(a) [tex]2.68\cdot 10^{-6} W/m^2[/tex]
The intensity of an electromagnetic wave is given by
[tex]I=\frac{P}{A}[/tex]
where
P is the power
A is the area of the surface considered
For the waves in the problem,
[tex]P=182 kW = 1.82\cdot 10^5 W[/tex] is the power
The area is a hemisphere of radius
[tex]r=104 km=1.04\cdot 10^5 m[/tex]
so
[tex]A=2\pi r^2=2\pi (1.04\cdot 10^5 m)^2=6.8\cdot 10^{10} m^2[/tex]
So, the intensity is
[tex]I=\frac{1.82\cdot 10^5 W}{6.8\cdot 10^{10}m^2}=2.68\cdot 10^{-6} W/m^2[/tex]
(b) [tex]5.9\cdot 10^{-7} W[/tex]
In this case, the area of the reflection is
[tex]A=0.22 m^2[/tex]
So, if we use the intensity of the wave that we found previously, we can calculate the power of the aircraft's reflection using the same formula:
[tex]P=IA=(2.68\cdot 10^{-6} W/m^2)(0.22 m^2)=5.9\cdot 10^{-7} W[/tex]
(c) [tex]8.7\cdot 10^{-18} W/m^2[/tex]
We said that the power of the waves reflected by the aircraft is
[tex]P=5.9\cdot 10^{-7} W[/tex]
If we assume that the reflected waves also propagate over a hemisphere of radius
[tex]r=104 km=1.04\cdot 10^5 m[/tex]
which has an area of
[tex]A=2\pi r^2=2\pi (1.04\cdot 10^5 m)^2=6.8\cdot 10^{10} m^2[/tex]
Then the intensity of the reflected waves at the radar site will be
[tex]I=\frac{P}{A}=\frac{5.9\cdot 10^{-7} W}{6.8\cdot 10^{10} m^2}=8.7\cdot 10^{-18} W/m^2[/tex]
(d) [tex]8.1\cdot 10^{-8} V/m[/tex]
The intensity of a wave is related to the maximum value of the electric field by
[tex]I=\frac{1}{2}c\epsilon_0 E_0^2[/tex]
where
c is the speed of light
[tex]\epsilon_0[/tex] is the vacuum permittivity
[tex]E_0[/tex] is the maximum value of the electric field vector
Solving the equation for [tex]E_0[/tex],
[tex]E_0=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(8.7\cdot 10^{-18} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12} F/m)}}=8.1\cdot 10^{-8} V/m[/tex]
(e) [tex]1.9\cdot 10^{-16} T[/tex]
The maximum value of the magnetic field vector is given by
[tex]B_0 = \frac{E_0}{c}[/tex]
Substituting the values,
[tex]B_0 = \frac{(8.1\cdot 10^{-8} V/m)}{3\cdot 10^8 m/s}=2.7\cdot 10^{-16} T[/tex]
And the rms value of the magnetic field is given by
[tex]B_{rms} = \frac{B_0}{\sqrt{2}}=\frac{2.7\cdot 10^{-16} T}{\sqrt{2}}=1.9\cdot 10^{-16} T[/tex]
