This module is part of the collection, A First Course in Electrical and Computer Engineering. The LaTeX source files for this collection were created using an optical character recognition technology, and because of this process there may be more errors than usual. Please contact us if you discover any errors.
We have established that vectors may be used to code complex numbers. Conversely, complex numbers may be used to code or represent the
orthogonal components of any two-dimensional vector. This makes them invaluable in electromagnetic field theory, where they are used to represent the
components of electric and magnetic fields.
The basic problem in electromagnetic field theory is to determine the
electric or magnetic field that is generated by a static or dynamic distribution
of charge. The key idea is to isolate an infinitesimal charge, determine the
field set up by this charge, and then to sum the fields contributed by all
such infinitesimal charges. This idea is illustrated in Figure 1, where the
charge
λλ, uniformly distributed over a line segment of length dxdx at point -x-x,
produces a field dE(x)dE(x) at the test point (0,h)(0,h). The field dE(x)dE(x) is a “vector”
field (as opposed to a “scalar” field), with components E1(x)E1(x) and E2(x)E2(x). The
intensity or field strength of the field dE(x)dE(x) is
|
d
E
(
x
)
|
=
λ
d
x
4
π
ϵ
0
(
h
2
+
x
2
)
|
d
E
(
x
)
|
=
λ
d
x
4
π
ϵ
0
(
h
2
+
x
2
)
(1)But the field strength is directed at angle θ(x)θ(x), as illustrated in Figure 1.
The field dE(x)dE(x) is real with components dE1(x)dE1(x) and dE2(x)dE2(x), but we code it
as a complex field. We say that the “complex” field at test point (0,h)(0,h) is
dE(x)=λdx4πϵ0(h2+x2)ejθ(x)dE(x)=λdx4πϵ0(h2+x2)ejθ(x)
(2)with components dE1(x)dE1(x) and dE2(x)dE2(x). That is,
d
E
(
x
)
=
d
E
1
(
x
)
+
j
d
E
2
(
x
)
d
E
(
x
)
=
d
E
1
(
x
)
+
j
d
E
2
(
x
)
(3)
d
E
1
(
x
)
=
λ
d
x
4
π
ϵ
0
(
h
2
+
x
2
)
c
o
s
θ
(
x
)
d
E
1
(
x
)
=
λ
d
x
4
π
ϵ
0
(
h
2
+
x
2
)
c
o
s
θ
(
x
)
(4)
d
E
2
(
x
)
=
λ
d
x
4
π
ϵ
0
(
h
2
+
x
2
)
s
i
n
θ
(
x
)
.
d
E
2
(
x
)
=
λ
d
x
4
π
ϵ
0
(
h
2
+
x
2
)
s
i
n
θ
(
x
)
.
(5)For charge uniformally distributed with density
λλ along the x-axis, the
total field at the test point (0,h)(0,h) is obtained by integrating dEdE:
∫
-
∞
∞
d
E
(
x
)
=
∫
-
∞
∞
λ
4
π
ϵ
0
(
h
2
+
x
2
)
[
cos
θ
(
x
)
+
j
s
i
n
θ
(
x
)
]
d
x
.
∫
-
∞
∞
d
E
(
x
)
=
∫
-
∞
∞
λ
4
π
ϵ
0
(
h
2
+
x
2
)
[
cos
θ
(
x
)
+
j
s
i
n
θ
(
x
)
]
d
x
.
(6)The functions cos θ(x)θ(x) and sinθ(x)sinθ(x) are
cosθ(x)=x(x2+h2)1/2;
sinθ(x)=h(x2+h2)1/2
cosθ(x)=x(x2+h2)1/2;sinθ(x)=h(x2+h2)1/2
(7)We leave it as a problem to show that the real component
E
1
E
1
of the field is
zero. The imaginary component
E
2
E
2
is
E
=
j
E
2
=
j
∫
-
∞
∞
λ
h
4
π
ϵ
0
d
x
(
x
2
+
h
2
)
3
/
2
E
=
j
E
2
=
j
∫
-
∞
∞
λ
h
4
π
ϵ
0
d
x
(
x
2
+
h
2
)
3
/
2
(8)
=
j
λ
h
4
π
ϵ
0
x
h
2
(
x
2
+
h
2
)
1
/
2
|
-
∞
∞
=
j
λ
h
4
π
ϵ
0
x
h
2
(
x
2
+
h
2
)
1
/
2
|
-
∞
∞
(9)
=
j
λ
h
4
π
ϵ
0
[
1
h
2
+
1
h
2
]
=
j
λ
2
π
ϵ
0
h
=
j
λ
h
4
π
ϵ
0
[
1
h
2
+
1
h
2
]
=
j
λ
2
π
ϵ
0
h
(10)
E
2
=
λ
2
π
ϵ
0
h
.
E
2
=
λ
2
π
ϵ
0
h
.
(11)We emphasize that the field at (0,h)(0,h) is a real field. Our imaginary answer
simply says that the real field is oriented in the vertical direction because
we have used the imaginary part of the complex field to code the vertical
component of the real field.
Show that the horizontal component of the field
E
E is zero.
Interpret this finding physically.
From the symmetry of this problem, we conclude that the field around
the infinitely long wire of Figure 1 is radially symmetric. So, in polar
coordinates, we could say
E
(
r
,
θ
)
=
λ
2
π
ϵ
0
r
E
(
r
,
θ
)
=
λ
2
π
ϵ
0
r
(12)which is independent of
θ
θ. If we integrated the field along a radial line perpendicular to the wire, we would measure the voltage difference
V
(
r
1
)
-
V
(
r
0
)
=
∫
r
0
r
1
λ
2
π
ϵ
0
r
d
r
=
λ
2
π
ϵ
0
[ logr1-log
r
0
]
.
V
(
r
1
)
-
V
(
r
0
)
=
∫
r
0
r
1
λ
2
π
ϵ
0
r
d
r
=
λ
2
π
ϵ
0
[logr1-log
r
0
].
(13)An electric field has units of volts/meter, a charge density
λλ has units of
coulombs/meter, and
ϵ
0
ϵ
0
has units of coulombs/volt-meter; voltage has units
of volts (of course).
(What does "Endorsed by" mean?)
"Reviewer's Comments: 'I recommend this book as a "required primary textbook." This text attempts to lower the barriers for students that take courses such as Principles of Electrical Engineering, […]"