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

2. Preparations for today - cover background reading pages of Chapter 7 and prepare questions for problems 1 through 3 on page 449.
In today’s class presentation we continue to develop key quantitative relationships in the use of magnetic methods, and as we do, we will tie them into homework problems 1 through 3.
Problems 7-1 through 7-3 will be Tuesday after Thanksgiving break (Dec. 3rd).

3. Potential versus Force
As noted the other day, the potential is the integral of the force over a displacement path.
From above, we obtain a basic definition of the potential (at right) for a unit positive test pole (mt).

4. Thus - F (sometimes also referred to as H) can be easily derived from the potential simply by taking the derivative of the potential

5. re: problem 7-2
Consider the field at a point along the axis of a dipole. 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.

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

7. 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

8. Considering the general case of the dipole field we have to take direction into account. However, in the case where the distance to the center of the dipole is much greater than the length of the dipole we can still treat the problem of computing the potential as one of scalar summation since the directions to each pole fall nearly along parallel lines.

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

10. r+ r r-

11. 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

12. Converting to common denominator yields
re: problem 7.1 -
Converting to common denominator yields
See Eqs where ml = M
From the previous discussion , the field intensity H (or F as in Berger) is just

13. H - monopole =
H - dipole
This yields the field intensity in the radial direction - i.e. in the direction toward the center of the dipole. However, we can also evaluate the horizontal and vertical components of the total field directly from the potential. 27

14. Vd represents the potential of the dipole.

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

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

19. 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.

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

21. 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.

22. Recall rule-of-thumb values for penetration depths have the vertical dipole exploration depth twice that of the horizontal dipole.

23. Gradients
& Problem 7-1
Berger discusses computation of vertical and horizontal gradients of magnetic field intensity.
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 horizontal component of the earth’s magnetic field?
Representing the earth’s horizontal field in dipole form as
The vertical gradient is just the variation with change of radius or

24. What is the horizontal gradient of the vertical component of the earth’s magnetic field (ZE) ? [see problem 7.1]
Recall

25. Recall the horizontal gradient operator?
Evaluate

26. An important point to carry away from the preceding discussion is the distinction between the horizontal and vertical gradient operators.
Thus if you were asked “what is the horizontal gradient of the horizontal component?” you are required to evaluate 46

27. Problem 7.3
For problem 7-3, the main field has inclination i of 50o. The following equations are needed for Problem 3. ZA is the vertical component of the anomalous field, and HA, the horizontal.
Eqns 7-36 and 7-37 are in general form and can be used for arbitrary inclination i.

28. Before we address problem 7-3 in more detail, let’s go back and look at the ideas discussed by Berger pertaining to equations 7-15 through In problem 7-3 you will want to make use of Eq. (7-18)

29. Consider the type of measurement provided by the proton precession magnetometer
The proton precession magnetometer measures the scalar magnitude of the earth’s main field.

30. In this diagram FET is is the vector sum of the earth’s main field and the anomalous field associated with a buried dipole field. The proton precession magnetometer measures the magnitude of FET.

31. Magnetic Elements for your location
F is known

32. In most applications the anomalous field FA is much smaller than the main field FE.

33. In this case, the magnetic anomaly is just the difference between the measured field (FET) at some point and the predicted value of the earth’s main field (FE) at that point. 53

34. FA
When FA (the anomalous field) is small, we consider the difference T = FET - FE to be equivalent to the projection of vector FA onto the direction of the main field.

35. In the case where FA is large the projection FAT is significantly different from T.

36. Let’s zoom in for a closer look at the tip of FE.

37. FET T Horizontal line parallel to earth’s surface
i is the inclination of the earth’s main magnetic field.  is the angle of FA relative to the earth’s main field FE.

38. FAT is the projection of FA onto the direction of the main field FE, and is considered equal to T, the scalar difference between FE and FET.

39. The horizontal and vertical projections of FA

40. The horizontal and vertical projections of FA appear in the expansion of FAT = FAcos(-i).

41. In summary - FAT is an approximation of T, the scalar difference obtained from measurements of the total field (FET) made by the proton precession magnetometer.
For the purposes of modeling we work backwards. Given a certain object, we compute the horizontal (HA) and vertical (ZA) components of the anomaly (see equations 7-36 and 7-37, for example) and combine them to obtain FAT - the anomaly we obtain from the proton precession magnetometer measurements.

42. Back to problem 7-3. In problem 7-3 you are asked to estimate the detectability of a buried magnetized stone wall. You are also instructed to approximate the effect of the wall using the equations for the sphere. The inclination in the area is given as 50o, but the total field is not given. As you can see in equations 7-36 and 7-37 we need the total field FE in order to solve for ZA and HA. How might you determine what FE is given the inclination?
inclination
Main field intensity
Guess FE ~ nT

43. Do we have everything we need
Do we have everything we need? If we are going to approximate the effect of the wall using a sphere we need to decide what the radius of the sphere will be. The cross sectional area of the wall is 1/2 m2 (0.5m x 1m). A cross section through the center of the sphere will have area R2. Let R2 = 0.5m2 and solve for the equivalent radius R. This yields an R of 0.4m. From the diagram below, we can see that z (depth to center of wall or sphere) should be 1.75m.
You now have everything you need to calculate ZA and HA.

44. Note that some texts
have
In the denominator
instead of

45. Note that some texts
have
In the denominator
instead of

46. (see Eq. 7-18)
The maximum anomaly associated with the sphere is over 6nT and easily detected by a proton precession magnetometer which can be read with an accuracy of 1nT. (Notice that at 0 meters FAT is 4.5 nT.)

47. Questions?

48. Vertically polarized sphere or dipole
Some comments on -
Simple Geometrical Objects - Magnetics Style
Vertically polarized sphere or dipole

49. surface
The orientation of H and Hr has changed. We now have to think in terms of their relation to a buried dipole rather than to the earth’s main dipole field. ZA and HA refer to components of the buried dipole field measured relative to the earth’s surface.

50. surface
ZA is the vertical component of the anomalous dipole field - vertical with respect to the earth’s surface.

51. What are cos and sin?

53. Capitol versus lower case Z

54. HA is the horizontal component - component oriented along the earth’s surface associated with the dipole field of a buried object.

55. What is M?
Recall from class last Thursday (see also Equation 7-5 (Burger) that the intensity of magnetization equals the magnetic dipole moment per unit volume or
and also,
. Thus

56. Is equivalent to equation 7-35 in the text.
Substitute M=IV into the above to obtain
Recall that the volume of the sphere is
hence,
where I = kFE

57. An easy approximation
Equations 7-36 and 7-37 are fairly cumbersome equations to solve. The foregoing approximations made assuming a vertically polarized sphere allow one to make a quicker estimate of the field. A comparison of results would be informative.
Solve these two equations at x=0 and compare ZA and HA to the values obtained in the preceding detailed analysis.

58. Some background figures
for discussion
Magnetics Lab
Due Thursday, Dec. 5th

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

60. 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.

61. 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.

63. You may find that the autocalculations become a little erratic
You may find that the autocalculations become a little erratic. If that happens, just run forward again.

64. The spurious peaks in the calculation disappear.

65. How many drums are represented by the triangular-shaped object you entered into your model?
Plot the corner coordinates for the triangular shaped object you derived at 1:1 scale and compute the area.

66. How many drums?

67. For Tuesday, December 3rd be prepared to discuss problems 7-6, and 7-8.
Finish reading Chapter 7; we’ll talk about simple geometrical representations in magnetic problem solving.
Also be prepared to ask questions about the final exam. (Reminder - the final exam will be from 3-5pm on Friday afternoon, Dec. 13th)