Question 6 of 8: Reading a Plane Wave from a Loop Probe
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2014 — 07-Elec-A7 Electromagnetics. Three hours, closed book, one approved Casio or Sharp calculator. Eight questions of equal value (20 marks each); the paper states that any five constitute a complete paper and that only the first five in the answer book are marked, but all eight are solved here so the set works as a complete study resource. Aids printed on the paper: $\epsilon_0 = 8.85\times10^{-12}$ F/m and $\mu_0 = 4\pi\times10^{-7}$ H/m, from which $\eta_0 = \sqrt{\mu_0/\epsilon_0} = 376.7\ \Omega$ and $c = 1/\sqrt{\mu_0\epsilon_0} = 3.00\times10^{8}$ m/s.
Reference texts. Sadiku, Elements of Electromagnetics, 7th ed.; Hayt & Buck, Engineering Electromagnetics, 9th ed.; Ulaby & Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; Pozar, Microwave Engineering, 4th ed.; Balanis, Antenna Theory: Analysis and Design, 4th ed.
Question 6: Reading a Plane Wave from a Loop Probe (20 marks)
Find. (i) the time-average power density of the wave, and (ii) the propagation vector $\vec{k}$ in east-north-up components.
Approach. Use Faraday's law for a small loop to convert the measured EMF into $H$, then $\eta_0$ and the Poynting product give the power density; for the direction, exploit the fact that $\vec{k}$ must be perpendicular to both E and H and reconstruct it as $\hat{E}\times\hat{H}$.
What the loop measures. For a loop small compared with the wavelength the flux is essentially uniform, and Faraday's law gives$$\mathcal{E} = -\frac{d\Phi}{dt},\qquad \Phi = \mu_0 H A\cos\psi,$$so the rms EMF at the optimum orientation (loop normal along H, $\psi = 0$) is $\mathcal{E} = \omega\mu_0 A H$. Note that the loop responds to the magnetic field, not the electric one.
Solve for the magnetic field. With $\omega = 2\pi\times10^{10}$ rad/s,$$\omega\mu_0 A = (6.283\times10^{10})(4\pi\times10^{-7})(10^{-4}) = 7.896\ \Omega\cdot\text{m}^{-1}\cdot\text{m}^{2},$$$$H = \frac{\mathcal{E}}{\omega\mu_0 A} = \frac{60}{7.896} = 7.60\ \text{A/m (rms)}.$$
Electric field and (i) power density. In free space the two fields are locked together by the intrinsic impedance:$$E = \eta_0 H = 376.7(7.60) = 2.86\times10^{3}\ \text{V/m (rms)},$$$$\boxed{S_{av} = EH = \eta_0 H^{2} = 2.18\times10^{4}\ \text{W/m}^{2} \approx 21.8\ \text{kW/m}^{2}.}$$Because both field values are rms, this product is already the time-average Poynting magnitude — no extra factor of one half is needed.
Set up the coordinate frame. Take $\hat{x}$ east, $\hat{y}$ north, $\hat{z}$ up. The magnetic field is horizontal along NE–SW, so$$\hat{H} = \tfrac{1}{\sqrt2}(\hat{x} + \hat{y}).$$
Locate the electric field. E must be perpendicular to H, so it lies in the vertical plane containing the NW–SE direction. Being inclined at $45^{\circ}$ to the vertical fixes it completely:$$\hat{E} = \cos45^{\circ}\,\hat{z} + \sin45^{\circ}\,\hat{n}_{NW} = (-0.5,\ 0.5,\ 0.707),$$where $\hat{n}_{NW} = (-\hat{x}+\hat{y})/\sqrt2$. A quick check confirms $\hat{E}\cdot\hat{H} = 0$ as required.
Direction of propagation. Energy flows along $\vec{E}\times\vec{H}$, so$$\hat{k} = \hat{E}\times\hat{H} = (-0.5,\ 0.5,\ -0.707),$$a ray heading north-west and descending at $45^{\circ}$ below the horizon. Its horizontal and vertical parts are equal in magnitude, which is the geometric mirror of E's $45^{\circ}$ tilt.
(ii) Magnitude and components. In free space$$|\vec{k}| = \frac{2\pi f}{c} = \frac{2\pi}{\lambda},\qquad \lambda = 3.00\ \text{cm},\qquad |\vec{k}| = 209.4\ \text{rad/m},$$$$\boxed{\vec{k} = -104.7\,\hat{x} + 104.7\,\hat{y} - 148.1\,\hat{z}\ \ \text{rad/m}.}$$
Figure 6.1 — The east-north-up triad with H along NE, E tilted 45° from vertical toward the north-west, and the resulting propagation vector.
Check: the data fix $\vec{k}$ only up to a four-fold sign ambiguity, because the question does not say which way E leans (north-west or south-east) nor which sense of the NE–SW axis H points along. All four possibilities lie in the vertical NW–SE plane at $45^{\circ}$ elevation, with components $(\pm104.7,\ \mp104.7,\ \pm148.1)$ rad/m; the descending north-west case is quoted above as the representative answer, and any of the four earns full marks provided E, H and $\vec{k}$ form a right-handed set.
The power density is enormous by broadcast standards — 21.8 kW/m2 is roughly sixteen times the solar constant — which is simply the consequence of demanding 60 V from a 1 cm2 loop. The result is nonetheless self-consistent: at 10 GHz the $\omega\mu_0 A$ factor is only about 7.9, so a large field is genuinely needed to reach that EMF.
