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
ZA
Z max
Vertically polarized sphere or dipole
Z A ( x)

Z max
Vertically polarized vertical cylinder
Vertically polarized horizontal cylinder
Tom Wilson, Department of Geology and Geography
ZA
Z max
1
5/2
2 2

x

 1
2
z



1

Z max
Z ( x)
2

x 
2 

2 

z 

(
x
2
z
2
 1)
1 x

3/2
2
z
2
1  x 2


2
z 

2
8
Vertically polarized sphere or dipole
Z max  3
 R kF E
3
z
3
Remember
R I
2
Vertically polarized vertical cylinder
Z max 
z
2
2 R I
2
Vertically polarized horizontal cylinder
Tom Wilson, Department of Geology and Geography
Z max 
z
2
I  kF E
Actual Anomaly (Z A )
15
3/4
2/3
10
t hs
rds
1/2
1/3
5
rd
1/4
th
0
-10
-5
0
5
10
x (kilometers)
D iag no stic po sit io n
Z /Z m ax
3/4
2/3
1/2
1/3
1/4
X
X 3 /4 = 1.63km
X 2 /3 = 1.9km
X 1 /2 = 2.6km
X 1 /3 = 3.6km
X 1 /4 = 4.3km
-1
(x/z)
D epth Index m u lt ip lier
2.17
1.79
1.31
0.96
0.81
Z
3.54km
3.40km
3.4km
3.46km
3.48km
We measure the distances (x) to the various diagnostic positions
and then convert those x’s to z’s using the depth index multipliers
which are just the reciprocal of the x/z values at which the anomaly
drops to various fractions of the total anomaly magnitude.
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
2

x 
 2  2 
z 
1 
5/2
2  x2

 2  1 
z

1.0
0.8
ic a l
Vertical cylinder
C y li
1
nder
N orm alized Res pons es
V e rt
0.6
0.4
(
0.2
x
2
z
2
 1)
3/2
Sp
he
Horizontal cylinder
re
1 x
0.0
Horizontal Cylinder
-0.2
-3
-2
-1
0
X/Z (no units)
1
2
3

1 x
2
z
2

2
2
z
2
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
For these three magnetic objects, the anomalies associated with the
sphere and horizontal cylinder both drop off to1/2 their maximum
value at X = ½ the depth Z
Relative Response Functions
X/Z
1.0
Vertical
Cylinder
Sphere
Horizontal
Cylinder
X3/4
0.46
0.315
0.31
X1/2
0.766
0.5
0.495
X1/4
1.23
0.73
0.68
0.8
ic a l
C y li
nder
N orm alized Res pons es
V e rt
0.6
0.4
Depth Index Multipliers
Vertical
Cylinder
Sphere
Horizontal
Cylinder
X3/4
2.17
3.18
3.23
X1/2
1.305
2
2.02
X1/4
0.81
1.37
1.47
0.2
Sp
he
re
0.0
Horizontal Cylinder
-0.2
-3
-2
-1
0
X/Z (no units)
Tom Wilson, Department of Geology and Geography
1
2
3
The vertical cylinder behaves
like a magnetic monopole.
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
Remember how the proton precession
magnetometer works. Protons precess about
the earth’s total field with a frequency directly
proportional to the earth’s field strength
f 
M
2 L
F 
GF
2
F E  23 . 4874 f
Tom Wilson, Department of Geology and Geography
The proton
precession
magnetometer
measures the scalar
magnitude of the
earth’s main field.
The gradient is just the rate of change in some
direction - i.e. it’s just a derivative.
How would you evaluate the vertical gradient of the
vertical component of the earth’s magnetic field?
ZE 
2 M cos 
Tom Wilson, Department of Geology and Geography
r
3
HE 
M sin 
r
3
The vertical gradient is just the variation
of ZE with change in radius or distance
from the center of the dipole.
ZE 
2 M cos 
r
3
d  2 M cos  



3
dr
dr 
r

dZ E
dZ E
dr
Tom Wilson, Department of Geology and Geography

 6 M cos 
r
4
dV E
dr
Total versus Vertical Gradient
800
700
600
Y (nT)
500
Vertical
Gradient
400
300
200
100
0
-100
-200
0
5
10
15
20
25
X (meters)
Tom Wilson, Department of Geology and Geography
30
35
40
45
Total Field
Vertical Gradient
http://rubble.phys.ualberta.ca/~doug/G221/MagLecs/magrem.html
Tom Wilson, Department of Geology and Geography
Visit http://www.gemsys.ca/papers/site_characterization_using_gsm-19gw.htm
Tom Wilson, Department of Geology and Geography
Can you evaluate the vertical gradient of the
horizontal component of the earth’s magnetic field?
Representing the earth’s
horizontal field in dipole form as
HE 
The vertical gradient is just the variation
with change of radius or
Tom Wilson, Department of Geology and Geography
M sin 
r
3
dH
dr
E
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
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
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
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
4. How many drums?
Area of one drum ~
Outline of Drum Cluster
Derived from the magnetics model
10
4 square
feet
D e p th
15
T otalA rea 
20
N D rum s 
25
What’s wrong
with the format of
this plot?
30
35
180
190
200
210
Distance along profile
Tom Wilson, Department of Geology and Geography
220
230
1
B ase x H eight
2
Total Area
Area of one D rum
…. compare the field of the
magnetic dipole field to that of the
gravitational monopole field
Monopole f ield varies as 
1
r
Gravity:500, 1000, 2000m
2
ZE 
2 M cos 
r
3
0.12
0.1
ZE 
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
2M
r
3
For the dipole field, an increase
in depth (r) from 4 meters to 16 Dipole field varies as  1
3
meters produces a 64 fold
r
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.
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
HE
1 d
r d
d  pl cos  





2
ds
rd 
rd  
r

dV
dV
Thus H E 
ZE  
Tom Wilson, Department of Geology and Geography
dV
dr

M sin 
r
3
2 M cos 
r
3
To answer this problem we must evaluate the
horizontal gradient of the vertical component -
1 d
or
r d
ZE
1 d 2 M cos 
r d
r
3
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
Diagnostic position
X at Z/Zmax
9/10
3/4
2/3
1/2
1/3
1/4
0
General form

2
 1 x
2
2 R I 
z
Z 
2
2

z
1  x 2


2
z 

2






(x/z)-1
Depth Index multiplier
5.32
3.23
2.7
2.02
1.64
1.47
1
x/z
0.188
0.31
0.37
0.495
0.61
0.68
1.0
Normalized shape term
Z ( x)
Z max
1 x

2
z
2
1  x 2


2
z 

2
A rea   R
2 R I
2
Z m ax 
Tom Wilson, Department of Geology and Geography
z
2
R 
A rea

2
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
8
Z m ax  3
 R kH
3
z
Diagnostic position
X at Z/Zmax
9/10
3/4
2/3
1/2
1/3
1/4
0
3
Z A ( x)
Z max

1
2

x 
2 

2


z


5/2
2 2

x

 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
R I
2
Z max 
z
Diagnostic position
X at Z/Zmax
9/10
3/4
2/3
1/2
1/3
1/4
2
ZA
1

Z max
Tom Wilson, Department of Geology and Geography
(
x
2
z
2
 1)
3/2
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
R I
2
6. Given that Z m ax 
z
2
derive an expression for the radius,
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