Electromagnetic Methods: Induction, Skin Depth, Natural-Field and Time-Domain Methods, Ground-Penetrating Radar

This chapter answers the Part B2 heading “Electromagnetic Methods” of the Geology and Geophysics (GG) paper: “Geo-electromagnetic spectrum; Biot Savart’s Law; Maxwell’s Equation, Helmholtz Equation, Basic concept of EM induction in the earth, Skin-depth, elliptic polarization, in-phase and quadrature components, phasor diagrams; Response function and response parameters; Measurements in different source receiver configurations; Earth’s natural electromagnetic methods- tellurics, geomagnetic depth sounding and magnetotellurics; Electromagnetic profiling and Sounding, Time domain EM method; EM scale modelling, processing of EM data and interpretation; Ground Penetrating Radar (GPR) Methods; Effect of conducting overburden.” The numericals are skin depth, the magnetotelluric (Cagniard) resistivity, the low-induction-number conductivity, decay in the time domain, scale-model frequency and GPR depth and velocity. Take μ0 = 4π × 10⁻⁷ H/m.

1. The geo-electromagnetic spectrum, Biot–Savart law, Maxwell’s and Helmholtz equations

EM methods use time-varying fields from about 10⁻⁴ Hz to 10⁹ Hz, and frequency sets what is sensed: low frequency reaches deep, high frequency resolves the shallow ground.

The geo-electromagnetic spectrum
MethodSourceFrequency or time
Magnetotellurics, tellurics, geomagnetic depth soundingNatural: magnetosphere (below about 1 Hz), lightning (above)about 10⁻⁴ to 10⁴ Hz
Frequency-domain loop EM (Slingram)Controlled loop transmitterabout 10² to 10⁵ Hz
VLFDistant radio transmittersabout 15 to 30 kHz
Time-domain EMSwitched loop currenttransient, microseconds to hundreds of ms
Ground-penetrating radarPulsed antennaabout 10⁷ to 10⁹ Hz

Biot–Savart law: a current element gives dB = (μ0/4π) I dl × r̂/r². For a circular loop of radius a, B = μ0 I/(2a) at the centre and B = μ0 I a²/(2(a² + z²)^(3/2)) on the axis, multiplied by N for N turns; at large distance the loop acts as a dipole of moment m = N I A. This is the primary field of a loop transmitter.

Maxwell’s equations: ∇×E = −∂B/∂t, ∇×H = J + ∂D/∂t, ∇·B = 0 and ∇·D = ρf, with J = σE, D = εE and B = μH (μ ≈ μ0 in rocks). Eliminating H gives the Helmholtz equation ∇²E + k²E = 0 with k² = ω²με − iωμσ for time dependence e^(iωt). The ratio of displacement to conduction current is ωε/σ, about 6 × 10⁻⁵ at 1 kHz for ρ = 100 Ω·m and εr = 10. So at EM-induction frequencies the quasi-static form k² ≈ −iωμσ holds and fields diffuse. At radar frequencies ωε ≳ σ, k ≈ ω√(με), and fields propagate as waves.

2. EM induction in the Earth: skin depth, phasors, in-phase and quadrature, elliptic polarisation

A changing primary field induces eddy currents in the ground (Faraday), which produce a secondary field; the receiver records the sum. In a uniform half-space the quasi-static wave falls as e^(−z/δ) e^(−iz/δ), with skin depth δ = √(2/(ωμσ)) = √(ρ/(π f μ0)) ≈ 503 √(ρ/f) m. At z = δ the amplitude is 1/e (37 percent); at 3δ it is about 5 percent; the phase lags by z/δ radians. Example: ρ = 100 Ω·m and f = 1 kHz give δ = 503 √0.1 = 159 m. Low frequency or high resistivity means deeper reach; the Bostick depth √(ρa/(ωμ0)) = δ/√2 is the usual depth assigned to a measurement.

Fields are written as phasors, a complex amplitude times e^(iωt). The secondary relative to the primary is a complex number: its real part is the in-phase component, along the primary, and its imaginary part is the quadrature component, 90° out of phase. In the closed-loop model of a conductor with inductance L and resistance R the secondary-to-primary ratio is −(α² + iα)/(1 + α²), times a geometric coupling factor, with response parameter α = ωL/R, and the secondary lags the primary by 90° + arctan α.

Response of the loop model
αIn-phaseQuadratureMeaning
≪ 1−α²−αPoor conductor or low frequency: mostly quadrature
1−0.5−0.5Quadrature at its maximum; lag 135°
≫ 1→ −1→ 0Good conductor or high frequency: all in-phase, field expelled

Elliptic polarisation. Over a dipping conductor the in-phase and quadrature secondary fields point in different directions, so the total field at a point traces an ellipse over a cycle. The tilt of its major axis and its ellipticity (minor over major axis, roughly the vertical quadrature over the in-phase field) are what tilt-angle methods such as VLF measure.

3. Response parameters, source–receiver configurations, profiling and sounding

The response depends on the response parameter (conductance, frequency and size of the target) and on geometry (spacing, dip, depth, coupling of transmitter, target and receiver). For a half-space the induction number is B = s/δ, with s the transmitter–receiver spacing. For low induction number, B ≪ 1, the quadrature ratio is (Hs/Hp)quad ≈ ωμ0σ s²/4, so σa = 4 (Hs/Hp)quad/(ωμ0 s²) reads conductivity directly. Example: s = 4 m, f = 10 kHz and (Hs/Hp)quad = 2 × 10⁻³ give σa = 4 × 2 × 10⁻³/(2π × 10⁴ × 4π × 10⁻⁷ × 16) = 6.33 mS/m.

  • Slingram (horizontal-loop EM): transmitter and receiver coils at fixed spacing are moved together along a line. At low induction number, coplanar horizontal coils (vertical dipoles) reach about 1.5 s deep, vertical coplanar coils (horizontal dipoles) about 0.75 s.
  • Fixed source, moving receiver: a large grounded or ungrounded loop with a roving receiver (Turam), a coaxial or perpendicular-coil arrangement, and airborne systems with the coils in a towed bird or on the aircraft.
  • Profiling and sounding: profiling keeps frequency and spacing fixed and moves along the line to find lateral conductors. Sounding varies the depth sensed: frequency sounding (change f at fixed s), geometric sounding (change s at fixed f) or time sounding (read the transient at successive times, next section).
⚠️ Quadrature is not “the smaller part”
Quadrature and in-phase are defined by phase relative to the primary, not by size. The conductivity formula above uses the quadrature component only, and only when B ≪ 1; it fails over good conductors, where the response is in-phase.

4. Natural-field methods: tellurics, geomagnetic depth sounding and magnetotellurics

Natural sources are solar-wind interactions with the magnetosphere (micropulsations, below about 1 Hz) and lightning travelling in the Earth–ionosphere cavity (above about 1 Hz, with the Schumann resonance near 7.8 Hz). Because the ground is far more conductive than air, these fields enter it as plane waves travelling almost vertically downward, so E and H at the surface are horizontal and orthogonal.

Tellurics measures only the horizontal electric field, with a roving pair compared against a base pair. Since E = ρJ, the telluric amplitude is higher over resistive basement highs and lower over conductive basin fill, which maps structure without any magnetic measurement.

Magnetotellurics (MT) measures orthogonal horizontal E and H. The impedance Z = Ex/Hy gives the Cagniard resistivity ρa = |Z|²/(ωμ0) = |Ex/Hy|²/(ωμ0), in Ω·m for E in V/m and H in A/m. In field units, E in mV/km, B in nT and period T in s, ρa = 0.2 T |E/B|². Example: T = 10 s, E = 10 mV/km and B = 2 nT give ρa = 0.2 × 10 × 25 = 50 Ω·m. The impedance phase is 45° over a uniform half-space; above 45° the resistivity falls with depth, below 45° it rises. Depth grows with √(ρT), and MT has weak-signal dead bands near 0.5 to 5 Hz and 1 to 5 kHz. Static shift, from near-surface heterogeneity, moves the ρa curve up or down by a constant factor on the log scale and leaves the phase unchanged.

Geomagnetic depth sounding (GDS) uses magnetic fields alone. Lateral conductivity contrasts produce a vertical field related to the horizontal one by the tipper, Hz = A Hx + B Hy, drawn as induction arrows: in the Parkinson convention they point toward the conductor (so toward the sea at a coast, the coast effect) and in the Wiese convention away from it. Global GDS uses periods of hours to days, such as the daily Sq variation and storms, to reach the mantle.

5. Time-domain EM and EM scale modelling

In time-domain EM (TDEM) a steady loop current is switched off, and the collapsing primary field induces eddy currents that diffuse downward and outward like a smoke ring. The receiver records the decay of the secondary field, usually dB/dt, in the off time. Early times sense shallow ground and late times deeper ground: the diffusion depth is about √(2ρt/μ0) ≈ 1260 √(ρt) m for t in s. Over a uniform half-space the late-time central-loop response falls as a power law, dB/dt ∝ t^(−5/2). A discrete good conductor decays exponentially, e^(−t/τ), where a long time constant τ means a good conductor. Example: τ = 5 ms leaves e^(−(15 − 5)/5) = e^(−2) = 0.14 of the 5 ms response at 15 ms.

TDEM measures when the primary field is off, so no primary has to be separated from the secondary; it sounds deep with a small receiver and is the preferred choice under conductive overburden (section 6).

EM scale modelling studies complex targets in the laboratory with a model scaled down by a factor n in length. The response depends on the induction number, so the product μσωL² must match: with μ unchanged, σm fm = n² σ f. A 1:500 model (n = 500) needs σm fm = 250,000 σ f, which is met with a highly conducting model material (graphite, metal sheet, brine) and an adjusted frequency. Models test the response of dipping plates, overburden and loop geometry and check interpretation charts, but only if the conductivity-frequency scaling is kept.

6. Ground-penetrating radar, conducting overburden, processing and interpretation

GPR sends short radar pulses of about 10 MHz to 1 GHz into the ground; they reflect at contrasts in relative permittivity εr. In low-loss, non-magnetic ground the velocity is v = c/√εr, with c = 0.3 m/ns, and a reflector at two-way time t lies at depth d = v t/2. Typical εr: air 1, ice 3–4, dry sand 3–5, saturated sand 20–30, fresh water about 80, clay 5–40. At normal incidence R = (√ε1 − √ε2)/(√ε1 + √ε2). Example: εr = 16 gives v = 0.075 m/ns, and t = 80 ns gives d = 3.0 m.

Attenuation grows with conductivity, α ≈ (σ/2)√(μ/ε), so range is short in clay and saline ground and long in dry sand, ice and granite. Vertical resolution is about a quarter of the wavelength in the ground, λ = v/f, so a higher frequency resolves finer detail but reaches less deep. A point reflector gives a diffraction hyperbola t² = t₀² + (2x/v)², which yields v from the curvature and so the depth and εr.

Conducting overburden of conductance S = σh hurts in two ways. It shields the target, since eddy currents in it weaken the field reaching the target and the field returning, so a deeper conductor looks weaker and shifted in phase; and it adds a smooth overburden response of its own that can mask a target. In TDEM the overburden transient dies quickly because its time constant is short, so late-time channels see only the target, and this is why TDEM is preferred there.

Processing and interpretation. Remove noise (50 Hz power-line harmonics by notch filters, sferics and cultural sources by stacking and editing), correct drift, and calibrate. For VLF the Fraser filter turns tilt-angle crossovers into peaks and current-density pseudosections image conductors. MT uses robust cross-spectral estimation, often with a remote reference. Interpretation is by forward modelling and inversion in 1D, 2D or 3D, constrained by boreholes. Ambiguity remains: a conductor’s conductance is better determined than its thickness and conductivity separately, much as in resistivity equivalence.

Key takeaways

  • Low frequency means deep and high frequency means shallow with fine resolution: MT about 10⁻⁴ to 10⁴ Hz, loop EM 10² to 10⁵ Hz, VLF 15 to 30 kHz, GPR 10⁷ to 10⁹ Hz. Quasi-static diffusion holds while ωε ≪ σ; radar works where ωε ≳ σ.
  • Skin depth δ = √(2/(ωμσ)) ≈ 503 √(ρ/f) m: amplitude 1/e at δ, about 5 percent at 3δ, phase lag z/δ. The Bostick depth is δ/√2.
  • The secondary has in-phase and quadrature parts. For the loop model the response is −(α² + iα)/(1 + α²), α = ωL/R: quadrature dominates at small α, equals in-phase at α = 1, and vanishes at large α. At low induction number σa = 4(Hs/Hp)quad/(ωμ0 s²).
  • MT: ρa = |Ex/Hy|²/(ωμ0) = 0.2 T |E/B|² (mV/km, nT, s); phase 45° for a uniform half-space, above 45° for resistivity falling with depth; static shift moves ρa but not phase. Tellurics maps E amplitude, GDS uses tipper and induction arrows.
  • TDEM reads the decay after turn-off: early times shallow, late times deep, half-space dB/dt ∝ t^(−5/2), a discrete conductor e^(−t/τ). It suits conductive overburden, which shields a target and adds its own response in the frequency domain. Scale models keep σm fm = n² σ f.
  • GPR: v = c/√εr with c = 0.3 m/ns and d = v t/2; high conductivity (clay, saline water) kills penetration, resolution is about λ/4, and a point reflector gives a hyperbola t² = t₀² + (2x/v)².

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Which condition justifies neglecting the displacement current in EM induction surveys on the ground?

    1. ωε ≫ σ, because the ground then behaves as a good dielectric and the fields propagate freely
    2. ωε ≪ σ, because the conduction current then far exceeds the displacement current
    3. The permeability of rocks is far above μ0, which suppresses the displacement current
    4. The frequency is above 1 GHz, where wavelengths in rock are very short and conduction is negligible
    Show answer

    Answer: B — ωε ≪ σ, because the conduction current then far exceeds the displacement current

    The ratio of displacement to conduction current is ωε/σ. At induction frequencies in rock it is tiny (about 6 × 10⁻⁵ at 1 kHz for 100 Ω·m), so the quasi-static form k² ≈ −iωμσ holds and fields diffuse. The opposite limit, ωε ≳ σ, is the propagation regime of radar. Rocks have μ ≈ μ0, so permeability plays no role.

  2. An EM measurement is repeated with the frequency raised by a factor of 4 over ground whose resistivity is also 4 times higher. How does the skin depth change?

    1. It doubles, because the higher resistivity makes the field penetrate twice as deep
    2. It halves, because the frequency enters through its square root and the resistivity does not
    3. It is unchanged, because δ depends on the ratio ρ/f and that ratio is the same
    4. It quadruples, because the two changes both increase the penetration of the field
    Show answer

    Answer: C — It is unchanged, because δ depends on the ratio ρ/f and that ratio is the same

    δ ≈ 503 √(ρ/f), so only the ratio ρ/f matters. Multiplying ρ by 4 and f by 4 leaves ρ/f, and therefore δ, unchanged. Higher resistivity alone would double δ, and higher frequency alone would halve it, and here the two cancel.

  3. A closed-loop conductor model has response function −(α² + iα)/(1 + α²), with response parameter α = ωL/R. At which α is the quadrature component largest in magnitude, and what is the in-phase component there, as a fraction of the primary?

    1. α = 1, where the in-phase component is −0.5
    2. α = 1, where the in-phase component is zero
    3. α ≫ 1, where the in-phase component is −1
    4. α ≪ 1, where the in-phase component is zero
    Show answer

    Answer: A — α = 1, where the in-phase component is −0.5

    The quadrature magnitude α/(1 + α²) peaks at α = 1, where it is 0.5. The in-phase magnitude α²/(1 + α²) is then also 0.5, so the response has equal in-phase and quadrature parts. The in-phase part is not zero at the quadrature peak, and at large α the quadrature part has already fallen toward zero.

  4. A magnetotelluric sounding has an impedance phase (E relative to H) near 45° at short periods that rises to about 65° at long periods. The simplest interpretation is:

    1. a static shift, because phase above 45° records a constant vertical shift of the ρa curve
    2. a uniform half-space whose resistivity changes only with the period of the source
    3. a resistive layer at depth, because the phase rises when the field meets a deeper insulator
    4. a conductor at depth, so the resistivity decreases with depth below the near-surface ground
    Show answer

    Answer: D — a conductor at depth, so the resistivity decreases with depth below the near-surface ground

    Over a uniform half-space the phase is 45°. A phase above 45° means the resistivity falls with depth, so a deeper conductor is indicated. A deeper resistor would pull the phase below 45°, and a static shift moves ρa but leaves the phase unchanged, so it cannot produce the phase change.

  5. At a coastline, the real induction arrows of geomagnetic depth sounding in the Parkinson convention generally point:

    1. away from the sea, along the direction of decreasing conductivity
    2. along the coast, parallel to the strike of the shoreline
    3. toward the sea, which is the more conductive side
    4. in no consistent direction, because the vertical field vanishes at a coast
    Show answer

    Answer: C — toward the sea, which is the more conductive side

    In the Parkinson convention the real arrows point toward the region of higher conductivity, which at a coast is the sea (the coast effect). The Wiese convention reverses the sign and points away. The vertical field is strongest, not zero, at a coast.

  6. Which statement about a time-domain EM sounding over a uniform half-space is correct?

    1. Early-time channels sense deeper ground because the eddy currents are strongest at depth at the start
    2. Late-time channels sense deeper ground, and the late-time dB/dt falls as t^(−5/2)
    3. The measured decay is exponential with a time constant that equals ρ/μ0 for the half-space
    4. The transient is recorded while the primary field is on, so the primary must be subtracted from it
    Show answer

    Answer: B — Late-time channels sense deeper ground, and the late-time dB/dt falls as t^(−5/2)

    The eddy currents diffuse downward with time, so late times sense deeper ground, and for a half-space the late-time central-loop dB/dt follows a power law t^(−5/2). An exponential decay with a time constant belongs to a discrete conductor, not the half-space. TDEM is recorded in the off-time, so no primary has to be removed.

  7. Beneath a thin, conductive overburden, why is time-domain EM generally preferred to frequency-domain loop EM for detecting a deeper good conductor?

    1. TDEM uses higher frequencies, which pass through the overburden without being attenuated
    2. TDEM transmits plane waves, which a thin conductive layer cannot disturb or attenuate
    3. TDEM records only the in-phase component, which a thin overburden does not produce
    4. The overburden transient dies out early, so late-time channels see the target’s longer decay
    Show answer

    Answer: D — The overburden transient dies out early, so late-time channels see the target’s longer decay

    The overburden is thin, so its time constant is short and its transient is gone by late times, while a good conductor at depth has a long time constant and still shows. A frequency-domain system receives the overburden response at every frequency. TDEM does not use plane waves, and it records a decaying total transient, not an in-phase component.

  8. A GPR survey on clay-rich, saline ground gives very shallow penetration. The main reason is that:

    1. the high conductivity causes strong attenuation, α ≈ (σ/2)√(μ/ε), of the radar wave
    2. the relative permittivity of wet clay is below that of air, so the pulse is not reflected at all
    3. the radar velocity in clay exceeds c, so the reflections return before the pulse has ended
    4. clay is magnetic enough to stop the wave, since the velocity formula contains the permeability
    Show answer

    Answer: A — the high conductivity causes strong attenuation, α ≈ (σ/2)√(μ/ε), of the radar wave

    Attenuation rises with conductivity, so clay and saline water absorb the radar energy within a short distance. The permittivity of wet clay is well above 1, not below it, and velocities are always below c. Clay is not magnetic enough to matter, and the low-loss velocity formula uses εr only.

  9. A single-turn circular loop of radius 0.5 m carries a current of 2 A. Using the Biot–Savart law with μ0 = 4π × 10⁻⁷ H/m, the magnetic field at the centre of the loop is:

    1. 0.80 μT
    2. 1.26 μT
    3. 2.51 μT
    4. 5.03 μT
    Show answer

    Answer: C — 2.51 μT

    At the centre of a circular loop B = μ0 I/(2a) = (4π × 10⁻⁷ × 2)/(2 × 0.5) = 2.51 × 10⁻⁶ T. The value 0.80 μT comes from the straight-wire form μ0 I/(2π a), 1.26 μT is half the correct value (the result for a semicircular arc, μ0 I/(4a)), and 5.03 μT comes from dropping the factor of 2 in the denominator.

  10. Which of the following statements about skin depth are correct?

    1. The phase of the field in the ground stays the same at every depth, equal to its surface value
    2. For a given resistivity the skin depth increases as the frequency decreases
    3. At a depth of one skin depth the amplitude has fallen to 50 percent of its surface value
    4. At a depth of three skin depths the amplitude has fallen to about 5 percent of its surface value
    Show answer

    Answer: B — For a given resistivity the skin depth increases as the frequency decreases; D — At a depth of three skin depths the amplitude has fallen to about 5 percent of its surface value

    δ ≈ 503 √(ρ/f), so lowering f deepens the penetration. The amplitude falls as e^(−z/δ), which is 37 percent at δ (not 50 percent) and e^(−3) ≈ 5 percent at 3δ. The phase lags by z/δ radians, so it changes with depth.

  11. Which of the following statements about the magnetotelluric method are correct?

    1. A static shift moves the ρa curve by a constant factor on a log scale and leaves the phase unchanged
    2. The Cagniard resistivity cannot be computed without the vertical magnetic field component
    3. Over a uniform half-space the impedance phase is 45° at every period
    4. The depth of penetration increases as the signal frequency is increased, because higher frequencies reach deeper
    Show answer

    Answer: A — A static shift moves the ρa curve by a constant factor on a log scale and leaves the phase unchanged; C — Over a uniform half-space the impedance phase is 45° at every period

    Static shift scales ρa but not the phase, and a half-space gives 45°. The Cagniard resistivity uses the horizontal components Ex and Hy only. Penetration grows as the frequency falls, since δ ∝ √(ρ/f).

  12. Which of the following statements about ground-penetrating radar are correct?

    1. In low-loss, non-magnetic ground the radar velocity is v = c/√εr, with c the vacuum speed of light
    2. The depth of a reflector is simply v t, where t is the two-way travel time of its echo
    3. Radar waves travel faster in water-saturated sand than in dry sand, since water fills the pores
    4. A higher antenna frequency gives finer vertical resolution but less penetration
    Show answer

    Answer: A — In low-loss, non-magnetic ground the radar velocity is v = c/√εr, with c the vacuum speed of light; D — A higher antenna frequency gives finer vertical resolution but less penetration

    v = c/√εr is the low-loss result, and a higher frequency gives a shorter wavelength (finer resolution) but greater attenuation. The depth is v t/2 because t is the two-way time. Saturated sand has a higher εr (20–30 against 3–5), so the velocity is lower, not higher.

  13. Use the skin-depth relation δ = 503 √(ρ/f) metres. In ground of resistivity 50 Ω·m, the frequency at which the skin depth is 500 m, in Hz to one decimal place, is ____ .

    Numerical answer — type the value.

    Show answer

    Answer: 50.6

    From δ = 503 √(ρ/f), f = ρ (503/δ)² = 50 × (503/500)² = 50 × 1.012 = 50.6 Hz. The usual slip is to forget to square the factor, giving 50 × 1.006 = 50.3.

  14. At a period T = 25 s a magnetotelluric station records |E| = 4 mV/km and |B| = 1 nT. Using ρa = 0.2 T |E/B|², with T in s, E in mV/km and B in nT, the Cagniard resistivity in Ω·m is ____ .

    Numerical answer — type the value.

    Show answer

    Answer: 80

    ρa = 0.2 × 25 × (4/1)² = 0.2 × 25 × 16 = 80 Ω·m. The practical-units constant 0.2 comes from ρa = μ0 T |E|²/(2π |B|²) after converting mV/km to V/m and nT to T. The ratio E/B must be squared.

  15. A low-induction-number conductivity meter has coil spacing s = 2 m and operates at f = 20 kHz. It reads a quadrature ratio (Hs/Hp)quad = 4 × 10⁻³. Using σa = 4 (Hs/Hp)quad/(ωμ0 s²) with μ0 = 4π × 10⁻⁷ H/m, the apparent conductivity in mS/m, to one decimal place, is ____ .

    Numerical answer — type the value.

    Show answer

    Answer: 25.3

    ω = 2π × 2 × 10⁴ = 1.2566 × 10⁵ rad/s, so ωμ0 s² = 1.2566 × 10⁵ × 1.2566 × 10⁻⁶ × 4 = 0.6317. Then σa = 4 × 4 × 10⁻³/0.6317 = 0.02533 S/m = 25.3 mS/m. Using f in place of ω, a factor 2π error, is the common slip.

  16. The secondary field of a discrete conductor decays as e^(−t/τ) with time constant τ = 2 ms. The ratio of the response at t = 8 ms to the response at t = 2 ms, to two decimal places, is ____ .

    Numerical answer — type the value.

    Show answer

    Answer: 0.05

    The ratio is e^(−(8 − 2)/2) = e^(−3) = 0.0498, which is 0.05. The common slip is to use e^(−8/2) = 0.018, which divides by the response at t = 0 instead of at 2 ms.

  17. A laboratory scale model is built 500 times smaller than the field target (n = 500), with the same permeability. The field has σ = 0.02 S/m and f = 400 Hz, and the model conductor has σm = 1 × 10⁵ S/m. The model frequency that preserves the induction number, in Hz, is ____ .

    Numerical answer — type the value.

    Show answer

    Answer: 20

    The induction number depends on μσωL², so σm fm Lm² = σ f L², which gives σm fm = n² σ f. Thus fm = 500² × 0.02 × 400/(1 × 10⁵) = 250,000 × 8/10⁵ = 20 Hz. Using n instead of n² gives 0.04 Hz, which is wrong.

  18. A GPR survey with co-located antennas over uniform ground records the diffraction hyperbola of a buried pipe, treated as a point reflector, with t² = t₀² + (2x/v)². The apex is at two-way time t₀ = 30 ns, and at a horizontal offset x = 2.0 m from the apex the arrival is at t = 50 ns. The depth of the pipe below the antennas, in metres to one decimal place, is ____ .

    Numerical answer — type the value.

    Show answer

    Answer: 1.5

    (2x/v)² = t² − t₀² = 2500 − 900 = 1600, so 2x/v = 40 ns and v = 4.0/40 = 0.1 m/ns. The depth is v t₀/2 = 0.1 × 30/2 = 1.5 m. The velocity corresponds to εr = (0.3/0.1)² = 9. Taking depth = v × 50 ns/2 = 2.5 m would use the offset arrival instead of the apex.