Environmental and Exploration Geophysics I
Magnetic Methods (V)
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
Problems we’ve been working on …
Tom Wilson, Department of Geology and Geography
Questions?
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 for
the potential.
1 d
r d
dV
dV
d  pl cos  
HE  




2
ds
rd
rd  r

M sin 
The negative sign
Thus H E 
r3
dV 2M cos 
ZE  

dr
r3
Tom Wilson, Department of Geology and Geography
is NOT needed
when computing
the gradient.
To answer this problem we must evaluate the
horizontal gradient of the vertical component or
1 d
ZE
r d
1 d 2 M cos 
r d
r3
See Equation 7.20
Tom Wilson, Department of Geology and Geography
Evaluate the horizontal gradient
dZ E dZ E
d 2 M cos 


dS
rd rd
r3
2 M sin 
2
2

  HE  
20, 000nT
3
8
r r
r
6.3x10 cm
Since  is co-latitude, the direction of increasing  is
southward (in the northern hemisphere). As we
travel from pole to equator ZE decreases, thus the
gradient is negative.
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
ZA
is a function of the unit-less variable x/z
Z max
Dipole/sphere
Relative Response Functions

x2 
 2  2 
z 
1 
2  x 2 5 / 2
 2  1
z

1.0
cal
Verti
0.6
Vertical cylinder
der
Cylin
Normalized Responses
0.8
0.4
1
x2
( 2  1) 3 / 2
z
0.2
h
Sp
er
Horizontal cylinder
e
0.0
Horizontal Cylinder
-0.2
-3
-2
-1
0
X/Z (no units)
1
2
3

1 x
2
2
1 x
z2

2
z2
The vertical field is often used to make a quick estimate of the magnitude
of an object. This is fairly accurate as long as i is 60 or greater
Tom Wilson, Department of Geology and Geography
Vertically polarized sphere or dipole
8 3
R kFE
Z max  3 3
z
RememberI  kFE
Vertically polarized horizontal cylinder
Area   R 2
R
Tom Wilson, Department of Geology and Geography
Area

Z max
2R 2 I

z2
Considerable difference in magnitude of
I  kFE  0.001x55, 000nT  55nT
R
Area


0.5m2

 0.4m
z  1.5  0.4  1.9m
For the dipole
Z max
8 3
 R kFE 8.38x0.064 x55 0.536
3 3


 0.078nT
3
z
1.9
6.86
For the horizontal cylinder
Z max
Tom Wilson, Department of Geology and Geography
2 R 2 kFE


z2
4. 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
The map view clearly indicates that consideration of two
possible origins may be appropriate - sphere or vertical cylinder.
Tom Wilson, Department of Geology and Geography
In general one will not make such extensive comparisons.
You may use only one of the diagnostic positions, for
example, the half-max (X1/2) distance for an anomaly to
quickly estimate depth if the object were a sphere or
buried vertical cylinder….
Burger limits his discussion to half-maximum relationships.
X1/2 = Z/2
X1/2 = 0.77Z
X1/2 = Z
X1/2 = Z/2
Breiner, 1973
Tom Wilson, Department of Geology and Geography
Just as an aside: The sample rate you use will depend
on the minimum depth of the objects you wish to find.
Your sample interval should probably be no greater than X1/2.
But don’t forget that equivalent solutions with
shallower origins do exist!
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
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  
z 2 
Z A ( x) 1 

5/ 2
Z max
2  x2

  1
 z2
 We’ll make quick work of it an use only


three diagnostic positions (red above)
Tom Wilson, Department of Geology and Geography
Z max 
R 2 I
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
ZA
1
 2
Z max
x
( 2  1)3 / 2
z
Again, we can get by with only three
diagnostic positions (red above)
Tom Wilson, Department of Geology and Geography
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
2.86
3.1
3.1
3.35
3.35
2.17
1.31
0.81
ZCylinder
1.95
1.95
2.03
2.03
2.00
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
X1/2 = 2.5 meters
2
X1/4 = 3.7 meters
1.37
Tom Wilson, Department of Geology and Geography
ZSphere
5.01
5.08
5.0
5
5.07
5.1
Multipliers
Cylinder
2.17
1.31
0.81
ZCylinder
3.47
3.28
3.00
Algebraic manipulation
5. 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
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
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
anomaly
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
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
30
35
180
190
200
210
220
Distance along profile
Tom Wilson, Department of Geology and Geography
230
Make sure the
scale of your
graph is 1:1
…. 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
Again - follow the recommended reporting format.
Specifically address listed points (1-5).
Tom Wilson, Department of Geology and Geography
Sampling issues – for leisure consideration …
Jump to last slide for reminders
You are asked to run a magnetic
survey to detect a buried drum.
What spacing do you use
between observation points?
Tom Wilson, Department of Geology and Geography
How often would you have to sample to detect this drum?
Tom Wilson, Department of Geology and Geography
…. how about this one?
The anomaly of the drum drops to ½ at
a distance = ½ the depth.
Tom Wilson, Department of Geology and Geography
Sampling does depend on available equipment!
As with the GEM2, newer
generation magnetometers can
sample at a walking pace.
Tom Wilson, Department of Geology and Geography
Remember, the field of a buried drum can be
approximated by the field of a dipole or buried
sphere. X1/2 for the sphere (the dipole) equals
one-half the depth z to the center of the dipole.
The half-width of the anomaly over any given
drum will be approximately equal to its depth
Or X1/2 =Z/2
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
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
General Review this coming Thursday
Turn in your magnetics lab report Thursday,
December 10th.
Exam, Friday December 17th; 3-5pm
Tom Wilson, Department of Geology and Geography