Question 7 of 8: E Field of a Plane Wave from Maxwell's Equations
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2013 — 04-BS-9 Basic Electromagnetics. Three-hour, closed-book exam (approved Casio/Sharp calculator only). Aids given: $\varepsilon_0=8.85\times10^{-12}$ F/m, $\mu_0=4\pi\times10^{-7}$ H/m, $e=1.6\times10^{-19}$ C. Format: eight questions offered; any five constitute a complete paper and only the first five appearing in the answer book are marked. All eight are solved below for completeness.
Reference texts: Sadiku, Elements of Electromagnetics / Hayt & Buck, Engineering Electromagnetics — electrostatics, Gauss's law, capacitance, magnetostatics, Faraday's law, Maxwell's equations and displacement current; Young & Freedman, University Physics with Modern Physics — induced EMF in a rotating loop, plane-wave impedance.
Question 7: E Field of a Plane Wave from Maxwell's Equations (20 marks)
Plane wave in vacuum: H field given, E field (perpendicular, in phase) found via the vacuum intrinsic impedance.
Approach. Apply Maxwell's source-free curl equation $\nabla\times\vec H=\varepsilon_0\,\partial\vec E/\partial t$ (using the supplied curl formula) to a plane wave of the assumed form, which yields the vacuum intrinsic impedance $\eta_0=E/H=\sqrt{\mu_0/\varepsilon_0}$; RMS values obey the same ratio as instantaneous ones for a sinusoid.
Set up the wave and apply the curl equation. With $\vec H=(H,0,0)\cos(\omega t-kz)$, only $\partial H_x/\partial z\ne0$ among the terms in the aid formula, so $\nabla\times\vec H=(0,\,\partial H_x/\partial z,\,0)=(0,\,kH\sin(\omega t-kz),\,0)$. Matching this to $\varepsilon_0\partial\vec E/\partial t$ with $\vec E=(0,E_y,0)\cos(\omega t-kz)$ (E perpendicular to H, transverse wave), whose time derivative gives $(0,-\omega\varepsilon_0E_y\sin(\omega t-kz),0)$,
yields $kH=-\omega\varepsilon_0 E_y$, i.e. $E_y=-kH/(\omega\varepsilon_0)$: $\vec E$ points along $-\hat y$ (so that $\vec E\times\vec H$ points along $+\hat z$, the propagation direction), with magnitude ratio $|E|/H=k/(\omega\varepsilon_0)$.
Reduce to the intrinsic impedance. In vacuum the dispersion relation is $\omega/k=c=1/\sqrt{\mu_0\varepsilon_0}$, so
$$\frac{|E|}{H}=\frac{k}{\omega\varepsilon_0}=\frac{\sqrt{\mu_0\varepsilon_0}}{\varepsilon_0}=\sqrt{\frac{\mu_0}{\varepsilon_0}}=\eta_0$$
$$\eta_0=\sqrt{\frac{4\pi\times10^{-7}}{8.85\times10^{-12}}}=\boxed{376.8\ \Omega}$$
RMS electric field. The $E/H$ ratio holds equally for RMS values (both scale by the same $1/\sqrt2$ factor from their peaks):
$$E_{\text{rms}}=\eta_0 H_{\text{rms}}=(376.8)(50\times10^{-6})$$
$$E_{\text{rms}}=\boxed{18.84\ \text{mV/m}}$$
