The geomagnetic elements are defined with Z positive downward, H the horizontal intensity, D the declination east of true north and I the inclination. Which set of relations is correct?

H = F cos I, Z = F sin I and the eastward component Y = H sin D

H = F sin I, Z = F cos I and the eastward component Y = H sin D

H = F cos I, Z = F sin I and the eastward component Y = H sin D

H = F cos I, Z = F sin I and the eastward component Y = H cos D

H = F cos I, Z = F tan I and the eastward component Y = H sin D

A lava flow, never tilted since it cooled, carries a stable primary thermoremanent magnetisation of inclination 0°. Assuming the geocentric axial dipole hypothesis, where did the flow cool?

At the paleo-equator, whichever the polarity was, because tan I = 2 tan λ gives λ = 0 for I = 0

At a paleo-pole, because the field of an axial dipole is horizontal there

At the paleo-equator only if the polarity was normal, since reversed polarity would give 90°

At an undetermined paleolatitude, since inclination alone cannot fix latitude without the declination

At the paleo-equator, whichever the polarity was, because tan I = 2 tan λ gives λ = 0 for I = 0

Orbiting spacecraft find that Mars has no global dipole field today, yet parts of its ancient southern highlands carry strong magnetisation. The most reasonable inference is that:

a core dynamo operated early in Mars’s history, magnetised the crust as it cooled, and later ceased

a core dynamo is active today but its field is completely screened by the thin atmosphere

the weak present field has slowly built a viscous remanence in the crust over billions of years

the iron-rich crust magnetises itself permanently without any external field, so no past dynamo is required

Which pair of corrections is routinely applied when reducing a ground magnetic survey to an anomaly map?

Removal of the diurnal variation using a base-station record and subtraction of the IGRF for the date

A Bouguer slab correction for station elevation and a free-air correction

A terrain correction with a density of 2.67 g/cm³ and a latitude correction from the normal-gravity formula

Removal of the lunar-solar tidal drift from a spring-gravimeter reading and an isostatic correction for crustal roots

Removal of the diurnal variation using a base-station record and subtraction of the IGRF for the date

Poisson’s relation connects the magnetic potential and the gravitational potential of one body. Which condition is essential for it to hold?

Density and magnetisation must both be uniform through the body, so M/ρ is constant

The body must be a sphere, since only a sphere has an equivalent point dipole and a closed-form potential

Density and magnetisation must both be uniform through the body, so M/ρ is constant

The magnetisation must be entirely induced by the present field, with no remanent part present

The observations must be made on the ground surface and not at flight height or any higher level

A small, deeply buried sphere carries only induced magnetisation with χ ≪ 1. At the magnetic pole the total-field anomaly directly over its centre is +A. The same sphere at the same depth is moved to the magnetic equator, where the Earth’s dipole field intensity is half that at the pole. Ignoring demagnetisation and regional gradients, the total-field anomaly directly over its centre is:

−A/4, because the antiparallel field over a horizontal dipole is half the axial value and the induced moment is also halved

+A/4, because the moment and the dipole-field factor are both halved, but the sign of the anomaly is unchanged

−A/2, because the induced moment is horizontal and the field over a horizontal dipole is antiparallel at half the axial strength

−A/4, because the antiparallel field over a horizontal dipole is half the axial value and the induced moment is also halved

−A, because the sign reverses while the magnitude is set by the sphere, its depth and its susceptibility alone

Which statement about the analytic-signal amplitude of magnetic anomaly data is correct?

For 2D sources its shape does not depend on the magnetisation direction, so no pole reduction is needed at low latitude

It is the first vertical derivative of the anomaly and therefore changes sign across the edge of a magnetised body

It can be computed only after a structural index has been chosen for the geometry of the source body

It is obtained by downward continuation to the source depth, so the source depth must be known in advance

A thin vertical dike, whose true structural index is N = 1, is analysed by Euler deconvolution with N = 3. At the anomaly peak, where the horizontal derivative is zero, the depth estimate compared with the true depth is:

about three times the true depth, because z₀ = N(T − B)/(∂T/∂z) scales with N

equal, because N cancels when the horizontal derivative is zero

one third of the true depth, because a larger N always means a shallower source

equal, because the Euler equation determines N from the data without any input

about three times the true depth, because z₀ = N(T − B)/(∂T/∂z) scales with N

Which of the following statements about the induction equation and the geodynamo are correct?

Frozen-flux behaviour needs a magnetic Reynolds number far above 1, so convection dominates over diffusion

The convection term ∇×(u×B) comes from Ohm’s law for a conductor moving with velocity u through the field

Frozen-flux behaviour needs a magnetic Reynolds number far above 1, so convection dominates over diffusion

A purely axisymmetric magnetic field can be maintained by dynamo action if the flow is vigorous enough

The convection term ∇×(u×B) comes from Ohm’s law for a conductor moving with velocity u through the field

The geomagnetic field is produced by the permanent magnetisation of the solid, iron-rich inner core

Which of the following statements about paleomagnetism are correct?

If magnetisation directions cluster more tightly once folded beds are restored to horizontal, the magnetisation predates the folding

In a reversal test, the normal and reverse mean directions should be antipodal once overprints are removed

Under a geocentric axial dipole the remanent declination gives the paleolatitude of the site through the relation tan D = 2 tan λ

If magnetisation directions cluster more tightly once folded beds are restored to horizontal, the magnetisation predates the folding

Paleomagnetic data fix the paleolongitude of a continent as well as its paleolatitude, because the paleo-pole marks one point on the globe

In a reversal test, the normal and reverse mean directions should be antipodal once overprints are removed

Which of the following statements about upward and downward continuation of magnetic data are correct?

Upward continuation multiplies the amplitude spectrum by e^(−kh), so it is a stable smoothing operation

Downward continuation multiplies the spectrum by e^(+kh) and so amplifies short-wavelength noise

Both operations rely on the field satisfying Laplace’s equation in the source-free region between the two levels

Upward continuation multiplies the amplitude spectrum by e^(−kh), so it is a stable smoothing operation

Downward continuation multiplies the spectrum by e^(+kh) and so amplifies short-wavelength noise

Upward continuation sharpens the anomalies of shallow sources more than those of deep sources, boosting short wavelengths

Both operations rely on the field satisfying Laplace’s equation in the source-free region between the two levels

Magnetic Methods: Geomagnetic Field, Paleomagnetism, Induction and Geodynamo, Susceptibility, Anomalies and Their Interpretation

This chapter answers the Part B2 heading “Magnetic Methods” of the Geology and Geophysics (GG) paper: “Geomagnetic field, paleomagnetism; Inclination, declination; Magnetic induction equation – convection, diffusion, Ohm’s law for a moving conductor, frozen flux theorem, magnetic fields of planets in the Solar System, Geodynamo and origin of Earth’s Magnetic field, magnetic susceptibility of rocks and measurements; Various corrections applied to magnetic data, IGRF, Poisson’s relation of gravity and magnetic potential field, upward and downward continuation, magnetic anomalies due to geometrical and irregular shaped bodies; Interpretation of processed magnetic anomaly data; Derivative, analytic signal and Euler Depth Solutions.” The numericals are components from F, I and D, paleolatitude, the magnetic Reynolds number, the anomaly of a sphere, continuation and Euler depth. Take μ0 = 4π × 10⁻⁷ H/m.

1. The geomagnetic field: elements, dipole formulas, IGRF and variations

The Earth’s field is a vector F at each point. Its elements are the horizontal component H, the vertical component Z (positive downward), the northward and eastward components X and Y, the declination D (angle of H east of true north) and the inclination I (angle of F below the horizontal, positive downward). They are tied by H = F cos I, Z = F sin I, tan I = Z/H, X = H cos D, Y = H sin D and F² = X² + Y² + Z². Surveys use the nanotesla (1 nT = 10⁻⁹ T, the old gamma); the main field runs from about 25,000 nT near the equator to about 65,000 nT near the poles.

To first order the field is that of a geocentric axial dipole, a bar magnet of moment m ≈ 8 × 10²² A·m² along the spin axis. With B₀ = μ0 m/(4π a³) ≈ 31,000 nT (a = 6371 km) and latitude φ: H = B₀ cos φ, Z = 2B₀ sin φ, F = B₀ √(1 + 3 sin² φ) and tan I = 2 tan φ. So F at the pole is twice F at the equator. The real field departs from this: the best-fitting dipole is tilted about 10° from the spin axis and meets the surface at the geomagnetic poles, which differ from the dip poles (where I = ±90°); the remainder is the non-dipole field.

Secular variation: slow change over years to centuries. Diurnal variation: tens of nT a day from ionospheric currents. Magnetic storms: hundreds of nT or more, when surveys stop.
IGRF (International Geomagnetic Reference Field): a spherical-harmonic model of the core-generated main field, revised every five years. Subtracting it leaves the crustal anomaly plus external variation.

⚠️ Two common slips

The law is tan I = 2 tan φ, not I = 2φ. And the dipole’s geomagnetic pole is not the dip pole where the measured field is vertical.

2. Paleomagnetism: remanence, paleolatitude, reversals and field tests

Rocks record the ambient field as natural remanent magnetisation. Thermoremanent (TRM) is acquired on cooling through the blocking temperature (at or a little below the Curie point) in igneous rock; the Curie point of magnetite is about 580 °C and of hematite about 675 °C. Detrital (DRM) is the alignment of grains in settling sediment, chemical (CRM) comes from grain growth or alteration, and viscous (VRM) is a soft overprint from the present field. Stepwise thermal or alternating-field demagnetisation strips the soft parts and isolates the characteristic direction. The Königsberger ratio Q = M_r/M_i (remanent over induced) says which part dominates the anomaly.

Paleolatitude. Averaged over tens of thousands of years the field is a geocentric axial dipole (the GAD hypothesis), so a rock’s remanent inclination gives the latitude λ where it formed: tan I = 2 tan λ. Example: I = 45° gives tan λ = 0.5, λ = 26.6°. Declination gives the direction to the paleo-pole but nothing about longitude, so paleomagnetism fixes paleolatitude and orientation, never paleolongitude. The mean of virtual geomagnetic poles is the paleomagnetic pole; its sequence with age for one continent is its apparent polar wander path, and paths of continents once joined coincide after the ocean between them is closed.

The field has reversed many times, recorded in lavas, in sediments and as linear marine magnetic stripes symmetric about spreading ridges (the Vine–Matthews–Morley hypothesis). Field tests: the fold test (directions clustering best after beds are restored to horizontal predate folding), the conglomerate test (random clast directions mean the magnetisation predates the conglomerate) and the reversal test (normal and reversed means antipodal, so overprints are removed).

⚠️ Remanence is not a bonus detail

Total magnetisation is the vector sum of induced and remanent parts. Where Q is above 1 (fresh basalt, many iron ores) an anomaly may be shifted or even reversed in sign compared with an induced-only model, which misplaces the source.

3. The induction equation, frozen flux, the geodynamo and planetary fields

Take Faraday’s law ∇×E = −∂B/∂t, Ampère’s law without displacement current ∇×B = μ0 J, and Ohm’s law for a moving conductor, J = σ(E + u×B), where u is the fluid velocity. Eliminating E and J and using ∇·B = 0 gives the induction equation ∂B/∂t = ∇×(u×B) + η ∇²B, with magnetic diffusivity η = 1/(μ0 σ). The first term is convection: the flow carries and stretches field lines. The second is diffusion: ohmic decay. Their ratio is the magnetic Reynolds number Rm = U L/η = μ0 σ U L.

Two limits of the induction equation
Regime	Dominates	Consequence
Rm ≪ 1	Diffusion	Field decays freely; for a conducting sphere of radius a the slowest mode dies in τ = μ0 σ a²/π²
Rm ≫ 1	Convection	Frozen-flux theorem: flux through any loop moving with the fluid stays constant, so field lines are carried with the flow

For the core τ is of the order of 10⁴ years, far less than the age of the Earth, so the field cannot be a relic and must be regenerated. Frozen flux lets secular variation be read as motion of the core fluid.

The geodynamo needs (i) a conducting fluid, the liquid iron-alloy outer core, (ii) vigorous motion, driven by thermal convection and by compositional convection as the inner core freezes and releases light elements, (iii) rotation, whose Coriolis force makes the flow helical, so differential rotation turns poloidal field into toroidal (ω-effect) and helical convection turns it back (α-effect), and (iv) Rm above a critical value, of order 10 to 100 or more; the core has hundreds to 10³. Cowling’s theorem forbids a purely axisymmetric field from being maintained, so a working dynamo is three-dimensional. A permanent magnet is no source: the core is far above the Curie temperature.

Magnetic fields of Solar System bodies
Body	Intrinsic field
Mercury	Weak dipole, surface field roughly 1 percent of the Earth’s, nearly aligned with the spin axis
Venus	None detected; very slow rotation
Moon, Mars	No global field today; magnetised ancient crust (strong on Mars) shows an early dynamo that has ceased
Jupiter	Strongest planetary field, moment of order 10⁴ times the Earth’s, made in liquid metallic hydrogen
Saturn	Almost perfectly axisymmetric dipole (tilt under 1°), commonly explained by a stable conducting layer screening the non-axisymmetric part
Uranus, Neptune	Strongly tilted (about 59° and 47°) and offset, rich in non-dipole terms; probably made in a shell of ionic water-rich fluid

4. Magnetic susceptibility of rocks and its measurement

A material in a field H acquires M = χH, where χ is the dimensionless SI volume susceptibility. The Earth’s field F (in tesla) means H = F/μ0, so M_induced = χF/μ0, valid for χ ≪ 1, which holds for nearly every rock. Diamagnetic minerals (quartz, calcite, halite) have χ ≈ −10⁻⁵; paramagnetic ones (biotite, olivine, pyroxene) about 10⁻⁵ to a few × 10⁻³; ferrimagnetic magnetite, titanomagnetite and pyrrhotite are strongly magnetic, magnetite with χ of order 1 SI or more. Hematite is weak.

Magnetite is the main carrier, so sediments and felsic rocks are usually very low (10⁻⁵ to 10⁻³), mafic igneous rocks higher (10⁻³ to 10⁻¹), and serpentinised ultramafics and iron formation higher still. The strong susceptibility of magnetite vanishes above the Curie temperature, so the magnetic crust ends at the Curie isotherm, some tens of km deep and shallower where the crust is hot.

Laboratory: an AC susceptibility (kappa) bridge places the sample in a coil, whose inductance change is proportional to χ; measuring in many orientations gives the anisotropy of susceptibility (rock fabric); χ against temperature gives the Curie point and so the carrier mineral. A hand-held meter reads outcrops and cores in the field.
Field magnetometers: the proton-precession (or Overhauser) type reads the total field F absolutely, to about 0.1 nT, from f = (γp/2π) F with γp/2π = 42.58 Hz per μT; optically pumped alkali-vapour magnetometers are more sensitive; the fluxgate reads components and is relative.

5. Corrections, IGRF, Poisson’s relation and continuation

Diurnal and storm variation: a base-station magnetometer records continuously and its time-synchronised variation is subtracted from the roving readings.
Main field: the IGRF value for the point and date is subtracted. This also removes the latitude trend, about 4 to 5 nT/km in mid-latitudes, so no separate latitude correction is made.
Airborne and cultural noise: tie-line levelling, heading and lag corrections, steady terrain clearance; fences, pipes and power lines are avoided or edited out.

There is no free-air or Bouguer analogue: the main-field vertical gradient is only about 0.02 nT/m. What remains is the residual anomaly ΔT = T(observed) − diurnal − IGRF.

Poisson’s relation. Let a body of uniform density ρ be uniformly magnetised with intensity M along direction α. With gravitational potential U = Gρ∫dV/r (g = ∇U, z downward), the magnetic scalar potential (H = −∇V_m) is V_m = −(M/(4πGρ)) ∂U/∂α. So each magnetic field component is a derivative of a gravity component, for example B_z = (μ0 M/(4πGρ)) ∂g_z/∂α. It needs ρ and M both uniform (constant M/ρ). Its use is the pseudo-gravity transformation: the gravity anomaly the body would give if its density were proportional to its magnetisation, to compare with measured gravity.

Continuation. Outside the sources the field obeys Laplace’s equation, so a field measured on one plane can be computed on another. For wavenumber k = 2π/λ, upward continuation by h multiplies the amplitude spectrum by e^(−kh): short wavelengths (shallow sources, noise) are suppressed, a stable smoothing used to isolate the regional field. Downward continuation multiplies by e^(+kh): it sharpens anomalies but amplifies short-wavelength noise without limit, and holds only down to just above the sources. Example: λ = 1 km, h = 200 m leaves e^(−2π × 0.2) = 0.28 of the amplitude.

6. Magnetic anomalies of geometric and irregular bodies

The total-field anomaly ΔT is, to a good approximation, the component of the anomalous field along the regional field. For induced magnetisation M = χF/μ0 is parallel to F. A dipole of moment m gives B_r = (μ0/4π) 2m cos θ/r³ and B_θ = (μ0/4π) m sin θ/r³, with θ from the moment axis. A uniformly magnetised sphere of radius R acts exactly as a dipole at its centre, m = (4/3)πR³M, so its anomaly falls as the inverse cube of distance.

Worked example. A sphere with R = 50 m, centre depth z = 150 m, χ = 0.05 and F = 55,000 nT at the magnetic pole has vertical M = χF/μ0. Above its centre ΔT = 2(μ0/4π) m/z³ = (2/3) χF (R/z)³ = (2/3)(0.05)(55,000)(1/27) = 67.9 nT; μ0 cancels. At the magnetic equator the same sphere gives −(1/3) χF_eq (R/z)³ directly above, a negative low: half the size for equal F, and a quarter of the polar peak because the Earth’s dipole makes F_eq half F_pole. At mid-latitudes the profile has a positive peak with its negative trough on the magnetic-pole side (north in the northern hemisphere).

Depth rule. Above a vertical dipole ΔZ ∝ (2z² − x²)/(x² + z²)^(5/2), which changes sign at x = ±√2 z: the depth is the zero-crossing distance divided by √2. Irregular bodies are modelled forward, a 2D body as a polygon cross-section and a 3D body as a stack of prisms with closed-form responses, adjusted until calculated and observed anomalies match, or inverted by least squares. Interpretation is non-unique: a shallow small body and a deeper larger one can give similar anomalies, so geological constraint is essential.

Fall-off with distance and Euler structural index N
Source	Field falls as	N
Sphere (dipole)	r⁻³	3
Horizontal cylinder; vertical pipe	r⁻²	2
Thin dike or sheet	r⁻¹	1
Thick contact (fault step)	about r⁰	0

7. Interpreting processed anomalies: derivatives, analytic signal and Euler depth

Reduction to the pole recomputes the anomaly as if the magnetisation were vertical, making anomalies symmetric over their sources; it is unstable at low inclination and fails if remanence is strong. Derivatives sharpen the data: the first vertical derivative multiplies the spectrum by k, favouring shallow, narrow sources and resolving close neighbours but amplifying noise; maxima of the total horizontal gradient of reduced-to-pole data lie over steep edges.

The analytic signal amplitude is |A| = √[(∂T/∂x)² + (∂T/∂y)² + (∂T/∂z)²]. It is positive, crests over each source, and for 2D sources its shape does not depend on the direction of magnetisation, so it needs no reduction to the pole and suits low latitudes. Over a 2D thin dike |A| ∝ 1/(x² + h²), so its full width at half maximum is 2h and the depth h is half that width; over a contact |A| ∝ 1/√(x² + h²).

Euler deconvolution uses the fact that the anomaly of a simple source is homogeneous of degree −N. For a source at (x₀, y₀, z₀) and background B, with z positive downward: (x − x₀) ∂T/∂x + (y − y₀) ∂T/∂y + (z − z₀) ∂T/∂z = N (B − T). Solved in a moving window for a chosen structural index N (table above), it gives position and depth. Worked example: N = 2, T = 62 nT, B = 20 nT, observation at the surface directly over the source (∂T/∂x = 0), ∂T/∂z = 0.7 nT/m: −z₀ (0.7) = 2(20 − 62), so z₀ = 120 m.

⚠️ Euler depth depends on the index you choose

Depth is proportional to N, so a thin dike (N = 1) analysed with N = 3 gives about three times the true depth. Choose N from geology or by testing, and take tight clustering of solutions as the check.

Key takeaways

H = F cos I, Z = F sin I, Y = H sin D. Axial dipole: tan I = 2 tan φ, F = B₀√(1 + 3 sin²φ), so F at the pole is twice that at the equator. IGRF is subtracted; there is no free-air analogue.
Remanence (TRM, DRM, CRM, VRM) is cleaned by demagnetisation. Paleolatitude comes from tan I = 2 tan λ, never paleolongitude; fold, conglomerate and reversal tests validate it; Q = M_r/M_i shows whether remanence matters.
∂B/∂t = ∇×(u×B) + η∇²B, η = 1/(μ0σ), follows from Ohm’s law J = σ(E + u×B). Rm = μ0σUL: Rm ≪ 1 diffusion, Rm ≫ 1 frozen flux. The geodynamo needs conducting fluid, convection, rotation and Rm above critical; Cowling forbids an axisymmetric dynamo.
Induced magnetisation is M = χF/μ0. Quartz and calcite are diamagnetic (χ < 0), magnetite controls the susceptibility of rocks, and the magnetic crust ends at the Curie isotherm. Proton-precession magnetometers read |F| from f = (γp/2π) F.
Poisson: V_m = −(M/(4πGρ)) ∂U/∂α for uniform ρ and M, giving the pseudo-gravity transform. Upward continuation multiplies by e^(−kh), a stable smoothing; downward by e^(+kh), which amplifies noise.
A sphere over the pole gives ΔT = (2/3)χF(R/z)³. Analytic signal |A| is independent of magnetisation direction (width at half maximum 2h for a thin dike). Euler: (x − x₀)T_x + (z − z₀)T_z = N(B − T), N = 3, 2, 1, 0 for sphere, cylinder or pipe, dike, contact.

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.

The geomagnetic elements are defined with Z positive downward, H the horizontal intensity, D the declination east of true north and I the inclination. Which set of relations is correct?
1. H = F sin I, Z = F cos I and the eastward component Y = H sin D
2. H = F cos I, Z = F sin I and the eastward component Y = H sin D
3. H = F cos I, Z = F sin I and the eastward component Y = H cos D
4. H = F cos I, Z = F tan I and the eastward component Y = H sin D
Show answer
Answer: B — H = F cos I, Z = F sin I and the eastward component Y = H sin D
F makes the angle I with the horizontal, so its horizontal projection is F cos I and its vertical projection is F sin I. H lies at the angle D from north, so the eastward part is H sin D and the northward part is H cos D. Swapping sine and cosine in H and Z (the tempting slip) would make the field vertical at the equator.
A lava flow, never tilted since it cooled, carries a stable primary thermoremanent magnetisation of inclination 0°. Assuming the geocentric axial dipole hypothesis, where did the flow cool?
1. At a paleo-pole, because the field of an axial dipole is horizontal there
2. At the paleo-equator only if the polarity was normal, since reversed polarity would give 90°
3. At an undetermined paleolatitude, since inclination alone cannot fix latitude without the declination
4. At the paleo-equator, whichever the polarity was, because tan I = 2 tan λ gives λ = 0 for I = 0
Show answer
Answer: D — At the paleo-equator, whichever the polarity was, because tan I = 2 tan λ gives λ = 0 for I = 0
For an axial dipole tan I = 2 tan λ, so I = 0 means λ = 0. A reversed field gives the opposite sign of inclination, not a different magnitude, and 0 stays 0. At a pole the field is vertical (I = 90°), not horizontal, which is the tempting confusion. Inclination alone fixes the latitude; declination is not needed for it.
Orbiting spacecraft find that Mars has no global dipole field today, yet parts of its ancient southern highlands carry strong magnetisation. The most reasonable inference is that:
1. a core dynamo operated early in Mars’s history, magnetised the crust as it cooled, and later ceased
2. a core dynamo is active today but its field is completely screened by the thin atmosphere
3. the weak present field has slowly built a viscous remanence in the crust over billions of years
4. the iron-rich crust magnetises itself permanently without any external field, so no past dynamo is required
Show answer
Answer: A — a core dynamo operated early in Mars’s history, magnetised the crust as it cooled, and later ceased
Strong crustal remanence needs a strong magnetising field when the rock formed or cooled. Since no such field exists now, one existed earlier, and an extinct dynamo is the standard reading. A weak present field cannot build a strong remanence, and no rock magnetises itself without an applied field. An atmosphere does not screen a planetary magnetic field.
Which of the following minerals has a negative volume susceptibility in SI units?
1. Magnetite
2. Biotite
3. Quartz
4. Pyrrhotite
Show answer
Answer: C — Quartz
Quartz is diamagnetic, with a small negative susceptibility of about −10⁻⁵ SI. Biotite is the tempting answer, but its iron makes it paramagnetic (small positive χ). Magnetite and pyrrhotite are ferrimagnetic with large positive susceptibility.
Which pair of corrections is routinely applied when reducing a ground magnetic survey to an anomaly map?
1. A Bouguer slab correction for station elevation and a free-air correction
2. A terrain correction with a density of 2.67 g/cm³ and a latitude correction from the normal-gravity formula
3. Removal of the lunar-solar tidal drift from a spring-gravimeter reading and an isostatic correction for crustal roots
4. Removal of the diurnal variation using a base-station record and subtraction of the IGRF for the date
Show answer
Answer: D — Removal of the diurnal variation using a base-station record and subtraction of the IGRF for the date
Time variation is removed with a base-station record, and the IGRF takes out the main field, including the latitude trend. The other pairs are gravity reductions: the vertical gradient of the magnetic field is only about 0.02 nT/m, so there is no routine free-air or Bouguer analogue.
Poisson’s relation connects the magnetic potential and the gravitational potential of one body. Which condition is essential for it to hold?
1. The body must be a sphere, since only a sphere has an equivalent point dipole and a closed-form potential
2. Density and magnetisation must both be uniform through the body, so M/ρ is constant
3. The magnetisation must be entirely induced by the present field, with no remanent part present
4. The observations must be made on the ground surface and not at flight height or any higher level
Show answer
Answer: B — Density and magnetisation must both be uniform through the body, so M/ρ is constant
Poisson’s relation holds for a body of any shape provided the density and the magnetisation (intensity and direction) are each uniform, so that M/ρ is constant. Remanent magnetisation is allowed as long as it is uniform. The observation level is irrelevant because the relation connects potentials everywhere outside the body.
A small, deeply buried sphere carries only induced magnetisation with χ ≪ 1. At the magnetic pole the total-field anomaly directly over its centre is +A. The same sphere at the same depth is moved to the magnetic equator, where the Earth’s dipole field intensity is half that at the pole. Ignoring demagnetisation and regional gradients, the total-field anomaly directly over its centre is:
1. +A/4, because the moment and the dipole-field factor are both halved, but the sign of the anomaly is unchanged
2. −A/2, because the induced moment is horizontal and the field over a horizontal dipole is antiparallel at half the axial strength
3. −A/4, because the antiparallel field over a horizontal dipole is half the axial value and the induced moment is also halved
4. −A, because the sign reverses while the magnitude is set by the sphere, its depth and its susceptibility alone
Show answer
Answer: C — −A/4, because the antiparallel field over a horizontal dipole is half the axial value and the induced moment is also halved
At the pole the induced moment is vertical and the field on the axis above it is 2(μ0/4π)m/z³. At the equator the induced moment is horizontal and the point lies on the perpendicular bisector, where the field is antiparallel with magnitude (μ0/4π)m/z³, half as large. The induced moment m ∝ χF is also halved because F is halved. So the anomaly is −(1/2)(1/2)A = −A/4. Forgetting the weaker inducing field gives the tempting −A/2.
Which statement about the analytic-signal amplitude of magnetic anomaly data is correct?
1. For 2D sources its shape does not depend on the magnetisation direction, so no pole reduction is needed at low latitude
2. It is the first vertical derivative of the anomaly and therefore changes sign across the edge of a magnetised body
3. It can be computed only after a structural index has been chosen for the geometry of the source body
4. It is obtained by downward continuation to the source depth, so the source depth must be known in advance
Show answer
Answer: A — For 2D sources its shape does not depend on the magnetisation direction, so no pole reduction is needed at low latitude
The amplitude √[(∂T/∂x)² + (∂T/∂z)²] is positive everywhere and, for 2D bodies, has the same shape whatever the magnetisation direction, which is its advantage at low latitudes. It needs no structural index (that is Euler deconvolution) and no depth input, and it is not a single derivative, so it does not change sign.
A thin vertical dike, whose true structural index is N = 1, is analysed by Euler deconvolution with N = 3. At the anomaly peak, where the horizontal derivative is zero, the depth estimate compared with the true depth is:
1. equal, because N cancels when the horizontal derivative is zero
2. one third of the true depth, because a larger N always means a shallower source
3. equal, because the Euler equation determines N from the data without any input
4. about three times the true depth, because z₀ = N(T − B)/(∂T/∂z) scales with N
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Answer: D — about three times the true depth, because z₀ = N(T − B)/(∂T/∂z) scales with N
With ∂T/∂x = 0 the Euler equation reduces to z₀ = N(T − B)/(∂T/∂z). The data fix T − B and ∂T/∂z, so the depth is proportional to the chosen N: N = 3 on a source with N = 1 triples the depth. N does not cancel, and it is an input, not something the equation solves for.
Which of the following statements about the induction equation and the geodynamo are correct?
1. Frozen-flux behaviour needs a magnetic Reynolds number far above 1, so convection dominates over diffusion
2. A purely axisymmetric magnetic field can be maintained by dynamo action if the flow is vigorous enough
3. The convection term ∇×(u×B) comes from Ohm’s law for a conductor moving with velocity u through the field
4. The geomagnetic field is produced by the permanent magnetisation of the solid, iron-rich inner core
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Answer: A — Frozen-flux behaviour needs a magnetic Reynolds number far above 1, so convection dominates over diffusion; C — The convection term ∇×(u×B) comes from Ohm’s law for a conductor moving with velocity u through the field
In a frame moving with the fluid the electric field is E + u×B, which puts the term ∇×(u×B) into the induction equation; when Rm ≫ 1 that term dominates and flux is frozen into the fluid. Cowling’s theorem rules out maintaining an axisymmetric field however vigorous the flow. The inner core is far above the Curie temperature, so it cannot be a permanent magnet.
Which of the following statements about paleomagnetism are correct?
1. Under a geocentric axial dipole the remanent declination gives the paleolatitude of the site through the relation tan D = 2 tan λ
2. If magnetisation directions cluster more tightly once folded beds are restored to horizontal, the magnetisation predates the folding
3. Paleomagnetic data fix the paleolongitude of a continent as well as its paleolatitude, because the paleo-pole marks one point on the globe
4. In a reversal test, the normal and reverse mean directions should be antipodal once overprints are removed
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Answer: B — If magnetisation directions cluster more tightly once folded beds are restored to horizontal, the magnetisation predates the folding; D — In a reversal test, the normal and reverse mean directions should be antipodal once overprints are removed
The fold test and the reversal test are described correctly. The latitude comes from the inclination (tan I = 2 tan λ), not the declination. A pole position is rotationally ambiguous about the Earth’s axis, so longitude cannot be recovered; only paleolatitude and orientation can.
Which of the following statements about upward and downward continuation of magnetic data are correct?
1. Upward continuation multiplies the amplitude spectrum by e^(−kh), so it is a stable smoothing operation
2. Downward continuation multiplies the spectrum by e^(+kh) and so amplifies short-wavelength noise
3. Upward continuation sharpens the anomalies of shallow sources more than those of deep sources, boosting short wavelengths
4. Both operations rely on the field satisfying Laplace’s equation in the source-free region between the two levels
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Answer: A — Upward continuation multiplies the amplitude spectrum by e^(−kh), so it is a stable smoothing operation; B — Downward continuation multiplies the spectrum by e^(+kh) and so amplifies short-wavelength noise; D — Both operations rely on the field satisfying Laplace’s equation in the source-free region between the two levels
Continuation is valid because the field is harmonic between the levels. Upward continuation damps short wavelengths (e^(−kh)); downward continuation boosts them (e^(+kh)), which is why it is unstable with noisy data. Upward continuation therefore suppresses, not sharpens, shallow-source anomalies.
At a station the total field is F = 50,000 nT, the inclination satisfies tan I = 4/3 with I positive, and the declination is 30° east of north. The eastward component Y, in nT, is ____ .
Numerical answer — type the value.
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Answer: 15000
tan I = 4/3 gives cos I = 3/5, so H = F cos I = 30,000 nT. Y = H sin D = 30,000 × sin 30° = 15,000 nT. The mistake to avoid is using F sin D, which ignores the dip.
A well-preserved lava flow carries a primary remanence of inclination 45°. Assuming the geocentric axial dipole hypothesis, its paleolatitude in degrees, to one decimal place, is ____ .
Numerical answer — type the value.
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Answer: 26.6
tan λ = tan I / 2 = tan 45° / 2 = 0.5, so λ = arctan 0.5 = 26.565°, which is 26.6° to one decimal place. Taking λ = 45°/2 = 22.5° is the common slip, because the law is on the tangents, not the angles.
In a model of the outer core take σ = 4 × 10⁵ S/m, flow speed U = 4 × 10⁻⁴ m/s, length scale L = 2 × 10⁶ m and μ0 = 4π × 10⁻⁷ H/m. The magnetic Reynolds number Rm = μ0 σ U L, to the nearest integer, is ____ .
Numerical answer — type the value.
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Answer: 402
Rm = μ0 σ U L = (4π × 10⁻⁷)(4 × 10⁵)(4 × 10⁻⁴)(2 × 10⁶). The product σUL = 3.2 × 10⁸ and times 1.2566 × 10⁻⁶ gives 402.1, so 402. A value far above 1 means convection dominates diffusion and flux is nearly frozen in.
A sphere of radius 50 m, with its centre at 150 m depth, has susceptibility χ = 0.05 SI and is magnetised only by induction. It lies at the magnetic pole, where F = 55,000 nT, and demagnetisation is ignored. The total-field anomaly directly above its centre, in nT to the nearest integer, is ____ .
Numerical answer — type the value.
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Answer: 68
M = χF/μ0 and m = (4/3)πR³M. Above the centre at the pole ΔT = 2(μ0/4π) m/z³ = (2/3) χF (R/z)³, since μ0 cancels. With R/z = 1/3: ΔT = (2/3)(0.05)(55,000)(1/27) = 67.9 nT, so 68 nT.
Euler deconvolution is applied with structural index N = 2 and background B = 20 nT. At a surface point directly above the source, so that ∂T/∂x = 0, the field is T = 62 nT and the vertical derivative is ∂T/∂z = 0.7 nT/m with z positive downward. The source depth z₀ in metres is ____ .
Numerical answer — type the value.
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Answer: 120
With the point at z = 0 and ∂T/∂x = 0 the equation (z − z₀) ∂T/∂z = N (B − T) becomes −z₀ (0.7) = 2 (20 − 62) = −84, so z₀ = 84/0.7 = 120 m. Equivalently z₀ = N (T − B)/(∂T/∂z).
A sinusoidal magnetic anomaly of wavelength 1.0 km measured at ground level is continued upward by 200 m. The fraction of its amplitude that remains, to two decimal places, is ____ .
Numerical answer — type the value.
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Answer: 0.28
Upward continuation multiplies the amplitude by e^(−kh) with k = 2π/λ. Here kh = 2π × 0.2/1.0 = 1.2566 and e^(−1.2566) = 0.2846, which is 0.28. Using k = 1/λ without the 2π (the slip) would give e^(−0.2) = 0.82.