Environmental and Exploration Geophysics I Magnetic Methods part

Environmental and Exploration Geophysics I Magnetic Methods (part II) tom. h. wilson tom. wilson@mail. wvu. edu Department of Geology and Geography West Virginia University Morgantown, WV

Due Tuesday, Dec. 5 th Abstract Background Results Discussion Abstract – a brief description of what you did and the results you obtained (in 200 -250 words).

Since the bedrock is magnetic, we have no way of differentiating between anomalies produced by bedrock and those produced by buried storage drums.

Acquisition of gravity data allows us to estimate variations in bedrock depth across the profile. With this knowledge, we can directly calculate the contribution of bedrock to the magnetic field observed across the profile.

Compare bedrock anomaly to total anomaly illustrate With the information on bedrock configuration we can clearly distinguish between the magnetic anomaly associated with bedrock and that associated with buried drums at the site.

How well do model and actual data agree? Location and extent

Improvements? Work using both forward and inverse calculations – iterate and adjust…

How many drums are represented by the triangular-shaped object you entered into your model? Use the magic eye to get the coordinates of the polygon defining the drums Plot the corner coordinates for the triangular shaped object you derived at 1: 1 scale and compute the area.

How many drums? Area of one drum ~ 4 square feet What’s wrong with the format of this plot? We’ll talk about the last bullet (1/r 3) on the results-tobe-discussed list a little later.

Don’t forget to include a nice summary of your effort. This includes a brief restatement of the problem with highlights of results. Also include perspectives on the validity of the results – the potential for ambiguity in the findings.

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www. sensoft. ca

The potential is the integral of the force (F) over a displacement path. From above, we obtain a basic definition of the potential (at right) for a unit positive test pole (mt). Note that we follow Berger et al. and assume the leading constants are ~1.

Thus - H (i. e. F/ptest, the field intensity) can be easily derived from the potential simply by taking the derivative of the potential

Consider the field at a point along the axis of a dipole as noted in problem 1. The dipole in this case could be a buried well casing. The field has vector properties however, in this case vectors are collinear and its easy to determine the net effect.

In terms of the potential we can write In the case at right, r+ is much greater than r- , thus in

Thus, the potential near either end of a long dipole behaves like the potential of an isolated monopole. If we are looking for abandoned wells, we expect to find anomalies similar to the gravity anomalies encountered over buried spherical objects. 21

Consider the case where the distance to the center of the dipole is much greater than the length of the dipole. This allows us to treat the problem of computing the potential of the dipole at an arbitrary point as one of scalar summation since the directions to each pole fall nearly along parallel lines.

If r is much greater than l (distance between the poles) then the angle between r+ and r- approaches 0 and r, r+ and r- can be considered parallel so that the differences in lengths r+ and r- from r equal to plus or minus the projections of l/2 into r.

r- r r+

Working with the potentials of both poles. . Recognizing that pole strength of the negative pole is the negative of the positive pole and that both have the same absolute value, we rewrite the above as

Converting to common denominator yields where pl = M – the magnetic moment From the previous discussion , the field intensity H is just

H - monopole = H - dipole This yields the field intensity in the radial direction i. e. in the direction toward the center of the dipole (along r). However, we can also evaluate the horizontal and vertical components of the total field directly from the potential.

H Toward dipole (Earth’s) center Vd represents the potential of the dipole.

HE is represented by the negative derivative of the potential along the earth’s surface or in the S direction.

Where M = pl and Let’s tie these results back into some observations made earlier in the semester with regard to terrain conductivity data. 32

Given What is HE at the equator? … first what’s ? is the angle formed by the line connecting the observation point with the dipole axis. So , in this case, is a colatitude or 90 o minus the latitude. Latitude at the equator is 0 so is 90 o and sin (90) is 1.

At the poles, is 0, so that What is ZE at the equator? is 90

ZE at the poles …. The variation of the field intensity at the pole and along the equator of the dipole may remind you of the different penetration depths obtained by the terrain conductivity meters when operated in the vertical and horizontal dipole modes.

Also compare the field of the magnetic dipole field to that of the gravitational monopole field Gravity: 500, 1000, 2000 m Increase r by a factor of 4 reduces g by a factor of 16

For the dipole field, an increase in depth (r) from 4 meters to 16 meters produces a 64 fold decrease in anomaly magnitude Thus the 7. 2 n. T anomaly (below left) produced by an object at 4 meter depths disappears into the background noise at 16 meters. 7. 2 n. T 0. 113 n. T

Recall you are 20 centimeters from the negative and positive poles of a dipole as shown below, and that each pole has pole strength of 1 ups. Observation point 20 cm - + What is the magnetic field intensity at the observation point in nano. Teslas?

With H in units of Oersteds, p in ups, and r in cm, we have units of - And the equivalent units for ups

3. What is the horizontal gradient in n. T/m of the Earth’s vertical field (ZE) in an area where the horizontal field (HE) equals 20, 000 n. T and the Earth’s radius is 6. 3 x 108 cm. 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

To answer this problem we must evaluate the horizontal gradient of the vertical component or Take a minute and give it a try.

4. A buried stone wall constructed from volcanic rocks has a susceptibility contrast of 0. 001 cgs emu with its enclosing sediments. The main field intensity at the site is 55, 000 n. T. 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 today’s handout. Background noise at the site is roughly 5 n. T.

Vertically Polarized Horizontal Cylinder General form Normalized shape term

Vertical Magnetic Anomaly Vertically Polarized Sphere Zmax and ZA refer to the anomalous field, i. e. the field produced by the object in consideration The notation can be confusing at times. In the above, consider H = FE= intensity of earth’s magnetic field at the survey location.

Vertically Polarized Vertical Cylinder

Sphere, Vertical Cylinder; Diagnostic positions Multipliers Sphere ZSph X 3/4 =0. 9 3. 18 2. 17 X 1/2 =1. 55 2 1. 31 X 1/4 =2. 45 1. 37 0. 81 ere Multipliers Cylinder z = _____ The depth ZCylinde r

Sphere or cylinder? Diagnostic positions Multipliers Sphere X 3/4 = 1. 6 ZSphere Multipliers Cylinder ZCylinder 3. 18 2. 17 2. 82 X 1/2 = 2. 5 2 1. 31 2. 69 X 1/4 = 3. 7 1. 37 0. 81 2. 34

6. Given that derive an expression for the radius, where I = k. HE. 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 =55000 n. T. Zmax (62. 2 n. T from graph below) is the maximum vertical component of the anomalous field produced by the vertical casing.

TODAY Tuesday Dec. 5 th Thursday Dec. 7 th