Environmental and Exploration Geophysics I

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

Potential versus Force 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 consider the 1/4 term =1 Tom Wilson, Department of Geology and Geography

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

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 Tom Wilson, Department of Geology and Geography

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

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. Tom Wilson, Department of Geology and Geography

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

arc - length relationship HE is represented by the negative derivative of the potential along the earth’s surface or in the S direction. Tom Wilson, Department of Geology and Geography

Evaluating the derivative along the surface Tom Wilson, Department of Geology and Geography

-dV/dS Tom Wilson, Department of Geology and Geography

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 Tom Wilson, Department of Geology and Geography

The field along the dipole equator 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. Tom Wilson, Department of Geology and Geography

 is 90 The field at the pole At the poles,  is 0, so that What is ZE at the equator? Tom Wilson, Department of Geology and Geography

Field at pole is twice that at the equator ZE 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. Tom Wilson, Department of Geology and Geography

Consider one of the Lab Questions …. 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 Tom Wilson, Department of Geology and Geography

A 4 fold increase in distance 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 Tom Wilson, Department of Geology and Geography

Some in-class problems for the last week of class 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 (ZE) in an area where the horizontal field (HE) equals 20,000 nT and the Earth’s radius is 6.3 x 108 cm. Tom Wilson, Department of Geology and Geography

Problem 3 (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. 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. Tom Wilson, Department of Geology and Geography

Can you find it? 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. Tom Wilson, Department of Geology and Geography

Problem 4 Vertically Polarized Horizontal Cylinder General form Normalized shape term Tom Wilson, Department of Geology and Geography

Detecting abandoned wells 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. Tom Wilson, Department of Geology and Geography

Question 5 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. Tom Wilson, Department of Geology and Geography

Question 5 Vertically Polarized Vertical Cylinder Tom Wilson, Department of Geology and Geography

diagnostic distance Sphere vs. Vertical Cylinder; z = __________ The depth 2.17 1.31 0.81 1.95 2.03 2.00 0.9 3.18 2 1.37 2.86 3.1 3.35 1.55 2.45 Tom Wilson, Department of Geology and Geography

Sphere or cylinder? Tom Wilson, Department of Geology and Geography

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. Tom Wilson, Department of Geology and Geography

non-uniqueness Magnetics lab, part 2: A perfect fit – but is it correct? Beware of the flattened drum solution Tom Wilson, Department of Geology and Geography

How many drums? Area of one drum ~ 4 square feet What’s wrong with the format of this plot? We’ll talk more about the last bullet (1/r3) on the results-to-be-discussed list a little later. Tom Wilson, Department of Geology and Geography

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

Environmental and Exploration Geophysics I