22-Elec-A7 Electromagnetics · December 2018

Question 7 of 8: Penetration depth of a 10 GHz wave into mercury

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

Notes on this paper

Paper format. National Examinations, December 2018, 16-Elec-A7 Electromagnetics. Three hours, closed book, one approved Casio or Sharp calculator. Eight questions, all of equal value; the rubric says any five constitute a complete paper, so a candidate answers five — but all eight are worked here, because the set is a study resource. Aids printed on the paper: $\varepsilon_0 = 8.85\times10^{-12}$ F/m, $\mu_0 = 4\pi\times10^{-7}$ H/m, $e = 1.6\times10^{-19}$ C.

Reference texts. M. N. O. Sadiku, Elements of Electromagnetics, 7th ed.; W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed.; F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; D. M. Pozar, Microwave Engineering, 4th ed.; C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed.

Constants used throughout. Question 6 hands us $\omega = 2\pi\times10^{10}$ rad/s together with $k = (2\pi/3)\ \text{cm}^{-1}$, so the setter’s phase velocity is $\omega/k = 3.000\times10^{8}$ m/s exactly; Question 3’s cut-off arithmetic only closes on round numbers under the same choice. Accordingly $c = 3\times10^{8}$ m/s and $\eta_0 = 120\pi = 376.99\ \Omega$ are used here. The paper’s own aid list would give $c = 2.99863\times10^{8}$ m/s and $\eta_0 = 376.83\ \Omega$ — a 0.05 % shift that changes no answer except at one deliberate boundary case, flagged in Question 3.

Question 7: Penetration depth of a 10 GHz wave into mercury (20 marks)

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.

Given. A centimetre-wave signal striking a liquid metal at normal incidence, with the metal characterised only by its resistivity.

Given data
Quantity	Symbol	Value
Frequency	$f$	10 GHz
Resistivity of mercury	$\rho$	$10^{-6}\ \Omega\cdot\text{m}$
Conductivity	$\sigma = 1/\rho$	$10^{6}$ S/m
Permeability	$\mu$	$\mu_0$ (mercury is non-magnetic)
Incidence	—	normal, from free space
Target attenuation	—	field reduced to 10 %

Find. The depth into the mercury at which the electric field has fallen to one tenth of its value at the surface.

Field amplitude inside the metal. The skin depth marks the 37 % point; ten per cent arrives 2.303 skin depths in, at 11.6 μm.

Approach. Test the loss tangent to confirm the good-conductor regime, take the skin depth from it, and scale by the natural logarithm of ten.

Classify the medium. The loss tangent at 10 GHz is $$\frac{\sigma}{\omega\varepsilon_0} = \frac{10^{6}}{2\pi\times10^{10}\times 8.85\times10^{-12}} = 1.80\times10^{6} ,$$ overwhelmingly greater than unity, so mercury is an excellent conductor at this frequency and the good-conductor approximations are exact to six figures. This check is worth a line: at the same conductivity but optical frequencies the conclusion would reverse.
Skin depth. In a good conductor the attenuation and phase constants are equal, $\alpha = \beta = \sqrt{\pi f\mu\sigma}$, so $$\delta = \frac{1}{\alpha} = \frac{1}{\sqrt{\pi f \mu_0 \sigma}} = \frac{1}{\sqrt{\pi \times 10^{10} \times 4\pi\times10^{-7} \times 10^{6}}} = 5.03\ \mu\text{m},$$ whence $\alpha = 1.987\times10^{5}$ Np/m. Solving the exact propagation constant $\gamma = \sqrt{j\omega\mu(\sigma + j\omega\varepsilon_0)}$ returns the same $\alpha$ to five figures, confirming the approximation.
Scale to the ten per cent point. The amplitude decays as $e^{-\alpha d}$, so $$e^{-\alpha d} = 0.10 \quad\Longrightarrow\quad d = \frac{\ln 10}{\alpha} = 2.303\,\delta = \frac{2.3026}{1.987\times10^{5}} = \boxed{1.16\times10^{-5}\ \text{m} = 11.6\ \mu\text{m}} .$$ For scale, that is about a seventh of a human hair, and the wavelength inside the mercury is only 31.6 $\mu$m against 3 cm in air — a compression of nearly a thousand to one.
How much actually gets in. The intrinsic impedance of the mercury is $\eta_m = (1+j)/(\sigma\delta) = 0.199(1+j)\ \Omega$, magnitude $0.281\ \Omega$ at $45^\circ$, against 377 $\Omega$ for free space. The reflection coefficient is therefore $$\Gamma = \frac{\eta_m - \eta_0}{\eta_m + \eta_0}, \qquad |\Gamma| = 0.9989, \qquad 1 - |\Gamma|^{2} = 0.0021 ,$$ so 99.79 % of the incident power bounces straight back off the surface and only about a fifth of one per cent enters the liquid at all. This is why a mercury surface is a mirror.
State the reference plane. Because of that reflection the field just inside the surface is already only $|\tau| = 1.49\times10^{-3}$ of the incident field — 0.149 %, far below the 10 % asked for. The question is therefore to be read, as intended, as the decay measured from the value at the surface inside the metal, and the answer stands at 11.6 $\mu$m. Read literally against the incident wave it has no positive solution, since the field never reaches 10 % of the incident value anywhere in the mercury.

Check: ‘10 % of its incident value’ is taken to mean 10 % of the transmitted field at the surface, which is the standard meaning of a penetration-depth question and the only reading with a finite answer. The surface reflection is quantified above rather than passed over, because it is the reason the two readings differ so sharply: the transmitted field starts at 0.149 % of the incident field, so on the literal reading the required depth would be negative.

Final results
Quantity asked	Result
Conductivity	$\sigma = 10^{6}$ S/m
Loss tangent at 10 GHz	$1.80\times10^{6}$ — a very good conductor
Skin depth	$\delta = 5.03\ \mu\text{m}$
Attenuation constant	$\alpha = 1.987\times10^{5}$ Np/m (1.726 dB/$\mu$m)
Depth for 10 % of the surface field	$d = \ln 10 / \alpha = 11.6\ \mu\text{m}$
Power reflected at the surface	$\|\Gamma\|^{2} = 99.79$ %

Topic. Plane waves in a good conductor: skin depth and surface reflection.

Key relations. $\tan\delta_{loss} = \sigma/\omega\varepsilon$; $\;\alpha = \beta = \sqrt{\pi f\mu\sigma}$; $\;\delta = 1/\alpha$; $\;\eta_m = (1+j)/(\sigma\delta)$; $\;\Gamma = (\eta_m - \eta_0)/(\eta_m + \eta_0)$; $\;d = \ln(1/\text{ratio})/\alpha$.

Why this works. Inside a conductor the displacement current is negligible beside the conduction current, and the wave equation degenerates into a diffusion equation whose solution decays and rotates at the same rate. That is why $\alpha$ and $\beta$ are equal there, and why the intrinsic impedance sits at $45^\circ$. Because that impedance is three orders of magnitude below the impedance of free space, the boundary behaves almost like a short circuit: nearly all the incident power is reflected, and the small fraction admitted is extinguished within a few micrometres.

Common pitfall. Two traps. Quoting the skin depth itself as the answer forgets that 37 % is not 10 %; the factor $\ln 10 = 2.303$ is the whole content of the last step. And using $\sigma$ in place of $\rho$, or vice versa, moves the answer by twelve orders of magnitude — always convert first and sanity-check that a metal has a conductivity in the millions, not the millionths.

Source. Sadiku, Elements of Electromagnetics, 7th ed., §10.3–10.6 (plane waves in lossy media, skin effect, reflection); Hayt and Buck, Engineering Electromagnetics, 9th ed., §11.2.

← Question 6 Next: Question 8 →