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18-Geol-A7 Applied Geophysics · December 2014

Question 5 of 8: HLEM Traverse Over a Conductive Sheet

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2014 — 04-Geol-A7 Applied Geophysics. Three-hour, open-book exam; any non-communicating calculator permitted. The NOTES state that SIX questions constitute a complete paper, but eight numbered questions are printed on the exam — all eight, and every lettered/numbered sub-part, are solved below. Two figures (the CMP gather of Q6 and the reversed-refraction time-distance plot of Q8) carry real numeric data that is only given graphically on the printed page; both were read from the printed figures, calibrated against each figure's own printed axes, and the reading tolerance is given in a check callout beside each calculation.

Reference texts: Telford, Geldart & Sheriff, Applied Geophysics (2nd ed.) — the primary reference for every method in this paper (gravity, magnetics, seismic refraction and reflection, electrical resistivity, induced polarization, electromagnetics); Kearey, Brooks & Hill, An Introduction to Geophysical Exploration (3rd ed.) — survey design and interpretation context; Blakely, Potential Theory in Gravity and Magnetic Applications — the dipole/sphere anomaly shapes used in Q2–Q3.

Question 5: HLEM Traverse Over a Conductive Sheet (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

This is the classic horizontal-loop EM (Slingram / "Tx-Rx pair") survey: the transmitter (Tx) coil generates a PRIMARY alternating magnetic field; where that primary field cuts across the near-vertical conductive sheet, it induces circulating EDDY CURRENTS in the sheet (Faraday's law, $\varepsilon=-d\Phi/dt$), and those eddy currents in turn generate their own SECONDARY magnetic field, which is what the receiver (Rx) coil actually senses superimposed on (a bucked-out) primary field. The three-panel sequence below shows the Tx-Rx pair traversing from well before the conductor, directly over it, to well after it, with the primary field (blue, dashed loops from Tx) and the induced eddy-current loop and its secondary field (green, only significant when the pair straddles the conductor) sketched at each position.

1. coils well before the sheetTxRxweak coupling, small secondary2. Tx and Rx straddle the sheetTxRxeddy currents: strong secondary field3. coils well past the sheetTxRxweak coupling, small secondaryBlue dashed: primary field of the Tx loop. Green: eddy current in the sheet and its secondary field.Response parameter α ≈ 10 (good conductor): in-phase 0.99, quadrature 0.100in-phase (solid): deep negative trough,small positive shouldersquadrature (dashed): same shape, 10× smallerResponse parameter α ≈ 0.1 (poor conductor): in-phase 0.01, quadrature 0.100quadrature (dashed) now dominates, Q/IP ≈ 10in-phase (solid) almost flatTx-Rx midpoint position (sheet at centre; grey lines: Tx or Rx directly over sheet)
Top: Tx-Rx pair at three stages of the traverse (primary field of the Tx loop in blue; eddy current and secondary field in green, strong only when the coils straddle the sheet). Bottom: In-phase (solid) and Out-of-phase (dashed) profiles for response parameter ≈10 and ≈0.1, scaled by the L-R circuit fractions.

(ii) Response profiles at response parameter ≈10. The response parameter $\alpha$ (formula sheet: $\tan\phi=\sigma\mu\omega tl$ for a sheet, the $\omega L/R$ of the equivalent eddy-current circuit) controls how the secondary field divides between the two phases. For that L-R circuit the secondary field, as a fraction of its perfect-conductor limit, has in-phase part $\alpha^2/(1+\alpha^2)$ and quadrature part $\alpha/(1+\alpha^2)$. At $\alpha\approx10$ these are $100/101\approx0.99$ and $10/101\approx0.10$. The profile SHAPE, which is the same for both components, comes from the coil-conductor geometry sketched above. Far from the sheet the response is zero. As the pair approaches there is a small positive shoulder, then the response passes through zero when the leading coil is directly over the sheet. It reaches a deep NEGATIVE trough when the sheet lies midway between Tx and Rx (strongest coupling), passes back through zero when the trailing coil is over the sheet, and ends with another small positive shoulder. So at $\alpha\approx10$ the In-phase profile is a large negative trough about one coil separation wide with small positive shoulders, and the Out-of-phase profile has the same shape at only about one tenth of that amplitude: the quadrature response peaks at $\alpha=1$ and has fallen back by $\alpha=10$.

(iii) Response profiles at response parameter ≈0.1. At $\alpha\approx0.1$ the in-phase fraction is $0.01/1.01\approx0.01$ and the quadrature fraction is $0.1/1.01\approx0.10$. The geometry, and so the profile shape (zero crossings with a coil over the sheet, central negative trough, small shoulders), is unchanged. What changes is the size of each component: the In-phase trough shrinks by a factor of about 100 to almost nothing, while the Out-of-phase trough stays about as large as it was at $\alpha\approx10$. The quadrature now dominates, with $Q/IP=1/\alpha\approx10$, the reverse of part (ii), where $IP/Q\approx10$. A large quadrature anomaly with little in-phase response marks a poor conductor; a strong in-phase anomaly with weak quadrature marks a good one.