This preview shows pages 1–3. Sign up to view the full content.
CHAPTER 12
12.1. A uniform plane wave in air,
E
+
x
1
=
E
+
x
10
cos
(
10
10
t
−
βz)
V/m, is normallyincident 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
η
c
=
r
jωµ
σ
=
(
1
+
j)
r
ωµ
2
σ
=
(
1
+
s
10
10
(
4
π
×
10
7
)
2
(
5
.
8
×
10
7
)
=
(
1
+
j)(.
0104
)
Note that the accuracy here is questionable, since we know the conductivity to only two significant
figures. We nevertheless proceed: Using
η
0
=
376
.
7288 ohms, we write
0
=
η
c
−
η
0
η
c
+
η
0
=
.
0104
−
376
.
7288
+
j.
0104
.
0104
+
376
.
7288
+
0104
=−
.
9999
+
0001
Now

0

2
=
.
9999, and so the transmitted power fraction is 1
−
0

2
=
.
0001, or about 0
.
01%
is
transmitted.
12.2. The plane
y
=
0 defines the boundary between two different dielectrics. For
y<
0,
±
0
R
1
=
1,
µ
1
=
µ
0
,
and
±
00
R
1
=
0; and for
y>
0,
±
0
R
2
=
5,
µ
2
=
µ
0
, and
±
00
R
2
=
0. Let
E
+
z
1
=
150 cos
(ωt
−
8
y)
V/m, and
find
a)
ω
:Have
β
=
8
=
ω/c
⇒
ω
=
8
c
=
2
.
4
×
10
9
sec
−
1
.
b)
H
+
1
: With
E
in the
z
direction, and propagation in the forward
y
direction,
H
will lie in the positive
x
direction, and its amplitude will be
H
x
=
E
y
/η
0
in region 1.
Thus
H
+
1
=
(
150
/η
0
)
cos
(ωt
−
8
a
x
=
0
.
40 cos
(
2
.
4
×
10
9
t
−
8
a
x
A
/
m
.
c)
H
−
1
: First,
E
−
z
1
=
0E
+
z
1
=
η
0
/
√
5
−
η
0
/
1
η
0
/
√
5
+
η
0
/
1
=
1
−
√
5
1
+
√
5
E
+
z
1
0
.
38
E
+
z
1
Then
H
−
x
1
=+
(
0
.
38
/η
0
)E
+
z
1
=
0
.
38
(
150
)
377
cos
(ωt
+
8
So finally,
H
−
x
1
=
0
.
15 cos
(
2
.
4
×
10
9
t
+
8
a
x
A
/
m
.
12.3. A uniform plane wave in region 1 is normallyincident on the planar boundary separating regions 1 and
2. If
±
00
1
=
±
00
2
=
0, while
±
0
R
1
=
µ
3
R
1
and
±
0
R
2
=
µ
3
R
2
, find the ratio
±
0
R
2
/±
0
R
1
if 20% of the energy in
the incident wave is reflected at the boundary. There are two possible answers. First, since

0

2
=
.
20,
and since both permittivities and permeabilities are real,
0
=±
0
.
447. we then set up
0
0
.
447
=
η
2
−
η
1
η
2
+
η
1
=
η
0
q
(µ
R
2
/±
0
R
2
)
−
η
0
q
(µ
R
1
/±
0
R
1
)
η
0
q
(µ
R
2
/±
0
R
2
)
+
η
0
q
(µ
R
1
/±
0
R
1
)
=
q
(µ
R
2
/µ
3
R
2
)
−
q
(µ
R
1
/µ
3
R
1
)
q
(µ
R
2
/µ
3
R
2
)
+
q
(µ
R
1
/µ
3
R
1
)
=
µ
R
1
−
µ
R
2
µ
R
1
+
µ
R
2
200
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document12.3. (continued)
Therefore
µ
R
2
µ
R
1
=
1
∓
0
.
447
1
±
0
.
447
=
(
0
.
382
,
2
.
62
)
⇒
±
0
R
2
±
0
R
1
=
±
µ
R
2
µ
R
1
²
3
=
(
0
.
056
,
17
.
9
)
12.4. The magnetic field intensity in a region where
±
00
=
0 is given as
H
=
5 cos
ωt
cos
βz
a
y
A/m, where
ω
=
5 Grad/s and
β
=
30 rad/m. If the amplitude of the associated electric field intensity is 2kV/m,
find
a)
µ
and
±
0
for the medium: In phasor form, the magnetic field is
H
ys
=
H
0
e
−
jβz
+
H
0
e
+
=
5 cos
⇒
H
0
=
2
.
5. The electric field will be
x
directed, and is
E
xs
=
η(
2
.
5
)e
−
−
2
.
5
)e
+
=
(
2
j)η(
2
.
5
)
sin
. Given the electric field amplitude of 2 kV/m, we write 2
×
10
3
=
5
η
,or
η
=
400
.Now
η
=
400
=
η
0
q
µ
r
/±
0
R
and we also have
β
=
30
=
(ω/c)
q
µ
R
±
0
R
.W
e
solve these two equations simultaneously for
µ
R
and
±
0
R
to find
µ
R
=
1
.
91 and
±
0
R
=
1
.
70.
Therefore
µ
=
1
.
91
×
4
π
×
10
−
7
=
2
.
40
µ
H
/
m
and
±
0
=
1
.
70
×
8
.
854
×
10
−
12
=
15
.
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Brown
 Electromagnet

Click to edit the document details