Question 4 of 8: Vertical component of H in an obliquely climbing plane wave
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 4: Vertical component of H in an obliquely climbing plane wave (20 marks)
Given. A single plane wave climbing at 45° above the horizon on a north-westerly track, with its electric vector lying flat.
Given data
Quantity
Symbol
Value
Frequency
$f$
10 GHz
Horizontal bearing of the ray
—
north-west
Elevation of the ray
$\psi$
45°
Average power density
$S$
$4\times10^{-8}$ W/m²
Polarisation
—
electric field horizontal
Intrinsic impedance of free space
$\eta_0$
$120\pi = 376.99\ \Omega$
Find. The rms magnitude of the vertical (upward) component of the magnetic field.
Left: the horizontal geometry — E is horizontal, so it must lie across the ground track. Right: the vertical plane containing the ray, where H is forced to lie and leans 45° from the vertical.
Approach. Get the total rms field from the power density, then settle the geometry with an explicit orthogonal triad rather than by inspection, because the two plausible trigonometric factors coincide at 45°.
Total fields from the power density. For a plane wave in free space the average Poynting magnitude is $S = E_{rms}H_{rms} = \eta_0 H_{rms}^2$, so $$H_{rms} = \sqrt{\frac{S}{\eta_0}} = \sqrt{\frac{4\times10^{-8}}{376.99}} = 1.030\times10^{-5}\ \text{A/m} = 10.30\ \mu\text{A/m},$$ and for completeness $E_{rms} = \sqrt{S\eta_0} = 3.883$ mV/m. The stated 10 GHz is a decoy: it licenses the plane-wave relation but never enters the arithmetic.
Fix a frame and write the propagation direction. Take east, north and up as $\hat{x},\hat{y},\hat{z}$. A north-westerly horizontal bearing is $(-1,1,0)/\sqrt{2}$, and lifting it to $45^\circ$ gives $$\hat{k} = \cos\psi\,\frac{(-1,1,0)}{\sqrt{2}} + \sin\psi\,\hat{z} = \left(-\tfrac12,\ \tfrac12,\ \tfrac{1}{\sqrt2}\right),$$ which is a unit vector as required.
Locate E. It must be perpendicular to $\hat{k}$ and, we are told, horizontal. The only horizontal direction satisfying $\hat{e}\cdot\hat{k} = 0$ is $\hat{e} = (1,1,0)/\sqrt{2}$ — the north-east / south-west line, at right angles to the ground track. This is the physical content of ‘the electric field is horizontal’.
Get H from the triad. With $\hat{h} = \hat{k}\times\hat{e}$, $$\hat{h} = \left(-\tfrac12,\ \tfrac12,\ \tfrac{1}{\sqrt2}\right) \times \frac{(1,1,0)}{\sqrt{2}} = \left(-\tfrac12,\ \tfrac12,\ -\tfrac{1}{\sqrt2}\right).$$ H therefore lies in the vertical plane containing the ray — it has to, being perpendicular to a horizontal E — and its vertical share is $|\hat{h}\cdot\hat{z}| = 1/\sqrt{2}$.
Identify the trigonometric factor properly. Repeating the construction for a general elevation gives $\hat{h} = \sin\psi\,\hat{x}' - \cos\psi\,\hat{z}$, so the vertical fraction is $\cos\psi$, not $\sin\psi$. The limits confirm it: a wave skimming the horizon ($\psi = 0$) with horizontal E has a purely vertical H, while one fired straight up ($\psi = 90^\circ$) has H entirely horizontal. At the paper’s $45^\circ$ the two functions happen to be equal, which is exactly why the construction is worth writing down.
Assemble the answer. $$H_{z,rms} = H_{rms}\cos\psi = 1.030\times10^{-5} \times 0.7071 = \boxed{7.28\ \mu\text{A/m}} .$$ The remaining horizontal component of H is the same size, $7.28\ \mu$A/m, directed along the ground track, and the two combine to the full $10.30\ \mu$A/m.
