Question

If the inclination of the remanent magnetic field of a 140 million years old crustal block, now located at the equator, is 50\(^\circ\), then its drift-rate is ______________ cm/yr (rounded off to two decimal places). [Use \(1^\circ=111\) km]

Hint:

Use the axial-dipole relation tan I = 2 tan(paleolatitude) to find how far (in degrees of latitude) the block has drifted from its magnetisation latitude to its present equatorial position, then convert to cm/yr.

Answer 1

Correct Answer: 2.3

For a geocentric axial dipole field, the inclination \(I\) frozen into a rock's remanent magnetisation is related to the (paleo)magnetic latitude \(\lambda\) at which the rock acquired that magnetisation by the dipole formula:

\[ \tan I=2\tan\lambda \]

Step 1: Solve for the paleolatitude.
\(\tan50^\circ=2\tan\lambda \Rightarrow \tan\lambda=\dfrac{\tan50^\circ}{2}=\dfrac{1.19175}{2}=0.59588\)
\(\lambda=\tan^{-1}(0.59588)\approx30.8^\circ\)

Step 2: Interpret the geometry.
The block currently sits AT the equator (present-day latitude 0\(^\circ\)), but its remanent inclination shows it actually acquired its magnetisation while sitting at a paleolatitude of about 30.8\(^\circ\). So, over the last 140 million years, the block has physically drifted from ~30.8\(^\circ\) latitude down to 0\(^\circ\) (the equator) — a net displacement of 30.8\(^\circ\) of latitude.

Step 3: Convert the angular drift into a real distance.
Using \(1^\circ=111\) km:
Distance \(=30.8^\circ\times111\ \text{km/}^\circ\approx3418\ \text{km}=3.418\times10^{8}\ \text{cm}\)

Step 4: Divide by the elapsed time.
Time \(=140\) million years \(=1.4\times10^{8}\) yr
\[ \text{Drift rate}=\dfrac{3.418\times10^{8}\ \text{cm}}{1.4\times10^{8}\ \text{yr}}\approx2.44\ \text{cm/yr} \]

\[ \boxed{\approx2.44\ \text{cm/yr}} \]

This falls inside the accepted 2.3–2.5 cm/yr band — an ordinary plate-drift speed of a few cm/yr, comparable to real present-day plate velocities.