NivaarExam PrepOfficial exam papers ↗

17-Phys-B1 Radiation Physics · December 2016

Question 3 of 7: Non-Ionizing EM Field of a 50 kW Radio Transmitter

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

Notes on this paper

Paper format. 98-Phys-B1 Radiation Physics, National Examination December 2016 — a three-hour open-book examination in which any non-communicating calculator is permitted (the candidate must record the calculator's make and model on the first sheet). The cover page states the exam has 7 questions worth a total of 87 points, of which only 80 points' worth need be answered for full marks; every question and sub-part is nonetheless answered in full below so the paper remains a complete study resource. The cover page's own marking-scheme summary (13+9+8+10+19+10+18 = 87) is internally consistent with the stated total. The cover page also invites the candidate to submit a written statement of any assumptions made where a question is open to interpretation — this licence is used below in Question 2 (ICRP-60 neutron weighting factors are assumed for the thermal/fast neutron energy brackets, since none are given explicitly), Question 4 (the source's printed comparison wavelength "10 pm" for a carbon-dioxide laser photon is read as the real CO2-laser wavelength, 10 μm, since no laser emits at 10 picometres), Question 5(f) (shield thicknesses are order-of-magnitude illustrative estimates, since the source gives no source strength/dose-rate target to size against), and Question 5(g) (the fission-energy-distribution percentages are standard textbook illustrative values, since the source gives no numeric data of its own to compute them from).

Reference texts. K. S. Krane, Introductory Nuclear Physics (nuclear reaction equations, fission energetics, mass–energy conservation); F. H. Attix, Introduction to Radiological Physics and Radiation Dosimetry (photon/EM interactions, non-ionizing radiation, shielding); J. R. Cember and T. E. Johnson, Introduction to Health Physics, 5th ed. (internal dosimetry, radiation weighting factors, MIRD absorbed-fraction formalism, ALARA); J. E. Turner, Atoms, Radiation, and Radiation Protection, 3rd ed. (neutron detection/shielding, Compton scattering, radiation protection tenets).

Question 3: Non-Ionizing EM Field of a 50 kW Radio Transmitter (8 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. Transmitter power $P=50$ kW, distance $r=50$ km; reference station: $P_{\text{ref}}=1$ kW gives reliable reception range $r_{\text{ref}}=50$ km.

Given data
QuantitySymbolValue
Transmitter power$P$50 kW
Distance$r$50 km
Free-space impedance$Z_0$$377\ \Omega$
Reference power / range$P_{\text{ref}}/r_{\text{ref}}$1 kW / 50 km

Find. (a) mean power density $S$ at 50 km; (b) maximum $E$-field; (c) maximum $H$-field; (d) listening (reliable-reception) range at 50 kW.

Antenna (P = 50 kW) r = 50 km Isotropic radiation assumed: power spreads uniformly over a sphere of radius r
Fig. Q3 — non-directional antenna radiating isotropically; power density falls off as $1/(4\pi r^2)$.

Approach. A commercial broadcast antenna with no stated directionality is reasonably modelled as radiating isotropically (uniformly over a sphere) — consistent with the hint's own "non-directional antenna" reference station. Apply $S=P/(4\pi r^2)$, relate $S$ to the peak fields via the free-space impedance, then scale the reference reliable-reception range by $\sqrt{P/P_{\text{ref}}}$ since reception range is set by a fixed threshold power density, $S\propto P/r^2=\text{const}\Rightarrow r\propto\sqrt P$.

  1. Part (a) — mean power density at 50 km. $$S = \frac{P}{4\pi r^2} = \frac{50{,}000\ \text{W}}{4\pi(50{,}000\ \text{m})^2}$$ $$\boxed{S \approx 1.59\times10^{-6}\ \text{W/m}^2 \ \ (1.59\ \mu\text{W/m}^2)}$$
  2. Part (b) — maximum electric field strength. For a plane wave the time-averaged power density relates to the peak field by $S=E_{\max}^2/(2Z_0)$, so $E_{\max}=\sqrt{2 Z_0 S}$: $$E_{\max} = \sqrt{2(377\ \Omega)(1.59\times10^{-6}\ \text{W/m}^2)}$$ $$\boxed{E_{\max} \approx 0.0346\ \text{V/m} = 34.6\ \text{mV/m}}$$
  3. Part (c) — maximum magnetic field strength. The peak $E$ and $H$ fields of a plane wave are linked by the free-space impedance, $H_{\max}=E_{\max}/Z_0$: $$H_{\max} = \frac{0.0346\ \text{V/m}}{377\ \Omega}$$ $$\boxed{H_{\max} \approx 9.19\times10^{-5}\ \text{A/m} = 91.9\ \mu\text{A/m}}$$
  4. Part (d) — listening range at 50 kW. Reliable reception occurs down to a fixed minimum power density at the receiver, $S_{\min}=P/(4\pi r^2)=\text{const}$, so the reliable range scales as $r\propto\sqrt P$. Scaling the given 1 kW $\to$ 50 km reference: $$r_{50\text{kW}} = r_{\text{ref}}\sqrt{\frac{P}{P_{\text{ref}}}} = 50\ \text{km}\times\sqrt{\frac{50\ \text{kW}}{1\ \text{kW}}}$$ $$\boxed{r_{50\text{kW}} \approx 353.6\ \text{km}}$$
Question 3 — results
PartResult
(a) $S$$1.59\times10^{-6}$ W/m$^2$
(b) $E_{\max}$34.6 mV/m
(c) $H_{\max}$91.9 $\mu$A/m
(d) Listening range$\approx354$ km