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In-Depth Information
With account for (6.2),
∞
H
x
H
x
H
x
o
(
e
ik
z
e
ik
z
)
−
=
−
=
b
m
sin
mzdm
,
(6
.
6)
0
where
(
k
)
2
(
k
)
2
2
mH
x
o
π
−
b
m
=
.
(6
.
7)
)
2
(
Substituting (6.4) and (6.6) in (6.5) and equating the terms with the same sub-
script, we obtain
a
m
−
a
m
=
a
m
+
a
m
=
b
m
,
0
,
(6
.
8)
whence
2
H
x
o
(
k
)
2
(
k
)
2
−
m
a
m
=
+
/
)
,
)
2
(1
π
(
(6
.
9)
2
H
x
o
(
k
)
2
(
k
)
2
−
m
a
m
=−
+
/
)
.
π
)
3
(1
(
Returning to (6.1) and (6.4), we write
⎧
⎨
me
y
sin
mz dm
2
H
x
o
(
k
)
2
(
k
)
2
0
−
∞
H
N
x
+
y
≤
0
π
)
2
(1
(
+
/
)
H
x
=
⎩
0
me
−
y
sin
mz dm
(
2
H
x
o
(
k
)
2
(
k
)
2
∞
−
H
x
−
y
≥
0
.
)
3
(1
π
+
/
)
(6
.
10)
Differentiation of the magnetic field
H
x
with respect to
z
gives the electric field
E
y
. On the Earth surface (
z
=
0):
⎧
⎨
m
2
e
y
dm
2
H
x
o
(
k
)
2
(
k
)
2
0
∞
−
E
y
+
y
≤
0
)
2
(1
π
(
+
/
)
E
y
=
H
x
=
z
⎩
(
k
)
2
(
k
)
2
m
2
e
−
y
dm
−
∞
E
N
y
−
2
H
x
o
y
≥
0
,
π
)
3
(1
(
+
/
)
0
(6
.
11)
where
E
N
E
N
y
y
,
are the normal electric fields within the left and right quarter-spaces.
