Environmental and Exploration Geophysics I
Magnetic Methods
(part II)
tom.h.wilson
wilson@geo.wvu.edu
Department of Geology and Geography
West Virginia University
Morgantown, WV
The potential is the integral of the
force (F) over a displacement path.
V    Fdr   
From above, we obtain a
basic definition of the
potential (at right) for a unit
positive test pole (mt).
p1 p2
dr
2
r
V
pobject
r
Note that we consider
the 1/4 term =1
Thus - H (i.e. F/ptest, the field
intensity) can be easily derived
from the potential simply by taking
the derivative of the potential
dV
p
H 
 2
dr r
Consider the field at a point along the axis of a dipole. The
dipole in this case could be a buried well casing.
The field has vector properties however, in this case
vectors are collinear and its easy to determine the net effect.
In terms of the
potential we can write
V
p p

r r
In the case at right, r+ is much
greater than r- , thus in
V
p p

r
r
V
p
r
Thus, the potential near either end of a long dipole behaves like
the potential of an isolated monopole. If we are looking for
abandoned wells, we expect to find anomalies similar to the
gravity anomalies encountered over buried spherical objects.
21
In spite of this special situation, the magnetic field of an
object is defined by the simple dipole field or combinations
of dipole fields. To examine the nature of the dipole field
consider the case where the distance to the center of the
dipole is much greater than the length of the dipole. This
allows us to treat the problem of computing the potential
as one of scalar summation since the directions to each
pole fall nearly along parallel lines.
If r is much much greater than l then the angle 
between r+ and r- approaches 0 and r, r+ and rcan be considered parallel and the differences in
lengths r+ and r- from r equal to plus or minus the
projections of l/2 into r.
r-
r
r+
Working with the potentials of both poles ..
Vdipole 
p p

r
r
Recognizing that pole strength of the
negative pole is the negative of the positive
pole and that both have the same absolute
value, we rewrite the above as
Vdipole  
p p

r r
Vdipole  
p
r  l cos 
2

p
r  l cos 
2
Converting to common denominator yields
Vdipole
pl cos 

r2
where pl = M – the
magnetic moment
From the previous discussion , the field intensity H is just
dV

dr
since V    Fdr, F  
dV
dr
H - monopole =
d  p p

   2
dr
dr  r  r
H - dipole
dVd
d  pl cos  

 

dr
dr  r 2 
dVp
2 pl cos 

r3
This yields the field intensity in the radial direction i.e. in the direction toward the center of the dipole
(along r). However, we can also evaluate the
horizontal and vertical components of the total field
directly from the potential.
H
Toward dipole
(Earth’s) center
ZE  
dVd
d  pl cos  
 

dr
dr  r 2 
Vd represents the
potential of the dipole.
HE is represented by the negative derivative of the
potential along the earth’s surface or in the S direction.
d  pl cos  



rd  r 2 
dV
M sin 

 HE 
ds
r3
Where M = pl
and
dV 2M cos
ZE  

dr
r3
Let’s tie these results back into some
observations made earlier in the semester
with regard to terrain conductivity data.
32
Given
M sin 
HE 
r3
What is HE at the equator? … first what’s ?
 is the angle formed by the line connecting the
observation point with the dipole axis. So , in
this case, is a colatitude or 90o minus the
latitude. Latitude at the equator is 0 so  is 90o
and sin (90) is 1.
M
HE  3
r
At the poles,  is 0, so that
HE  0
What is ZE at the equator?
2M cos
ZE 
r3
 is 90
ZE  0
2M cos
ZE 
r3
ZE at the poles ….
2M
ZE  3
r
The variation of the field intensity at the pole and
along the equator of the dipole may remind you of
the different penetration depths obtained by the
terrain conductivity meters when operated in the
vertical and horizontal dipole modes.
I=kH
I  kH
I is the intensity of magnetization and H can be considered
the ambient (for example - Earth’s) magnetic field
intensity. k is the magnetic susceptibility.
Also recall Equation 7-5 (Burger). The intensity
of magnetization is equivalent to the magnetic
moment per unit volume or
M
I
V
and also, I  kH .
M pl

I
V
V
Magnetic dipole
moment per unit volume
Thus
p
 kH yielding
and
A
p  kAH
p  kAH
Recall from our earlier discussions that magnetic
field intensity
p
H  2 so that
r
p  Hr 2
The anomalous field induced in a magnetic
object is simply …
HA 
kHA
r2
Let’s consider this in the context of a potential application
Field intensity can be described in a variety of different
units, including, for example: Oersteds, nanoTeslas, and
gammas. The units you often encounter in exploration are
nanoTeslas (nT).
Pole strengths have a variety of units as well, but the one we
will use is the “unit pole strength” or ups.
In our basic definition for magnetic field
intensity associated with a single pole,
we have ..
p
H 2
r
p
H 2
r
With H in units of Oersteds, p in ups, and r in
cm, we have units of -
ups
1 Oersted  1
2
cm
And the equivalent units for ups
ups  Oersteds-cm
2
5
1 Oersted also equals 10 nT .
Problem: Imagine you are 20 centimeters from the
negative and positive poles of a dipole as shown below,
and that each pole has pole strength of 1 ups.
Observation
point
20cm
-
+
What is the magnetic field intensity at the observation point?
Vertical Magnetic Anomaly
Vertically Polarized Sphere
Z max
8 3
 R kH
3 3
z
Zmax and ZA refer to the
anomalous field, i.e.
the field produced by the
object in consideration
Diagnostic position
X at Z/Zmax
9/10
3/4
2/3
1/2
1/3
1/4
0
4 3
x2
 R kH (2  2 )
z
ZA  3 3
2
x
z
( 2  1)5/ 2
z
x/z
0.19
0.315
0.377
0.5
0.643
0.73
1.41
(x/z)-1
Depth Index multiplier
5.26
3.18
2.65
2
1.56
1.37
0.71
2

x
2  
2

z
Z A ( x) 1 


Z max
2  x 2 5 / 2
  1
 z2



The notation can be confusing at times. In
the above, consider H = FE= intensity of
earth’s magnetic field at the survey location.
Vertically Polarized Vertical Cylinder
Z max 
R I
2
z2
Diagnostic position
X at Z/Zmax
9/10
3/4
2/3
1/2
1/3
1/4
x/z
0.27
.046
0.56
0.766
1.04
1.23
(x/z)-1
Depth Index multiplier
3.7
2.17
1.79
1.31
0.96
0.81
zkHA
 R 2 zI
Z A  2 2 3/ 2 , or 2 2 3/ 2
(x  z )
(x  z )
ZA 
R 2 I
1
z2
x2
( 2  1)3 / 2
z
ZA
1
 2
Z max
x
( 2  1)3 / 2
z
Anomaly A: Sphere, Vertical Cylinder;
Diagnostic
positions
Multiplier
s
Sphere
ZSphere
Multipliers
Cylinder
X3/4 =0.9
3.18
2.17
X1/2 =1.55
2
1.31
X1/4 =2.45
1.37
0.81
z = __________
The depth
ZCylind
er
Sphere or cylinder?
Diagnostic
positions
Multipliers
Sphere
ZSphere
Multiplier
s Cylinder
ZCylinder
X3/4 = 1.3
3.18
4.13
2.17
2.82
X1/2 = 2.05
2
4.1
1.31
2.69
X1/4 = 2.9
1.37
3.97
0.81
2.34
Due Thursday, Dec. 9th
Since the bedrock is
magnetic, we have no
way of differentiating
between anomalies
produced by bedrock
and those produced by buried storage drums.
Acquisition of gravity data allows us to estimate variations
in bedrock depth across the profile. With this knowledge,
we can directly calculate the contribution of bedrock to the
magnetic field observed across the profile.
With the information on bedrock configuration we can clearly
distinguish between the magnetic anomaly associated with
bedrock and that associated with buried drums at the site.
Work with the inversions – iterate and adjust…
How many drums are represented by the
triangular-shaped object you entered into your
model?
Use the magic eye to get the coordinates of the
polygon defining the drums
Plot the corner coordinates for the
triangular shaped object you derived at
1:1 scale and compute the area.
How many drums?