Question

A communications truck with a 44-cm-diameter dish receiver on the roof starts out 10 km from its base station. It drives directly away from the base station at 50 km/h for 1.0 h, keeping the receiver pointed at the base station. The base station antenna broadcasts continuously with 2.5 kW of power, radiated uniformly in all directions. How much electromagnetic energy does the truck’s dish receive during that 1.0 h?

Short answer

0.000181J

Step-by-step solution

Given Parameters

radius of dish(r): 44/2 cm= 0.22m

$s_0=10km=10000m$

v = 50km/h = 50000m/h = 13.88m/s

$P_{source}=2.5kW=2500W$

Time = 1hr

Concept

Energy received by dish is power received over the course of 1hr time period.

$E={\int }_0^{1hr}{P}_{dish}dt$ ____ eq (1)

Now, Intensity received by dish is $I=\frac{P}{A}$,

Where, P = power

A = Area

So, P = IA

Substituting this in eq(1) we getE=∫01hrIdishAdt____ eq(2)Now intensity received by dish = Power emitted by sourcetotal surface area=Psource4πr2Putting this in eq(2) we get;

$E={\int }_0^{1hr}\frac{P_{source}}{4{\mathrm{\pi r}}^2}Adt$____ eq(3)

Now we know that the truck is moving away from the station at a constant speed. So truck's distance from station at any time t is

$r=s_0+vt$

Substituting this in equation 3 we get,

$$\begin{gathered} E={\int }_0^{1hr}\frac{P_{source}A}{4\mathrm{\pi }{({\mathrm{s}}_0+\mathrm{vt})}^2}dt \\ =\frac{P_{source}A}{4\mathrm{\pi }}{\int }_0^{1hr}\frac{1}{{({\mathrm{s}}_0+\mathrm{vt})}^2}dt \\ =\frac{P_{source}A}{4\mathrm{\pi v}}{\left[\right(-1\left){({\mathrm{s}}_0+\mathrm{vt})}^{-1}\right]}_0^{1hr}=\frac{-P_{source}A}{4\mathrm{\pi v}}[\frac{1}{s_0+v\left(1hr\right)}-\frac{1}{s_0}] \end{gathered}$$

Here A = ${\mathrm{\pi r}}^2$

Substituting this value in above equation we get:

$E=\frac{-P_{Source}{r}^2}{4\mathrm{v}}[\frac{1}{S_0+v\left(1hr\right)}-\frac{1}{S_0}]$

Substituting all the values in equation now, we get;

$$\begin{gathered} E=\frac{-\left(2500W\right){(0.22m)}^2}{4(13.88m/s)}[\frac{1}{10000+50000}-\frac{1}{10000}] \\ E=-2.179\times (-8.333\times {10}^{-5)} \\ E=0.000181J \end{gathered}$$