Question 4 of 8: RMS Vertical Component of E in an Obliquely Travelling Wave
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2018 — 16-Elec-A7, Electromagnetics. Closed book, three hours, approved Casio or Sharp calculator only. Eight questions, all of equal value; the paper states that any five constitute a complete paper, so each question carries 20 of the 100 marks that are actually graded. All eight are solved here, because this set is a study resource rather than an examination script.
Aids given on the paper. $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$, $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$, $e = 1.6\times10^{-19}\ \text{C}$. Those two constants fix $c = 1/\sqrt{\mu_0\varepsilon_0} = 2.9986\times10^{8}\ \text{m/s}$ and $\eta_0 = \sqrt{\mu_0/\varepsilon_0} = 376.82\ \Omega$ — which is why Question 1 quotes a 377 Ω line: it is free space itself, rendered as a cable.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission-line transients, stubs, waveguide modes); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. (Biot–Savart, motional EMF, plane waves); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed.; F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short-element fields and the reactive near-field boundary).
Constants used throughout: the paper's own aid list is taken as authoritative, so $c = 2.9986\times10^{8}$ m/s and $\eta_0 = 376.82\ \Omega$. Question 1 supplies its own propagation velocity ($3\times10^{8}$ m/s) and that stated value is used there. Question 5 is the one place where the difference between $2.9986\times10^{8}$ and $3.00\times10^{8}$ changes the wording of the answer, and both readings are worked through explicitly.
Question 4: RMS Vertical Component of E in an Obliquely Travelling Wave (20 marks)
Given. A plane wave in free space, climbing steeply, whose magnetic field is known to lie horizontally.
Given data
Quantity
Symbol
Value
Frequency
$f$
10 GHz
Power density (Poynting magnitude)
$S$
0.01 W/m$^{2}$
Direction of propagation
$\theta$
30° from the vertical
Magnetic field
$\mathbf{H}$
horizontal
Medium
—
free space, $\eta_0 = 376.82\ \Omega$
Find. The rms value of the vertical component of the electric field.
The vertical plane containing the ray. H is horizontal and normal to the page; E lies in the page, square to the ray, and therefore leans 30° from the horizontal (60° from the vertical).
Approach. Get the total rms field from the power density, then settle the geometry: because $\mathbf{H}$ is horizontal and perpendicular to the ray, $\mathbf{E}$ must lie in the vertical plane of propagation, and being perpendicular to the ray it leans the complementary angle.
Total field from the power density. For a plane wave in free space $S = E_{\text{rms}}^{2}/\eta_0$, so $$E_{\text{rms}} = \sqrt{S\,\eta_0} = \sqrt{(0.01)(376.82)} = 1.941\ \text{V/m}.$$ This is the only place the stated 10 GHz matters — it guarantees the wave is a well-formed plane wave, and then drops out.
Fix the geometry with a coordinate frame. Take $\hat{z}$ vertical and put the ray in the $x$–$z$ plane: $$\hat{k} = \sin 30^\circ\,\hat{x} + \cos 30^\circ\,\hat{z} .$$ “$\mathbf{H}$ horizontal” and $\mathbf{H}\perp\hat{k}$ force $\hat{h} = \hat{y}$, the one horizontal direction square to the ray.
Get the direction of E. For a plane wave $\mathbf{E}\times\mathbf{H}$ points along $\hat{k}$, so $$\hat{e} = \hat{h}\times\hat{k} = \hat{y}\times(\sin 30^\circ\,\hat{x} + \cos 30^\circ\,\hat{z}) = \cos 30^\circ\,\hat{x} - \sin 30^\circ\,\hat{z} .$$ So $\mathbf{E}$ lies in the same vertical plane as the ray, and its vertical direction cosine is $\sin 30^\circ$, not $\cos 30^\circ$.
Take the vertical component. $$E_{z,\text{rms}} = E_{\text{rms}}\sin 30^\circ = (1.941)(0.5) \;\Rightarrow\; \boxed{E_{z,\text{rms}} = 0.971\ \text{V/m}} .$$
Check by putting the components back together. The horizontal part is $E_{\text{rms}}\cos 30^\circ = 1.681$ V/m, and $\sqrt{0.971^{2} + 1.681^{2}} = 1.941$ V/m, recovering the total. It is also worth noting $H_{\text{rms}} = E_{\text{rms}}/\eta_0 = 5.15$ mA/m, entirely horizontal, and $E_{\text{rms}}H_{\text{rms}} = 0.01$ W/m$^2$ as required.
The step that earns the marks is recognising that the angle is quoted from the vertical. A ray leaning 30° off vertical is climbing at 60° elevation, and the electric field — locked square to it — leans 30° off the horizontal (60° off the vertical), so only half of it is vertical. Had the same 30° been measured from the horizontal instead, the vertical component would have been $1.681$ V/m, a factor of 1.73 different.
Final results
Quantity asked
Result
Free-space intrinsic impedance (from the paper’s aids)
