Environmental and Exploration Geophysics I
Magnetic Methods (IV)
tom.h.wilson
wilson@geo.wvu.edu
Department of Geology and Geography
West Virginia University
Morgantown, WV
Could a total-field magnetic
survey detect the illustrated
burial chamber (spherical void) in
a region where FE = 55,000nT
and i = 70o?
To do this we can compute FAT directly
using equations 7-36 and 7-37.
ZA
HA
4
 R3kFE sin i
 3
2
2 3/ 2
x

z


4
 R3kFE cos i
 3
2
2 3/ 2
x

z






3z 2
3xz



cot
i

1
 2



2
2
2
(x  z )
x

z









 

3x 2
3xz




1

tan
i




2
2
  x2  z 2 
  (x  z )






-40
-30
-20
-10
0
10
20
30
40
2.00
Depth to sphere center (m)
Sphere radius (m)
Susceptibility (cgs emu)
Horizontal increment (m)
3
1
0.0001
2
Earth's field (FE) (nT)
Earth's field inclination (?)
55000
70
Magnetic Field Intensity (nT)
1.50
1.00
0.50
0.00
-0.50
-1.00
Position (m)
We’ve got an FAT over 1.5 nT. A typical proton precession
magnetometer reads differences of 1 nT. This is much too
small of an anomaly to detect. Also, consider that our
magnetometer usually sits atop a 2 meter rod. How will that
change the maximum value of FAT.
ZA
HA
FAT
At 5 meters the inverse cube
relationship produces a pretty
significant drop in the magnitude of FAT.
-40
-30
-20
-10
0
10
20
30
40
0.40
Magnetic Field Intensity (nT)
0.30
Depth to sphere center (m)
Sphere radius (m)
Susceptibility (cgs emu)
Horizontal increment (m)
5
1
0.0001
2
Earth's field (FE) (nT)
Earth's field inclination (?)
55000
70
0.20
0.10
0.00
-0.10
-0.20
Position (m)
At 0.35 nT this is much
too small to be detected.
ZA
HA
FAT
In class the other day, we
recorded the following
numbers at approximately
5 minute intervals.
55400
55300
55200
55100
F (nT)
Time
0
5
10
15
20
25
30
35
40
60
FE
FE
54792
54784
54794
54792
54484
55310
54781
54783
54754
54787
54600
54500
54400
0
54785
F (nT)
20
30
40
50
60
70
The smaller background
variations have a standard
deviation of about ±13nT.
54790
54780
54775
FE
54770
54765
54760
54755
54750
30
10
Time (minutes)
54795
20
FE
54700
54800
10
54900
54800
FE
0
55000
40
Time (minutes)
50
60
70
58500
58000
time
12:02
12:05
12:07 PM
12:14 PM
12:22 PM
12:27 PM
12:33 PM
12:42 PM
12:46 PM
12:53 PM
FET
54962
57764
56539
57783
57797
55023
54944
54965
55602
55176
F
2802
-1225
1244
14
-2774
-79
21
637
-426
T
1547
4349
3124
4368
4382
1608
1529
1550
2187
1761
57500
57000
56500
56000
55500
55000
54500
11:52 12:00 12:07 12:14 12:21 12:28 12:36 12:43 12:50 12:57
Here we have much larger deviations of ±1283nT. Of course
these measurements were made in the building so they are not
representative of what might be happening at a field site, but you
shouldn’t loose sight of the potential influence of background
noise on conclusions drawn from the finest of models.
Is there a quicker way to estimate the possibility
that we might observe this anomaly?
We could use the simple geometrical object
representations.
Which object would you use to approximate this situation?
The vertical field of a simple
sphere or dipole
Z max
8 3
R kFE
3 3
z
Remember where this equation comes from?
The magnetic response of a sheet of dipoles is
obtained by carrying out integrations over two
sheets: one consisting of the negative poles and
the other of the positive poles.
Z Atopsheet
 /2
Idy
 2
cos    2 Id  2 I
 / 2
r
where
I m
area
These individual integrations are very
similar to the ones used to derive the
Bouguer plate effects.
See Equation 7-45, on page 429 of Berger.
Z Atopsheet  2I
The effect of the bottom sheet will also equal
Z Abottomsheet  2I
The negative
Sign comes from the convention that defines upward pointing
vectors (from the positive pole) as negative. So the net result
….. is
Z Ainfinitesheet  2I  2I  0
The process yields an intermediate more useful result.
2 I top and
The contribution from the top of the rod is
the contribution from the base of the rod is  2Ibot
The total field of this infinitely long intrusive (dike) will be
Z A  2Itop  2Ibot
or just

Z A  2I top  bot

(7.46)
In Problem 7.6 we are asked to determine the vertical field anomaly
(ZA) over the intrusive shown in the diagram (see text) at a point
directly over the center of the intrusive. The intrusive has a very long
strike-length. FE is vertical and equal to 55,000nT. Use equation 7-46
and compute ZA for two cases. Case 1: assume that the base of the
intrusive is located at 12km beneath the surface. Case 2: assume the
base is located at infinity. Compare the two results.

Z A  2I top  bot


Z A  2I top  bot
Z A  2 I  top 

Determining 
In general for the top,
0.0031
55000
5
8.5
7
Calculate bot in a similar fashion and take
their difference at each point x along a profile.
160
140
Anomalous Field Intensity (nT)
Susceptability
Main Field Intensity
Depth to Top
Depth to Base
Horizontal Width
top
 xw 
 xw 
2   tan 1 
2
 tan 1 
 z 
 z 




120
100
80
60
Z
40
20
0
-20
-40
-40
-30
-20
-10
0
10
Distance from center (meters)
20
30
40
Is there a quicker way to estimate the possibility
that we might observe this anomaly?
Again, those simple geometrical object
representations might get us in the ballpark.
Which object would you use to approximate this situation?
The vertical field of a
horizontal cylinder
Z max
2R 2 I

z2
The magnetic data graphed below represent vertical field
measurements (ZA) in an area where shallow crystalline
basement is overlain by non-magnetic sediments. The
basement gneisses are intruded by numerous thin
kimberlite pipes. Both gneisses and kimberlite pipes are
eroded to a common level surface. Determine the likely
depth to basement. FE=58,000nT and i=80o.
Problem 7-7
6
Intensity (nT)
5
4
3
2
1
0
-200
-150
-100
-50
0
50
Distance in m eters
100
150
200
Problem 7-7
6
Intensity (nT)
5
4
3
2
1
0
-200
-150
-100
-50
0
50
Distance in m eters
Do you remember what to do?
What simple geometrical object should be
used in this case and what property of the
curve do you need to measure?
100
150
200
Determine the depth z to the center of the
basalt flow. Also indicate whether you
think the flow is faulted (two offset semiinfinite sheets) or just terminates (a semiinfinite sheet). What evidence do you
have to support your answer? Refer to
illustration on page 477 and associated
discussion.
This problem relies primarily on a
qualitative understanding of equation 7-47.
 1 x 
1 x 
Z A  2kFE  tan    tan 

z
 z  t 

Field of the semi-infinite plate
X = 0 at the surface point directly over the edge of the
plate. The field at a point X is derived from the two angles
shown below - 1 and 2 - used in the text.
1
z
2
t
This is just a special case of the preceding example

The angle subtended by the top of the sheet at x is
2

The angle subtended by the bottom of the sheet at x is
2
 top
 bot
Z A ( x) half  sheet  2 I 1   2 
 top  1 

 x
 tan 1  
2
z
 bot   2 

 x 
 tan 1 

2
 z t 

 x
 x 
Z A  2 I  tan 1   tan 1

z
 z  t 

 1 x 
1 x 
Z A  2kFE  tan    tan 
 Eqn. 7 - 47
z
 z  t 

z
Faulted basaltic sheet or isolated sill?


1  x 
1  x  
Z A  2kFE   tan     tan 

2
z
2
z

t
 



Simple-geometrical-object representation
Vertically Polarized Faulted Horizontal Slab
Z  It
x
x2  z 2


It  x
Z 2
z 
x2
1 2

z
Z max






z
 Z ( x  z )  It 2
z  z2
It
Z max 
2z
Horizontal Slab
Z max 
8
6
It
 6.14nT
2z
4
The edge of the fault is located at the
2
inflection
point.
Z (nT)
Xmax=z
0
z = 1.75m
-2
t = 0.5m
-4
-6
-8
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
x (meters)
z
t
Vertically Polarized Faulted Horizontal Slab
or Semi Infinite sheet
Diagnostic position
Z/Zmax
0
1/4
1/2
3/4
1
(x/z)-1
Depth Index multiplier
7.87
3.73
2.21
1
x/z
0
0.127
0.268
0.452
1
Z
Z max


2 x

z
x2
1 2

z







 x
 x 
Z A  2kFE  tan 1   tan 1

z
z

t





Variations of Z across edge of isolated sheet.
4
3
2
1
Z (nT)
0
-1
-2
-3
-4
-30
-20
-10
0
10
20
30
x (meters)
z
t
----------------------++++++++++++++++
Half-plate (the Slab, semi
infinite plate, the half-sheet …)
Sem i-Infinite Sheets
4
3
upper
sheet
2
1
Z (nT)
combined
response
0
-1
lower
sheet
-2
-3
-4
-30
-20
-10
0
10
20
30
x (meters)
Look carefully at the anomaly profile shown in Problem 7-8
and consider the overall shape of the anomaly and how it
may allow you to discriminate between the faulted versus
terminated flow interpretations.
The yellow curve at left is derived
from the full computation (terms
shown below)
8
Anomalous Field Intensity (nT)
6
4
2
SGOA
0
Sum ZA + ZB
-2
-4


x 
1  x 
1 
Z A  2kFE   tan     tan 

2
z
2
z

t
 A
 A 



x 
1  x 
1 
Z B  2kFE   tan     tan 

2
z
2
z

t
 B
 B 

-6
-40
-20
0
20
40
-8
Distance from fault (meters)
The blue curve is obtained from the
SGO approximation
x
Z  It 2
x  z2
Peter’s half slope
Anomalous Field Intensity (nT)
300
250
200
150
Z
100
50
0
-50
-20
-15
-10
-5
0
5
10
15
Distance from center (km)
We’ll come back to this on Thursday
20
The final will be comprehensive, but the focus
will be on material covered since the last exam.
Come prepared to ask questions