Environmental and Exploration Geophysics I

tom.h.wilsonwilson@geo.wvu.edu

Department of Geology and GeographyWest Virginia University

Morgantown, WV

Magnetic Magnetic Methods (IV)Methods (IV)

Could a total-field magnetic survey detect the illustrated burial chamber (spherical void) in a region where FE = 55,000nT and i = 70o?

32

3/ 2 2 22 22 2

4cos 3 33 1 tan

( )

E

A

R kF i x xzH i

x zx zx z

32

3/ 2 2 22 22 2

4sin 3 33 cot 1

( )

E

A

R kF i z xzZ i

x zx zx z

To do this we can compute FAT directly using equations 7-36 and 7-37.

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.

Depth to sphere center (m) 3Sphere radius (m) 1Susceptibility (cgs emu) 0.0001Horizontal increment (m) 2

Earth's field (FE) (nT) 55000Earth's field inclination (?) 70

-1.00

-0.50

0.00

0.50

1.00

1.50

2.00-40 -30 -20 -10 0 10 20 30 40

Position (m)

Ma

gn

eti

c F

ield

Inte

ns

ity

(n

T)

ZAHA

FAT

At 5 meters the inverse cube relationship produces a pretty

significant drop in the magnitude of FAT.

At 0.35 nT this is much too small to be detected.

-0.20

-0.10

0.00

0.10

0.20

0.30

0.40-40 -30 -20 -10 0 10 20 30 40

Position (m)

Ma

gn

eti

c F

ield

Inte

ns

ity

(n

T)

ZAHAFAT

Depth to sphere center (m) 5Sphere radius (m) 1Susceptibility (cgs emu) 0.0001Horizontal increment (m) 2

Earth's field (FE) (nT) 55000Earth's field inclination (?) 70

FE

54400

54500

54600

54700

54800

54900

55000

55100

55200

55300

55400

0 10 20 30 40 50 60 70

Time (minutes)

F (

nT

)

FETime FE0 547925 5478410 5479415 5479220 5448425 5531030 5478135 5478340 5475460 54787

In class the other day, we recorded the following

numbers at approximately 5 minute intervals.

FE

54750

54755

54760

54765

54770

54775

54780

54785

54790

54795

54800

0 10 20 30 40 50 60 70

Time (minutes)

F (

nT

)

FE

The smaller background variations have a standard deviation of about ±13nT.

time FET F T12:02 54962 2802 154712:05 57764 -1225 4349

12:07 PM 56539 1244 312412:14 PM 57783 14 436812:22 PM 57797 -2774 438212:27 PM 55023 -79 160812:33 PM 54944 21 152912:42 PM 54965 637 155012:46 PM 55602 -426 2187

12:53 PM 55176 1761

54500

55000

55500

56000

56500

57000

57500

58000

58500

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?

3

3

max38

z

kFRZ

E

The vertical field of a simple sphere or dipole

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.

/ 2

/ 22 cos 2 2

topsheetA

IdyZ Id I

r

where mI area

See Equation 7-45, on page 429 of Berger.

These individual integrations are very similar to the ones used to derive the Bouguer plate effects.

IZtopsheetA 2

The effect of the bottom sheet will also equal

IZsheetbottomA 2

The negative

Sign comes from the convention that defines upward pointing vectors (from the positive pole) as negative. So the net result ….. is

022 nfinite

IIZsheetiA

The process yields an intermediate more useful result.

The contribution from the top of the rod is topI2 and

the contribution from the base of the rod is botI2

The total field of this infinitely long intrusive (dike) will be

bottopA IIZ 22

or just

bottopA IZ 2

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.

(7.46)

bottopA IZ 2

bottopA IZ 2

2A topZ I

Determining

Susceptability 0.0031Main Field Intensity 55000Depth to Top 5Depth to Base 8.5Horizontal Width 7

-40

-20

0

20

40

60

80

100

120

140

160

-40 -30 -20 -10 0 10 20 30 40

Distance from center (meters)

An

om

alo

us

Fie

ld I

nte

nsi

ty (

nT

)

Z

1 12 2tan tantop

w wx x

z z

In general for the top,

Calculate bot in a similar fashion and take their difference at each point x along a profile.

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?

2

2

max

2

z

IRZ

The vertical field of a

horizontal cylinder

Problem 7-7

0

1

2

3

4

5

6

-200 -150 -100 -50 0 50 100 150 200

Distance in meters

Inte

nsi

ty (

nT

)

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.

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?

Problem 7-7

0

1

2

3

4

5

6

-200 -150 -100 -50 0 50 100 150 200

Distance in meters

Inte

nsi

ty (

nT

)

Determine the depth z to the center of the basalt flow. Also indicate whether you think the flow is faulted (two offset semi-infinite sheets) or just terminates (a semi-infinite sheet). What evidence do you have to support your answer? Refer to illustration on page 477 and associated discussion.

tantan2 11

tz

x

z

xkFZ EA

This problem relies primarily on a qualitative understanding of equation 7-47.

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

2

z

t

This is just a special case of the preceding example

The angle subtended by the top of the sheet at x is top

2

The angle subtended by the bottom of the sheet at x is bot

2

212)( IxZ sheethalfA

11 tan

2top

x

z

12 tan

2bot

x

z t

tantan2 11

tz

x

z

xIZA

47-7 Eqn. tantan2 11

tz

x

z

xkFZ EA

z

Faulted basaltic sheet or isolated sill?

1 12 tan tan2 2A E

x xZ kF

z z t

Vertically Polarized Faulted Horizontal Slab

22 zx

xItZ

2

22

1z

x

x

z

ItZ

22max )(zz

zItzxZZ

z

ItZ

2max

Simple-geometrical-object representation

Horizontal Slab

-8

-6

-4

-2

0

2

4

6

8

Z (nT)

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

x (meters)

t

nTz

ItZ 14.6

2max

Xmax=z

z

z = 1.75m

t = 0.5m

The edge of the fault is located at the inflection point.

Vertically Polarized Faulted Horizontal Slab or Semi Infinite sheet

2

2max 1

2

z

x

x

zZ

Z

-4

-3

-2

-1

0

1

2

3

4

Z (nT)

Variations of Z across edge of isolated sheet.

-30 -20 -10 0 10 20 30

x (meters)

Half-plate (the Slab, semi infinite plate, the half-sheet …)

- - - - - - - - - - - - - - - - - - - - - - -+ + + + + + + + + + + + + + + +

z

t

tantan2 11

tz

x

z

xkFZ EA

combinedresponse

uppersheet

lowersheet

-4

-3

-2

-1

0

1

2

Z (nT)

3

4

Semi-Infinite Sheets

-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.

-8

-6

-4

-2

0

2

4

6

8

-40 -20 0 20 40

Distance from fault (meters)

An

om

alo

us

Fie

ld I

nte

nsi

ty (

nT

)

SGOA

Sum ZA + ZB

1 1

1 1

2 tan tan2 2

2 tan tan2 2

A EA A

B EB B

x xZ kF

z z t

x xZ kF

z z t

The yellow curve at left is derived from the full computation (terms

shown below)

22 zx

xItZ

The blue curve is obtained from the

SGO approximation

Peter’s half slope

-50

0

50

100

150

200

250

300

-20 -15 -10 -5 0 5 10 15 20

Distance from center (km)

An

om

alo

us

Fie

ld I

nte

nsi

ty (

nT

)

Z

We’ll come back to this on Thursday

The final will be comprehensive, but the focus will be on material covered since the last exam.

Come prepared to ask questions