧 𝑑𝑎 = −
∫0
2𝜋
∫0
𝐿
−𝜎𝜌𝑉02
2𝐿2
𝐚𝜌 ⋅ 𝐚𝜌 𝑎𝑑𝜙𝑑𝑧 =
𝜋𝑎2 𝜎 2
𝑉0 = 𝑉02 ∕𝑅 W
𝐿
where the wire resistance is 𝑅 = 𝐿∕(𝜋𝑎2 𝜎) 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:
−
∮𝑠
𝐒 ⋅ 𝐧 𝑑𝑎 =
∫𝑣
𝐄 ⋅ 𝐉 𝑑𝑣
The right hand side evaluates as
∫𝑣
𝐿
𝐄 ⋅ 𝐉 𝑑𝑣 =
∫0 ∫0
2𝜋
∫0
𝑎
𝜎𝑉02
𝐿2
𝜌 𝑑𝜌 𝑑𝜙 𝑑𝑧 =
𝜋𝑎2 𝜎 2
𝑉0 = 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
𝜇0 . Determine:
a) 𝜖 ′ : 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 𝜎 = 𝜔𝜖 ′′ . Using these, we take the
ratio, 𝛽∕𝛼, to find
√
]
[
( ′)
( )
𝜔 𝜇𝜖 ′ 1 + (1∕8)(𝜖 ′′ ∕𝜖 ′ )2
𝛽
𝜖
20
1 𝜖 ′′
= 2 ′′ +
=
=
√
𝛼 0.5
𝜖
4 𝜖′
(𝜔𝜖 ′′ ∕2) 𝜇∕𝜖 ′
This becomes the quadratic equation:
(
𝜖 ′′
𝜖′
)2
(
− 160
𝜖 ′′
𝜖′
)
+8=0
(
)
The solution to the quadratic is 𝜖 ′′ ∕𝜖 ′ = 0.05, which means that we can neglect the second
√
. √
term in Eq. (60b), so that 𝛽 = 𝜔 𝜇𝜖 ′ = (𝜔∕𝑐) 𝜖𝑟′ . With the given frequency of 100 MHz,
√
and with 𝜇 = 𝜇0 , we find 𝜖𝑟′ = 20(3∕2𝜋) = 9.55, so that 𝜖𝑟′ = 91.3, and finally 𝜖 ′ = 𝜖𝑟′ 𝜖0 =
8.1 × 10−10 F∕m.
b) 𝜖 ′′ : Using Eq. (60a), the set up is
√
√
2(0.5)
𝜔𝜖 ′′ 𝜇
𝜖′
10−8 √
′′
𝛼 = 0.5 =
91.3 = 4.0 × 10−11 F∕m
⇒
𝜖
=
=
2
𝜖′
2𝜋(377)
2𝜋 × 108 𝜇
c) 𝜂: Using Eq. (62b), we find
√ [
( )]
.
𝜇
1 𝜖 ′′
377
1
+
𝑗
=√
𝜂=
(1 + 𝑗.025) = (39.5 + 𝑗0.99) ohms
𝜖′
2 𝜖′
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
𝜂
(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
𝑒
cos(0.025) 𝐚𝑧 = 0.202𝑒−𝑧 𝐚𝑧 W∕m2
< 𝐒 >= 𝑒{𝐄𝑠 × 𝐇∗𝑠 } =
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 × 10−6 W∕m2 . The incident power on
the given area is then 𝑃 = 9.2 × 10−6 × (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 𝜖 = 10−9 ∕4𝜋 F/m and 𝜇𝑟 = 1. If 𝐄 in this region
is (500∕𝜌) cos(𝜔𝑡 − 4𝑧)𝐚𝜌 V/m, find:
a) 𝜔, with the help of Maxwell’s equations in cylindrical coordinates: We use the two curl equations, beginning with ∇ × 𝐄 = −𝜕𝐁∕𝜕𝑡, where in this case,
∇×𝐄=
So
𝐵𝜙 =
𝜕𝐸𝜌
𝜕𝑧
𝐚𝜙 =
𝜕𝐵𝜙
2000
sin(𝜔𝑡 − 4𝑧)𝐚𝜙 = −
𝐚
𝜌
𝜕𝑡 𝜙
2000
2000
sin(𝜔𝑡 − 4𝑧)𝑑𝑡 =
cos(𝜔𝑡 − 4𝑧) T
∫
𝜌
𝜔𝜌
Then
𝐻𝜙 =
𝐵𝜙
𝜇0
=
2000
cos(𝜔𝑡 − 4𝑧) A∕m
(4𝜋 × 10−7 )𝜔𝜌
We next use ∇ × 𝐇 = 𝜕𝐃∕𝜕𝑡, where in this case
∇×𝐇=−
𝜕𝐻𝜙
𝜕𝑧
𝐚𝜌 +
1 𝜕(𝜌𝐻𝜙 )
𝐚𝑧
𝜌 𝜕𝜌
where the second term on the right hand side becomes zero when substituting our 𝐻𝜙 . So
∇×𝐇=−
𝜕𝐻𝜙
𝜕𝑧
𝐚𝜌 = −
𝜕𝐷𝜌
8000
𝐚
sin(𝜔𝑡
−
4𝑧)𝐚
=
𝜌
𝜕𝑡 𝜌
(4𝜋 × 10−7 )𝜔𝜌
And
𝐷𝜌 =
∫
−
8000
8000
cos(𝜔𝑡 − 4𝑧) C∕m2
sin(𝜔𝑡 − 4𝑧)𝑑𝑡 =
(4𝜋 × 10−7 )𝜔𝜌
(4𝜋 × 10−7 )𝜔2 𝜌
Finally, using the given 𝜖,
𝐷𝜌
𝐸𝜌 =
𝜖
=
8000
cos(𝜔𝑡 − 4𝑧) V∕m
(10−16 )𝜔2 𝜌
This must be the same as the given field, so we require
500
8000
=
⇒ 𝜔 = 4 × 108 rad∕s
−16
2
𝜌
(10 )𝜔 𝜌
b) 𝐇(𝜌, 𝑧, 𝑡): From part 𝑎, we have
𝐇(𝜌, 𝑧, 𝑡) =
4.0
2000
cos(4 × 108 𝑡 − 4𝑧)𝐚𝜙 A∕m
cos(𝜔𝑡 − 4𝑧)𝐚𝜙 =
−7
𝜌
(4𝜋 × 10 )𝜔𝜌
c) 𝐒(𝜌, 𝜙, 𝑧): This will be
𝐒(𝜌, 𝜙, 𝑧) = 𝐄 × 𝐇 =
=
500
4.0
cos(4 × 108 𝑡 − 4𝑧)𝐚𝜌 ×
cos(4 × 108 𝑡 − 4𝑧)𝐚𝜙
𝜌
𝜌
2.0 × 10−3
cos2 (4 × 108 𝑡 − 4𝑧)𝐚𝑧 W∕m2
𝜌2
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11.19
d) the average power passing through every cross-section 8 < 𝜌 < 20 mm, 0 < 𝜙 < 2𝜋. Using
the result of part 𝑐, we find < 𝐒 >= (1.0 × 103 )∕𝜌2 𝐚𝑧 W∕m2 . The power through the given
cross-section is now
P=
∫0
2𝜋
.020
∫.008
( )
20
1.0 × 103
3
= 5.7 kW
𝜌
𝑑𝜌
𝑑𝜙
=
2𝜋
×
10
ln
2
8
𝜌
11.20. Voltage breakdown in air at standard temperature and pressure occurs at an electric field strength
of approximately 3 × 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 10𝜇m 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
𝜂 = 𝜂0 , 𝜃𝜂 = 𝛼 = 0):
|<𝐒>|=
𝐸02
2𝜂0
=
(3 × 106 )2
= 1.2 × 1010 W∕m2
2(377)
To avoid breakdown, the power in a 10-𝜇m radius cylinder is then bounded by
𝑃 < (1.2 × 1010 )(𝜋 × (10−5 )2 ) = 3.75 W
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11.21. The cylindrical shell, 1 cm < 𝜌 < 1.2 cm, is composed of a conducting material for which 𝜎 = 106
S/m. The external and internal regions are non-conducting. Let 𝐻𝜙 = 2000 A/m at 𝜌 = 1.2 cm.
a) Find 𝐇 everywhere: Use Ampere’s circuital law, which states:
∮
𝐇 ⋅ 𝑑𝐋 = 2𝜋𝜌(2000) = 2𝜋(1.2 × 10−2 )(2000) = 48𝜋 A = 𝐼𝑒𝑛𝑐𝑙
Then in this case
𝐉=
48
𝐼
𝐚 =
𝐚 = 1.09 × 106 𝐚𝑧 A∕m2
𝐴𝑟𝑒𝑎 𝑧 (1.44 − 1.00) × 10−4 𝑧
With this result we again use Ampere’s circuital law to find 𝐇 everywhere within the shell as a
function of 𝜌 (in meters):
𝐻𝜙1 (𝜌) =
1
2𝜋𝜌 ∫0
2𝜋
𝜌
∫.01
1.09 × 106 𝜌 𝑑𝜌 𝑑𝜙 =
54.5 4 2
(10 𝜌 − 1) A∕m (.01 < 𝜌 < .012)
𝜌
Outside the shell, we would have
𝐻𝜙2 (𝜌) =
48𝜋
= 24∕𝜌 A∕m (𝜌 > .012)
2𝜋𝜌
Inside the shell (𝜌 < .01 m), 𝐻𝜙 = 0 since there is no enclosed current.
b) Find 𝐄 everywhere: We use
𝐄=
1.09 × 106
𝐉
=
𝐚𝑧 = 1.09 𝐚𝑧 V∕m
𝜎
106
which is valid, presumeably, outside as well as inside the shell.
c) Find 𝐒 everywhere: Use
𝐏 = 𝐄 × 𝐇 = 1.09 𝐚𝑧 ×
=−
54.5 4 2
(10 𝜌 − 1) 𝐚𝜙
𝜌
59.4 4 2
(10 𝜌 − 1) 𝐚𝜌 W∕m2 (.01 < 𝜌 < .012 m)
𝜌
Outside the shell,
𝐒 = 1.09 𝐚𝑧 ×
26
24
𝐚𝜙 = − 𝐚𝜌 W∕m2 (𝜌 > .012 m)
𝜌
𝜌
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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 𝛿. The dielectric is lossless and the operating
frequency is 400 MHz. Calculate the resistance per meter length of the:
a) inner conductor: First
1
1
𝛿=√
=√
= 3.3 × 10−6 m = 3.3𝜇m
𝜋𝑓 𝜇𝜎
𝜋(4 × 108 )(4𝜋 × 10−7 )(5.8 × 107 )
Now, using (90) with a unit length, we find
𝑅𝑖𝑛 =
1
1
=
= 0.42 ohms∕m
−3
2𝜋𝑎𝜎𝛿 2𝜋(2 × 10 )(5.8 × 107 )(3.3 × 10−6 )
b) outer conductor: Again, (90) applies but with a different conductor radius. Thus
2
𝑎
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 × 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) =
1
𝐿
= 1.4 × 10−3 Ω∕m
=
7
𝜎𝐴 (1.2 × 10 )𝜋(.012 − .0092 )
b) 20 MHz: Now the skin effect will limit the effective cross-section. At 20 MHz, the skin depth
is
𝛿(20MHz) = [𝜋𝑓 𝜇0 𝜎]−1∕2 = [𝜋(20 × 106 )(4𝜋 × 10−7 )(1.2 × 107 )]−1∕2 = 3.25 × 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) =
1
1
𝐿
=
=
= 4.1 × 10−2 Ω∕m
7
𝜎𝐴 2𝜋𝑏𝛿 (1.2 × 10 )(2𝜋(.01))(3.25 × 10−5 )
c) 2 GHz: Using the same formula as in part 𝑏, we find the skin depth at 2 GHz to be 𝛿 = 3.25×10−6
m. The resistance (using the other formula) is 𝑅(2GHz) = 4.1 × 10−1 Ω∕m.
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11.24
a) Most microwave ovens operate at 2.45 GHz. Assume that 𝜎 = 1.2 × 106 S/m and 𝜇𝑟 = 500 for
the stainless steel interior, and find the depth of penetration:
1
1
𝛿=√
=√
= 9.28 × 10−6 m = 9.28𝜇m
9
−7
6
𝜋𝑓 𝜇𝜎
𝜋(2.45 × 10 )(4𝜋 × 10 )(1.2 × 10 )
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 𝛼 = 𝛽 = 𝜋𝑓 𝜇𝜎 = 1∕𝛿. So, assuming that the direction into the conductor
is 𝑧, the depth-dependent field is written as
𝐸𝑠 (𝑧) = 50𝑒−𝛼𝑧 𝑒−𝑗𝛽𝑧 = 50𝑒−𝑧∕𝛿 𝑒−𝑗𝑧∕𝛿 = 50 exp(−𝑧∕9.28) exp(−𝑗 𝑧∕9.28 )
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟
⏟⏟⏟
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 𝜇m) 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 × 105 m/s. Assuming the conductor is non-magnetic, determine the frequency
and the conductivity: First, we use
𝑓=
𝑣
3 × 105
= 109 Hz = 1 GHz
=
𝜆 3 × 10−4
Next, for a good conductor,
𝛿=
𝜆
4𝜋
4𝜋
1
=
= 1.1 × 105 S∕m
⇒ 𝜎= 2
=√
−8 )(109 )(4𝜋 × 10−7 )
2𝜋
𝜆
𝑓
𝜇
(9
×
10
𝜋𝑓 𝜇𝜎
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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 𝜎 = 1.6 × 107 S/m.
a) Find 𝑅, the resistance per unit length, at an operating frequency of 2.4 GHz: First
𝛿=√
1
=√
= 2.57 × 10−6 m = 2.57𝜇m
8
−7
7
𝜋𝑓 𝜇𝜎
𝜋(2.4 × 10 )(4𝜋 × 10 )(1.6 × 10 )
1
Then, using (90) with a unit length, we find
𝑅𝑖𝑛 =
1
1
= 4.84 ohms∕m
=
−3
2𝜋𝑎𝜎𝛿 2𝜋(0.8 × 10 )(1.6 × 107 )(2.57 × 10−6 )
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=
2𝜋𝜖0
2𝜋(8.854 × 10−12 )
=
= 3.46 × 10−11 F∕m
ln(𝑏∕𝑎)
ln(4∕.8)
𝜇0
4𝜋 × 10−7
ln(𝑏∕𝑎) =
ln(4∕.8) = 3.22 × 10−7 H∕m
2𝜋
2𝜋
√
c) Find 𝛼 and 𝛽 if 𝛼 + 𝑗𝛽 = 𝑗𝜔𝐶(𝑅 + 𝑗𝜔𝐿): Taking real and imaginary parts of the given
expression, we find
𝐿=
]1∕2
[√
√
)
} 𝜔 𝐿𝐶
(
{√
𝑅 2
−1
𝑗𝜔𝐶(𝑅 + 𝑗𝜔𝐿) = √
1+
𝛼 = Re
𝜔𝐿
2
and
𝛽 = Im
{√
[√
]1∕2
√
(
)
𝜔 𝐿𝐶
𝑅 2
𝑗𝜔𝐶(𝑅 + 𝑗𝜔𝐿) = √
1+
+1
𝜔𝐿
2
}
These can be found by writing out
}
{√
√
𝑗𝜔𝐶(𝑅 + 𝑗𝜔𝐿) = (1∕2) 𝑗𝜔𝐶(𝑅 + 𝑗𝜔𝐿) + 𝑐.𝑐.
𝛼 = Re
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
𝛼 = 3.0 × 10−2 Np∕m and 𝛽 = 50.3 rad∕m.
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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 𝜔 = 4 × 1010 rad/s:
a) 𝛼Tef ∕𝛼brass : From the appendix we find 𝜖 ′′ ∕𝜖 ′ = .0003 for Teflon, making the material a good
dielectric. Also, for Teflon, 𝜖𝑟′ = 2.1. For brass, we find 𝜎 = 1.5×107 S/m, making brass a good
conductor at the stated frequency. For a good dielectric (Teflon) we use the approximations:
√
( ′′ ) ( ) √
( ) √
. 𝜎 𝜇
𝜖
1
1 𝜖 ′′ 𝜔
′
𝜔 𝜇𝜖 =
=
𝛼=
𝜖𝑟′
2 𝜖′
𝜖′
2
2 𝜖′ 𝑐
( )]
[
√
. √
. √ ′
1 𝜖 ′′
𝜔
=
𝜔
𝛽 = 𝜔 𝜇𝜖 1 +
𝜇𝜖′
=
𝜖𝑟′
8 𝜖′
𝑐
For brass (good conductor) we have
√ ( )
.
. √
1
(4 × 1010 )(4𝜋 × 10−7 )(1.5 × 107 ) = 6.14 × 105 m−1
𝛼 = 𝛽 = 𝜋𝑓 𝜇𝜎brass = 𝜋
2𝜋
Now
√
(
)
√
1∕2 𝜖 ′′ ∕𝜖 ′ (𝜔∕𝑐) 𝜖𝑟′
𝛼Tef
(1∕2)(.0003)(4 × 1010 ∕3 × 108 ) 2.1
= 4.7 × 10−8
=
=
√
5
𝛼brass
6.14
×
10
𝜋𝑓 𝜇𝜎brass
b)
√
𝑐 𝜋𝑓 𝜇𝜎brass
(2𝜋∕𝛽Tef )
𝛽brass
𝜆Tef
(3 × 108 )(6.14 × 105 )
=
=
=
=
= 3.2 × 103
√
√
𝜆brass
(2𝜋∕𝛽brass )
𝛽Tef
(4 × 1010 ) 2.1
𝜔 𝜖𝑟′ Tef
c)
𝑣Tef
(𝜔∕𝛽Tef )
𝛽
=
= brass = 3.2 × 103 as before
𝑣brass
(𝜔∕𝛽brass )
𝛽Tef
11.28. A uniform plane wave in free space has electric field given by 𝐄𝑠 = 10𝑒−𝑗𝛽𝑥 𝐚𝑧 + 15𝑒−𝑗𝛽𝑥 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 𝜙 = tan−1 (10∕15) = 33.7◦ to the 𝑦 axis.
b) Find 𝐇𝑠 : With propagation in forward 𝑥, we would have
𝐇𝑠 =
15 −𝑗𝛽𝑥
−10 −𝑗𝛽𝑥
𝑒
𝐚𝑦 +
𝑒
𝐚𝑧 A∕m = −26.5𝑒−𝑗𝛽𝑥 𝐚𝑦 + 39.8𝑒−𝑗𝛽𝑥 𝐚𝑧 mA∕m
377
377
c) determine the average power density in the wave in W∕m2 : Use
[
]
} 1 (10)2
(15)2
1 {
𝐏𝑎𝑣𝑔 = Re 𝐄𝑠 × 𝐇∗𝑠 =
𝐚𝑥 +
𝐚𝑥 = 0.43𝐚𝑥 W∕m2 or 𝑃𝑎𝑣𝑔 = 0.43 W∕m2
2
2 377
377
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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−𝑗𝛽𝑧 . 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−𝑗𝛽𝑧
𝜂0
b) Determine an expression for the average power density in the wave in W∕m2 by direct application of Eq. (77): We have
)
(
𝐸0
1
1
+𝑗𝛽𝑧
∗
−𝑗𝛽𝑧
(𝐚 − 𝑗𝐚𝑥 )𝑒
×
𝐏𝑧,𝑎𝑣𝑔 = 𝑅𝑒(𝐄𝑠 × 𝐇𝑠 ) = 𝑅𝑒 𝐸0 (𝐚𝑥 + 𝑗𝐚𝑦 )𝑒
2
2
𝜂0 𝑦
=
𝐸02
𝜂0
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 𝑒𝑗Δ𝛽𝑧 ) 𝑒−𝑗𝛽𝑧
where Δ𝛽 = 𝛽𝑥 − 𝛽𝑦 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 Δ𝛽) at which the field is:
a) Linearly-polarized: We want the 𝑥 and 𝑦 components to be in phase, so therefore
𝑚𝜋
, (𝑚 = 1, 2, 3...)
Δ𝛽𝑧𝑙𝑖𝑛 = 𝑚𝜋 ⇒ 𝑧𝑙𝑖𝑛 =
Δ𝛽
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−𝑗𝛽𝑧 . Therefore,
Δ𝛽𝑧𝑐𝑖𝑟𝑐 =
(2𝑛 + 1)𝜋
(2𝑛 + 1)𝜋
⇒ 𝑧𝑐𝑖𝑟𝑐 =
, (𝑛 = 0, 1, 2, 3...)
2
2Δ𝛽
c) Assume intrinsic impedance 𝜂 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 𝑒𝑗Δ𝛽𝑧 𝑒−𝑗𝛽𝑧 and
𝜂
{
} 𝐸2
1
< 𝐒 >= 𝑒 𝐄𝑠 × 𝐇∗𝑠 = 0 𝐚𝑧 W∕m2
2
𝜂
where it is assumed that 𝐸0 is real. 𝜂 is real because the medium is evidently lossless.
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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 (𝜖𝑟𝑦 ) differs from that seen by waves polarized along 𝑥 (𝜖𝑟𝑥 ). Suppose 𝜖𝑟𝑥 = 2.15, 𝜖𝑟𝑦 = 2.10,
and the wave electric field at input is polarized at 45◦ to the positive 𝑥 and 𝑦 axes. Assume free space
wavelength 𝜆.
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 𝜋∕2. Our requirement over length 𝐿 is thus 𝛽𝑥 𝐿 − 𝛽𝑦 𝐿 = 𝜋∕2, or
𝐿=
𝜋
𝜋𝑐
=
√
√
2(𝛽𝑥 − 𝛽𝑦 ) 2𝜔( 𝜖𝑟𝑥 − 𝜖𝑟𝑦 )
With the given values, we find,
𝐿=
(58.3)𝜋𝑐
𝜆
= 58.3 = 14.6 𝜆
2𝜔
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 𝜋 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−𝑗𝛽𝑥 𝐿 𝐚𝑥 + 𝑒−𝑗𝛽𝑦 𝐿 𝐚𝑦 ] = 𝐸0 𝑒−𝑗𝛽𝑥 𝐿 [𝐚𝑥 + 𝑒−𝑗(𝛽𝑦 −𝛽𝑥 )𝐿 𝐚𝑦 ]
where (𝛽𝑦 − 𝛽𝑥 )𝐿 = −𝜋 (since 𝛽𝑥 > 𝛽𝑦 ), and so 𝐄𝑠 (𝐿) = 𝐸0 𝑒−𝑗𝛽𝑥 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−𝑗𝛽𝑧 𝐚𝑥 + 18𝑒−𝑗𝛽𝑧 𝑒𝑗𝜙 𝐚𝑦 V/m, propagating in a medium characterized
by complex intrinsic impedance, 𝜂.
a) Find 𝐇𝑠 : With the wave propagating in the forward 𝑧 direction, we find:
𝐇𝑠 =
]
1[
−18𝑒𝑗𝜙 𝐚𝑥 + 15𝐚𝑦 𝑒−𝑗𝛽𝑧 A∕m
𝜂
b) Determine the average power density in W∕m2 : We find
{
}
{ }
} 1
(15)2 (18)2
1 {
1
𝑃𝑧,𝑎𝑣𝑔 = Re 𝐄𝑠 × 𝐇∗𝑠 = Re
+
=
275
Re
W∕m2
2
2
𝜂∗
𝜂∗
𝜂∗
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11.34. Given the general elliptically-polarized wave as per Eq. (93):
𝐄𝑠 = [𝐸𝑥0 𝐚𝑥 + 𝐸𝑦0 𝑒𝑗𝜙 𝐚𝑦 ]𝑒−𝑗𝛽𝑧
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−𝑗𝜙 𝐚𝑦 ]𝑒−𝑗𝛽𝑧 𝑒𝑗𝛿
where 𝛿 is a constant: Adding the two fields gives
[
(
)
(
) ]
𝐄𝑠,𝑡𝑜𝑡 = 𝐸𝑥0 1 + 𝑒𝑗𝛿 𝐚𝑥 + 𝐸𝑦0 𝑒𝑗𝜙 + 𝑒−𝑗𝜙 𝑒𝑗𝛿 𝐚𝑦 𝑒−𝑗𝛽𝑧
⎤
⎡
⎢
⎥
(
)
(
)
= ⎢𝐸𝑥0 𝑒𝑗𝛿∕2 𝑒−𝑗𝛿∕2 + 𝑒𝑗𝛿∕2 𝐚𝑥 + 𝐸𝑦0 𝑒𝑗𝛿∕2 𝑒−𝑗𝛿∕2 𝑒𝑗𝜙 + 𝑒−𝑗𝜙 𝑒𝑗𝛿∕2 𝐚𝑦 ⎥ 𝑒−𝑗𝛽𝑧
⎢
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⎥
⏟⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏟
⎢
⎥
2 cos(𝛿∕2)
2 cos(𝜙−𝛿∕2)
⎣
⎦
[
] 𝑗𝛿∕2 −𝑗𝛽𝑧
This simplifies to 𝐄𝑠,𝑡𝑜𝑡 = 2 𝐸𝑥0 cos(𝛿∕2)𝐚𝑥 + 𝐸𝑦0 cos(𝜙 − 𝛿∕2)𝐚𝑦 𝑒 𝑒
, which is
linearly polarized.
b) Find 𝛿 in terms of 𝜙 such that the resultant wave is polarized along 𝑥: By inspecting the part 𝑎
result, we achieve a zero 𝑦 component when 2𝜙 − 𝛿 = 𝜋 (or odd multiples of 𝜋).
260
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Engineering Electromagnetics 9th solution
Engineering Electromagnetics (Chungnam National University)
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CHAPTER 12 – 9th Edition
+
+
12.1. A uniform plane wave in air, 𝐸𝑥1
= 𝐸𝑥10
cos(1010 𝑡 − 𝛽𝑧) 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
√
√
√
𝑗𝜔𝜇
𝜔𝜇
1010 (4𝜋 × 107 )
𝜂𝑐 =
= (1 + 𝑗)
= (1 + 𝑗)
= (1 + 𝑗)(.0104)
𝜎
2𝜎
2(5.8 × 107 )
Note that the accuracy here is questionable, since we know the conductivity to only two significant
figures. We nevertheless proceed: Using 𝜂0 = 376.7288 ohms, we write
Γ=
𝜂𝑐 − 𝜂0
.0104 − 376.7288 + 𝑗.0104
=
= −.9999 + 𝑗.0001
𝜂𝑐 + 𝜂0
.0104 + 376.7288 + 𝑗.0104
Now |Γ|2 = .9999, and so the transmitted power fraction is 1 − |Γ|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, 𝜖𝑟1 = 9, 𝜖𝑟1
′ = 3, 𝜖 ′′ = 0, and 𝜇 = 𝜇 . Let 𝐸 + = 10 cos(𝜔𝑡 − 15𝑧) V/m and find
𝜇1 = 𝜇0 . For 𝑧 > 0, 𝜖𝑟2
2
0
𝑟2
𝑥1
√
√
√
√
′ ∕𝑐 = 15. So 𝜔 = 15𝑐∕ 𝜖 ′ = 15 × (3 × 108 )∕ 9 =
a) 𝜔: We have 𝛽 = 𝜔 𝜇0 𝜖1′ = 𝜔 𝜖𝑟1
𝑟1
1.5 × 109 s−1 .
√
√
√
′ = 𝜂 ∕ 𝜖 ′ = 377∕ 9 = 126 ohms. Next we apply
b) < 𝐒+
>:
First
we
need
𝜂
=
𝜇
∕𝜖
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 ∕𝜂1 𝐚𝑧 =
1
(1∕2)(10)2 ∕126 𝐚𝑧 = 0.40 𝐚𝑧 W∕m2 .
c) < 𝐒−
>: First, we need to evaluate the reflection coefficient:
1
√
√
√
√
√
√
′ − 𝜂 ∕ 𝜖′
′ −
′
𝜖
𝜖
𝜖𝑟2
𝜂
∕
0
0
𝑟2
𝑟1
𝑟1
𝜂2 − 𝜂1
9− 3
=
=√
=√
Γ=
√
√
√
√ = 0.27
𝜂2 + 𝜂1
′ + 𝜂 ∕ 𝜖′
′ +
′
9
+
3
𝜂0 ∕ 𝜖𝑟2
𝜖
𝜖
0
𝑟1
𝑟1
𝑟2
Then < 𝐒−
>= −|Γ|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 .
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12.3. A uniform plane wave in region 1 is normally-incident on the planar boundary separating regions 1
′ = 𝜇 3 and 𝜖 ′ = 𝜇 3 , find the ratio 𝜖 ′ ∕𝜖 ′ if 20% of the energy in the
and 2. If 𝜖1′′ = 𝜖2′′ = 0, while 𝜖𝑟1
𝑟1
𝑟2
𝑟2
𝑟2 𝑟1
incident wave is reflected at the boundary. There are two possible answers. First, since |Γ|2 = .20,
and since both permittivities and permeabilities are real, Γ = ±0.447. we then set up
√
√
′ )−𝜂
′ )
(𝜇
∕𝜖
(𝜇𝑟1 ∕𝜖𝑟1
𝜂
0
𝑟2 𝑟2
0
𝜂2 − 𝜂1
Γ = ±0.447 =
= √
√
𝜂2 + 𝜂1
′ )+𝜂
′ )
𝜂0 (𝜇𝑟2 ∕𝜖𝑟2
(𝜇𝑟1 ∕𝜖𝑟1
0
√
√
3
3
(𝜇𝑟2 ∕𝜇𝑟2
) − (𝜇𝑟1 ∕𝜇𝑟1
) 𝜇 −𝜇
𝑟2
=√
= 𝑟1
√
𝜇
+
𝜇
3
3
𝑟1
𝑟2
(𝜇𝑟2 ∕𝜇𝑟2
) + (𝜇𝑟1 ∕𝜇𝑟1
)
Therefore
′
𝜖𝑟2
𝜇𝑟2
1 ∓ 0.447
=
= (0.382, 2.62) ⇒ ′ =
𝜇𝑟1
1 ± 0.447
𝜖𝑟1
(
𝜇𝑟2
𝜇𝑟1
)3
= (0.056, 17.9)
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 𝜖2′′ ∕𝜖2′ = 0.05, 𝜖𝑟2
𝜇2 = 𝜇0 . 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 𝜖2′′ ∕𝜖2′ = 0.05 << 1, we recognize region 2 as a good dielectric. Its intrinsic
impedance is therefore approximated well by Eq. (62b), Chapter 11:
]
√ [
′′
𝜖
𝜇0
1
377
𝜂2 =
1 + 𝑗 2′ = √ [1 + 𝑗0.025]
2 𝜖2
𝜖2′
5
The reflection coefficient encountered by the incident wave from region 1 is therefore
√
√
𝜂2 − 𝜂1
(377∕ 5)[1 + 𝑗.025] − 377 (1 − 5) + 𝑗.025
Γ=
=
= −0.383 + 𝑗0.011
=
√
√
𝜂2 + 𝜂1
(377∕ 5)[1 + 𝑗.025] + 377 (1 + 5) + 𝑗.025
The fraction of the incident power that is reflected is then |Γ|2 = 0.147, and thus the fraction
of the power that is transmitted into region 2 is 1 − |Γ|2 = 0.853. Still using the good dielectric
approximation, the attenuation coefficient in region 2 is found from Eq. (60a), Chapter 11:
′′ √
𝜇0
. 𝜔𝜖2
377
𝛼=
= (2𝜋 × 107 )(0.05 × 5 × 8.854 × 10−12 ) √ = 2.34 × 10−2 Np∕m
2
𝜖2′
2 5
Now, the power that propagates into region 2 is expressed in terms of the incident power through
−2 )𝑧
< 𝑆2 > (𝑧) = 5(1 − |Γ|2 )𝑒−2𝛼𝑧 = 5(.853)𝑒−2(2.34×10
= 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 × 10−2 )
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12.5. The region 𝑧 < 0 is characterized by 𝜖𝑟′ = 𝜇𝑟 = 1 and 𝜖𝑟′′ = 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, 𝛽 = 𝑘0 = 10 = 𝜔∕𝑐 = 𝜔∕3 × 108 . Thus,
𝜔 = 3 × 109 s−1 , or 𝑓 = 𝜔∕2𝜋 = 4.7 × 108 Hz.
b) Specify the intrinsic impedance of the region 𝑧 > 0 that would provide the appropriate reflected
wave: Use
◦
𝜂 − 𝜂0
𝐸
◦
50𝑒𝑗20
1
Γ= 𝑟 =
= 𝑒𝑗20 = 0.31 + 𝑗0.11 =
𝐸𝑖𝑛𝑐
150
3
𝜂 + 𝜂0
Now
(
𝜂 = 𝜂0
(
)
)
1 + 0.31 + 𝑗0.11
1+Γ
= 377
= 691 + 𝑗177 Ω
1−Γ
1 − 0.31 − 𝑗0.31
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 𝜙 = 20◦ = .349rad, and we use
𝑧𝑚𝑎𝑥 =
−𝜙 −.349
=
= −0.017 m = −1.7 cm
2𝛽
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
Γ𝑓 =
𝜂𝑔 − 𝜂0
𝜂𝑔 + 𝜂0
=
𝜂0 ∕𝑛𝑔 − 𝜂0
𝜂0 ∕𝑛𝑔 + 𝜂0
=
1 − 𝑛𝑔
1 + 𝑛𝑔
Taking the glass index, 𝑛𝑔 , as 1.45, we find Γ𝑓 = −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 Γ𝑓 ; i.e.,
Γ𝑏 = −Γ𝑓 .
Now, the wave power that makes it through (assuming total reflection at the 45◦ interface)
will be
𝑃𝑜𝑢𝑡 = 𝑃𝑖𝑛 (1 − |Γ𝑓 |2 )(1 − |Γ𝑏 |2 ) = 𝑃𝑖𝑛 (1 − |0.18|2 )2 = 0.93𝑃𝑖𝑛
So we have 93% net transmission.
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12.7. The semi-infinite regions 𝑧 < 0 and 𝑧 > 1 m are free space. For 0 < 𝑧 < 1 m, 𝜖𝑟′ = 4, 𝜇𝑟 = 1, and
𝜖𝑟′′ = 0. A uniform plane wave with 𝜔 = 4 × 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,
√
𝜔 𝜖𝑟′
2(4 × 108 )
= 2.67 rad∕m
𝛽2 =
=
𝑐
3 × 108
Then, with the middle layer √
thickness of 1 m, 𝛽2 𝑑 = 2.67 rad. Also, the intrinsic impedance of
the middle layer is 𝜂2 = 𝜂0 ∕ 𝜖𝑟′ = 𝜂0 ∕2. We now find the input impedance:
[
𝜂𝑖𝑛 = 𝜂2
]
[
]
𝜂0 cos(𝛽2 𝑑) + 𝑗𝜂2 sin(𝛽2 𝑑)
377 2 cos(2.67) + 𝑗 sin(2.67)
=
= 231 + 𝑗141
𝜂2 cos(𝛽2 𝑑) + 𝑗𝜂0 sin(𝛽2 𝑑)
2 cos(2.67) + 𝑗2 sin(2.67)
Now, at the first interface,
Γ12 =
𝜂𝑖𝑛 − 𝜂0
231 + 𝑗141 − 377
=
= −.176 + 𝑗.273 = .325∠123◦
𝜂𝑖𝑛 + 𝜂0
231 + 𝑗141 + 377
The standing wave ratio measured in region 1 is thus
𝑠1 =
1 + |Γ12 | 1 + 0.325
=
= 1.96
1 − |Γ12 | 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:
Γ23 =
𝜂0 − 𝜂0 ∕2 1 − 1∕2 1
=
=
𝜂0 + 𝜂0 ∕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 Γ12 is 𝜙 = 123◦ = 2.15 rad. Thus
𝑧𝑚𝑎𝑥 =
−𝜙
−2.15
=
= −.81 m
2𝛽
2(4∕3)
12.8. A wave starts at point 𝑎, propagates 1m through a lossy dielectric rated at 𝛼𝑑𝐵 = 0.1dB/cm, reflects
at normal incidence at a boundary at which Γ = 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 10−0.1𝛼𝑑𝐵 𝐿 |Γ|2 10−0.1𝛼𝑑𝐵 𝐿 ⇒
𝑃𝑓
𝑃𝑖
= |0.3 + 𝑗0.4|2 10−0.2(0.1)100 = 2.5 × 10−3
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12.9. Region 1, 𝑧 < 0, and region 2, 𝑧 > 0, are both perfect dielectrics (𝜇 = 𝜇0 , 𝜖 ′′ = 0). A uniform plane
wave traveling in the 𝐚𝑧 direction has a radian frequency of 3 × 1010 rad/s. Its wavelengths in the two
regions are 𝜆1 = 5 cm and 𝜆2 = 3 cm. What percentage of the energy incident on the boundary is
a) reflected; We first note that
(
′
𝜖𝑟1
=
2𝜋𝑐
𝜆1 𝜔
)2
(
′
and 𝜖𝑟2
=
2𝜋𝑐
𝜆2 𝜔
)2
′ ∕𝜖 ′ = (𝜆 ∕𝜆 )2 . Then with 𝜇 = 𝜇 in both regions, we find
Therefore 𝜖𝑟1
2
1
0
𝑟2
√
Γ=
=
𝜂2 − 𝜂1
=
𝜂2 + 𝜂1
𝜂0
𝜂0
√
′ −𝜂
1∕𝜖𝑟2
0
′ +𝜂
1∕𝜖𝑟2
0
√
√
′
1∕𝜖𝑟1
′
1∕𝜖𝑟1
√
′ ∕𝜖 ′ − 1
𝜖𝑟1
𝑟2
(𝜆 ∕𝜆 ) − 1
=√
= 2 1
(𝜆2 ∕𝜆1 ) + 1
′ ∕𝜖 ′ + 1
𝜖𝑟1
𝑟2
𝜆2 − 𝜆1
3−5
1
=
=−
𝜆2 + 𝜆1
3+5
4
The fraction of the incident energy that is reflected is then |Γ|2 = 1∕16 = 6.25 × 10−2 .
b) transmitted? We use part 𝑎 and find the transmitted fraction to be
1 − |Γ|2 = 15∕16 = 0.938.
c) What is the standing wave ratio in region 1? Use
𝑠=
1 + |Γ| 1 + 1∕4 5
=
= = 1.67
1 − |Γ| 1 − 1∕4 3
′′ = 0, and 𝜖 ′ is unknown. Find 𝜖 ′ if
12.10. In Fig. 12.1, let region 2 be free space, while 𝜇𝑟1 = 1, 𝜖𝑟1
𝑟1
𝑟1
: Since region 2 is free space, the reflection coeffia) the amplitude of 𝐄−
is one-half that of 𝐄+
1
1
cient is
√
√
′
′ −1
−
𝜖
𝜖𝑟1
𝜂
−
𝜂
∕
0
0
|𝐄1 | 𝜂0 − 𝜂1
𝑟1
1
′
Γ= + =
=√
=
=
⇒ 𝜖𝑟1
= 9.
√
2
|𝐄1 | 𝜂0 + 𝜂1
′
′
𝜂0 + 𝜂0 ∕ 𝜖𝑟1
𝜖𝑟1 + 1
b) < 𝐒−
> is one-half of < 𝐒+
>: This time
1
1
|2
|√ ′
| 𝜖 − 1|
|
|
𝑟1
| = 1 ⇒ 𝜖 ′ = 34
|Γ|2 = || √
|
𝑟1
2
| 𝜖′ + 1 |
|
|
𝑟1
|
|
c) |𝐄1 |𝑚𝑖𝑛 is one-half |𝐄1 |𝑚𝑎𝑥 : Use
√
′
𝜖𝑟1 − 1
|𝐄1 |𝑚𝑎𝑥
1 + |Γ|
1
′
=𝑠=
= 2 ⇒ |Γ| = Γ = = √
⇒ 𝜖𝑟1
=4
|𝐄1 |𝑚𝑖𝑛
1 − |Γ|
3
′ +1
𝜖𝑟1
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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 = − (𝜙 + 𝜋) = −0.3𝜆 ⇒ 𝜙 = 0.2𝜋
2𝛽
where 𝛽 = 2𝜋∕𝜆 has been used. Next,
|Γ| =
𝑠−1 3−1 1
=
=
𝑠+1 3+1 2
So we now have
Γ = 0.5𝑒𝑗0.2𝜋 =
𝜂𝑢 − 𝜂0
𝜂𝑢 + 𝜂0
We solve for 𝜂𝑢 to find
𝜂𝑢 = 𝜂0 (1.70 + 𝑗1.33) = 641 + 𝑗501 Ω
12.12. A 50MHz uniform plane wave is normally incident from air onto the surface of a calm ocean.
For seawater, 𝜎 = 4 S/m, and 𝜖𝑟′ = 78.
a) Determine the fractions of the incident power that are reflected and transmitted: First we find
the loss tangent:
4
𝜎
=
= 18.4
𝜔𝜖 ′
2𝜋(50 × 106 )(78)(8.854 × 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
√
𝜂𝑠 = 𝜋𝑓 𝜇∕𝜎(1 + 𝑗), and the reflection coefficient becomes
√
Γ= √
where
𝜋𝑓 𝜇∕𝜎 (1 + 𝑗) − 𝜂0
𝜋𝑓 𝜇∕𝜎 (1 + 𝑗) + 𝜂0
√
√
𝜋𝑓 𝜇∕𝜎 = 𝜋(50 × 106 )(4𝜋 × 10−7 )∕4 = 7.0. The fraction of the power reflected is
√
[
𝜋𝑓 𝜇∕𝜎 − 𝜂0 ]2 + 𝜋𝑓 𝜇∕𝜎
𝑃𝑟
[7.0 − 377]2 + 49.0
= 0.93
=
= |Γ|2 = √
𝑃𝑖
[7.0 + 377]2 + 49.0
[ 𝜋𝑓 𝜇∕𝜎 + 𝜂0 ]2 + 𝜋𝑓 𝜇∕𝜎
The transmitted fraction is then
𝑃𝑡
= 1 − |Γ|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.
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12.13. A right-circularly-polarized plane wave is normally incident from air onto a semi-infinite slab of plexiglas (𝜖𝑟′ = 3.45, 𝜖𝑟′′ = 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 𝜂 = 𝜂0 ∕ 3.45 = 203 Ω. Then
Γ=
203 − 377
= −0.30
203 + 377
The reflected power fraction is thus |Γ|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 − |Γ|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, Γ = −1, the incident
and reflected fields will add to give the total field:
𝐄𝑡𝑜𝑡 = 𝐄𝑖 + 𝐄𝑟 = 𝐸0 (𝐚𝑥 + 𝑗𝐚𝑦 )𝑒−𝑗𝛽𝑧 − 𝐸0 (𝐚𝑥 + 𝑗𝐚𝑦 )𝑒+𝑗𝛽𝑧
⎤
⎡
⎢(
[
]
)
(
) ⎥
= 𝐸0 ⎢ 𝑒−𝑗𝛽𝑧 − 𝑒𝑗𝛽𝑧 𝐚𝑥 + 𝑗 𝑒−𝑗𝛽𝑧 − 𝑒𝑗𝛽𝑧 𝐚𝑦 ⎥ = 2𝐸0 sin(𝛽𝑧) 𝐚𝑦 − 𝑗𝐚𝑥
⎢⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟
⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟ ⎥
⎥
⎢
−2𝑗 sin(𝛽𝑧)
⎦
⎣ −2𝑗 sin(𝛽𝑧)
b) Determine the real instantaneous form of the result of part 𝑎:
{
}
[
]
𝐄(𝑧, 𝑡) = Re 𝐄𝑡𝑜𝑡 𝑒𝑗𝜔𝑡 = 2𝐸0 sin(𝛽𝑧) cos(𝜔𝑡)𝐚𝑦 + sin(𝜔𝑡)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(𝛽𝑧).
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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 𝜃𝑐 =
√
𝑛𝑠
1
= sin(45◦ ) = √ ⇒ 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
√
√
√
𝜂𝑓 = 𝜂𝑠 𝜂𝑔 ⇒ 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 𝜃𝑐 with the reduced surrounding index. At the first normal incidence boundary
(from air to film to glass), the input intrinsic impedance is
( )
𝜂𝑓2
𝜂02 ∕𝑛2𝑓
𝑛𝑔
𝜂𝑖𝑛1 =
=
= 𝜂0
𝜂𝑔
𝜂0 ∕𝑛𝑔
𝑛2𝑓
At the second normal incidence boundary at the prism exit (glass to film to air), the input intrinsic impedance is
( )
𝜂02 ∕𝑛2𝑓
𝜂𝑓2
1
=
= 𝜂0
𝜂𝑖𝑛2 =
𝜂0
𝜂0
𝑛2𝑓
The reflection coefficients at the two boundaries will be
𝑛𝑔 − 𝑛2𝑓
𝜂𝑖𝑛1 − 𝜂0
2.55 − (2.14)2
Γ1 =
= −0.285
=
=
𝜂𝑖𝑛1 + 𝜂0
2.55 + (2.14)2
𝑛𝑔 + 𝑛2𝑓
Γ2 =
𝜂𝑖𝑛2 − 𝜂𝑔
𝜂𝑖𝑛2 + 𝜂𝑔
=
𝑛𝑔 − 𝑛2𝑓
𝑛𝑔 + 𝑛2𝑓
= Γ1
The power ratio will be:
)(
) (
)2
𝑃𝑜𝑢𝑡 (
= 1 − |Γ1 |2 1 − |Γ2 |2 = 1 − (0.285)2 = 0.845
𝑃𝑖𝑛
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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 𝜂𝑖𝑛,𝑏 : Since region 3 is a quarter-wave layer, 𝛽3 𝑙𝑏 = 𝜋∕2, and (47) reduces to
𝜂𝑖𝑛,𝑏 =
𝜂32
𝜂4
b) Find 𝜂𝑖𝑛,𝑎 : Again, with region 2 of quarter-wave thickness, 𝛽2 𝑙𝑎 = 𝜋∕2 and (48) becomes
𝜂𝑖𝑛,𝑎 =
𝜂22
𝜂𝑖𝑛,𝑏
=
𝜂22 𝜂4
𝜂32
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:
𝜂𝑖𝑛,𝑎 = 𝜂0 ⇒ 𝜂22 𝜂4 = 𝜂32 𝜂0
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
( 2)( ) ( 2)
𝜂0
𝜂0
𝑛23
𝜂0
=
𝜂
⇒
𝑛
=
0
4
𝜂4
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, 𝜂𝑖𝑛,𝑏 = 𝜂𝑖𝑛,𝑎 = 𝜂4 . The reflection coefficient at the front surface is
therefore
𝜂𝑖𝑛,𝑎 − 𝜂0
𝜂 − 𝜂0
1 − 𝑛4
1 − 1.45
= 4
=
=
= −0.184
Γ𝑖𝑛 =
𝜂𝑖𝑛,𝑎 + 𝜂0
𝜂4 + 𝜂0
1 + 𝑛4
1 + 1.45
The transmitted power fraction is then
𝑃𝑡
= 1 − |Γ𝑖𝑛 |2 = 1 − (0.184)2 = 0.97
𝑃𝑖𝑛
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12.17. A uniform plane wave in free space is normally-incident onto a dense dielectric plate of thickness
𝜆∕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, 𝜂1 = 𝜂3 = 𝜂0 , and the quarter-wave section input impedance is therefore
𝜂𝑖𝑛 =
𝜂22
𝜂3
=
𝜂02 ∕𝑛2
𝜂0
=
𝜂0
𝑛2
The reflection coefficient at the front surface is then
Γ𝑖𝑛 =
𝜂𝑖𝑛 − 𝜂0
1 − 𝑛2
=
𝜂𝑖𝑛 + 𝜂0
1 + 𝑛2
√
For half-power reflection, we require that |Γ𝑖𝑛 |2 = 0.5, or Γ𝑖𝑛 = ±1∕ 2. Since 𝑛 must be
greater than 1, we choose the minus sign option and write:
1 − 𝑛2
1
= −√ ⇒ 𝑛 =
1 + 𝑛2
2
[√
]1∕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) 𝜆∕2; (b) 𝜆∕4; (c) 𝜆∕8.
With region 3 being a perfect conductor, 𝜂3 = 0, and Eq. (36) gives the input impedance to the
structure as 𝜂𝑖𝑛 = 𝑗𝜂2 tan 𝛽𝓁. The reflection coefficient is then
𝜂22 tan2 𝛽𝓁 − 𝜂02 + 𝑗2𝜂0 𝜂2 tan 𝛽𝓁
𝜂𝑖𝑛 − 𝜂0
𝑗𝜂2 tan 𝛽𝓁 − 𝜂0
Γ=
= Γ𝑟 + 𝑗Γ𝑖
=
=
𝜂𝑖𝑛 + 𝜂0
𝑗𝜂2 tan 𝛽𝓁 + 𝜂0
𝜂 2 tan2 𝛽𝓁 + 𝜂 2
2
0
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
[
]
( )
]
[
Γ
2𝜂
𝜂
tan
𝛽𝓁
𝑖
2 0
−1
−1
−1 (2.90) tan 𝛽𝓁
𝜙 = tan
= tan
= tan
Γ𝑟
tan 𝛽𝓁 − 2.10
𝜂 2 tan2 𝛽𝓁 − 𝜂 2
2
0
where 𝜂2 = 𝜂0 ∕1.45 has been used. We can now evaluate the phase shift for the three given
cases. First, when 𝓁 = 𝜆∕2, 𝛽𝓁 = 𝜋, and thus 𝜙(𝜆∕2) = 0. Next, when 𝓁 = 𝜆∕4, 𝛽𝓁 = 𝜋∕2,
and
𝜙(𝜆∕4) → tan−1 [2.90] = 71◦
as 𝓁 → 𝜆∕4. Finally, when 𝓁 = 𝜆∕8, 𝛽𝓁 = 𝜋∕4, and
]
[
2.90
= −69.2◦ (or 291◦ )
𝜙(𝜆∕8) = tan−1
1 − 2.10
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12.19. You are given four slabs of lossless dielectric, all with the same intrinsic impedance, 𝜂, known to
be different from that of free space. The thickness of each slab is 𝜆∕4, where 𝜆 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 𝜆, or ii.)
𝑑1 = 𝑑3 = 0, 𝑑2 = 𝜆∕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 = 𝜆∕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
𝜂𝑖𝑛,1 = 𝜂 2 ∕𝜂0 . We transform this back to the back surface of slab 3, moving through a distance of 𝜆∕4 in free space: 𝜂𝑖𝑛,2 = 𝜂02 ∕𝜂𝑖𝑛,1 = 𝜂03 ∕𝜂 2 . We next transform this impedance to
the front surface of slab 3, producing 𝜂𝑖𝑛,3 = 𝜂 2 ∕𝜂𝑖𝑛,2 = 𝜂 4 ∕𝜂03 . We continue in this manner
until reaching the front surface of slab 1, where we find 𝜂𝑖𝑛,7 = 𝜂 8 ∕𝜂07 . Assuming 𝜂 < 𝜂0 , the
ratio 𝜂 𝑛 ∕𝜂0𝑛−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.
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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:
𝜎
4
= 18.4
=
′
6
𝜔𝜖
2𝜋(50 × 10 )(78)(8.854 × 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), and with 𝜇 = 𝜇0 , the intrinsic
impedance is 𝜂𝑠 = 𝜋𝑓 𝜇∕𝜎(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
√
𝑛𝑠𝑒𝑎 𝜔 . √
.
𝜇𝜎
𝛽=
= 𝜋𝑓 𝜇𝜎 ⇒ 𝑛𝑠𝑒𝑎 = 𝑐
= 26.8
𝑐
4𝜋𝑓
Then, using Snell’s law, the angle of refraction is found:
sin 𝜃2 =
𝑛𝑠𝑒𝑎
sin 𝜃1 = 26.8 sin(60◦ ) ⇒ 𝜃2 = 1.9◦
𝑛1
.
This angle is small enough so that cos 𝜃2 = 1. Therefore, for s polarization,
7.0(1 + 𝑗) − 377∕ cos 60◦
. 𝜂 − 𝜂𝑠1
=
= −0.98 + 𝑗0.018 = 0.98∠179◦
Γ𝑠 = 𝑠2
𝜂𝑠2 + 𝜂𝑠1
7.0(1 + 𝑗) + 377∕ cos 60◦
The fraction of the power reflected is now |Γ𝑠 |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
. 𝜂𝑝2 − 𝜂𝑝1
7.0(1 + 𝑗) − 377 cos 60◦
=
= −0.93 + 𝑗0.069 = 0.93∠176◦
Γ𝑝 =
𝜂𝑝2 + 𝜂𝑝1
7.0(1 + 𝑗) + 377 cos 60◦
The fraction of the power reflected is now |Γ𝑝 |2 = 0.86. The fraction transmitted is then 0.14.
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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 (𝜖𝑟′ = 3.45, 𝜖𝑟′′ = 0, 𝜇 = 𝜇0 ).
a) Determine the fractions of the incident power that are reflected
and transmitted: In plexiglas,
√
′
′
−1
−1
Brewster’s angle is 𝜃𝐵 = 𝜃1 = tan (𝜖𝑟2 ∕𝜖𝑟1 ) = tan ( 3.45) = 61.7◦ . Then the angle of
refraction is 𝜃2 = 90◦ − 𝜃𝐵 (see Example 12.9), or 𝜃2 = 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
𝜂 − 𝜂1𝑠
.614𝜂0 − 2.11𝜂0
=
= −0.549
Γ𝑠 = 2𝑠
𝜂2𝑠 + 𝜂1𝑠
.614𝜂0 + 2.11𝜂0
◦ ) = 2.11𝜂 ,
where 𝜂1𝑠 = 𝜂1 sec 𝜃1 = 𝜂0 sec(61.7
0
√
◦
and 𝜂2𝑠 = 𝜂2 sec 𝜃2 = (𝜂0 ∕ 3.45) sec(28.3 ) = 0.614𝜂0 . Now, the reflected power fraction
is |Γ|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 𝜙 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 𝜙 such that total confinement will occur, with 𝑛0 = 1. The quantity sin 𝜙
is known as the numerical aperture of the guide.
From the illustration we see that 𝜙1 maximizes when 𝜃1 is at its minimum value. This minimum will
be the critical angle for the 𝑛1 − 𝑛2 interface, where sin 𝜃𝑐 = sin 𝜃1 = 𝑛2 ∕𝑛1 . Let the refracted angle
to the right of the vertical interface (not shown) be 𝜙2 , where 𝑛0 sin 𝜙1 = 𝑛1 sin 𝜙2 . Then we see that
𝜙2 + 𝜃1 = 90◦ , and so sin 𝜃1 = cos 𝜙2 . Now, the numerical aperture becomes
√
√
√
𝑛1
sin 𝜙1𝑚𝑎𝑥 =
sin 𝜙2 = 𝑛1 cos 𝜃1 = 𝑛1 1 − sin2 𝜃1 = 𝑛1 1 − (𝑛2 ∕𝑛1 )2 = 𝑛21 − 𝑛22
𝑛0
)
(√
−1
2
2
Finally, 𝜙1𝑚𝑎𝑥 = sin
𝑛1 − 𝑛2 is the numerical aperture angle.
12.23. Suppose that 𝜙1 in Fig. 12.16 is Brewster’s angle, and that 𝜃1 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◦ − 𝜙1 . Therefore, 𝜙1 = 𝜃1 , and thus sin 𝜙1 = sin 𝜃1 .
Thus
√
𝑛2
𝑛1
⇒ 𝑛0 = (𝑛1 ∕𝑛2 ) 𝑛21 − 𝑛22
= sin 𝜃1 =
sin 𝜙1 = √
𝑛1
2
2
𝑛0 + 𝑛1
Alternatively, we
√ could have used the result of Problem 12.22, in which it was found that
sin 𝜙1 = (1∕𝑛0 ) 𝑛21 − 𝑛22 , which we then set equal to sin 𝜃1 = 𝑛2 ∕𝑛1 to get the same result.
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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, 𝛼: 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 𝛼∕2. Since
this angle is also Brewster’s angle, we may write:
)
)
(
(
1
1
−1
−1
= 1.21 rad = 69.2◦
= 2 sin
𝛼 = 2 sin
√
√
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 Γ𝑠 = (𝜂𝑠2 − 𝜂𝑠1 )∕(𝜂𝑠2 + 𝜂𝑠1 ), where
𝜂𝑠2 =
𝜂 √
𝜂
𝜂2
= √2
= 02 1 + 𝑛2
cos(𝜃𝐵2 ) 𝑛∕ 1 + 𝑛2
𝑛
𝜂𝑠1 =
√
𝜂
𝜂1
= √1
= 𝜂0 1 + 𝑛2
cos(𝜃𝐵1 ) 1∕ 1 + 𝑛2
and
Thus, at the first interface, Γ = (1 − 𝑛2 )∕(1 + 𝑛2 ). At the second interface, Γ will be equal but
of opposite sign to the above value. The power transmission coefficient through each interface
is 1 − |Γ|2 , so that for both interfaces, we have, with 𝑛 = 1.45:
[
) 2 ]2
( 2
(
)
𝑃𝑡𝑟
𝑛 −1
2 2
= 0.76
= 1 − |Γ|
= 1−
𝑃𝑖𝑛𝑐
𝑛2 + 1
The insertion loss is now
𝓁𝑖 (dB) = 10 log10 (0.76) = −1.19 dB
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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, 𝜃𝐵 = 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 𝜃𝑐 = 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:
√
[
]
√
𝑑𝛽
𝜔𝜇𝜎
1 𝜔𝜇𝜎 −1∕2 𝜇𝜎
𝛽 = 𝜋𝑓 𝜇𝜎 =
→
=
2
𝑑𝜔 2
2
2
Thus
𝑑𝜔
=
𝑑𝛽
(
𝑑𝛽
𝑑𝜔
)−1
√
2𝜔
= 𝑣𝑔
=2
𝜇𝜎
and
𝜔
𝜔
=
𝑣𝑝 = = √
𝛽
𝜔𝜇𝜎∕2
√
2𝜔
𝜇𝜎
Therefore 𝑣𝑔 ∕𝑣𝑝 = 2.
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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(𝜆) = 𝑛𝑎 + 𝑛𝑏 (𝜆 − 𝜆𝑎 ), where 𝑛𝑎 , 𝑛𝑏 , and 𝜆𝑎 are constants, and where 𝜆 is the
free space wavelength.
a) Show that 𝑑∕𝑑𝜔 = −(2𝜋𝑐∕𝜔2 )𝑑∕𝑑𝜆: With 𝜆 as the free space wavelength, we use 𝜆 = 2𝜋𝑐∕𝜔,
from which 𝑑𝜆∕𝑑𝜔 = −2𝜋𝑐∕𝜔2 . Then 𝑑∕𝑑𝜔 = (𝑑𝜆∕𝑑𝜔) 𝑑∕𝑑𝜆 = −(2𝜋𝑐∕𝜔2 ) 𝑑∕𝑑𝜆.
b) Using 𝛽(𝜆) = 2𝜋𝑛∕𝜆, determine the wavelength-dependent (or independent) group delay over
a unit distance: This will be
[
]
[
]]
𝑑𝛽
𝑑 2𝜋𝑛(𝜆)
2𝜋𝑐 𝑑 2𝜋 [
1
𝑡𝑔 =
=
=
𝑛𝑎 + 𝑛𝑏 (𝜆 − 𝜆𝑎 )
=− 2
𝑣𝑔
𝑑𝜔 𝑑𝜔
𝜆
𝜔 𝑑𝜆 𝜆
[
] 2𝜋 ]
2𝜋 [
2𝜋𝑐
= − 2 − 2 𝑛𝑎 + 𝑛𝑏 (𝜆 − 𝜆𝑎 ) +
𝑛
𝜆 𝑏
𝜔 [ 𝜆
]
2𝜋𝑛𝑏 𝜆𝑎
2𝜋𝑛
1
𝜆2
= (𝑛𝑎 − 𝑛𝑏 𝜆𝑎 ) s∕m
=−
− 2𝑎 +
2𝜋𝑐
𝑐
𝜆
𝜆2
c) Determine 𝛽2 from your result of part 𝑏: 𝛽2 = 𝑑 2 𝛽∕𝑑𝜔2 |𝜔0 . Since the part 𝑏 result is independent
of wavelength (and of frequency), it follows that 𝛽2 = 0.
d) Discuss the implications of these results, if any, on pulse broadening: A wavelength-independent
group delay (leading to zero 𝛽2 ) 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 𝜆 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 𝛽2 = 10 ps2 ∕km.
Over √
what distance will the pulse spread
√ to twice its initial width? After propagation, the width is
′
2
2
𝑇 = 𝑇 + (Δ𝜏) = 2𝑇 . Thus Δ𝜏 = 3𝑇 , where Δ𝜏 = 𝛽2 𝑧∕𝑇 . Therefore
√
√
𝛽2 𝑧 √
3(5 ps)2
3𝑇 2
=
= 3𝑇 or 𝑧 =
= 4.3 km
𝑇
𝛽2
10 ps2 ∕km
12.30. A 𝑇 = 20 ps transform-limited pulse propagates through 10 km of a dispersive channel for which
𝛽2 = 12 ps2 ∕km. The pulse then propagates through a second 10 km channel for which 𝛽2 =
−12 ps2 ∕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 𝛽2 down the length of the channel. In fact,
we may write in general:
𝐿
1
Δ𝜏 =
𝛽 (𝑧) 𝑑𝑧
𝑇 ∫0 2
Having 𝛽2 change sign at the midpoint, yields a zero Δ𝜏, 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 𝛽2 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 𝛽2 𝐿 = −𝛽2′ 𝐿′ , assuming dispersion terms of higher
order than 𝛽2 do not exist.
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Engineering Electromagnetics 9th solution
Engineering Electromagnetics (Chungnam National University)
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CHAPTER 13 – 9th Edition
13.1. The conductors of a coaxial transmission line are copper (𝜎𝑐 = 5.8 × 10−7 S/m) and the dielectric is
polyethylene (𝜖𝑟′ = 2.26, 𝜎∕𝜔𝜖 ′ = 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 Ω: Us
1
𝑍0 =
2𝜋
√
√
( ) 2𝜋 𝜖 ′ (50)
𝜇 (𝑏)
𝑏
𝑟
= 50 ⇒ ln
=
ln
= 1.25
𝜖′
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=
( )
𝑏
2𝜋𝜖 ′
= 2𝜋(2.26)(8.854 × 10−2 ) = 1.257
= 10−10 ⇒ ln
ln(𝑏∕𝑎)
𝑎
So 𝑏∕𝑎 = 𝑒1.257 = 3.51, or 𝑎 = 4∕3.51 = 1.138 mm.
c) 𝐿 = 0.2 𝜇H∕m: Use
𝐿=
( ) 2𝜋(0.2 × 10−6 )
𝜇0 ( 𝑏 )
𝑏
= 0.2 × 10−6 ⇒ ln
=
ln
=1
2𝜋
𝑎
𝑎
4𝜋 × 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, 𝜖𝑟 = 2.0,
𝜇𝑟 = 1, 𝜎𝑐 = 1.0 × 107 S/m, 𝜎 = 1.0 × 10−5 S/m, and 𝑓 = 300 MHz.
First, we note that the metal is a good conductor, as confirmed by the loss tangent:
𝜎𝑐
1.0 × 107
=
= 3.0 × 108 >> 1
8
−12
𝜔𝜖 ′
2𝜋(3.00 × 10 )(2)(8.854 × 10 )
So the skin depth into the metal is
√
√
2
2
= 9.2 𝜇m
=
𝛿=
8
𝜔𝜇𝜎𝑐
2𝜋(3.00 × 10 )(4𝜋 × 10−7 )(1.0 × 107 )
The current can be said to exist in layers of thickness 𝛿 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 𝛿. 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:
(
)
)
(
1
1
1
1
1 1
𝑅=
+
=
= 7.6 Ω∕m
+
2𝜋𝛿𝜎𝑐 𝑎 𝑏
2𝜋(9.2 × 10−6 )(1.0 × 107 ) 0.25 × 10−3 2.5 × 10−3
using Eq. (12).
𝐿 = 𝐿𝑒𝑥𝑡 =
)
( )
(
𝜇
4𝜋 × 10−7
𝑏
2.5
=
= 0.46 𝜇H∕m
ln
ln
2𝜋
𝑎
2𝜋
0.25
using Eq. (11).
𝐶=
2𝜋(2)(8.854 × 10−12 )
2𝜋𝜖 ′
=
= 48 pF∕m
ln(𝑏∕𝑎)
ln(2.5∕0.25)
using Eq. (9).
𝐺=
2𝜋(1.0 × 10−5 )
2𝜋𝜎
=
= 27 𝜇S∕m
ln(𝑏∕𝑎)
ln(2.5∕0.25)
using Eq. (10).
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13.3. Two aluminum-clad steel conductors are used to construct a two-wire transmission line. Let
𝜎𝐴𝑙 = 3.8 × 107 S/m, 𝜎𝑆𝑡 = 5 × 106 S/m, and 𝜇𝑆𝑡 = 100 𝜇H∕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, 𝛿, will tell us. We have,
for aluminum,
1
1
=√
𝛿=√
= 2.58 × 10−5 m
7
−7
7
𝜋𝑓 𝜇0 𝜎𝐴𝑙
𝜋(10 )(4𝜋 × 10 )(3.8 × 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
1
=
= 0.023 Ω∕m
𝜋𝑎𝛿𝜎𝐴𝑙
𝜋(.014)(2.58 × 10−5 )(3.8 × 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=
𝜋𝜖0
cosh−1 (𝑑∕2𝑎)
=
𝜋 × 8.85 × 10−12
= 1.42 × 10−11 F∕m = 14.2 pF∕m
cosh−1 (4∕(2 × 0.55))
Finally, the inductance per unit length will be
𝐿=
𝜇0
4𝜋 × 10−7
cosh(𝑑∕2𝑎) =
cosh (4∕(2 × 0.55)) = 7.86 × 10−7 H∕m = 0.786 𝜇H∕m
𝜋
𝜋
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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 𝜖𝑟 = 2.26 and 𝜎∕(𝜔𝜖 ′ ) = 4.0 ×
10−4 .
From the loss tangent, we find the conductivity of polyethylene:
𝜎 = (4.0 × 10−4 )(2𝜋 × 8.00 × 108 )(2.26)(8.854 × 10−12 ) = 4.0 × 10−5 S∕m
As polyethylene is a good dielectric, its penetration depth is (using Eq. (60a), Chapter 11):
√
√
2𝜖𝑟′
2 2.26
1
2 𝜖′
= 199 m
𝛿𝑝 =
=
=
=
𝛼𝑝
𝜎 𝜇
𝜎𝜂0
(4.0 × 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
𝜎𝑐 = 5.8 × 107 ) the skin depth will be (from Eq. (82), Chapter 11):
𝛿𝑐 =
1
1
= 2.3 𝜇m
=
8
𝜋𝑓 𝜇𝜎𝑐
𝜋(8.00 × 10 )(4𝜋 × 10−7 )(5.8 × 107 )
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=
𝜋(2.26)(8.854 × 10−12 )
𝜋𝜖 ′
=
= 22.7 pF∕m
cosh−1 (𝑑∕2𝑎)
cosh−1 (8)
𝐿 = 𝐿𝑒𝑥𝑡 =
𝜇
4𝜋 × 10−7
cosh−1 (𝑑∕2𝑎) =
cosh−1 (8) = 1.11 𝜇H∕m
𝜋
𝜋
𝐺=
𝑅=
𝜋(4.0 × 10−5 )
𝜋𝜎
=
= 46 𝜇S∕m
cosh−1 (𝑑∕2𝑎)
cosh−1 (8)
1
1
= 4.8 ohms∕m
=
−4
𝜋𝑎𝛿𝑐 𝜎𝑐
𝜋(5.0 × 10 )(2.3 × 10−6 )(5.8 × 107 )
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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 𝜎 and 𝜎𝑐 are zero. Find the dielectric constant of the
insulating medium if
a) 𝑍0 = 300 Ω: Use
√
(
)
√
( )
𝜇0
120𝜋
𝑑
8
1
−1
−1
′
𝜖𝑟 =
⇒
cosh
cosh
300 =
= 1.107 ⇒ 𝜖𝑟′ = 1.23
𝜋 𝜖𝑟′ 𝜖0
2𝑎
300𝜋
2(.5)
b) 𝐶 = 20 pF/m: Use
20 × 10−12 =
20 × 10−12
𝜋𝜖 ′
′
cosh−1 (8) = 1.99
⇒
𝜖
=
𝑟
−1
𝜋𝜖
cosh (𝑑∕2𝑎)
0
c) 𝑣𝑝 = 2.6 × 108 m/s:
1
𝑐
=√
𝑣𝑝 = √
=√
⇒ 𝜖𝑟′ =
′
′
𝜇
𝜖
𝜖
𝜖
𝐿𝐶
0 0 𝑟
𝑟
1
(
3.0 × 108
2.6 × 108
)2
= 1.33
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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
√
√
( )
𝜂
. 1
.
1 𝑅
𝑏
𝐶
𝐿
where 𝑍0 =
=
= 0 ln
𝛼= 𝑅
2
𝐿 2 𝑍0
𝐶
2𝜋
𝑎
Then the skin effect resistance for the coax line is
)
(
(
)
1
1 1
𝑏
1
=
+
+
1
𝑅=
2𝜋𝛿𝜎𝑐 𝑎 𝑏
2𝜋𝑏𝛿𝜎𝑐 𝑎
So
(𝑏∕𝑎) + 1
𝜋
2𝜋𝑏𝛿𝜎𝑐 𝜂0 ln(𝑏∕𝑎)
.
𝛼=
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 ⇒ 𝑥 = 3.6
so that
𝑍0 =
377
ln(3.6) = 77 ohms
2𝜋
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(𝜌) =
𝑉0,𝑚𝑎𝑥
𝑉0
𝐚𝜌 ⇒ 𝐄𝑚𝑎𝑥 =
𝐚 = 𝐄𝑏𝑑
𝜌 ln(𝑏∕𝑎)
𝑎 ln(𝑏∕𝑎) 𝜌
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)
2𝜋𝑎𝐸𝑏𝑑
=
(𝜂0 ∕2𝜋) ln(𝑏∕𝑎)
𝜂0
The maximum line power (under dc operation) is
𝑃𝑚𝑎𝑥 = 𝑉0,𝑚𝑎𝑥 𝐼0,𝑚𝑎𝑥 =
2 ln(𝑏∕𝑎)
2𝜋𝑎2 𝐸𝑏𝑑
𝜂0
=
2𝜋𝑏2 2 ln(𝑏∕𝑎)
𝐸
𝜂0 𝑏𝑑 (𝑏∕𝑎)2
Defining 𝑥 = 𝑏∕𝑎, we need to maximize the function ln(𝑥)∕𝑥2 :
(
)
𝑥 − 2𝑥 ln(𝑥)
𝑑 ln(𝑥)
1
⇒ 𝑥 = 𝑒1∕2 = 1.65
= 0 ⇒ ln(𝑥) =
=
2
4
𝑑𝑥
2
𝑥
𝑥
So
𝑍0 =
𝜂0 ( 𝑏 ) 𝜂0
ln
= 30 ohms
=
2𝜋
𝑎
4𝜋
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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 Ω, find 𝜖𝑟′ : We use
√
𝑍0 =
(
)
𝜇 (𝑑 )
377 2 .04
′
=
15
⇒
𝜖
=
= 2.8
𝑟
𝜖′ 𝑏
15
9
b) Assume copper conductors and operation at 2 × 108 rad/s. If 𝑅𝐶 = 𝐺𝐿, determine the loss
tangent of the dielectric: For copper, 𝜎𝑐 = 5.8 × 107 S/m, and the skin depth is
√
√
2
2
=
𝛿=
= 1.2 × 10−5 m
8
𝜔𝜇0 𝜎𝑐
(2 × 10 )(4𝜋 × 10−7 )(5.8 × 107 )
Then
2
2
=
= 0.98 Ω∕m
7
𝜎𝑐 𝛿𝑏 (5.8 × 10 )(1.2 × 10−5 )(.003)
𝑅=
Now
𝐶=
𝜖 ′ 𝑏 (2.8)(8.85 × 10−12 )(3)
=
= 3.7 × 10−10 F∕m
𝑑
0.2
and
𝐿=
𝜇0 𝑑
(4𝜋 × 10−7 )(0.2)
=
= 8.4 × 10−8 H∕m
𝑏
3
Then, with 𝑅𝐶 = 𝐺𝐿,
𝐺=
𝜎 𝑏
(.98)(3.7 × 10−10 )
𝑅𝐶
= 4.4 × 10−3 S∕m = 𝑑
=
−8
𝐿
𝑑
(8.4 × 10 )
Thus 𝜎𝑑 = (4.4 × 10−3 )(0.2∕3) = 2.9 × 10−4 S∕m. The loss tangent is
𝑙.𝑡. =
𝜎𝑑
2.9 × 10−4
= 5.85 × 10−2
=
′
𝜔𝜖
(2 × 108 )(2.8)(8.85 × 10−12 )
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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 Ω: With the maximum dimension of 8mm, we have, using (24):
√
)
)
(
(
8 − 2𝑎
300𝜋
1 𝜇
−1 8 − 2𝑎
=
300
⇒
= 6.13
cosh
=
cosh
𝑍0 =
𝜋 𝜖′
2𝑎
2𝑎
120𝜋
Solve for 𝑎 to find 𝑎 = 0.56 mm. Then 𝑑 = 8 − 2𝑎 = 6.88 mm.
b) a planar line with 𝑍0 = 15 Ω: In this case our maximum dimension dictates that
So, using (8), we write
√ √
√
𝜇
64 − 𝑏2
15
𝑍0 =
64 − 𝑏2 =
=
15
⇒
𝑏
𝜖′
𝑏
377
√
𝑑 2 + 𝑏2 = 8.
Solving, we find 𝑏 = 7.99 mm and 𝑑 = 0.32 mm.
c) a 72 Ω 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
√
( ) 2𝜋(72)
𝜇 (𝑏)
𝑏
1
=
72
⇒
ln
=
ln
= 1.20 ⇒ 𝑎 = 𝑏𝑒−1.20 = 4𝑒−1.20 = 1.2
𝑍0 =
′
2𝜋 𝜖
𝑎
𝑎
120𝜋
Summarizing, 𝑎 = 1.2 mm and 𝑏 = 4 mm.
13.9. A microstrip line is to be constructed using a lossless dielectric for which 𝜖𝑟′ = 7.0. If the line is to
have a 50-Ω characteristic impedance, determine:
a) 𝜖𝑟,𝑒𝑓 𝑓 : We use Eq. (34) (under the assumption that 𝑤∕𝑑 > 1.3) to find:
]−1
[
= 5.0
𝜖𝑟,𝑒𝑓 𝑓 = 7.0 0.96 + 7.0(0.109 − 0.004 × 7.0) log10 (10 + 50) − 1
b) 𝑤∕𝑑: Still under the assumption that 𝑤∕𝑑 > 1.3, we solve Eq. (33) for 𝑑∕𝑤 to find
[
(
) ]1∕0.555
𝜖𝑟′ − 1
𝜖𝑟′ + 1 −1
𝑑
= (0.1)
𝜖𝑟,𝑒𝑓 𝑓 −
− 0.10
𝑤
2
2
[
]1∕0.555
𝑤
= 1.60
= (0.1) 3.0 (5.0 − 4.0)−1
− 0.10 = 0.624 ⇒
𝑑
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13.10. Two microstrip lines are fabricated end-to-end on a 2-mm thick wafer of lithium niobate (𝜖𝑟′ = 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:
𝜖𝑟1,𝑒𝑓 𝑓 =
[
( )]−0.555
5.8 3.8
1
1 + 10
= 3.60
+
2
2
2
For line 2:
[
( )]−0.555
1
5.8 3.8
+
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:
[
]
√
( )2
( )
1
1
𝑎𝑖𝑟
𝑍01
= 60 ln 4
+ 16
+ 2 = 89.6 ohms
2
2
𝜖𝑟2,𝑒𝑓 𝑓 =
and for line 2:
𝑎𝑖𝑟
𝑍02
⎡ ( )
1
+
= 60 ln ⎢4
⎢ 2.5
⎣
√
16
(
1
2.5
)2
⎤
+ 2⎥ = 79.1 ohms
⎥
⎦
The actual line impedances are given by Eq. (31). Using our results, we find
𝑍 𝑎𝑖𝑟
𝑍 𝑎𝑖𝑟
89.6
79.1
=√
=√
= 47.2 Ω and 𝑍02 = √ 02
= 41.2 Ω
𝑍01 = √ 01
𝜖𝑟1,𝑒𝑓 𝑓
𝜖
𝑟2,𝑒𝑓 𝑓
3.60
3.68
The reflection coefficient at the junction is now
Γ=
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 − |Γ|2 ) = −10 log10 (0.995) = 0.02 dB
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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 𝜆𝑐1 = 0.4 cm. The guide is operated at wavelength 𝜆 = 1 mm. How many modes propagate? The
cutoff wavelength for mode 𝑚 is 𝜆𝑐𝑚 = 2𝑛𝑑∕𝑚, where 𝑛 is the refractive index of the guide interior.
For the first mode, we are given
𝜆𝑐1 =
2𝑛𝑑
0.4 0.2
= 0.4 cm ⇒ 𝑑 =
=
cm
1
2𝑛
𝑛
Now, for mode 𝑚 to propagate, we require
𝜆≤
0.4
0.4 0.4
2𝑛𝑑
=
⇒ 𝑚≤
=
=4
𝑚
𝑚
𝜆
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 (𝜖𝑟′ = 2.1). Determine the
maximum allowable plate separation, 𝑑: We require that 𝑓 < 𝑓𝑐1 , which, using (41), becomes
𝑓<
𝑐
3 × 108
𝑐
= 3.45 cm
= √
⇒ 𝑑𝑚𝑎𝑥 =
2𝑛𝑑
2𝑛𝑓𝑚𝑎𝑥
2 2.1 (3 × 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 × 1010
=
= 1010 ⇒ 𝑛 = 3 or 𝜖𝑟 = 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 × 109 )(1.45)(.01)
𝑚𝑐
⇒ 𝑚<
=
= 3.09
2𝑛𝑑
𝑐
3 × 108
The maximum allowed 𝑚 in this case is thus 3, and the propagating modes will be TM1 , TE1 , TM2 ,
TE2 , TM3 , and TE3 .
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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
√
)
(
𝑓𝑐3 2
3(3 × 1010 )
𝑐
where 𝑓𝑐3 =
= 3.1 × 1010 = 31 GHz
1−
𝑣𝑔3 =
𝑛
𝑓
2(1.45)(1)
Thus
𝑣𝑔3
3 × 1010
=
1.45
√
(
1−
31
32
)2
= 5.13 × 109 cm∕s
For the TEM mode (assuming no material dispersion) 𝑣𝑔,𝑇 𝐸𝑀 = 𝑐∕𝑛 = 3 × 1010 ∕1.45 = 2.07 × 1010
cm/s. The group delay difference is now
)
(
(
)
1
1
1
1
−
= 10
Δ𝑡𝑔 = 𝑧
−
= 1.5 ns
𝑣𝑔3 𝑣𝑔,𝑇 𝐸𝑀
5.13 × 109 2.07 × 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 𝜆 = 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∕𝜆 = 3 × 108 ∕.015 = 20
GHz. Now, using (57),
√
√
(
)
( )2
𝑓𝑐2 2
15
𝑐
𝑐
1−
1−
𝑣𝑔2 =
= 2 × 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 𝜖𝑟1
′
𝜖𝑟2 = 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
√
2.1
−1
𝜃𝐵 = tan
= 35.9◦
4.0
The ray angle is thus 𝜃 = 90 − 35.9 = 54.1◦ . The cutoff frequency for the 𝑚 = 1 mode is
𝑓𝑐1 =
𝑐
3 × 1010
= 7.5 GHz
=
√
2(1)(2)
′
2𝑑 𝜖𝑟1
The frequency is thus 𝑓 = 𝑓𝑐1 ∕ cos 𝜃 = 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.
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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 𝜖𝑟2
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,
′ ∕𝜖 ′ )1∕2 . The maximum mode ray angle is then 𝜃
◦
𝜃𝑐 , where sin 𝜃𝑐 = (𝜖𝑟2
1 𝑚𝑎𝑥 = 90 − 𝜃𝑐 . Now,
𝑟1
using (39), we write
(
)
(
)
)
(
𝑐
𝜋
𝜋𝑐
−1
◦
−1
−1
90 − 𝜃𝑐 = cos
=
cos
= cos
√
𝑘𝑚𝑎𝑥 𝑑
4𝑑𝑓𝑚𝑎𝑥
2𝜋𝑓𝑚𝑎𝑥 𝑑 4
√
Now
cos(90 − 𝜃𝑐 ) = sin 𝜃𝑐 =
′
𝜖𝑟2
′
𝜖𝑟1
=
𝑐
4𝑑𝑓𝑚𝑎𝑥
√
√
Therefore 𝑓𝑚𝑎𝑥 = 𝑐∕(2 2.1𝑑) = (3 × 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):
√
( ) 2 ( 𝑝 )2
𝑚
𝑐
+
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 =
3 × 108
2.5 × 109
𝑐
=
=
2𝑛𝑎
2𝑛(.06)
𝑛
𝑓𝑐01 =
𝑐
3 × 108
3.75 × 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
√
( )2 ( ) 2
30𝑐
1
4.5 × 109
𝑐
1
+
=
=
𝑓𝑐11 =
2𝑛
.06
.04
2𝑛
𝑛
The allowed frequency range is then
3.75
4.5
GHz < 𝑓 <
GHz
𝑛
𝑛
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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 𝜖𝑟′ .
a) Determine the maximum allowable value of 𝜖𝑟′ 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):
√
)
( 𝑝𝑐 )2
(
𝑚𝑐 2
𝑓𝑐𝑚𝑝 =
+
4𝑛𝑏
2𝑛𝑏
√
where 𝑛 = 1 in guide 1 and 𝑛 = 𝜖𝑟′ 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 𝜖𝑟′ ). For this range to be viable, it is required that 𝜖𝑟′ < 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 𝜖𝑟′ , guide dimensions as needed, and other
known constants: This was already found in part 𝑎:
𝑐
𝑐
<𝑓 < √
2𝑎
𝜖𝑟′ 𝑎
where 𝜖𝑟′ < 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
√
( )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
𝑐
3 × 1010
3 × 1010
= 1.1 cm and 𝑏 =
= 0.90 cm
=
=
9
2𝑓𝑐10
2𝑓𝑐01
2(13.6 × 10 )
2(16.7 × 109 )
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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
[
]
𝑏
𝐶
−1
𝑦 = cos
(0 < 𝑥 < 𝑎, 0 < 𝑦 < 𝑏)
𝜋
cos(𝜋𝑥∕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 ⇒ ln cos(𝑘𝑚 𝑥) = − ln cos(𝑘𝑝 𝑦) + ln(𝐶)
Combining terms:
[
ln cos(𝑘𝑚 𝑥) = − ln
cos(𝑘𝑝 𝑦)
𝐶
]
[
𝐶
= + ln
cos(𝑘𝑝 𝑦)
]
⇒ cos(𝑘𝑚 𝑥) =
𝐶
cos(𝑘𝑝 𝑦)
Then, for TE11 , we have 𝑘𝑝 = 𝜋∕𝑏 and 𝑘𝑚 = 𝜋∕𝑎. Incorporating these leads to
[
[
]
]
1
𝐶
𝑏
𝐶
−1
−1
𝑦=
cos
= cos
Q.E.D
𝑘𝑝
cos(𝑘𝑚 𝑥)
𝜋
cos(𝜋𝑥∕𝑎)
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13.23
a) Using the relation 𝑃𝑎𝑣 = (1∕2)Re{𝐄𝑠 × 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 =
𝛽10 2 2
𝐸 sin (𝜅10 𝑥) 𝐚𝑧
2𝜔𝜇 0
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
}
𝛽
1 {
𝑃𝑎𝑣 = Re 𝐄𝑦𝑠 × 𝐇∗𝑥𝑠 = 10 𝐸02 sin2 (𝜅10 𝑥) 𝐚𝑧
2
2𝜔𝜇
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 =
𝛽10 𝑎𝑏 2 𝑎𝑏 2
𝐸 = 𝐸 sin 𝜃10 W
4𝜔𝜇 0 4𝜂 0
√
where 𝜂 = 𝜇∕𝜖, and 𝜃10 is the wave angle associated with the TE10 mode. Interpret. First,
the integration:
𝑏
𝑎
𝛽10 2 2
𝛽 𝑎𝑏
𝐸0 sin (𝜅10 𝑥) 𝐚𝑧 ⋅ 𝐚𝑧 𝑑𝑥 𝑑𝑦 = 10 𝐸02
∫0 ∫0 2𝜔𝜇
4𝜔𝜇
√
Next, from (54), we have 𝛽10 = 𝜔 𝜇𝜖 sin 𝜃10 , which, on substitution, leads to
𝑃 =
𝑎𝑏
𝑃 = 𝐸02 sin 𝜃10 W with 𝜂 =
4𝜂
√
𝜇
𝜖
The sin 𝜃10 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 𝛽∕𝑑𝜔2 , for given mode in a parallel-plate or rectangular
waveguide is given by
( ) [
( 𝜔 )2 ]−3∕2
𝑑 2𝛽
𝑛 𝜔𝑐 2
𝑐
=−
1−
𝜔𝑐 𝜔
𝜔
𝑑𝜔2
where 𝜔𝑐 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
[
( 𝜔 )2 ]−1∕2
𝑑𝛽
𝑛
𝑐
=
1−
𝑑𝜔 𝑐
𝜔
Taking the derivative of this equation with respect to 𝜔 leads to
(
)
( )[
( ) [
( 𝜔 )2 ]−3∕2 2𝜔2
( 𝜔 )2 ]−3∕2
𝑑 2𝛽
1
𝑛
𝑛 𝜔𝑐 2
𝑐
𝑐
𝑐
−
1
−
1
−
=
=
−
2
3
𝑐
2
𝜔
𝜔𝑐
𝜔
𝜔
𝑑𝜔
𝜔
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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 + (Δ𝜏)2 , where Δ𝜏 = 𝛽2 𝐿∕𝑇 for a
transform limited pulse (assumed gaussian). 𝛽2 is the Problem 13.24 result evaluated at the operating
frequency, or
( )2 ]−3∕2
( )2 [
𝑑 2𝛽
1
1
1
𝛽2 =
1−
|𝜔=10 GHz = −
2
10
8
1.1
𝑑𝜔
(2𝜋 × 10 )(3 × 10 ) 1.1
= 6.1 × 10−19 s2 ∕m = 0.61 ns2 ∕m
Now Δ𝜏 = 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 ⇒ 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 𝛽2 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 𝜇m, with 𝑛1 = 1.48 and
𝑛2 = 1.45. If the operating wavelength √
is 𝜆 = 1.3 𝜇m, what modes will propagate? We use the
condition expressed through (141): 𝑘0 𝑑 𝑛21 − 𝑛22 ≥ (𝑚 − 1)𝜋. Since 𝑘0 = 2𝜋∕𝜆, the condition
becomes
√
2(10) √
2𝑑
𝑛21 − 𝑛22 ≥ (𝑚 − 1) ⇒
(1.48)2 − (1.45)2 = 4.56 ≥ 𝑚 − 1
𝜆
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 𝜆 = 1.55 𝜇m. If the slab thickness is 5 𝜇m, what is the maximum value of 𝑛1 if 𝑛2 = 3.30?
Using (142) we have
√
√
(
)2
√
( )2
1.55
𝜆
2𝜋𝑑
2
2
2
𝑛1 − 𝑛 2 < 𝜋 ⇒ 𝑛1 <
+ 𝑛2 =
+ (3.30)2 = 3.304
𝜆
2𝑑
2(5)
13.28. In a symmetric slab waveguide, 𝑛1 = 1.50, 𝑛2 = 1.45, and 𝑑 = 10 𝜇m.
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 × 108 ∕1.45 = 2.07 × 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 × 108 m∕s.
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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,
𝜃1 , 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 𝜃𝑚𝑖𝑛 = 𝜃𝑐,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 = 𝜔∕𝛽𝑚𝑖𝑛 , where 𝛽𝑚𝑖𝑛 = 𝑛1 𝑘0 sin 𝜃1,𝑚𝑖𝑛 =
𝑛1 𝑘0 𝑛3 ∕𝑛1 = 𝑛3 𝑘0 . Thus 𝑣𝑝,𝑚𝑎𝑥 = 𝜔∕(𝑛3 𝑘0 ) = 𝑐∕𝑛3 .
13.30. A step index optical fiber is known to be single mode at wavelengths 𝜆 > 1.2 𝜇m. Another fiber is
to be fabricated from the same materials, but is to be single mode at wavelengths 𝜆 > 0.63 𝜇m. 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):
√
2𝜋𝑎
𝑛21 − 𝑛22
𝜆 > 𝜆𝑐 =
2.405
With 𝜆 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
× 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
𝜌0
2.879
1.619
+
= 1.10
= 0.65 +
3∕2
𝑎
(2.405)
(2.405)6
Therefore, 𝜌0 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 𝜇m at free space wavelength 𝜆 = 1.30
𝜇m. If the cutoff wavelength is specified as 𝜆𝑐 = 1.20 𝜇m, find the expected mode field radius at
𝜆 = 1.55 𝜇m.
In this problem it is helpful to use the relation 𝑉 = 2.405(𝜆𝑐 ∕𝜆), and rewrite Eq. (164) to read:
( )6
( )3∕2
𝜌0
𝜆
𝜆
+ 0.015
= 0.65 + 0.434
𝑎
𝜆𝑐
𝜆𝑐
At 𝜆 = 1.30 𝜇m, 𝜆∕𝜆𝑐 = 1.08, and at 1.55 𝜇m, 𝜆∕𝜆𝑐 = 1.29. Using these values, along with
our new equation, we write
[
]
0.65 + 0.434(1.29)3∕2 + 0.015(1.29)6
𝜌0 (1.55) = 4.5
= 5.3 𝜇m
0.65 + 0.434(1.08)3∕2 + 0.015(1.08)6
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Sm ch (14) Engineering Electromagnetics 9th solution
Engineering Electromagnetics (Chungnam National University)
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CHAPTER 14 – 9th Edition
14.1. A short dipole carrying current 𝐼0 cos 𝜔𝑡 in the 𝐚𝑧 direction is located at the origin in free space.
a) If 𝑘 = 1 rad/m, 𝑟 = 2 m, 𝜃 = 45◦ , 𝜙 = 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):
)
(
𝐼0 𝑑
1
1
−𝑗𝑘𝑟
𝐸𝑟𝑠 =
+
(13𝑎)
𝜂 cos 𝜃 𝑒
2𝜋
𝑟2 𝑗𝑘𝑟3
)
(
𝐼0 𝑑
1
1
−𝑗𝑘𝑟 𝑗𝑘
𝐸𝜃𝑠 =
(13𝑏)
𝜂 sin 𝜃 𝑒
+ 2+
4𝜋
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 𝜃 = 45 , sin 𝜃 = cos 𝜃 = 1∕ 2. Also, with
𝑘 = 1 = 2𝜋∕𝜆, it follows that 𝜆 = 2𝜋 m. The above two equations can be simplified by these
substitutions, while dropping all amplitude terms that are common to both. Obtain
(
)
1
1
𝐴𝑟 =
+
𝑒−𝑗𝑟
𝑟2 𝑗𝑟3
)
(
1
1
1
1
𝐴𝜃 =
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
𝑒−𝑗2 = (0.90)𝑒𝑗56.3 𝑒−𝑗2
𝐴𝜃 = 𝑗 + − 𝑗
4 8
16
4
The total vector is now 𝐀 = 𝐴𝑟 𝐚𝑟 + 𝐴𝜃 𝐚𝜃 . We can normalize the vector by first finding the
magnitude:
√
1√
(1.12)2 + (0.90)2 = 0.359
|𝐀| = 𝐀 ⋅ 𝐀∗ =
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 𝐚𝜃
𝐴𝑟 =
In real instantaneous form, this becomes
(
)
𝐀𝑁 (𝑡) = Re 𝐀𝑁𝑠 𝑒𝑗𝜔𝑡 = 0.780 cos(𝜔𝑡 − 141.2◦ )𝐚𝑟 + 0.627 cos(𝜔𝑡 − 58.3◦ )𝐚𝜃
We evaluate this at 𝑡 = 0 to find
𝐀𝑁 (0) = 0.780 cos(141.2◦ )𝐚𝑟 + 0.627 cos(58.3◦ )𝐚𝜃 = −0.608𝐚𝑟 + 0.330𝐚𝜃
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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 + 0.477𝐚𝜃 . We next convert this to rectangular components:
1
𝑎𝑁𝑥 = 𝐚𝑁 (0) ⋅ 𝐚𝑥 = −0.879 sin 𝜃 cos 𝜙 + 0.477 cos 𝜃 cos 𝜙 = √ (−0.879 + 0.477) = −0.284
2
𝑎𝑁𝑦 = 𝐚𝑁 (0) ⋅ 𝐚𝑦 = −0.879 sin 𝜃 sin 𝜙 + 0.477 cos 𝜃 sin 𝜙 = 0 since 𝜙 = 0
1
𝑎𝑁𝑧 = 𝐚𝑁 (0) ⋅ 𝐚𝑧 = −0.879 cos 𝜃 − 0.477 sin 𝜃 = √ (−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◦ < 𝜃 < 100◦ ? We use the
far-zone phasor fields, (22) and (23), with 𝑘 = 2𝜋∕𝜆, and first find the average power density:
𝐼 2𝑑 2𝜂
1
∗
𝑃𝑎𝑣𝑔 = Re[𝐸𝜃𝑠 𝐻𝜙𝑠
] = 0 2 2 sin2 𝜃 W∕m2
2
8𝜆 𝑟
We integrate this over the given belt, an at radius 𝑟:
𝑃𝑏𝑒𝑙𝑡 =
∫0
2𝜋
100◦
∫80◦
𝐼02 𝑑 2 𝜂
8𝜆2 𝑟2
sin2 𝜃 𝑟2 sin 𝜃 𝑑𝜃 𝑑𝜙 =
𝜋𝐼02 𝑑 2 𝜂
4𝜆2
100◦
∫80◦
sin3 𝜃 𝑑𝜃
Evaluating the integral, we find
𝑃𝑏𝑒𝑙𝑡 =
𝜋𝐼02 𝑑 2 𝜂
𝜋𝐼02 𝑑 2 𝜂 [ 1
( 2
)]100
−
=
(0.344)
cos
𝜃
sin
𝜃
+
2
3
80
4𝜆2
4𝜆2
The total power is found by performing the same integral over 𝜃, where 0 < 𝜃 < 180◦ . Doing
this, it is found that
𝜋𝐼 2 𝑑 2 𝜂
𝑃𝑡𝑜𝑡 = (1.333) 0 2
4𝜆
The fraction of the total power in the belt is then 𝑓 = 0.344∕1.333 = 0.258.
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14.2. For the Hertzian dipole, prepare a curve, 𝑟 vs. 𝜃 in polar coordinates, showing the locus in the 𝜙 = 0
plane where:
a) The radiation field |𝐸𝜃𝑠 | is one-half of its value at 𝑟 = 104 m, 𝜃 = 𝜋∕2: Assuming the far field
approximation, we use (22) with 𝑘 = 2𝜋∕𝜆 to set up the equation:
|𝐸𝜃𝑠 | =
𝐼0 𝑑𝜂
𝐼0 𝑑𝜂
1
sin 𝜃 = ×
⇒ 𝑟 = 2 × 104 sin 𝜃
2𝜆𝑟
2 2 × 104 𝜆
b) The average radiated power density, 𝑃𝑟,𝑎𝑣 , is one-half of its value at 𝑟 = 104 m, 𝜃 = 𝜋∕2. To
find the average power, we use (22) and (23) in
2 2
2 2
√
1
1 1 𝐼0 𝑑 𝜂
1 𝐼0 𝑑 𝜂 2
∗
𝑃𝑟,𝑎𝑣 = Re{𝐸𝜃𝑠 𝐻𝜙𝑠
sin
𝜃
=
⇒
𝑟
=
2 × 104 sin 𝜃
×
}=
2
2 4𝜆2 𝑟2
2 2 4𝜆2 (108 )
√
The polar plots for field (𝑟 = 2 × 104 sin 𝜃) and power (𝑟 = 2 × 104 sin 𝜃) are shown below.
Both are circles.
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14.3. Two short antennas at the origin in free space carry identical currents of 5 cos 𝜔𝑡 A, one in the 𝐚𝑧
direction, one in the 𝐚𝑦 direction. Let 𝜆 = 2𝜋 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 𝜃 = 90◦ , only the
𝐸𝜃𝑠 component will be present (as (136) and (137) show). Since we are in the far zone, (138)
applies. We use 𝜃 = 90◦ , 𝑑 = 0.1, 𝜆 = 2𝜋, 𝜂 = 𝜂0 = 120𝜋, and 𝑟 = 1000 to write:
𝐼0 𝑑𝜂
5(0.1)(120𝜋) −𝑗1000
sin 𝜃𝑒−𝑗2𝜋𝑟∕𝜆 𝐚𝜃 = 𝑗
𝑒
𝐚𝜃
2𝜆𝑟
4𝜋(1000)
= 𝑗(1.5 × 10−2 )𝑒−𝑗1000 𝐚𝜃 = −𝑗(1.5 × 10−2 )𝑒−𝑗1000 𝐚𝑧 V∕m
𝐄𝑠 = 𝐸𝜃𝑠 𝐚𝜃 = 𝑗
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 × 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 × 10−2 )(𝐚𝑦 + 𝐚𝑧 ).
d) Find 𝐄 at (1000, 0, 0) at 𝑡 = 0: This is found through
(
)
𝐄(𝑡) = Re 𝐄𝑠 𝑒𝑗𝜔𝑡 = (1.5 × 10−2 ) sin(𝜔𝑡 − 1000)(𝐚𝑦 + 𝐚𝑧 )
Evaluating at 𝑡 = 0, we find
𝐄(0) = (1.5 × 10−2 )[− sin(1000)](𝐚𝑦 + 𝐚𝑧 ) = −(1.24 × 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 × 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, 𝛿𝑟 and 𝛿𝜃 , 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
.
.
𝜋
𝜋
𝜋
= −
and 𝛿𝜃 = tan−1 (−∞) = −
𝛿𝑟 = tan−1 (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 =
.
𝐸𝜃𝑠 =
𝐼0 𝑑
2𝜋𝑟2 (𝑘𝑟)
𝜂 cos 𝜃 exp(−𝑗𝜋∕2) = −𝑗
𝐼0 𝑘𝑑
4𝜋𝑟(𝑘𝑟)2
where, in free space
𝜂 sin 𝜃 exp(−𝑗𝜋∕2) = −𝑗
𝐼0 𝜂𝑑
2𝜋𝑘𝑟3
𝐼0 𝜂𝑑
4𝜋𝑘𝑟3
cos 𝜃
sin 𝜃
√
𝜇0 ∕𝜖0
𝜂
1
=
= √
𝑘 𝜔 𝜇0 𝜖0
𝜔𝜖0
The field vector can now be constructed as
𝐄𝑠 =
]
(−𝑗𝐼0 ∕𝜔)𝑑 [
2 cos 𝜃 𝐚𝑟 + sin 𝜃 𝐚𝜃
3
4𝜋𝜖0 𝑟
This expression is equivalent to the static dipole field (Eq. (35) in Chapter 4), provided that
𝐼0 = 𝑗𝜔𝑄, which is just a statement that the current is the time derivative of the time-harmonic
charge.
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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 − 𝛿𝜙 )] = 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
sin 𝜃 exp(𝑗0) = 0 2 sin 𝜃
𝐻𝜙 = 0
4𝜋𝑟 (𝑘𝑟)
4𝜋𝑟
Now, using the Biot-Savart law for a short element of assumed differential length, 𝑑, we have
𝐇=
𝐼0 𝑑𝐋 × 𝐚𝑅
4𝜋𝑅2
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 × 𝐚𝑟 )
4𝜋𝑟2
=
𝐼0 𝑑
4𝜋𝑟2
sin 𝜃 𝐚𝜙
(done)
{
}
14.6. Evaluate the time-average Poynting vector, < 𝐒 >= (1∕2)𝑒 𝐄𝑠 × 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 >= 𝑒 𝐸𝜃𝑠 𝐻𝜙𝑠
𝐚𝑟 − 𝐸𝑟𝑠 𝐻𝜙𝑠
𝐚𝜃
2
Substituting (10), (13a) and (13b), this becomes:
{(
]
[ 2
)
𝐼0 𝑑 2
𝑘
1
𝑘
1
𝑘
𝑘
1
𝜂 sin2 𝜃 2 − 𝑗 3 − 4 + 𝑗 3 + 4 − 𝑗 5 𝐚𝑟
< 𝐒 > = 𝑒
2
4𝜋
𝑟
𝑟
𝑟
𝑟
𝑟
𝑟
}
]
[
(𝐼 𝑑)2
𝑘
1
1
𝑘
− 0 2 sin 𝜃 cos 𝜃 −𝑗 3 − 4 + 4 − 𝑗 5 𝐚𝜃
8𝜋
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
(
𝐼0 𝑑 𝑘
4𝜋𝑟
)2
𝜂 sin𝜃 𝐚𝑟 W∕m2
which we see to be identical to Eq. (26).
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14.7. A short current element has 𝑑 = 0.03𝜆. Calculate the radiation resistance for each of the following
current distributions:
a) Uniform: In this case, (30) applies directly and we find
𝑅𝑟𝑎𝑑 = 80𝜋 2
( )2
𝑑
= 80𝜋 2 (.03)2 = 0.711 Ω
𝜆
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 Ω.
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 Ω.
{
}
14.8. Evaluate the time-average Poynting vector, < 𝐒 >= (1∕2)𝑒 𝐄𝑠 × 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:
𝜔𝜇 𝜋𝑎2 𝐼0 𝑘
.
sin 𝜃 exp[−𝑗(𝑘𝑟 − 𝛿𝜙 )]
𝐸𝜙𝑠 = −𝑗 0
4𝜋𝑟
. 𝜔𝜇 𝜋𝑎2 𝐼0 𝑘 1
𝐻𝜃𝑠 = 𝑗 0
sin 𝜃 exp[−𝑗(𝑘𝑟 − 𝛿𝜃 )]
4𝜋𝑟
𝜂
.
where, in the approximation, 𝛿𝜙 = 𝛿𝜃 = tan−1 (𝑘𝑟). So now
{
}
{
}
1
∗
< 𝐒 >= (1∕2)𝑒 𝐄𝜙𝑠 × 𝐇∗𝜃𝑠 = − 𝑒 𝐸𝜙𝑠 𝐻𝜃𝑠
𝐚𝑟
2
Using the above field expressions, we find
| < 𝐒 > | = 𝑆𝑟(𝑚𝑑)
1
=
2
(
𝐼0 𝑘
4𝜋𝑟
)2
[𝜔𝜇0 𝜋𝑎2 ]2
1 2
sin 𝜃
𝜂
which we now compare to the power density from the Hertzian dipole, Eq. (26):
(
)2
1 𝐼0 𝑘
𝑑 2 𝜂 sin2 𝜃
𝑆𝑟(𝐻𝑑) =
2 4𝜋𝑟
For equal power densities from the two structures, we thus require (assuming free space):
( )
√
𝜔𝜇0
1
2𝜋
[𝜔𝜇0 𝜋𝑎2 ]2 = 𝑑 2 𝜂 ⇒ 𝑑 = 𝜋𝑎2
= 𝜋𝑎2 (𝜔 𝜇0 𝜖0 ) = 𝜋𝑎2
𝜂
𝜂
𝜆
or
𝑑=
(2𝜋𝑎)2
2𝜆
(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 0.02𝜆, what value of
𝐼0 is required to:
a) provide a radiation-field amplitude of 100 mV/m at a distance of one mile, at 𝜃 = 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 𝑑𝜂
𝐼0 (0.02)(120𝜋)
|𝐸𝜃 | = 0 0 sin(90◦ ) =
= 0.1 ⇒ 𝐼0 = 85.4 A
4𝜆𝑟
(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 = 10𝜋 2 𝐼02 (0.02)2 = 1 ⇒ 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 𝜃, 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 − 𝜃) = sin 𝜃 ⇒ 𝑐 = 𝑏 sin 𝜃 (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 0.01𝜆, what value of 𝐼0 is required to
a) provide a radiation field amplitude of 100 mV/m at a distance of 1 mi, at 𝜃 = 90◦ : The image
antenna below the plane provides a radiation pattern that is identical to a dipole antenna of
length 0.02𝜆. The radiation field is thus given by (22) in free space, where 𝑘 = 2𝜋∕𝜆, 𝜃 = 90◦ ,
and with an additional factor of 1∕2 included to account for the linear current distribution:
|𝐸𝜃 | =
4𝑟|𝐸𝜃 |
4(5289)(12 × .0254)(100 × 10−3 )
1 𝐼0 𝑑𝜂0
=
⇒ 𝐼0 =
= 85.4 A
2 2𝜆𝑟
(𝑑∕𝜆)𝜂0
(.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 = 10𝜋 2 (𝑑∕𝜆)2 . The current magnitude is now
√
]
[
[
]1∕2
2𝑃𝑎𝑣 1∕2
2
2(1)
=
𝐼0 =
= 7.1 A
=√
2
2
𝑅𝑟𝑎𝑑
10𝜋 (𝑑∕𝜆)
10 𝜋(.02)
14.12. Find the zeros in 𝜃 for the E-plane pattern of a dipole antenna for which (using Fig. 14.8 as a guide):
a) 𝓁 = 𝜆: We look for zeros in the pattern function, Eq. (59), for which 𝑘𝓁 = (2𝜋∕𝜆)𝜆 = 2𝜋.
] [
]
[
cos(2𝜋 cos 𝜃) − cos(2𝜋)
cos(𝑘𝓁 cos 𝜃) − cos(𝑘𝓁)
=
𝐹 (𝜃) =
sin 𝜃
sin 𝜃
Zeros will occur whenever
cos(2𝜋 cos 𝜃) = 1 ⇒ 𝜃 = (0, 90◦ , 180◦ , 270◦ )
You are encouraged to show (using small argument approximations for cosine and sine) that
𝐹 (𝜃) does approach zero when 𝜃 approaches zero or 180◦ (despite the sin 𝜃 term in the
denominator).
b) 2𝓁 = 1.3𝜆: In this case 𝑘𝓁 = (2𝜋∕𝜆)(1.3𝜆∕2) = 1.3𝜋. The pattern function becomes
[
]
cos(1.3𝜋 cos 𝜃) − cos(1.3𝜋)
𝐹 (𝜃) =
sin 𝜃
Zeros of this function will occur at 𝜃 = (0, 57.42◦ , 122.58◦ , 180◦ ).
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14.13. The radiation field of a certain short vertical current element is 𝐸𝜃𝑠 = (20∕𝑟) sin 𝜃 𝑒−𝑗10𝜋𝑟 V/m if it
is located at the origin in free space.
a) Find 𝐸𝜃𝑠 at 𝑃 (𝑟 = 100, 𝜃 = 90◦ , 𝜙 = 30◦ ): Substituting these values into the given formula,
find
20
sin(90◦ )𝑒−𝑗10𝜋(100) = 0.2𝑒−𝑗1000𝜋 V∕m
𝐸𝜃𝑠 =
100
b) Find 𝐸𝜃𝑠 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:
𝐸𝜃𝑠 = 0.2𝑒−𝑗1000𝜋 𝑒𝑗10𝜋(0.1 sin 30 = 0.2𝑒−𝑗1000𝜋 𝑒𝑗0.5𝜋 V∕m
c) Find 𝐸𝜃𝑠 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:
(
)
𝐸𝜃𝑠 = 0.2𝑒−𝑗1000𝜋 𝑒𝑗0.5𝜋 + 𝑒−𝑗0.5𝜋 = 0.2𝑒−𝑗1000𝜋 2 cos(0.5𝜋) = 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 = 𝜆, evaluate the maximum directivity in dB, and the
half-power beamwidth.
𝐷𝑚𝑎𝑥 is found using Eq. (64) with 𝑘𝓁 = 𝜋, 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 𝜃 = 90◦ .
With 𝑘𝓁 = 𝜋, the pattern function, Eq. (59), reaches a maximum value of 2 at 𝜃 √
= 90◦ . The
half-power beamwidth is found numerically by setting the pattern function equal to 2 and thus
solving
[
] √
cos(𝜋 cos 𝜃) − cos(𝜋)
𝐹 (𝜃) =
= 2
sin 𝜃
The result is 𝜃 = 66.1◦ , which means that the deviation angle from the direction along the
maximum is 𝜃1∕2 ∕2 = 90 − 66.1 = 23.9◦ . The half-power beamwidth is therefore 𝜃1∕2 =
2 × 23.9 = 47.8◦ .
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14.15. For a dipole antenna of overall length 2𝓁 = 1.3𝜆, determine the locations in 𝜃 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 = (2𝜋∕𝜆)(1.3𝜆∕2) = 1.3𝜋. The angular intensity distribution is given by the square
of the pattern function, Eq. (59), with 𝑘𝓁 = 1.3𝜋 substituted:
[
cos(1.3𝜋 cos 𝜃) − cos(1.3𝜋)
| < 𝐒 > | ∝ [𝐹 (𝜃)] =
sin 𝜃
]2
2
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 𝜃 = 90◦ , and has
value [𝐹 (𝜃)]2 = 2.52. The secondary maxima occur at 𝜃 = 33.8◦ and 146.2◦ , and both peaks
have values [𝐹 (𝜃)]2 = 0.47. The sidelobe level in decibels is thus
(
)
2.52
𝑆𝑠 [𝑑𝐵] = 10 log10
= 7.3 dB
0.47
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14.16. For a dipole antenna of overall length, 2𝓁 = 1.5𝜆:
a) Evaluate the locations in 𝜃 at which the zeros and maxima in the E-plane pattern occur: In this
case, 𝑘𝓁 = (2𝜋∕𝜆)(1.5𝜆∕2) = 1.5𝜋. The angular intensity distribution is given by the square of
the pattern function, Eq. (59), with 𝑘𝓁 = 1.3𝜋 substituted:
[
cos(1.5𝜋 cos 𝜃) − cos(1.5𝜋)
| < 𝐒 > | ∝ [𝐹 (𝜃)] =
sin 𝜃
2
]2
[
cos(1.5𝜋 cos 𝜃)
=
sin 𝜃
]2
Zeros occur whenever cos(1.5𝜋 cos 𝜃) = 0, which in turn occurs whenever 1.5𝜋 cos 𝜃 is an odd
multiple of 𝜋∕2 (positive or negative). All zeros over the range 0 ≤ 𝜃 ≤ 𝜋 are found with just
the first two odd multiples, or
1
cos 𝜃𝑧 = ± , ±1 ⇒ 𝜃𝑧 = (0, 70.5◦ , 109.5◦ , 180◦ )
3
The function does indeed zero at 𝜃 = 0 and 180◦ , despite the presence of the sin 𝜃 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 𝜃𝑚𝑎𝑥,𝑚 = (42.6◦ , 137.4◦ ) (main lobes) and 𝜃𝑚𝑎𝑥,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
(𝜃 = 42.6◦ ), [𝐹 (𝜃)]2 evaluates as 1.96. The sidelobe at 𝜃 = 90◦ gives a value of 1.00. The
sidelobe level in decibles is thus
)
(
1.96
= 2.9 dB
𝑆𝑠 [𝑑𝐵] = 10 log10
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.
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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 = 4𝜋𝐾𝑚𝑎𝑥 ∕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 × 𝐴𝑟𝑒𝑎 =
𝐷𝑚𝑎𝑥 𝑃𝑟
4𝜋𝑟2
× 𝐴𝑟𝑒𝑎 =
1.64(36.5) 2
(1) = 4.77 𝜇W
4𝜋(103 )2
14.18. Repeat Problem 14.17, but with a full-wave antenna (2𝓁 = 𝜆). Numerical integrals may be necessary.
In this case, 𝑘𝓁 = 𝜋, and the radiation resistance is found by applying Eq. (65):
𝑅𝑟𝑎𝑑 = 60
∫0
𝜋
[
cos(𝜋 cos 𝜃) − cos(𝜋)
sin 𝜃
]2
sin 𝜃 𝑑𝜃 = 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 = 𝜋), and the result was found to be 𝐷𝑚𝑎𝑥 = 2.41. The received power is now
found as in Problem 14.17:
𝑃𝑟𝑒𝑐 =
𝐷𝑚𝑎𝑥 𝑃𝑟
4𝜋𝑟2
× 𝐴𝑟𝑒𝑎 =
2.41(99.5) 2
(1) = 19.1 𝜇W
4𝜋(103 )2
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14.19. Design a two-element dipole array that will radiate equal intensities in the 𝜙 = 0, 𝜋∕2, 𝜋, and 3𝜋∕2
directions in the H-plane. Specify the smallest relative current phasing, 𝜉, and the smallest element
spacing, 𝑑.
The array function is given by Eq. (81), with 𝑛 = 2:
|𝐴2 (𝜓)| =
1 || sin(𝜓) ||
2 || sin(𝜓∕2) ||
This has periodic maxima occurring at 𝜓 = 0, ±2𝑚𝜋, and in the H-plane, 𝜓 = 𝜉 + 𝑘𝑑 cos 𝜙.
Now, for broadside operation, we have maxima at 𝜙 = ±𝜋∕2, at which 𝜓 = 𝜉, so we set
𝜉 = 0 to get the array function principal maximum at 𝜓 = 0. Having done this, we now have
𝜓 = 𝑘𝑑 cos 𝜙, which we need to have equal to ±2𝜋 when 𝜙 = 0, 𝜋, in order to get endfire
operation. This will happen when 𝑘𝑑 = 2𝜋𝑑∕𝜆 = 2𝜋, so we set 𝑑 = 𝜆.
14.20. A two-element dipole array is configured to provide zero radiation in the broadside (𝜙 = ±90◦ ) and
endfire (𝜙 = 0, 180◦ ) directions, but with maxima occurring at angles in between. Consider such a
set-up with 𝜓 = 𝜋 at 𝜙 = 0 and 𝜓 = −3𝜋 at 𝜙 = 𝜋, 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 (𝜓)| =
1 || sin(𝜓) ||
2 || sin(𝜓∕2) ||
This is clearly zero at 𝜓 = 𝜋 and at 𝜓 = −3𝜋.
b) Determine the required relative current phase, 𝜉: We know that 𝜓 = 𝜉 +𝑘𝑑 cos 𝜙 in the H-plane.
Setting up the two given conditions, we have:
𝜋 = 𝜉 + 𝑘𝑑 cos(0) = 𝜉 + 𝑘𝑑
and
−3𝜋 = 𝜉 + 𝑘𝑑 cos(𝜋) = 𝜉 − 𝑘𝑑
Adding these two equations gives 𝜉 = −𝜋, 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:
𝜓(𝜙 = 0) = 𝜋 = −𝜋 + 𝑘𝑑 cos(0) = −𝜋 + 𝑘𝑑
Therefore, 𝑘𝑑 = 2𝜋∕𝜆 = 2𝜋, or 𝑑 = 𝜆.
d) Determine the values of 𝜙 at which maxima in the radiation pattern occur: With the values as
found, the array function now becomes
|𝐴2 (𝜙)| =
1 || sin[𝜋(2 cos 𝜙 − 1)] ||
|
|
2 || sin[ 𝜋 (2 cos 𝜙 − 1)] ||
2
This function maximizes at 𝜙 =
60◦ , 120◦ , 240◦ ,
and 300◦ , as found numerically.
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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, 𝜉 = −𝜋∕2. Determine the values of 𝜙 at which the maxima occur when the frequency is
changed to
a) 𝑓 = 1.5𝑓0 : The original phase and length from the example are 𝜉 = −𝜋∕2 and 𝑑 = 𝜆0 ∕4,
where 𝜆0 = 𝑐∕𝑓0 . Thus 𝑘𝑑 = (2𝜋∕𝜆0 )(𝜆0 ∕4) = 𝜋∕2. The H-plane array function in that case
was found to be
[
]
[𝜓 ]
[
]
𝜉 𝑘𝑑
𝜋 𝜋
|
𝐴(𝜋∕2, 𝜙)| = cos
= cos
+
cos 𝜙 = cos − + cos 𝜙
|𝜆0
2
2
2
4 4
Now, if we change the frequency to 1.5𝑓0 , while leaving 𝜉 fixed (and the antenna physical length
is unchanged), we obtain 𝜆 = 𝜆0 ∕1.5, and 𝑘𝑑 = (2𝜋∕𝜆0 )(1.5)(𝜆0 ∕4) = 3𝜋∕4. The new array
function is then:
[
]
[
]
[
]
𝜉 𝑘𝑑
𝜋 3𝜋
𝜋
|
𝐴(𝜋∕2, 𝜙)|
= cos
+
cos 𝜙 = cos − +
cos 𝜙 = cos (3 cos 𝜙 − 2)
|𝜆0 ∕1.5
2
2
4
8
8
This function amplitude maximizes when 3 cos 𝜙 − 2 = 0, or 𝜙 = cos−1 (2∕3) = ±48.2◦ .
b) 𝑓 = 2𝑓0 : Now, 𝜆 = 𝜆0 ∕2 and 𝑘𝑑 = (2𝜋∕𝜆0 )(2)(𝜆0 ∕4) = 𝜋. The array function becomes
]
[
[
]
[
]
𝜉 𝑘𝑑
𝜋 𝜋
𝜋
|
+
cos 𝜙 = cos − + cos 𝜙 = cos (2 cos 𝜙 − 1)
= cos
𝐴(𝜋∕2, 𝜙)|
|𝜆0 ∕2
2
2
4 2
4
This function amplitude maximizes when 2 cos 𝜙 − 1 = 0, or 𝜙 = cos−1 (1∕2) = ±60.0◦ .
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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 𝜉 ′ = 𝜉 𝑓 ∕𝑓0 , where 𝑓0 is the original (design) frequency.
Under this condition, radiation will maximize in the 𝜙 = 0 direction regardless of frequency (show
this). Backward radiation (along 𝜙 = 𝜋) 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 𝜙 = 0 and 𝜙 = 𝜋, 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
( )
( )
𝑓
𝑓
′
𝜉+
𝑘𝑑 cos 𝜙
𝜓 =
𝑓0
𝑓0
With 𝜉 = −𝜋∕2 and 𝑑 = 𝜆0 ∕4, The H-plane array function at the original frequency, 𝑓0 , is
[
]
[𝜓 ]
[
]
𝜉 𝑘𝑑
𝜋
|
𝐴(𝜋∕2, 𝜙)| = cos
= cos
+
cos 𝜙 = cos (cos 𝜙 − 1)
|𝑓0
2
2
2
4
The effect on this of changing the frequency is then
)]
[( )
]
[ (
[ ′]
𝑓 𝜋
𝑓 𝜉 𝑘𝑑
𝜓
|
+
cos 𝜙
= cos
(cos 𝜙 − 1)
= cos
𝐴(𝜋∕2, 𝜙)| = cos
|𝑓
2
𝑓0 2
2
𝑓0 4
The condition for the zero argument (which maximizes 𝐴) is (cos 𝜙 − 1) = 0 This term is
unaffected by changing the frequency, and so the array function will always maximize in the
𝜙 = 0 direction.
In the forward direction (𝜙 = 0), the value of |𝐴|2 (proportional to the radiation intensity) is
unity. In the backward direction (𝜙 = 180◦ ), we find
)
)
(
(
𝜋𝑓
𝑓 𝜋
|𝐴(𝜙 = 𝜋)|2 = cos2 −
= cos2
𝑓0 2
2𝑓0
The front-to-back ratio is then
[
𝑅𝑓 𝑏 = 10 log10
[
]
]
|𝐴(𝜙 = 0)|2
1
= 10 log10
(
)
|𝐴(𝜙 = 𝜋)|2
cos2 𝜋𝑓 ∕2𝑓0
Evaluate the front-to-back ratio for a) 𝑓 = 1.5𝑓0 , b) 𝑓 = 2𝑓0 , c) 𝑓 = 0.75𝑓0 . Substituting these
values, we find:
]
1
= 3 dB
𝑓 = 1.5𝑓0 ∶ 𝑅𝑓 𝑏 = 10 log10
cos2 (1.5𝜋∕2)
]
[
1
= 0 dB
𝑓 = 2𝑓0 ∶ 𝑅𝑓 𝑏 = 10 log10
cos2 (𝜋)
]
[
1
= 8.3 dB
𝑓 = 0.75𝑓0 ∶ 𝑅𝑓 𝑏 = 10 log10
cos2 (𝜋∕2.67)
[
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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, 𝜉, 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 𝜃 = 𝜋∕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 𝜉:
𝐄(𝑧, 𝜃 = 𝜋∕2) = 𝐽
[
]
𝐼0 𝜂
[1 − cos(𝑘𝓁)] 𝑒−𝑗𝑘𝑧 𝐚𝑥 + 𝐚𝑦 𝑒𝑗𝜉
2𝜋𝑧
a) To achieve left circular polarization for propagation in the forward 𝑧 direction, the vector array
function must be
[
] [
]
𝐚𝑥 + 𝐚𝑦 𝑒𝑗𝜉 = 𝐚𝑥 + 𝑗𝐚𝑦 ⇒ 𝜉 = 𝜋∕2
b) To achieve 45◦ polarization, the vector array function must be
[
] [
]
𝐚𝑥 + 𝐚𝑦 𝑒𝑗𝜉 = 𝐚𝑥 ± 𝐚𝑦 ⇒ 𝜉 = 0, 𝜋
14.24. Consider a linear endfire array, designed for maximum radiation intensity at 𝜙 = 0, using 𝜉 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 𝜉 = −𝜋∕2 and 𝑑 = 𝜆∕4, so that 𝑘𝑑 = 𝜋∕2. Then in the H-plane
𝜓 = 𝜉 + 𝑘𝑑 cos 𝜙 =
𝜋
(cos 𝜙 − 1)
2
The array function is then:
[
]
[
]
1 sin(𝑛𝜓∕2)
1 sin[(𝑛𝜋∕4)(cos 𝜙 − 1)]
𝐴(𝜓) =
=
𝑛 sin(𝜓∕2)
𝑛 sin[(𝜋∕4)(cos 𝜙 − 1)]
In the forward direction (𝜙 = 0), 𝐴 = 1. In the backward direction (𝜙 = 𝜋), 𝐴 = ±1∕𝑛, where
𝑛 is an odd integer. The front-to-back ratio is therefore:
)
(
|𝐴(𝜙 = 0)|2
= 10 log10 (𝑛2 ) dB
𝑅𝑓 𝑏 = 10 log10
|𝐴(𝜙 = 𝜋)|2
in which performance is clearly better as 𝑛 increases.
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14.25. A six-element linear dipole array has element spacing 𝑑 = 𝜆∕2.
a) Select the appropriate current phasing, 𝜉, to achieve maximum radiation along 𝜙 = ±60◦ : With
𝑑 = 𝜆∕2, we have 𝑘𝑑 = 𝜋 and in the H-plane,
𝜓 = 𝜉 + 𝑘𝑑 cos 𝜙 = 𝜉 + 𝜋 cos 𝜙
We set the principal maximum to occur at 𝜓 = 0, and with 𝜙 = 60◦ , the condition for this
becomes
𝜋
𝜋
⇒ 𝜉=−
𝜓 = 0 = 𝜉 + 𝜋 cos(60◦ ) = 𝜉 +
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
[
]
[
]
sin[3(𝜉 + 𝑘𝑑 cos 𝜙)]
1
1 sin[(3𝜋∕2)(2 cos 𝜙 − 1)]
𝐴(−𝜋∕2, 𝜙) =
=
6 sin[(1∕2)(𝜉 + 𝑘𝑑 cos 𝜙)]
6 sin[(𝜋∕4)(2 cos 𝜙 − 1)]
In the broadside direction (𝜙 = 90◦ ) the array function becomes
[
] √
2
1 sin[−3𝜋∕2]
=
𝐴(−𝜋∕2, 𝜋∕2) =
6 sin[−𝜋∕4]
6
The intensity in this direction (relative to that along 𝜙 = 60◦ ) is the square of this or 1∕18
In the endfire direction (𝜙 = 0◦ ) the array function becomes
[
] √
2
1 sin[3𝜋∕2]
𝐴(−𝜋∕2, 0) =
=
6 sin[𝜋∕4]
6
The intensity in this direction (relative to that along 𝜙 = 60◦ ) is the square of this or 1∕18, as
before. The same result is found for 𝜙 = 180◦ and for 𝜙 = −90◦ . The intensity pattern is shown
below:
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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:
)
(
2𝜋𝑑 𝜋
+
𝜉=±
𝜆
𝑛
where the plus or minus sign choices give maximum radiation along 𝜙 = 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 𝜆: We first construct the expression for 𝜓
in the H-plane, using the given current phase expression, and in which the minus sign is chosen:
𝜓 = 𝜉 + 𝑘𝑑 cos 𝜙 = −
2𝜋𝑑 𝜋 2𝜋𝑑
− +
cos 𝜙
𝜆
𝑛
𝜆
Note that 𝜓(𝜙 = 0) = −𝜋∕𝑛. The array function, Eq. (81) is:
[
]
[
]
1 sin(𝑛𝜓∕2)
1 sin[(𝑛𝜋𝑑∕𝜆)(cos 𝜙 − 1) − 𝜋∕2]
𝐴(𝜉, 𝜙) =
=
𝑛 sin(𝜓∕2)
𝑛 sin[(𝜋𝑑∕𝜆)(cos 𝜙 − 1) − (𝜋∕2𝑛)]
This function is equal to 1 when 𝜙 = 0, regardless of the choice of 𝑑. When 𝜙 = 𝜋, the array
function becomes
[
]
1 sin[(−2𝑛𝜋𝑑∕𝜆) − (𝜋∕2)]
𝐴(𝜉, 𝜋) =
𝑛 sin[(−2𝜋𝑑∕𝜆) − (𝜋∕2𝑛)]
which we require to be zero. This will happen when 𝜓(𝜙 = 𝜋) = −𝜋. Using the above formula
for 𝜓, we set up the condition:
)
(
𝑛−1 𝜆
4𝜋𝑑 𝜋
− = −𝜋 ⇒ 𝑑 =
𝜓(𝜙 = 𝜋) = −
𝜆
𝑛
𝑛
4
b) Show that the spacing as found in part 𝑎 approaches 𝜆∕4 for a large number of elements: It is
readily seen from the part 𝑎 result that when 𝑛 → ∞, 𝑑 → 𝜆∕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 𝜙 = 𝜋. The result is
[
]
1 sin(−𝑛𝜋∕2)
𝐴(𝜉, 𝜋) =
𝑛 sin(−𝜋∕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:
[
]
1 sin(𝑛𝜓∕2)
𝐴(𝜉, 𝜙) =
𝑛 sin(𝜓∕2)
wherein if 𝜓 = −𝜋, we will get a zero result only if 𝑛 is even.
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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 𝜙 between the zeros on
either side of the principal maximum at 𝜙 = 90◦ . Show that for large 𝑛 this separation is approximated
.
.
by Δ𝜙 = 2𝜆∕𝐿, where 𝐿 = 𝑛𝑑 is the overall length of the array.
For broadside operation, 𝜉 = 0, and, with 𝑘𝑑 = 2𝜋𝑑∕𝜆, we have
𝜓 = 𝜉 + 𝑘𝑑 cos 𝜙 =
2𝜋𝑑
cos 𝜙
𝜆
With this condition, the array function, Eq. (81) is:
[
]
[
]
1 sin(𝑛𝜓∕2)
1 sin[(𝑛𝜋𝑑∕𝜆) cos 𝜙]
𝐴(𝜉, 𝜙) =
=
𝑛 sin(𝜓∕2)
𝑛 sin[(𝜋𝑑∕𝜆) cos 𝜙]
Zeros in this function will occur whenever
𝑛𝜋𝑑
cos 𝜙 = ±𝑚𝜋
𝜆
where 𝑚 is an integer. Now, on either side of the principal maximum at 𝜓 = 0, zeros will occur
at +𝜋 and −𝜋 (giving a 2𝜋 separation), which we can associate with the two angles 𝜙+ and 𝜙−
respectively, which lie on either side of 90◦ . We can therefore write:
)
𝑛𝜋𝑑 (
cos 𝜙+ − cos 𝜙− = 2𝜋
𝜆
Using a trig identity, this becomes:
[
]
]
[
1
1
2𝑛𝜋𝑑
sin (𝜙+ + 𝜙− ) sin (𝜙− − 𝜙+ ) = 2𝜋
𝜆
2
2
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
1
Because we are considering the broadside direction, we have (𝜙+ + 𝜙− )∕2 = 𝜋∕2 and so the
first sine term is just unity. We find
[
]
𝜆
1
sin (𝜙− − 𝜙+ ) =
2
𝑛𝑑
Now as 𝑛 gets large, both sides of the equation become << 1, so that we may approximate the
sine term by just its argument:
. 𝜆
1 −
(𝜙 − 𝜙+ ) =
2
𝑛𝑑
−
+
Defining Δ𝜙 = 𝜙 − 𝜙 , the above expression becomes
. 2𝜆 . 2𝜆
=
(done)
Δ𝜙 =
𝑛𝑑
𝐿
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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:
|𝑍21 |2
|𝑍12 |2
𝑃
𝑃𝐿2
=
=
= 𝐿1
𝑃𝑟1
𝑅11 𝑅22
𝑅11 𝑅22
𝑃𝑟2
We have
𝑃𝐿1
𝑃𝐿2
10−3
−7
=
=
10
=
𝑃𝑟1
𝑃𝑟2
104
So
𝑃𝐿1 = 𝑃𝑟2 × 10−7 = 100 × 10−7 = 10−5 = 10 𝜇W
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 𝜆2
𝐷1 (𝜃1 𝜙1 ) 𝐷2 (𝜃2 𝜙2 )
(4𝜋𝑟)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 𝜇W
3
2
(4𝜋 × 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
| cos[(𝜋∕2) cos(45◦ )] |2
| = 0.643
𝐷2 (𝜃2 = 45◦ ) = 𝐷𝑚𝑎𝑥 × |𝐹 (45◦ )|2 = 1.64 × ||
|
sin(45◦ )
|
|
So now,
𝑃𝐿2 (45◦ ) =
100(1)2
(1.64)(0.643) = 672 nW
(4𝜋 × 103 )2
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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 𝜆: From
Eq. (104), it follows that
4𝜋
𝐷𝑚𝑎𝑥 = 2 𝐴𝑚𝑎𝑥
𝜆
b) Express the current amplitude, 𝐼0 , needed to radiate total power, 𝑃𝑟 , in terms of 𝑃𝑟 , 𝐴𝑚𝑎𝑥 , and
𝜆: Combining Eqs. (64) and (65), we can write
2𝑃𝑟
𝐼02
=
120[𝐹 (𝜃)]2𝑚𝑎𝑥
𝐷𝑚𝑎𝑥
where, for a half-wave dipole, [𝐹 (𝜃)]2𝑚𝑎𝑥 = 1. Then, using the part 𝑎 result, we find
√
𝐼0 =
𝜋𝑃𝑟 𝐴𝑚𝑎𝑥
15𝜆2
c) At what values of 𝜃 and 𝜙 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 (𝜃) for the
antenna reaches its maximum. We know from Example 14.2 that this happens at 𝜃 = 𝜋∕2, and
at all values of 𝜙.
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