Environmental and Exploration Geophysics I
Magnetic Methods (IV)
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
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
From the
bedrock
Tom Wilson, Department of Geology and Geography
anomaly
Tom Wilson, Department of Geology and Geography
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
…. 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
Follow the recommended reporting format.
Specifically address points mentioned in the results section, above.
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
The first problem relates to our discussions of the
dipole field and their derivatives.
7.1. 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
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
1 d
r d
dV
dV
d  pl cos  
HE  




ds
rd
rd  r 2 
M sin 
Thus H E 
r3
dV 2M cos 
ZE  

3
dr
r
Tom Wilson, Department of Geology and Geography
To answer this problem we must evaluate the
horizontal gradient of the vertical component -
1 d
ZE
r d
or
1 d 2 M cos 
r d
r3
Take a minute and give it a try.
Hint: See Equation 7.20
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.
What is z?
What is I?
Background noise at
the site is roughly 5nT.
Tom Wilson, Department of Geology and Geography
Vertically Polarized Horizontal Cylinder
General form


2
 1 x 2 
2R I 
z 
Z
2 
2
z
1  x 2  
 

z2  

2
Diagnostic position
X at Z/Zmax
9/10
3/4
2/3
1/2
1/3
1/4
0
x/z
0.188
0.31
0.37
0.495
0.61
0.68
1.0
Normalized shape term
1 x
2
Z ( x)
z2

Z max  x 2 2
1 

z2 

Z max
Area   R 2
2 R 2 I

z2
Tom Wilson, Department of Geology and Geography
R
Area

(x/z)-1
Depth Index multiplier
5.32
3.23
2.7
2.02
1.64
1.47
1
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
Z max
8 3
 R kH
3 3
z
Diagnostic position
X at Z/Zmax
9/10
3/4
2/3
1/2
1/3
1/4
0
2

x
2  
z 2 
Z A ( x) 1 

5/ 2
Z max
2  x2

  1
 z2



Tom Wilson, Department of Geology and Geography
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
Z max 
R 2 I
z2
Diagnostic position
X at Z/Zmax
9/10
3/4
2/3
1/2
1/3
1/4
ZA
1
 2
Z max
x
( 2  1)3 / 2
z
Tom Wilson, Department of Geology and Geography
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
Determine depths (z) assuming a sphere or a cylinder
and see which assumption yields consistent estimates.
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
It’s all about using diagnostic positions and
the depth index multipliers for each geometry.
Tom Wilson, Department of Geology and Geography
Unknow n Anom aly
16
X3/4
Intensity (nT)
14
12
X1/2
10
8
X1/4
6
4
2
0
-4
-3
-2
-1
0
1
2
3
4
Distance in m eters
distance
Sphere vs. Vertical Cylinder; z = diagnostic
__________
The depth
Diagnostic
positions
X3/4 =
X1/2 =
X1/4 =
0.9
1.55
2.45
Tom Wilson, Department of Geology and Geography
Multipliers
Sphere
ZSphere
Multipliers
Cylinder
3.18
2
1.37
2.86
3.1
3.35
2.17
1.31
0.81
ZCylinder
1.95
2.03
2.00
Another Unknown Anomaly
Intensity (nT)
5
4
gmax
3
g3/4
2
g1/2
1
g1/4
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 meters
3.18
5.01
2.17
3.47
X1/2 = 2.5 meters
2
5.0
1.31
3.28
X1/4 = 3.7 meters
1.37
5.07
0.81
2.99
Tom Wilson, Department of Geology and Geography
ZSphere
Multipliers
Cylinder
ZCylinder
Algebraic manipulation
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
Feel free to discuss these problems in groups, but realize that you
will have to work through problems independently on the final.
Tom Wilson, Department of Geology and Geography
Problems 1 & 2 are due today, 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