H Field Amplitude of a Plane EM Wave: 04-BS-9, December 2014, Q6

Question 6: H Field Amplitude of a Plane EM Wave (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.

Quantity	Value
$\vec E=(0,0,E)\cos(\omega t-kx)$, plane wave in free space	propagating along $+x$
Frequency $f$	$10^{10}$ Hz
Peak $E$ amplitude $E_0$	$10^{-6}$ V/m (as printed; a weak, received-signal-level field)
$c$	$3\times10^8$ m/s

Find. The RMS amplitude of $\vec H$.

Approach. Apply Faraday's law in point form, $\nabla\times\vec E=-\partial\vec B/\partial t$, using the given curl identity, to find $\vec B(x,t)$ directly from the stated $\vec E$ field; convert to $H=B/\mu_0$, then to RMS. Note the printed amplitude is $10^{-6}$ V/m (a microvolt-per-metre field, typical of a received radio signal), so $H$ comes out in nanoamperes per metre.

Evaluate $\nabla\times\vec E$. With $E_x=E_y=0$ and $E_z=E_0\cos(\omega t-kx)$ (function of $x,t$ only), the curl identity gives a single nonzero component: $$(\nabla\times\vec E)_y=\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}=0-\left[E_0 k\sin(\omega t-kx)\right]=-E_0 k\sin(\omega t-kx)$$ so $\nabla\times\vec E=(0,\,-E_0k\sin(\omega t-kx),\,0)$.
Integrate Faraday's law for $B_y$. $$-\frac{\partial B_y}{\partial t}=-E_0k\sin(\omega t-kx)\ \Rightarrow\ \frac{\partial B_y}{\partial t}=E_0k\sin(\omega t-kx)$$ $$B_y=-\frac{E_0k}{\omega}\cos(\omega t-kx)=-\frac{E_0}{c}\cos(\omega t-kx)\quad(\text{since }k/\omega=1/c)$$ The peak magnitude is $B_0=E_0/c$, i.e. the familiar plane-wave relation $E_0/B_0=c$.
Convert to $H$ and to RMS. $$H_0=\frac{B_0}{\mu_0}=\frac{E_0}{c\,\mu_0}=\frac{10^{-6}}{(3\times10^8)(4\pi\times10^{-7})}$$ $$H_0=\boxed{2.653\times10^{-9}\ \text{A/m (peak)}}$$ $$H_{\text{rms}}=\frac{H_0}{\sqrt2}=\frac{2.653\times10^{-9}}{\sqrt2}$$ $$H_{\text{rms}}=\boxed{1.876\times10^{-9}\ \text{A/m}}$$

Quantity	Result
$B_0$ (peak)	$E_0/c=3.33\times10^{-15}$ T
$H_0$ (peak)	$2.653\times10^{-9}$ A/m
$H_{\text{rms}}$	$1.876\times10^{-9}$ A/m

Concept & Theory

Topic. Faraday's law applied to a uniform plane wave in free space, relating the $E$ and $H$ field amplitudes.

Key relations. $\nabla\times\vec E=-\partial\vec B/\partial t$; $k=\omega/c$; intrinsic impedance of free space $\eta_0=c\mu_0=377.0\ \Omega$ with the stated $c$ ($\sqrt{\mu_0/\varepsilon_0}=376.8\ \Omega$ with the aid $\varepsilon_0$; the 0.05 % gap is rounding in the given constants), so $H_0=E_0/\eta_0$.

Why this works. A uniform plane wave has $E$, $H$, and the direction of propagation mutually perpendicular, with $B$ locked to $E$ by Faraday's law exactly as derived here; the resulting ratio $E_0/H_0=c\mu_0=\eta_0$ is a universal property of plane waves in a given medium, independent of frequency.

Common pitfall. Reporting the peak value $H_0$ when the question asks for the RMS amplitude, or omitting the $1/\sqrt2$ factor.

Source. Sadiku, Elements of Electromagnetics, Ch. 10 (Maxwell's equations, uniform plane waves).

Question 6 of 8: H Field Amplitude of a Plane EM Wave

Question 6: H Field Amplitude of a Plane EM Wave (20 marks)