Question 8 of 8: Crossed Current Elements — Plane of Linear Polarization

Question 8: Crossed Current Elements — Plane of Linear Polarization (equal value)

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. Two co-located Hertzian current elements ($l=1$ m, $f=5$ MHz), one vertical ($\hat z$), one horizontal ($\hat x$), equal current amplitude, $90^\circ$ relative phase; each ALONE gives maximum power density $S_{max}=10^{-7}$ W/m$^2$ (at its own broadside) on a 1 km sphere.

Find. The locus on the sphere where the total field is linearly polarized, and its RMS amplitude there.

Approach. A Hertzian dipole of moment direction $\hat p$ radiates $\mathbf E\propto[\hat p-(\hat p\cdot\hat r)\hat r]$ (the transverse projection of $\hat p$ onto the sphere at $\hat r$). Superpose the $\hat z$- and $\hat x$-directed elements (in phase quadrature) and find where the two transverse-projection vectors are parallel (linear polarization) rather than generally elliptical.

1. Vector field pattern. With $\hat r=(\sin\Theta\cos\Phi,\sin\Theta\sin\Phi,\cos\Theta)$ (standard spherical angles about $z$), the two (equal-amplitude) transverse vectors are $$\mathbf V_1=\hat z-(\hat z\cdot\hat r)\hat r,\qquad \mathbf V_2=\hat x-(\hat x\cdot\hat r)\hat r,$$ and the total phasor field is $\mathbf E\propto \mathbf V_1+j\mathbf V_2$ (the $90^\circ$ phase carried as $j$).
2. Where are $\mathbf V_1,\mathbf V_2$ parallel? Working out $\mathbf V_1\times\mathbf V_2$ in closed form gives $$\mathbf V_1\times\mathbf V_2=\sin\Phi\,\bigl(\sin^2\Theta\cos\Phi,\ \sin^2\Theta\sin\Phi,\ \sin\Theta\cos\Theta\bigr),$$ which vanishes identically (for every $\Theta$) only when $\sin\Phi=0$, i.e. $\Phi=0^\circ$ or $180^\circ$.
3. Identify the locus. $\Phi=0^\circ/180^\circ$ is the full vertical plane containing BOTH current elements (the plane through the $z$-axis and the $x$-axis — i.e. the plane that contains the vertical element and the direction the horizontal element points along). So $$\boxed{\text{the field is linearly polarized everywhere in the vertical plane containing the two elements, for every elevation angle}}.$$ (Elsewhere on the sphere, e.g. anywhere in the $y$-$z$ plane broadside to the horizontal element, the two contributions are unequal in magnitude and generally out of the special quadrature alignment, giving elliptical polarization; on the $\pm y$ axis specifically the two contributions are equal and in quadrature, giving CIRCULAR polarization — the complementary special case to this one.)
4. Amplitude in that plane. In the $\Phi=0$ plane, $|\mathbf V_1|=\sin\Theta$ and $|\mathbf V_2|=|\cos\Theta|$ (each dipole's own $\sin\theta$ pattern relative to its own axis), and they point along a common direction. The instantaneous total is then a single sinusoid $$E(t)\propto E_0[\sin\Theta\cos\omega t+\cos\Theta\sin\omega t]=E_0\sin(\omega t+\Theta),$$ whose amplitude is $E_0\sqrt{\sin^2\Theta+\cos^2\Theta}=E_0$ — CONSTANT, independent of $\Theta$, and equal to each element's own peak broadside amplitude.
5. Numerical amplitude. From $S_{max}=E_0^2/(2\eta_0)$, $$E_0=\sqrt{2\eta_0S_{max}}=\sqrt{2\times377\times10^{-7}}=8.68\times10^{-3}\ \text{V/m (peak)},$$ $$\boxed{E_{rms}=\frac{E_0}{\sqrt2}=6.14\times10^{-3}\ \text{V/m}=6.14\ \text{mV/m}}\quad\text{(constant throughout the plane).}$$