Equal Mean Power Density Between Two Radio Stations: 17-Phys-B1 Radiation Physics, December 2017, Q3

Question 3: Equal Mean Power Density Between Two Radio Stations (5 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. Town midway between two cities 100 km apart ($r_1=r_2=50$ km initially); station 1 power $P_1=70$ kW; station 2 power $P_2=50$ kW; both antennas non-directional (isotropic).

Find. The distance to travel from the town, and the direction, so the two stations' mean power densities at the traveller's new location are equal.

Approach. An isotropic antenna's mean power density falls off as $S=P/(4\pi r^2)$; since $P_1>P_2$, the town (equidistant from both) already sees a stronger field from station 1, so the traveller must move away from station 1 and toward station 2 until the two inverse-square terms balance.

Set up the equal-density condition. Let $x$ be the distance travelled toward the weaker (50 kW) station, so the new distances are $r_1=50+x$ and $r_2=50-x$ (km). Equal power density requires $$\frac{P_1}{4\pi r_1^2} = \frac{P_2}{4\pi r_2^2} \quad\Longrightarrow\quad \frac{r_1}{r_2} = \sqrt{\frac{P_1}{P_2}}$$
Solve for the travel distance. Substituting $r_1=50+x$, $r_2=50-x$ and $\sqrt{P_1/P_2}=\sqrt{70/50}=1.1832$: $$\frac{50+x}{50-x} = 1.1832 \;\Longrightarrow\; 50+x = 1.1832(50-x) \;\Longrightarrow\; x = \frac{50(1.1832-1)}{1.1832+1}$$ $$\boxed{x \approx 4.20\ \text{km, travelling away from the 70 kW station and toward the 50 kW station}}$$
Check. With $r_1=54.20$ km and $r_2=45.80$ km, $S_1=P_1/(4\pi r_1^2)$ and $S_2=P_2/(4\pi r_2^2)$ both evaluate to $1.90\times10^{-6}\ \text{W/m}^2$, confirming the two densities are indeed equal at that point.

Question 3 — results
Quantity	Result
Travel distance $x$	≈ 4.20 km
Direction	Away from the 70 kW station, toward the 50 kW station
Equal power density $S_1=S_2$	≈ $1.90\times10^{-6}$ W/m$^2$

Topic. Inverse-square law for isotropic radiators and non-ionizing (RF) power density.

Key relations. $S=P/(4\pi r^2)$ (isotropic mean power density); $r_1/r_2=\sqrt{P_1/P_2}$ at the equal-density locus.

Why this works. Equal power density is a single ratio condition independent of the absolute powers, so the traveller need not know the actual density value at all — only the power ratio and the fixed 100 km baseline are needed to locate the point.

Common pitfall. Assuming the equal-density point is simply proportionally closer to the weaker station along a linear (not inverse-square) rule — that would place it at $x=100\times50/120\approx41.7$ km from the midpoint, far past even the stronger station, which is not the physically correct locus for a $1/r^2$ field.

Source. F. H. Attix, Introduction to Radiological Physics and Radiation Dosimetry, Ch. 1 (inverse-square law); J. E. Turner, Atoms, Radiation, and Radiation Protection, 3rd ed., Ch. 15 (non-ionizing radiation, RF power density).

Question 3 of 7: Equal Mean Power Density Between Two Radio Stations

Question 3: Equal Mean Power Density Between Two Radio Stations (5 marks)