Environmental and Exploration Geophysics I

1. 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

2. More background discussion & Questions Magnetics Lab Due Tuesday, Dec. 5th Abstract Background Results Discussion Abstract – a brief description of what you did and the results you obtained (in 200-250 words).

3. What are you looking for? Since the bedrock is magnetic, we have no way of differentiating between anomalies produced by bedrock and those ? produced by buried storage drums.

4. Why gravity? 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.

5. 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.

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

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

8. 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.

9. 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/r3) on the results-to-be-discussed list a little later.

10. Conclusions! 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.

11. www.sensoft.ca

12. www.sensoft.ca

13. Consider Problem 1 in terms of the potential 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.

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

15. Problem 1 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.

16. In terms of the potential we can write In the case at right, r+ is much greater than r- , thus in 0

17. 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

18. The Dipole Field 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.

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

20. r- r r+

21. 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

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

23. Thus .. 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.

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

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

28. 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

29. 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 90o minus the latitude. Latitude at the equator is 0 so  is 90o and sin (90) is 1.

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

31. 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.

32. Final Lab Question Also compare the field of the magnetic dipole field to that of the gravitational monopole field Gravity:500, 1000, 2000m A more rapid decay Increase r by a factor of 4 reduces g by a factor of 16

33. 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 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

34. Observation point 20cm + - Problem 2 - Part 1 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. What is the magnetic field intensity at the observation point in nanoTeslas?

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

36. 3. 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. 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

37. Problem 3 To answer this problem we must evaluate the horizontal gradient of the vertical component - or Take a minute and give it a try.

38. 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 today’s handout. Background noise at the site is roughly 5nT.

39. Problem 4 Vertically Polarized Horizontal Cylinder General form Normalized shape term

40. Vertical Magnetic Anomaly Vertically Polarized Sphere Question 5 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.

41. Question 5 Vertically Polarized Vertical Cylinder

42. Sphere, Vertical Cylinder; z = __________ The depth

43. Sphere or cylinder?

44. 6. Given that 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.

45. Take some time to look over the assignment. Bring questions to class next time .....

46. Review of Due Dates Part 1 Magnetics Problem set TODAY Magnetics Lab Tuesday Dec. 5th Paper Summaries Thursday Dec. 7th Part 2 Magnetics Problem set Thursday Dec. 7th