Question 5 of 8: Rms EMF induced in a loop by an obliquely arriving plane wave
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2017 — 16-Elec-A7 Electromagnetics. Three hours, closed book (one Casio or Sharp approved calculator). Eight questions of equal value; the rubric states that any five questions constitute a complete paper and that only the first five answered will be marked. All eight are solved here, because the set is a study resource rather than a sitting. Aids printed on the paper: $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}$ and $\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}$.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission-line transients, stub networks, waveguides); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (plane waves, polarisation, Biot–Savart, inductance); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. (magnetostatics, guided waves); F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed. (wave polarisation, radiation); C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short-element fields).
Constants used throughout. The aid sheet's $\varepsilon_0$ and $\mu_0$ give $c = 2.9986 \times 10^{8}\ \text{m/s}$ and $\eta_0 = 376.82\ \Omega$, while every round figure the paper quotes (10 km at $3 \times 10^{8}\ \text{m/s}$, 300 and 600 MHz, 10 GHz giving a 3 cm wavelength) is built on $c = 3.00 \times 10^{8}\ \text{m/s}$ and $\eta_0 = 120\pi = 376.99\ \Omega$. Those are the values used below; the two sets differ by less than 0.05 % and no answer or conclusion changes.
Question 5: Rms EMF induced in a loop by an obliquely arriving plane wave
Given. A horizontally polarised wave climbing steeply out of the north-west is sampled by a small multi-turn loop lying in the vertical east–west plane.
Find. The rms EMF appearing at the loop terminals.
Figure 5.1 — plan view of the ground track and the loop trace (left), and an elevation along the track showing E out of the page with H perpendicular to the ray (right).
Approach. Convert the power density into a magnetic field, work out the direction of $\mathbf{H}$ from the stated polarisation and geometry, project it onto the loop normal, then apply Faraday's law to the $N$-turn loop.
Get the field magnitudes. For a plane wave in free space $S = E_{rms}H_{rms} = \eta_0 H_{rms}^{2}$, so$$H_{rms} = \sqrt{\frac{S}{\eta_0}} = \sqrt{\frac{2}{376.99}} = 0.07284\ \text{A/m}, \qquad E_{rms} = \sqrt{S\eta_0} = 27.46\ \text{V/m}$$and the product $E_{rms}H_{rms}$ does return $2\ \text{W/m}^{2}$.
Write the propagation direction. In east–north–up coordinates the ray travels away from the south-east toward the north-west at $60^\circ$ elevation, so with $\hat{g} = (-1/\sqrt{2}, 1/\sqrt{2}, 0)$ the horizontal ground track,$$\hat{k} = \cos\psi\,\hat{g} + \sin\psi\,\hat{z} = (-0.3536,\ 0.3536,\ 0.8660)$$
Fix the polarisation vectors. The electric field is horizontal and must be transverse, so $\hat{e} \propto \hat{z} \times \hat{k}$, which lies along the north-east / south-west horizontal line. The magnetic field then follows from $\hat{h} = \hat{k} \times \hat{e} = (0.6124,\ -0.6124,\ 0.5)$. Reading that vector: its horizontal part has magnitude $\cos 30^\circ = \sin\psi = 0.866$ and lies along the ground track, while its vertical part is $\cos\psi = 0.5$. In other words, with $\mathbf{E}$ horizontal the magnetic field must lie in the vertical plane of propagation, leaning $90^\circ - \psi = 30^\circ$ above the horizontal.
Project onto the loop normal. A loop whose plane is the vertical east–west plane has its normal along north–south, $\hat{n} = \hat{y}$. Only the component of $\mathbf{H}$ along $\hat{n}$ threads the loop:$$|\hat{h} \cdot \hat{n}| = \sin\psi \cos 45^\circ = 0.866 \times 0.7071 = 0.6124$$the $45^\circ$ being the angle between the north-west ground track and the north–south loop normal. Hence $H_{n,rms} = 0.6124 \times 0.07284 = 0.04460\ \text{A/m}$.
Apply Faraday's law. For an electrically small loop the field is taken uniform over the area, so the flux linkage is $\lambda = N\mu_0 H_n A$ and $e = -\,d\lambda/dt$ has rms value$$\mathcal{E}_{rms} = N\,\omega\,\mu_0\,A\,H_{n,rms}$$With $\omega = 2\pi(3 \times 10^{9}) = 1.885 \times 10^{10}\ \text{rad/s}$ the prefactor is $N\omega\mu_0 A = 10 \times 1.885 \times 10^{10} \times 1.2566 \times 10^{-6} \times 2.5 \times 10^{-3} = 592.2$, so$$\mathcal{E}_{rms} = 592.2 \times 0.04460 = \boxed{26.4\ \text{V rms}}$$
Sanity-check the size of the answer. A 26 V signal from a $2\ \text{W/m}^{2}$ wave looks large until the frequency is noticed: the EMF scales as $f$, and at 3 GHz the factor $\omega\mu_0 A$ alone is $59.2\ \text{V per A/m}$ per turn. The same loop at 3 MHz would deliver $26\ \text{mV}$.
Check: the answer above uses the standard electrically small loop model, which is what the data are designed for. At 3 GHz, however, $\lambda = 10\ \text{cm}$ while a square loop of $25\ \text{cm}^{2}$ is $5\ \text{cm}$ on a side — a full $\lambda/2$. The uniform-field assumption is therefore marginal: a rigorous treatment would integrate the phase taper of $H_n$ across the loop, which reduces the net flux linkage and so the EMF. The value quoted is the upper bound the question intends; it should be read as "of order 26 V" rather than as three-figure truth.
