(a) 2.97 m
The wavelength of an electromagnetic wave is given by:
[tex]\lambda = \frac{c}{f}[/tex]
where
[tex]\lambda[/tex] is the wavelength
c is the speed of light
f is the frequency
For the radio wave in the problem, the frequency is
[tex] f = 101 MHz = 101 \cdot 10^6 Hz[/tex]
Therefore, the wavelength is
[tex]\lambda = \frac{3\cdot 10^8 m/s}{101\cdot 10^6 Hz}=2.97 m[/tex]
(b) [tex]1.05\cdot 10^{-4} W/m^2[/tex]
The intensity of the radio signal is given by
[tex]I=\frac{P}{A}[/tex]
where
P is the power of the signal
A is the area over which the signal is radiated
In this situation:
P = 50,000 W is the power
the area is a hemisphere with a radius of
r = 8.70 km = 8700 m
So the area to be considered is
[tex]A=2\pi r^2 = 2\pi (8700 m)^2=4.76\cdot 10^8 m^2[/tex]
Therefore, the intensity of the signal is
[tex]I=\frac{50000 W}{4.76\cdot 10^8 m^2}=1.05\cdot 10^{-4} W/m^2[/tex]
(c) 0.281 V/m
The intensity of an electromagnetic wave can be written as
[tex]I=\frac{1}{2}c\epsilon_0 E^2[/tex]
where
c is the speed of light
[tex]\epsilon_0[/tex] is the vacuum permittivity
E is the amplitude of the electric field
Re-arranging the equation, we get
[tex] E=\sqrt{\frac{2I}{c \epsilon_0}}[/tex]
And substituting
[tex]I=1.05\cdot 10^{-4} W/m^2[/tex]
we find
[tex] E=\sqrt{\frac{2(1.05\cdot 10^{-4} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12} F/m)}}=0.281 V/m[/tex]
