Advanced Electrodynamics

Solar constant at Earth: $I = 1361$ W/m$^2$.

Radiation pressure on a perfectly absorbing surface:

$P = \frac{I}{c} = \frac{1361}{3 \times 10^8} = 4.54 \times 10^{-6}\ \text{N}/m^2 = 4.54\ \mu\text{Pa}$

For a perfect reflector: $P = 9.07\,\mu\text{Pa}$.

This is tiny compared to atmospheric pressure ($10^5$ Pa), but is significant for:

• Solar sails: A 100 m $\times$ 100 m sail with 90% reflectivity experiences $F \approx 0.12$ N, producing acceleration $a \approx 0.6$ mm/s$^2$ for a 100 kg sail.
• Asteroid deflection: Sustained radiation pressure can perturb asteroid orbits over years.
• Atom optics: Laser cooling uses radiation pressure to slow atoms to microkelvin temperatures.

For a rectangular waveguide ($a \times b$) operating in TE$_{10}$ mode at frequency $f$:

(a) Write the complete expressions for all six field components ($E_x, E_y, E_z, B_x, B_y, B_z$).

(b) Sketch the field pattern: show the direction and relative magnitude of $\mathbf{E}$ and $\mathbf{B}$ in the $xy$-plane at $z = 0$.

(c) Find the positions of maximum surface current density on the walls and explain why the waveguide loss is minimised by making the broad wall dimension $a$ as large as possible (for a given $f$).

Solution:

(a) For TE$_{10}$: $B_z = B_0\cos(\pi x/a)\,e^{i(\beta z - \omega t)}$.

$E_x = 0, \quad E_y = \frac{i\omega\mu_0 a}{\pi}B_0\sin\!\left(\frac{\pi x}{a}\right)e^{i(\beta z - \omega t)}$

$E_z = 0$

$B_x = -\frac{i\beta a}{\pi}B_0\sin\!\left(\frac{\pi x}{a}\right)e^{i(\beta z - \omega t)}, \quad B_y = 0$

$B_z = B_0\cos\!\left(\frac{\pi x}{a}\right)e^{i(\beta z - \omega t)}$

(b) The electric field $E_y$ is purely vertical, with a $\sin(\pi x/a)$ profile: zero at the side walls ($x = 0, a$) and maximum at the centre ($x = a/2$). The magnetic field forms closed loops in the $xz$-plane.

(c) Surface current $\mathbf{K} = \hat{\mathbf{n}} \times \mathbf{H}$. On the broad walls ($y = 0, b$): $\mathbf{K}$ has components from $B_x$ and $B_z$With maximum at $x = a/2$ (where $\sin(\pi x/a) = 1$). The power loss per unit length is:

$P_{\text{loss} = \frac{R_s}{2}\oint |\mathbf{K}|^2\, dl}$

Where $R_s = \sqrt{\omega\mu_0/(2\sigma)}$ is the surface resistance. For fixed $f$Increasing $a$ reduces the current density on the broad walls and increases the power-handling capacity.

A half-wave dipole antenna of length $\ell = \lambda/2$ carries a sinusoidal current distribution:

$I(z) = I_0\cos(kz), \quad -\lambda/4 \leq z \leq \lambda/4$

(a) Calculate the radiation fields $\mathbf{E}$ and $\mathbf{B}$ in the far field.

(b) Find the angular distribution of radiated power $dP/d\Omega$.

(c) Calculate the total radiated power and the radiation resistance. Compare with the short-dipole result $R_{\text{rad} = 197(\ell/\lambda)^2\,\Omega}$.

Solution:

(a) The vector potential in the far field:

$A_z = \frac{\mu_0}{4\pi}\frac{e^{ikr}}{r}\int_{-\lambda/4}^{\lambda/4}I_0\cos(kz')\,e^{-ikz'\cos\theta}\,dz'$

The integral evaluates to:

$A_z = \frac{\mu_0 I_0}{4\pi}\frac{e^{ikr}}{r}\frac{2\cos\!\left(\frac{\pi}{2}\cos\theta\right)}{k\sin^2\theta}$

The radiation fields:

$E_\theta = ikA_z\sin\theta = \frac{i\mu_0 c I_0}{4\pi}\frac{e^{ikr}}{r}\frac{\cos\!\left(\frac{\pi}{2}\cos\theta\right)}{\sin\theta}$

$B_\phi = E_\theta/c$

(b) The angular distribution:

$\frac{dP}{d\Omega} = \frac{r^2}{2Z_0}|E_\theta|^2 = \frac{Z_0 I_0^2}{32\pi^2}\frac{\cos^2\!\left(\frac{\pi}{2}\cos\theta\right)}{\sin^2\theta}$

(c) Total power:

$P = \frac{Z_0 I_0^2}{32\pi^2}\int_0^{2\pi}\!\!\int_0^\pi \frac{\cos^2\!\left(\frac{\pi}{2}\cos\theta\right)}{\sin^2\theta}\sin\theta\,d\theta\,d\phi$

$= \frac{Z_0 I_0^2}{16\pi}\int_0^\pi \frac{\cos^2\!\left(\frac{\pi}{2}\cos\theta\right)}{\sin\theta}\,d\theta$

With the substitution $u = \cos\theta$: $\int_{-1}^{1}\frac{\cos^2(\pi u/2)}{1-u^2}\,du = 1.2188$ (the Siegel integral).

$P = \frac{377 \times 1.2188}{16\pi}I_0^2 = 9.16\,I_0^2$

Radiation resistance: $R_{\text{rad} = 2P/I_0^2 = 18.3\,\Omega}$.

For comparison, a short dipole ($\ell \ll \lambda$) of length $\lambda/2$ would give $R_{\text{rad} = 197 \times 0.25 = 49.3\,\Omega}$. The half-wave dipole has lower radiation resistance because the current distribution (cosine) has less total effective acceleration than a uniform current.

The directivity of the half-wave dipole is $D = 1.64$ (2.15 dBi), slightly higher than the short dipole ($D = 1.5$).