Question

What is the value of magnetotelluric impedance phase over a homogeneous half-space?

• A. 0°
• B. 90°
• C. 45°
• D. 180°

Hint:

For a homogeneous half-space, impedance \(Z=\sqrt{i\omega\mu_0\rho}\); taking the square root of \(i=e^{i\pi/2}\) halves its phase angle.

Answer 1

The Correct Option is C

For a uniform (1-D) conducting half-space, the electric and magnetic fields of a plane EM wave diffusing downward obey the quasi-static Maxwell equations (displacement currents neglected in the Earth):

\( \dfrac{\partial E_x}{\partial z} = -i\omega\mu_0 H_y \), \( \dfrac{\partial H_y}{\partial z} = -\sigma E_x \)

Combining these gives the diffusion equation \( \dfrac{\partial^2 E_x}{\partial z^2} = i\omega\mu_0\sigma E_x = k^2 E_x \), so the field decaying into the half-space is \( E_x(z) = E_x(0)e^{-kz} \) with

\[ k = \sqrt{i\omega\mu_0\sigma} = \sqrt{\omega\mu_0\sigma}\,e^{i\pi/4} \]

since \( i = e^{i\pi/2} \) and \( \sqrt{i} = e^{i\pi/4} \). From Faraday's law, \( H_y(z) = \dfrac{k}{i\omega\mu_0}E_x(z) \), so the magnetotelluric impedance is

\[ Z = \frac{E_x}{H_y} = \frac{i\omega\mu_0}{k} = \sqrt{\frac{i\omega\mu_0}{\sigma}} \]

The phase of the numerator \(i\omega\mu_0\) is \(90^\circ\) and the phase of \(k\) is \(45^\circ\), so the phase of \(Z\) is \(90^\circ - 45^\circ = 45^\circ\). This \(45^\circ\) phase is independent of frequency, resistivity or permeability — it is a universal signature of a homogeneous (1-D) half-space, and MT interpreters use departures from \(45^\circ\) as a diagnostic of lateral resistivity heterogeneity.

\(\boxed{\text{Impedance phase} = 45^\circ}\)