22-Elec-A7 Electromagnetics · December 2018

Question 6 of 8: Electric field and power density of a circularly polarised wave

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 6: Electric field and power density of a circularly polarised 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. A magnetic field specified component by component, with the two transverse components in phase quadrature and equal in amplitude.

Find. The electric field of the wave, and its average power density.

Approach. Identify the medium from the phase velocity implied by the data, then apply the supplied curl to Ampère’s law to recover E, and form the Poynting vector directly.

1. Read off frequency, wavelength and velocity. $$f = \frac{\omega}{2\pi} = 10\ \text{GHz}, \qquad \lambda = \frac{2\pi}{k} = 3\ \text{cm}, \qquad v_p = \frac{\omega}{k} = \frac{2\pi\times10^{10}}{2\pi/0.03} = 3.00\times10^{8}\ \text{m/s}.$$ The phase velocity is $c$, so the wave is in free space and $\eta_0 = 120\pi = 376.99\ \Omega$ applies. Both components depend on $(\omega t - kz)$ only, so the wave travels in the $+z$ direction.
2. Apply Ampère’s law. There is no conduction current, so $\nabla\times\mathbf{H} = \varepsilon_0\,\partial\mathbf{E}/\partial t$. Since nothing depends on $x$ or $y$, the supplied curl collapses to $$\nabla\times\mathbf{H} = \left(-\frac{\partial H_y}{\partial z},\ \frac{\partial H_x}{\partial z},\ 0\right) = \left(kH_0\cos(\omega t - kz),\ kH_0\sin(\omega t - kz),\ 0\right).$$ Incidentally the aid as printed has a slip in its third component — it should read $\partial B/\partial x - \partial A/\partial y$ — but here both candidate terms vanish, so nothing turns on it.
3. Integrate in time. Dividing by $\varepsilon_0$ and integrating with respect to $t$ (so each cosine becomes a sine over $\omega$), $$\mathbf{E} = \frac{kH_0}{\varepsilon_0\omega}\left(\sin(\omega t - kz),\ -\cos(\omega t - kz),\ 0\right).$$ The prefactor simplifies at once, because $k/\omega = 1/v_p = \sqrt{\mu_0\varepsilon_0}$ and therefore $k/(\omega\varepsilon_0) = \sqrt{\mu_0/\varepsilon_0} = \eta_0$.
4. The electric field. $$\boxed{\ \mathbf{E} = \eta_0 H_0\left(\sin(\omega t - kz)\,\hat{x} - \cos(\omega t - kz)\,\hat{y}\right), \qquad |\mathbf{E}| = \eta_0 H_0 = 0.377\ \text{V/m}\ }$$ This is the same construction as $\mathbf{E} = \eta_0\,\mathbf{H}\times\hat{z}$, and it can be checked independently against Faraday’s law, $\nabla\times\mathbf{E} = -\mu_0\,\partial\mathbf{H}/\partial t$, which it satisfies identically.
5. Identify the polarisation. At a fixed plane, $|\mathbf{H}| = H_0$ and $|\mathbf{E}| = \eta_0 H_0$ for every $t$, while the vectors rotate from $\hat{x}$ towards $\hat{y}$. Pointing the right thumb along the direction of travel $+\hat{z}$, the fingers curl the same way, so the wave is $\boxed{\text{right-hand circularly polarised}}$ on the IEEE convention. Because the magnitudes never vary, they are already the rms values — there is no second factor of $\sqrt{2}$ to take out.
6. Average power density. With E and H perpendicular and both of constant magnitude, the Poynting vector is steady rather than pulsating: $$\mathbf{S} = \mathbf{E}\times\mathbf{H} = \eta_0 H_0^{2}\left(\sin^{2} + \cos^{2}\right)\hat{z} = \eta_0 H_0^{2}\,\hat{z},$$$$\boxed{S_{avg} = \eta_0 H_0^{2} = 376.99 \times (10^{-3})^{2} = 3.77\times10^{-4}\ \text{W/m}^{2} = 377\ \mu\text{W/m}^{2}} .$$ That is twice what a linearly polarised wave of peak amplitude $H_0$ would deliver, which makes sense: this wave carries two orthogonal linear components of amplitude $H_0$ apiece.