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

2. Tom Wilson, Department of Geology and Geography From above, we obtain a basic definition of the potential (at right) for a unit positive test pole (m t ). The potential is the integral of the force (F) over a displacement path. Note that we consider the 1/4  term =1

3. Tom Wilson, Department of Geology and Geography Thus - H (i.e. F/p test, the field intensity) can be easily derived from the potential simply by taking the derivative of the potential The reciprocal relationship between potential and field intensity

4. Tom Wilson, Department of Geology and Geography 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 Working with the potentials of both poles..

5. Tom Wilson, Department of Geology and Geography Converting to common denominator yields From the previous discussion, the field intensity H is just where pl = M – the magnetic moment

6. Tom Wilson, Department of Geology and Geography 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.

7. Tom Wilson, Department of Geology and Geography V d represents the potential of the dipole. H Toward dipole center (i.e. center of Earth’s dipole field

8. Tom Wilson, Department of Geology and Geography H E is represented by the negative derivative of the potential along the earth’s surface or in the S direction.

9. Tom Wilson, Department of Geology and Geography

11. 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

12. Tom Wilson, Department of Geology and Geography Given What is H E 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.

13. Tom Wilson, Department of Geology and Geography At the poles,  is 0, so that What is Z E at the equator?  is 90

14. Tom Wilson, Department of Geology and Geography Z E at the poles …. The variation of the field intensity at the poles 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.

15. Tom Wilson, Department of Geology and Geography …. compare the field of the magnetic dipole field to that of the gravitational monopole field Gravity:500, 1000, 2000m Increase r by a factor of 4 reduces g by a factor of 16

16. Tom Wilson, Department of Geology and Geography For the dipole field, an increase in depth (r) from 4 meters to 16 meters produces a 64 fold decrease in anomaly magnitude 7.2 nT 0.113 nT Thus the 7.2 nT anomaly (below left) produced by an object at 4 meter depths disappears into the background noise at 16 meters.

17. Tom Wilson, Department of Geology and Geography On Tuesday during the last week of class, we’ll work through some problems that will help you review materials we’ve covered on magnetic fields. Some of the problems are not too much different from those we worked for gravitational fields and so will help initiate some review of gravity methods. The first problem relates to our discussions of the dipole field and their derivatives. 3. What is the horizontal gradient in nT/m of the Earth’s vertical field (Z E ) in an area where the horizontal field (H E ) equals 20,000 nT and the Earth’s radius is 6.3 x 10 8 cm.

18. Tom Wilson, Department of Geology and Geography (see problem 7.1) 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.

19. Tom Wilson, Department of Geology and Geography To answer this problem we must evaluate the horizontal gradient of the vertical component - or Take a minute and give it a try.

20. Tom Wilson, Department of Geology and Geography 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 following formula for Zmax. Background noise at the site is roughly  5nT.

21. Tom Wilson, Department of Geology and Geography Vertically Polarized Horizontal Cylinder General formNormalized shape term

22. Tom Wilson, Department of Geology and Geography 5. In your survey area you encounter two magnetic anomalies, both of which form nearly circular patterns in map view. These anomalies could be produced by a variety of objects, but you decide to test two extremes: the anomalies are due to 1) a concentrated, roughly equidemensional shaped object (a sphere); or 2) to a long vertically oriented cylinder.

23. Tom Wilson, Department of Geology and Geography Vertical Magnetic Anomaly Vertically Polarized Sphere The notation can be confusing at times. In the above, consider H = F E = intensity of earth’s magnetic field at the survey location. Z max and Z A refer to the anomalous field, i.e. the field produced by the object in consideration

24. Tom Wilson, Department of Geology and Geography Vertically Polarized Vertical Cylinder

25. Tom Wilson, Department of Geology and Geography Sphere vs. Vertical Cylinder; z = __________ Diagnostic positionsMultipliers Sphere Z Sphere Multipliers CylinderZ Cylinder X 3/4 = X 1/2 = X 1/4 = The depth 0.9 1.55 2.45 2.86 3.1 3.35 1.95 2.03 2.00 2.17 1.31 0.81 3.18 2 1.37 diagnostic distance

26. Tom Wilson, Department of Geology and Geography Diagnostic positionsMultipliers Sphere Z Sphere Multipliers Cylinder Z Cylinder X 3/4 = 1.63.182.17 X 1/2 = 2.521.31 X 1/4 = 3.71.370.81 Sphere or cylinder?

27. Tom Wilson, Department of Geology and Geography 6. Given that derive an expression for the radius, where I = kH E. 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 H E =55000nT. Z max (62.2nT from graph below) is the maximum vertical component of the anomalous field produced by the vertical casing.

28. Tom Wilson, Department of Geology and Geography Magnetics lab, part 2: A perfect fit – but is it correct?

29. Tom Wilson, Department of Geology and Geography How many drums? 4 square feet Area of one drum ~ We’ll talk more about the last bullet (1/r 3 ) on the results-to- be-discussed list a little later. What’s wrong with the format of this plot?

30. Tom Wilson, Department of Geology and Geography Problems 1 & 2 will be due this Thursday, December 3 rd Next week will be spent in review Problems 3-6 are due next Tuesday, Dec 8 th Magnetics lab, Magnetics paper summaries are due Thursday December 10 th Exam, Thursday December 17 th ; 3-5pm