Roy K.K. Potential theory in applied geophysics
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13.8 Long Line Cable Placed in an Infinite and Homogenous Medium 419
We can write
H
ψ
=
σbγ
γρ
=
σb
ρ
=
I
2πρ
when ω → 0. (13.255)
Therefore
‘b’ is =
I
2πσ
. (13.256)
In a time varying field
b=
1
2π (σ +iω ∈)
. (13.257)
Therefore
Π
z
=bK
0
(γρ)=
1
2π (σ +iω ∈)
.K
0
(γρ) (13.258)
=
I
2π
.
iωµ
γ
2
.K
0
(γρ) . (13.259)
Theexistingelectricandmagnetic field components are
E
z
= −iωµ
I
2π
K
0
(γρ) (13.260)
H
ψ
= −
I
2π
.
∂
∂ρ
K
0
(γρ) . (13.261)
All other field components are zero i.e.,
Eρ =Eψ =0Hρ =H
z
=0. (13.262)
In the radiation zone, where γρ >> 1
K
0
(z) → K
1
(z) →
0
π
2z
e
−z
. (13.263)
the asymptotic values of E
z
and H
ψ
are given as
E
z
= −iωµ.
I
2π
0
π
2γρ
.e
−γρ
(13.264)
and
H
ψ
= −
Iγ
2π
.K
1
(γρ)
=
Iγ
2π
0
π
2γρ
e
−γρ
. (13.265)
Now substituting γ = α +iβ, we can separate the real and imaginary parts
1
√
γ
=
1
α +iβ
=
1
4
α
2
+ β
2
.e
−
i
2
tan
−1
(
β
α
)
.
420 13 Electromagnetic Wave Propagation
Therefore the expressions for E
z
and H
ψ
can be written as
E
z
=
µIω
2π
.e
−iπ
2
.
0
π
2ρ
.
1
4
α
2
+ β
2
.e
−
i
2
tan
−1
β
α
.e
−(α+iβ)ρ
(13.266)
H
ψ
=
I
2π
α
2
+ β
2
e
itan
−1
β
α
.
0
π
2ρ
.
1
4
α
2
+ β
2
.e
−
i
2
tan
−1
β
α
.e
−(α+iβ)ρ
. (13.267)
Real components of the electric and magnetic fields ar e given by
Re
E
z
=
μIω
√
8π
.
e
−αρ
√
ρ
.
1
4
α
2
+ β
2
cos
ωt −
1
2
tan
−1
β
α
−
π
2
− βρ
(13.268)
Re
H
ψ
=
I
√
8π
.
e
−αρ
√
ρ
.
1
4
α
2
+ β
2
cos
ωt −βρ +
1
2
tan
−1
β
α
. (13.269)
For negligible displacement cur rent
α = β =
0
ωµσ
2
,
hence
E
z
H
ψ
= µω.
1
α
2
+ β
2
=
0
ωµ
σ
=Z. (13.270)
So we can measure conductivity of the ground by measuring E
Z
and H
ψ
.Since
γ =
√
iωµσ,
E
z
= −iωµ
I
2π
K
0
P
√
i
(13.271)
and
H
ψ
= −
I
2π
.
∂
∂ρ
K
0
P
√
i
(13.272)
where P =
√
ωµσ.HereK
0
P
√
i
and K
′
0
P
√
i
have both real and imagi-
nary components as
K
0
P
√
i
=kerP+keipP=K
0
Pe
iπ/4
(13.273)
and
K
′
0
p
√
i
=ker
′
p+kei
′
p. (13.274)
The phase varies from - π/2to+π/2
13.9 Long Cable on the Surface of a Homogeneous Earth 421
13.9 Electromagnetic Field due to a Long Cable on the
Surface of a Homogeneous Earth
An infinitely long cable is placed along the x direction (Fig. 13.12) therefore
the starting wave equation is
∇
2
Π
x
= γ
2
Π
x
(13.275)
where
Π
x
is the vector potential.
In this problem
Π
y
=0and
H
z
= 0. Since there is no variation along
the x – direction, the Helmholtz equation reduces to the form in Cartisian
coordinate as
∂
2
Π
x
∂z
2
+
∂
2
Π
x
∂y
2
= γ
2
Π
x
. (13.276)
Applying the method of separation of variables, i.e.
Π
x
= Y (y) Z (z)
we get
1
Y
d
2
Y
dy
2
= −n
2
. (13.277)
The solution are e
iny
,e
−iny
.Ifwetake
1
Z
d
2
Z
dz
2
= γ
2
+ n
2
(13.278)
The solution are
Z=e
±
z
√
γ
2
+n
2
.
Therefore the expression for the perturbation potential is
Π
x
=
α
0
f
n
e
−z
√
γ
2
+n
2
g
n
e
z
√
γ
2
+n
2
e
−iny
dn. (13.279)
Fig. 13.12. An infinitely long line electrode carrying alternating current on the
surface of the earth
422 13 Electromagnetic Wave Propagation
The expressions for the vector po tentials in the first and second media are
respectively given by
Π
0x
=
Π
0
+
α
0
f
n
e
−Z
√
n
2
+γ
2
0
.e
−iny
dn (13.280)
[in the first medium] and
Π
1x
=
α
0
g
n
e
Z
√
n
2
+γ
2
1
e
−iny
dn (13.281)
[in the second medium] .
Here the source potential
Π
0
is
Π
0
=
I
2π (σ + iω ∈)
.K
0
(γρ)=
iωμI
2πγ
2
0
K
0
(γρ) . (13.282)
Substituting the value of Π
0
the source potential, the (13.280) becomes
Π
0x
=
iωµ
0
I
2πγ
2
0
K
0
(γ
0
ρ)+
α
0
f
n
e
−
√
n
2
+z
2
.Z
e
−iny
dn (13.283)
and the vector potential in the second medium is
Π
1x
=
α
0
g
n
e
z
√
n
2
+γ
2
1
e
−iny
dn. (13.284)
Now
E
x
= −γ
2
Π
x
+graddiv
Π
x
= −γ
2
Π
x
. (13.285)
Since
∂
∂x
=0
Hy =
γ
2
iωµ
curlΠ
x
=
−γ
2
iωµ
.
∂
Π
x
∂z
(13.286)
H
z
= −
γ
2
iωµ
∂
Π
x
∂y
(13.287)
because
Π
y
and
Π
z
are also zero.
The Sommerfeld formula
e
−γ
√
ρ
2
+z
2
ρ
2
+z
2
=
α
0
n
n
2
+ γ
2
e
−nz
J
0
(nρ) dn (13.288)
13.9 Long Cable on the Surface of a Homogeneous Earth 423
is used in time varying field to bring the source potential in the same format
as perturbation potential.
It changes to the Webers formula for γ = 0 i.e.,
1
ρ
2
+z
2
=
α
0
e
−nz
J
0
(nρ)dn=
2
π
α
0
K
0
(γρ)e
inz
dn. (13.289)
In this problem
1
n
2
+ γ
2
0
=
1
π
α
−α
K
0
(γ
0
y) e
−iny
dy. (13.290)
Here the changes are (i) from ρ to γ
0
, (ii) from n to y and (iii) from z to n.
The Sommerfeld formula can change to the form
e
−z
√
n
2
+γ
2
0
n
2
+ γ
2
0
=
1
π
α
−α
K
0
γ
0
y
2
+z
2
e
iny
dy. (13.291)
Now let
K
0
(γ
0
ρ)=K
0
γ
0
y
2
+z
2
=
∞
−∞
A
n
e
−iny
dn
y
,
then
A
n
=
1
2π
∞
−∞
K
0
γ
0
y
2
+ z
2
e
iny
.dy. (13.292)
Applying the Fourier Transform, we shall have
A
n
=
1
2π
.π.
e
−
√
n
2
+γ
2
0
n
2
+ γ
2
0
. (13.293)
Therefore
K
0
γ
0
y
2
+z
2
=
1
2
∞
−∞
e
−z
√
n
2
+γ
2
0
n
2
+ γ
2
0
e
−iny
.dn. (13.294)
The vector potential
Π
0x
is
Π
0x
=P
∞
−∞
e
−|z−h|
√
n
2
+γ
2
0
n
2
+ γ
2
0
e
−iny
dn
+
∞
0
f
n
e
−z
√
n
2
+γ
2
0
.e
−iny
dn, where P =
iωμ
0
I
2πγ
2
0
. (13.295)
424 13 Electromagnetic Wave Propagation
Applying the boundary conditions
γ
2
0
Π
0x
= γ
2
1
Π
1x
(13.296)
γ
2
0
µ
0
.
∂
Π
0x
∂z
=
γ
2
1
µ
1
.
∂π
1x
∂z
(13.297)
and substituting
P =
i̟μ
0
I
2πγ
2
0
,
we get
γ
2
0
.P e
−h
√
n
2
+γ
2
0
n
2
+ γ
2
0
+ γ
2
0
f =γ
2
1
g (13.298)
γ
2
0
Pe
h
√
n
2
+γ
2
0
.
n
2
+ γ
2
0
n
2
+ γ
2
0
+ γ
2
0
(−1) .
n
2
+ γ
2
0
.f
n
=γ
2
1
n
2
+ γ
2
1
.g
n
.
(13.299)
Substituting
n
2
0
=n
2
+ γ
2
0
n
2
1
=n
2
+ γ
2
1
,
the values of f
n
and g
n
are found out to be equal to
f
n
=P.
n
0
− n
1
n
0
(n
0
+ n
1
)
e
−n
0
h
(13.300)
g
n
=2P.
γ
2
0
γ
2
1
.e
−n
0
h
.
1
n
0
+ n
1
. (13.301)
When the line source is brought on the surface of the earth. i.e., when h = 0,
the vector pot ential on the surface of the earth is given by
Π
0x
=P
∞
0
e
−n
0
z
n
0
+ n
1
e
−iny
dn + P
∞
0
n
0
− n
1
n
0
(n
0
+ n
1
)
e
−n
0
z
e
−iny
dn
=2P
∞
0
e
−n
0
z
n
0
+ n
1
e
−iny
dn. (13.302)
Therefore, the electric and magnetic fields are
E
0x
= −2Pγ
2
0
∞
0
e
−n
0
z
n
0
+ n
1
.e
−iny
dn (13.303)
13.9 Long Cable on the Surface of a Homogeneous Earth 425
H
0y
= −2P
γ
2
0
iωμ
∞
0
n
0
e
−n
0
z
n
0
+ n
1
.e
−iny
dn (13.304)
H
0z
= −2P
γ
2
0
iωμ
∞
0
(−in) e
−n
0
z
n
0
+ n
1
.e
−iny
dn. (13.305)
On the surface of the earth at z = 0 and substituting the value of P ,weget
E
x
=
iωµI
π
∞
0
e
−iny
n
0
+n
1
.dn (13.306)
H
y
= −
µI
π
∞
0
n
0
e
−iny
n
0
+n
1
.dn (13.307)
.
H
z
= −
µI
π
∞
0
−in
n
0
+n
1
.e
−iny
dn.. (13.308)
Since
K
0
γ
0
y
2
+z
2
=
∞
0
e
−n
0
z
n
0
.e
−iny
dn (13.309)
we get
E
x
= −
iωµI
π
∞
0
(n
1
− n
0
)e
−iny
(n
2
1
− n
2
0
)
dn (13.310)
= −
iωµI
π (γ
2
1
− γ
2
0
)
⎡
⎣
∞
0
n
1
e
−iny
dn −
∞
0
n
0
e
−iny
dn
⎤
⎦
. (13.311)
If we differentiate with respect to z twice, we shall get
∂
2
∂z
2
K
0
γ
0
y
2
+z
2
=
∞
0
n
0
e
−n
0
z
.e
−iny
dn. (13.312)
Now,
z
Lim
→ 0=
∞
0
n
0
e
−iny
dn
426 13 Electromagnetic Wave Propagation
and
K
0
γ
1
y
2
+z
2
=
∞
0
e
−n
1
z
n
1
.e
−iny
dn (13.313)
∂
2
∂z
2
K
0
γ
1
y
2
+z
2
z=0
=
∞
0
n
1
e
−iny
dn.
Therefore, the expression for
E
x
can be written as
E
x
= −
iωμI
π (γ
2
1
− γ
2
0
)
Lt
z→0
∂
2
∂z
2
K
0
γ
1
y
2
+ z
2
− Lt
z→0
∂
2
∂z
2
K
0
γ
0
y
2
+ z
2
. (13.314)
Differentiating K
0
γ
0
y
2
+z
2
with respect to z, we get
∂
∂z
K
0
γ
0
y
2
+ z
2
=
γ
0
y
2
+ z
2
K
′
0
γ
0
y
2
+ z
2
(13.315)
∂
2
∂z
2
K
0
γ
0
y
2
+ z
2
=
γ
2
0
z
2
y
2
+ z
2
K
′′
0
γ
0
y
2
+ z
2
+γ
0
y
2
+ z
2
+
z
2
√
y
2
+z
2
y
2
+ z
2
K
′
0
γ
0
y
2
+ z
2
.
At
z=0,
∂
2
∂z
2
K
0
γ
0
y
2
+ z
2
=
γ
0
y
K
′
0
(γ
0
y) . (13.316)
Therefore ,
E
x
=
−iωμI
π (γ
2
1
− γ
2
0
)
γ
1
y
K
′
0
(γ
1
y) −
γ
0
y
K
′
0
(γ
0
y)
. (13.317)
Since
K
′
0
(x)=−K
1
(x)
=
E
x
−
iωµI
π (γ
2
1
− γ
2
1
)
.
1
y
2
[γ
0
yK
1
(γ
0
y) − γ
1
yK
1
(γ
1
y)] . (13.318)
When γ
0
y << 1
Lt
x→0
K
1
(x) = 1
Here, x = γ
0
y
13.9 Long Cable on the Surface of a Homogeneous Earth 427
Fig. 13.13. Variation of amplitude and phase with the electromagnetic
parameter P
E
x
= −
iωµI
πγ
2
1
y
2
[1 − γ
1
yK
1
(γ
1
y)] . (13.319)
Since
E
0
x
=
−iωµI
π
,
E
x
E
0
x
=
1
γ
2
1
y
2
[1 −γ
1
yK
1
(γ
1
y)] . (13.320)
The general expressions for the electric field and the magnetic field are given
by
E
x
= −
iωμI
π (γ
2
1
− γ
2
0
)
.
γ
1
y
2
[γ
0
yK
1
(γ
0
y) − γ
1
yK
1
(γ
1
y)] (13.321)
and
H
z
= −
I
π (γ
2
1
− γ
2
0
)
.
1
y
3
[2γ
0
y/K
1
(γ
0
y) +
γ
2
0
y
2
K
0
(γ
0
y) − 2γ
1
yK
1
(γ
1
y) − γ
2
1
y
2
K
0
(γ
1
y)
(
. (13.322)
When γ
0
→ 0
H
z
= −
I
πγ
2
1
.
1
y
3
'
2 −2γ
1
yK
1
(γ
1
y) − γ
2
1
y
2
K
0
(γ
1
y)
(
(13.323)
H
0
z
= −
I
2π
.
∂
∂y
K
0
(γ
1
y) ≈
1
2πy
. (13.324)
Then
H
z
H
0
z
=
2
γ
2
1
y
2
'
2 −2γ
1
yK
1
(γ
1
y) − γ
2
y
2
K
0
(γ
1
y)
(
. (13.325)
428 13 Electromagnetic Wave Propagation
Since
γ
1
=
iωµσ,
Therefore
γ
1
y = y
√
ωμσ.
√
i = P
√
i
H
z
H
0
z
=
2
P
2
2 −P
√
iK
1
P
√
i
− P
2
iK
0
P
√
i
. (13.326)
13.10 Electromagnetic Induction due to an Infinite
Cylinder in an Uniform Field
An infinitely long cylinder has its axis horizontal and is directed along the z
direction. The field is vertical due to an oscillating magnetic dipole of larger
radius such that one gets uniform vertical field at the center of the loop
(Fig. 13.14) The starting Helmholtz equation is
∇
2
A=γ
2
A (13.327)
where
A is a vector potential. Here the field is assumed to be harmonic, i.e.
H=He
iωt
. Since the cylinder is infinitely long,
∂
∂z
=0.
From Maxwell’s equation, we can write
curl
H =
J +
∂
D
∂t
=(σ + iω ∈)
E. (13.328)
We define
Fig. 13.14. An infinitely conducting cylinder buried inside a earth in the presence
of a vertical uniform field