Vertical component of E in a wave climbing at 45 degrees: 22-Elec-A7 Electromagnetics, May 2017, Q5

Question 5: Vertical component of E in a wave climbing at 45 degrees (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 uniform linearly polarised plane wave in air, climbing at a known angle, with its power density and the orientation of its magnetic field specified.

Given data
Quantity	Symbol	Value
Frequency	$f$	$10$ GHz
Elevation of the propagation direction	$\psi$	$45^\circ$ above horizontal
Time-average power density	$S$	$0.1$ W/m$^{2}$
Magnetic-field polarisation	$\hat{H}$	horizontal
Intrinsic impedance of free space	$\eta_0$	$376.8\ \Omega$

Find. The rms magnitude of the vertical (upward) component of $\mathbf{E}$.

Figure 5.1 — H is horizontal and transverse, so it points out of the vertical plane; E must then lie in that plane, perpendicular to k, leaning 45 degrees from the vertical.

Approach. Get the total rms field from the power density, then use the strict orthogonality of $\mathbf{E}$, $\mathbf{H}$ and $\mathbf{k}$ to fix the direction of $\mathbf{E}$ and project it onto the vertical.

Total field from the power density. For a plane wave in free space $S = E_{\text{rms}}^{2}/\eta_0$, so $$E_{\text{rms}} = \sqrt{S\,\eta_0} = \sqrt{(0.1)(376.8)} = \boxed{6.14\ \text{V/m}} ,$$ with the companion $H_{\text{rms}} = \sqrt{S/\eta_0} = 16.3$ mA/m.
Fix the direction of E. Take $\hat{x}$ east, $\hat{y}$ north, $\hat{z}$ up and let the wave climb in the vertical plane, $\hat{k} = (\cos\psi)\,\hat{x} + (\sin\psi)\,\hat{z}$. The magnetic field must be transverse and is stated to be horizontal, so it is the horizontal direction perpendicular to that vertical plane, $\hat{H} = \hat{y}$. Then $\hat{E} = \hat{H}\times\hat{k}$: $$\hat{E} = \hat{y}\times\left[(\cos\psi)\hat{x} + (\sin\psi)\hat{z}\right] = (\sin\psi)\,\hat{x} - (\cos\psi)\,\hat{z} .$$ So $\mathbf{E}$ lies in the vertical plane of propagation and is tilted $\psi$ away from the vertical — it points forward and downward, at right angles to the ray.
Project onto the vertical. The vertical share of the unit vector is $|\hat{E}\cdot\hat{z}| = \cos\psi$, hence $$E_{z,\text{rms}} = E_{\text{rms}}\cos\psi = 6.139 \times \cos 45^\circ = \boxed{4.34\ \text{V/m}} .$$
Why it is $\cos\psi$ and not $\sin\psi$. At $\psi = 45^\circ$ the two functions are equal, which hides the distinction; the general rule is that the field leans away from the vertical by the same angle the ray leans away from the horizontal, so the vertical component always carries $\cos\psi$. A wave climbing steeply ($\psi \to 90^\circ$) has an almost horizontal $\mathbf{E}$ and hence almost no vertical component, which is the correct limit.
The frequency is not needed. Nothing in the answer uses 10 GHz. Its role is to certify that we are dealing with a uniform plane wave in a lossless medium (free-space $\eta_0$ applies, and any observation point is many wavelengths from the source), so the power-density-to-field conversion is legitimate. Recognising a decoy datum is part of the mark.

Final results
Quantity asked	Result
Total rms electric field	$E_{\text{rms}} = 6.14$ V/m
Total rms magnetic field	$H_{\text{rms}} = 16.3$ mA/m
Direction of E	in the vertical plane of propagation, $45^\circ$ from the vertical
Vertical component (answer)	$E_{z,\text{rms}} = E_{\text{rms}}\cos 45^\circ = 4.34$ V/m
Horizontal component of E	$4.34$ V/m as well (equal only because $\psi = 45^\circ$)

Topic. Plane-wave orthogonality and the projection of a linear polarisation onto a chosen axis.

Key relations. $S = E_{\text{rms}}^{2}/\eta_0 = \eta_0 H_{\text{rms}}^{2}$; $\eta_0 = \sqrt{\mu_0/\varepsilon_0} = 376.8\ \Omega$; $\hat{E} = \hat{H}\times\hat{k}$ and $\hat{k} = \hat{E}\times\hat{H}$; $E_z = E\cos\psi$ for a ray at elevation $\psi$ with horizontal $\mathbf{H}$.

Why this works. For a uniform plane wave the three vectors form a right-handed orthogonal triad, so naming the propagation direction and either field direction fixes the third completely — no integration and no material data beyond $\eta_0$ are required. Stating that $\mathbf{H}$ is horizontal is the essential clue: it forces the plane of polarisation of $\mathbf{E}$ to be the vertical plane containing the ray, which is what makes a vertical component exist at all. Had $\mathbf{E}$ been the horizontal one, the vertical component would be exactly zero.

Common pitfall. Multiplying by $\sin\psi$ instead of $\cos\psi$ — invisible at $45^\circ$, wrong everywhere else — and using the peak rather than the rms relation ($S = E_{\text{pk}}^{2}/2\eta_0$) so the answer comes out $\sqrt{2}$ too large.

Source. Ulaby, Fundamentals of Applied Electromagnetics ch. 7 (plane-wave propagation, polarisation); Hayt & Buck, Engineering Electromagnetics 9th ed. ch. 11.

Question 5 of 8: Vertical component of E in a wave climbing at 45 degrees

Question 5: Vertical component of E in a wave climbing at 45 degrees (20 marks)