Environmental and Exploration Geophysics I
Magnetic Methods (III)
tom.h.wilson
tom.wilson@mail.wvu.edu
Department of Geology and Geography
West Virginia University
Morgantown, WV
Tom Wilson, Department of Geology and Geography
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
Tom Wilson, Department of Geology and Geography
The reciprocal relationship between
potential and field intensity
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
Tom Wilson, Department of Geology and Geography
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  
Tom Wilson, Department of Geology and Geography
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
Tom Wilson, Department of Geology and Geography
since V    Fdr, F  
dV
dr
H - monopole =
H - dipole

dVp
dr

d  p p
  2
dr  r  r
dVd
d  pl cos  

 

dr
dr  r 2 
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.
Tom Wilson, Department of Geology and Geography
H
Toward dipole
center (i.e. center of
Earth’s dipole field
dVd
d  pl cos  
ZE  
 

dr
dr  r 2 
Tom Wilson, Department of Geology and Geography
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.
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
d  pl cos  



rd  r 2 
Tom Wilson, Department of Geology and Geography
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
Tom Wilson, Department of Geology and Geography
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
Tom Wilson, Department of Geology and Geography
At the poles,  is 0, so that
HE  0
What is ZE at the equator?
2M cos
ZE 
r3
 is 90
ZE  0
Tom Wilson, Department of Geology and Geography
2M cos
ZE 
r3
ZE at the poles ….
2M
ZE  3
r
The variation of the field intensity at the poles 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.
Tom Wilson, Department of Geology and Geography
…. compare the field of the
magnetic dipole field to that of the
gravitational monopole field
1
Monopole f ield varies as  2
r
Gravity:500, 1000, 2000m
2M cos
ZE 
3
r
0.12
2M
ZE  3
r
0.1
0.08
0.06
0.04
0.02
0
-1500
-1000
-500
0
500
1000
1500
Increase r by a factor of 4
reduces g by a factor of 16
Tom Wilson, Department of Geology and Geography
For the dipole field, an increase
in depth (r) from 4 meters to 16 Dipole field varies as  1
meters produces a 64 fold
r3
decrease in anomaly magnitude
Thus the 7.2 nT anomaly (below left) produced by an object at 4
meter depths disappears into the background noise at 16 meters.
0.113 nT
7.2 nT
8
0.15
7
Intensity (nT)
Intensity (nT)
6
5
4
3
2
0.1
0.05
1
0
-1
-5
-3
-1
1
Distance in m eters
Tom Wilson, Department of Geology and Geography
3
5
0
-10
-5
0
Distance in m eters
5
10
On Tuesday during the last week of class, we’ll work
through some problems that will help you review
materials we’ve covered on magnetic fields. Some of
the problems are not too much different from those
we worked for gravitational fields and so will help
initiate some review of gravity methods.
The first problem relates to our discussions of the
dipole field and their derivatives.
3. What is the horizontal gradient in nT/m of the Earth’s
vertical field (ZE) in an area where the horizontal field (HE)
equals 20,000 nT and the Earth’s radius is 6.3 x 108 cm.
Tom Wilson, Department of Geology and Geography
(see problem 7.1) Recall that horizontal
gradients refer to the derivative evaluated along
the surface or horizontal direction and we use
the form of the derivative discussed earlier.
dV
dV
d  pl cos  
HE  




ds
rd
rd  r 2 
M sin 
Thus H E 
r3
dV 2M cos
ZE  

dr
r3
Tom Wilson, Department of Geology and Geography
1 d
r d
To answer this problem we must evaluate the
horizontal gradient of the vertical component -
1 d
ZE
r d
or
1 d 2M cos 
r d
r3
Take a minute and give it a try.
Tom Wilson, Department of Geology and Geography
4. A buried stone wall constructed from volcanic rocks has
a susceptibility contrast of 0.001cgs emu with its enclosing
sediments. The main field intensity at the site is 55,000nT.
Determine the wall's detectability with a typical proton
precession magnetometer. Assume the magnetic field
produced by the wall can be approximated by a vertically
polarized horizontal cylinder. Refer to figure below, and see
following formula for Zmax.
Background noise at
the site is roughly 5nT.
Tom Wilson, Department of Geology and Geography
Vertically Polarized Horizontal Cylinder
Z max
2 R 2 I

z2
Diagnostic position
X at Z/Zmax
9/10
3/4
2/3
1/2
1/3
1/4
0
General form


2
x


1

2
2
2R I 
z 
Z
z 2   x 2 2 
 1 
2 
z  

Tom Wilson, Department of Geology and Geography
x/z
0.188
0.31
0.37
0.495
0.61
0.68
1.0
(x/z)-1
Depth Index multiplier
5.32
3.23
2.7
2.02
1.64
1.47
1
Normalized shape term
2
x
1
2
Z ( x)
z

Z max  x 2 2
1 
2
z 

5. In your survey area you encounter two magnetic
anomalies, both of which form nearly circular
patterns in map view. These anomalies could be
produced by a variety of objects, but you decide to
test two extremes: the anomalies are due to 1) a
concentrated, roughly equidemensional shaped
object (a sphere); or 2) to a long vertically oriented
cylinder.
Tom Wilson, Department of Geology and Geography
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
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.
Tom Wilson, Department of Geology and Geography
(x/z)-1
Depth Index multiplier
5.26
3.18
2.65
2
1.56
1.37
0.71
Vertically Polarized Vertical Cylinder
Z max 
Diagnostic position
X at Z/Zmax
9/10
3/4
2/3
1/2
1/3
1/4
R I
2
z2
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
Tom Wilson, Department of Geology and Geography
ZA
1
 2
Z max
x
( 2  1)3 / 2
z
Unknow n Anom aly
16
Intensity (nT)
14
12
10
8
6
4
2
0
-4
-3
-2
-1
0
1
2
3
4
Distance in m eters
Sphere vs. Vertical Cylinder; z =
Diagnostic positions
X3/4 =
X1/2 =
X1/4 =
0.9
1.55
2.45
Tom Wilson, Department of Geology and Geography
Multipliers
Sphere
3.18
2
1.37
ZSphere
2.86
3.1
3.35
diagnostic distance
__________
The depth
Multipliers Cylinder
2.17
1.31
0.81
ZCylinder
1.95
2.03
2.00
Another Unknown Anomaly
5
Intensity (nT)
4
3
2
1
0
-1
-5
-4
-3
-2
-1
0
1
Distance in meters
2
3
4
5
Sphere or cylinder?
Diagnostic positions
Multipliers
Sphere
X3/4 = 1.6
3.18
2.17
X1/2 = 2.5
2
1.31
X1/4 = 3.7
1.37
0.81
Tom Wilson, Department of Geology and Geography
ZSphere
Multipliers
Cylinder
ZCylinder
6. Given that Z max 
 R2 I
2
derive an expression for the radius,
z
where I = kHE. Compute the depth to the top of the casing for
the anomaly shown below, and then estimate the radius of the
casing assuming k = 0.1 and HE =55000nT. Zmax (62.2nT from
graph below) is the maximum vertical component of the
anomalous field produced by the vertical casing.
Abandoned well
70
Intensity (nT)
60
50
40
30
20
10
0
-15
-10
-5
0
Distance in m eters
Tom Wilson, Department of Geology and Geography
5
10
15
Magnetics lab, part 2: A perfect fit – but is it correct?
Tom Wilson, Department of Geology and Geography
How many drums?
Outline of Drum Cluster
Derived from the magnetics model
10
Area of one drum ~
4 square
feet
Depth
15
1
TotalArea  Base x Height
2
Total Area
N Drums 
Area of one Drum
20
25
What’s wrong
with the format of
this plot?
30
35
180
190
200
210
220
Distance along profile
Tom Wilson, Department of Geology and Geography
230
We’ll talk more about the last
bullet (1/r3) on the results-tobe-discussed list a little later.
Problems 1 & 2 will be due this Thursday,
December 3rd
Next week will be spent in review
Problems 3-6 are due next Tuesday, Dec 8th
Magnetics lab, Magnetics paper summaries are due
Thursday December 10th
Exam, Thursday December 17th; 3-5pm
Tom Wilson, Department of Geology and Geography