Answer:
[tex]7.516\cdot 10^9 \mu V/m[/tex]
Explanation:
The relationship between the power of an electromagnetic wave and the amplitude of its electric field is given by the equation:
[tex]P=\epsilon_0 E^2 c[/tex]
where
P is the power of the wave
[tex]\epsilon_0[/tex] is the vacuum permittivity
E is the amplitude of the electric field
c is the speed of light
In this problem we have:
[tex]P=150 kW = 1.5\cdot 10^5 W[/tex] is the power of the microwaves
[tex]\epsilon_0 = 8.85\cdot 10^{-12}} F/m[/tex] is the vacuum permittivity
[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light
Solving for E, we find:
[tex]E=\sqrt{\frac{P}{\epsilon_0 c}}=\sqrt{\frac{1.5\cdot 10^5}{(8.85\cdot 10^{-12})(3.0\cdot 10^8)}}=7516 V/m =7.516\cdot 10^9 \mu V/m[/tex]
The strength of the amplitude electric field received back at the airport is [tex]1.9407 \times 10^ 9\ \mu V/m[/tex].
The given parameters;
The relationship between wave Intensity (I), Power (P), area (A), amplitude electric field strength (E₀), and speed of light(c) is given as follows;
[tex]I = \frac{P}{A} = \frac{c\times \epsilon _0\times E_o^2 }{2} \\\\c\times \epsilon _o\times E_o^2 \times A = 2P\\\\E_o^2 = \frac{2P}{A \times c \times \epsilon _o} \\\\E_o = \sqrt{\frac{2P}{A \times c \times \epsilon _o}} \\\\E_o = \sqrt{\frac{2\times 150,000}{(30) \times (3\times 10^8) \times (8.85\times 10^{-12})}} \\\\E_o = 1940.74 \ V/m\\\\E_o = 1.9407 \times 10^ 9\ \times 10^{-6}\ V/m\\\\E_o = 1.9407\times 10^ 9 \ \mu \ V/m[/tex]
Thus, the strength of the amplitude electric field received back at the airport is [tex]1.9407 \times 10^ 9\ \mu V/m[/tex]
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