22-Elec-A7 Electromagnetics · December 2018

Question 8 of 8: Scaling a short-element field in frequency, range and angle

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

Notes on this paper

Paper format. National Examinations, December 2018, 16-Elec-A7 Electromagnetics. Three hours, closed book, one approved Casio or Sharp calculator. Eight questions, all of equal value; the rubric says any five constitute a complete paper, so a candidate answers five — but all eight are worked here, because the set is a study resource. Aids printed on the paper: $\varepsilon_0 = 8.85\times10^{-12}$ F/m, $\mu_0 = 4\pi\times10^{-7}$ H/m, $e = 1.6\times10^{-19}$ C.

Reference texts. M. N. O. Sadiku, Elements of Electromagnetics, 7th ed.; W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed.; F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; D. M. Pozar, Microwave Engineering, 4th ed.; C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed.

Constants used throughout. Question 6 hands us $\omega = 2\pi\times10^{10}$ rad/s together with $k = (2\pi/3)\ \text{cm}^{-1}$, so the setter’s phase velocity is $\omega/k = 3.000\times10^{8}$ m/s exactly; Question 3’s cut-off arithmetic only closes on round numbers under the same choice. Accordingly $c = 3\times10^{8}$ m/s and $\eta_0 = 120\pi = 376.99\ \Omega$ are used here. The paper’s own aid list would give $c = 2.99863\times10^{8}$ m/s and $\eta_0 = 376.83\ \Omega$ — a 0.05 % shift that changes no answer except at one deliberate boundary case, flagged in Question 3.

Question 8: Scaling a short-element field in frequency, range and angle (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. One calibrated measurement of the lower-frequency radiator, and the statement that the two elements are physically identical and identically driven.

Given data
Quantity	Symbol	Value
Reference frequency	$f_1$	10 MHz
Second frequency	$f_2$	30 MHz
Reference sphere radius	$r_1$	10 km
Target sphere radius	$r_2$	5 km
Maximum power density at $r_1$, 10 MHz	$S_1$	$10^{-8}$ W/m²
Angle of the target point from the zenith	$\theta$	30°
Element current and length	—	identical for the two radiators

Find. The rms vertical component of the 30 MHz electric field at that point.

The pattern and the two spheres. The maximum lies on the horizon, 90° from the element axis; the asked-for point sits 30° from the zenith on the inner sphere.

Approach. Convert the calibrated power density into a reference field, scale it by the three ratios that the short-element formula contains, and then project onto the vertical — which contributes a second sine.

Reference field from the reference power density. On the 10 km sphere the maximum sits at $\theta = 90^\circ$, broadside to the vertical element, and there $$E_1 = \sqrt{S_1\,\eta_0} = \sqrt{10^{-8} \times 376.99} = 1.942\times10^{-3}\ \text{V/m} = 1.942\ \text{mV/m},$$ an rms value, because the given power density is an average. This one measurement absorbs every antenna constant — the current, the element length and all the numerical factors — so none of them ever needs to be unpacked.
Write the scaling law. For a short element the far field is $$E_\theta = \frac{\eta_0 k I \ell \sin\theta}{4\pi r} \;\propto\; \frac{f\,I\,\ell\,\sin\theta}{r} ,$$ and since $I$ and $\ell$ are stated to be the same for both radiators, only three ratios survive: frequency, range and pattern angle.
Apply the three ratios. Frequency triples, range halves (so the field doubles), and $\sin 30^\circ = 0.5$: $$E_\theta = E_1 \times \frac{f_2}{f_1} \times \frac{r_1}{r_2} \times \sin\theta = 1.942 \times 3 \times 2 \times 0.5 = 5.825\ \text{mV/m}.$$ Note the direction of the range factor: moving closer increases the field, which is the easiest sign to invert under exam pressure.
Project onto the vertical. The far field of a vertical element lies along $\hat{\theta}$, and $\hat{z}\cdot\hat{\theta} = -\sin\theta$, so the vertical component carries a SECOND sine: $$E_z = E_\theta \sin\theta = 5.825 \times 0.5 = \boxed{2.91\ \text{mV/m (rms)}} .$$ Equivalently $E_z \propto \sin^{2}\theta$, and here $\sin^{2}30^\circ = 0.25$.
Cross-check through the power density. The power density at the target point is $$S_2 = \frac{E_\theta^{2}}{\eta_0} = S_1\left(\frac{f_2}{f_1}\right)^{2}\left(\frac{r_1}{r_2}\right)^{2}\sin^{2}\theta = 10^{-8} \times 9 \times 4 \times 0.25 = 9\times10^{-8}\ \text{W/m}^{2},$$ which reproduces $E_\theta = \sqrt{S_2\eta_0} = 5.825$ mV/m independently.
Confirm the assumptions hold at the new frequency. At 30 MHz the free-space wavelength is 10 m, so the 5 km sphere is 500 wavelengths out, comfortably in the far field; and the element must still satisfy $\ell \ll \lambda$ at the higher frequency for the $\sin\theta$ pattern to apply, which is what the stated equality of element lengths quietly requires.

Final results
Quantity asked	Result
Reference field at 10 km, 10 MHz, on the horizon	$E_1 = 1.942$ mV/m (rms)
Scaling applied	$\times 3$ in frequency, $\times 2$ in range, $\times \sin 30^\circ$ in pattern
Total field at the target point	$E_\theta = 5.825$ mV/m (rms)
Vertical component (a second $\sin\theta$)	$E_z = 2.91$ mV/m (rms)
Power density there	$S_2 = 9\times10^{-8}$ W/m²
Far-field check at 30 MHz	$\lambda = 10$ m; $r_2 = 500\lambda$

Topic. Far field of a short current element, and the difference between $E_\theta$ and $E_z$.

Key relations. $E_\theta = \eta_0 k I \ell \sin\theta / (4\pi r)$; $\;S = E_{rms}^{2}/\eta_0$; $\;\hat{z}\cdot\hat{\theta} = -\sin\theta$, so $E_z = E_\theta\sin\theta \propto \sin^{2}\theta$; far field for $r \gg \lambda$.

Why this works. A short vertical element radiates nothing along its own axis and most strongly broadside to it, and the field it produces is proportional to frequency because radiation depends on the rate of change of the current, not its size alone. Once one field value is measured, every antenna constant is fixed, and any other point follows from ratios. The extra subtlety here is that the radiated field points along $\hat{\theta}$, a direction that tilts away from the vertical as the observation point rises; resolving it onto the vertical introduces the same sine a second time, so the vertical component peaks on the horizon and falls off twice as fast in angle as the total field does.

Common pitfall. The near-universal error is to stop at $E_\theta$ and report 5.825 mV/m, answering ‘the field’ rather than ‘the vertical component of the field’. A close second is measuring $\theta$ from the horizon instead of from the axis; the question says ‘away from directly overhead’, which is measured from the element axis, and reading it the other way would give $\sin^{2}60^\circ = 0.75$ and an answer three times too large.

Source. Balanis, Antenna Theory: Analysis and Design, 4th ed., §4.2 (infinitesimal dipole); Sadiku, Elements of Electromagnetics, 7th ed., §13.2.

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