22-Elec-A7 Electromagnetics · May 2013

Question 8 of 8: Short Radiating Element Moved onto a Conducting Plane

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

Question 8: Short Radiating Element Moved onto a Conducting Plane (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 short vertical current element of fixed length radiates first into free space and then from the surface of a horizontal perfect conductor, with the frequency reduced and the drive current doubled.

Find. (i) the maximum rms electric field on the 20 km hemisphere in case B, and (ii) the location on that hemisphere at which it occurs.

Approach. Write the far field of a short current element, extract the current moment from case A, apply image theory to account for the conducting plane, and scale the four independent factors — image, current, wavelength and distance — into case B; then read the location of the maximum from the pattern function.

1. Write the far field of a short element. For a Hertzian element of length $l$ carrying current $I$, $$E_{\theta}=\frac{\eta_0 I l\,\beta\sin\theta}{4\pi r}=\frac{\eta_0 I l\sin\theta}{2\lambda r}$$ so the field is proportional to the current moment $Il$, inversely proportional to both wavelength and distance, and maximum at $\theta=90^\circ$ — broadside to the element, which for a vertical element means the horizontal plane.
2. Extract the current moment from case A. Using the free-space maximum at $\theta=90^\circ$, $$Il=\frac{2\lambda_1 r_1 E_1}{\eta_0}=\frac{2\times10.0\times10^{4}\times500\times10^{-6}}{376.99}=0.2653\ \text{A}\cdot\text{m}$$
3. Apply image theory to the conducting plane. A vertical current element standing on a horizontal perfect conductor has an image directly beneath it carrying current in the same direction and phase. Above the plane, the element and its image are indistinguishable from a single element of length $2l$, so at any given current the field in the upper half space is doubled. Consistently, the radiation resistance also doubles — the same current radiates twice the power into half the solid angle — while retaining the $f^{2}$ dependence the question states.
4. Scale case A into case B. Four independent factors act on the field: $$\frac{E_2}{E_1}=\underbrace{2}_{\text{image}}\times\underbrace{2}_{\text{current}}\times\underbrace{\frac{\lambda_1}{\lambda_2}=\frac{10}{30}}_{\text{frequency}}\times\underbrace{\frac{r_1}{r_2}=\frac{10}{20}}_{\text{distance}}=\frac{2}{3}$$ The image and current gains of four are outweighed by the threefold wavelength increase and the doubled range.
5. Evaluate the field. $$E_2=\tfrac{2}{3}\times500\ \mu\text{V/m}=\boxed{333\ \mu\text{V/m (rms)}}$$ Rebuilding it directly from the current moment gives the same value: $$E_2=\frac{\eta_0\,(2Il)\times2}{2\lambda_2 r_2}=\frac{376.99\times0.5305\times2}{2\times30.0\times20.0\times10^{3}}=333\ \mu\text{V/m}$$
6. Locate the maximum. The pattern function is unchanged by the move: $|E|\propto\sin\theta$, which peaks at $\theta=90^\circ$, the direction perpendicular to the element. For a vertical element on a horizontal plane, $\theta=90^\circ$ is the horizontal direction, i.e. grazing along the conductor. Hence $$\boxed{\text{the maximum occurs at zero elevation, on the 20 km circle where the hemisphere meets the plane}}$$ and, because the element is rotationally symmetric about its own axis, it occurs equally at every azimuth around that circle. Nothing is radiated straight up along the element axis.

The reduction from 500 to 333 μV/m is a useful reminder that a short element is a frequency-sensitive radiator: at fixed physical length its electrical length $l/\lambda$ shrinks as the frequency falls, and the field falls with it. Tripling the wavelength alone would have cut the field to a third; only the image and the doubled current recovered most of that loss.