Question

M radio station KRTH in Los Angeles broadcasts on an assigned frequency of 101 MHz with a power of 50,000 W. (a) What is the wavelength of the radio waves produced by this station? Answer 1 m (b) Estimate the average intensity of the wave at a distance of 8.70 km from the radio transmitting antenna. Assume for the purpose of this estimate that the antenna radiates equally in all directions, so that the intensity is constant over a hemisphere centered on the antenna. Answer 2 W/m2 (c) Estimate the amplitude of the electric field at this distance.

Answer 1

(a) 2.97 m

The wavelength of an electromagnetic wave is given by:

`\lambda = (c)/(f)`

where

`\lambda` is the wavelength

c is the speed of light

f is the frequency

For the radio wave in the problem, the frequency is

`f = 101 MHz = 101 \cdot 10^6 Hz`

Therefore, the wavelength is

`\lambda = (3\cdot 10^8 m/s)/(101\cdot 10^6 Hz)=2.97 m`

(b)
`1.05\cdot 10^(-4) W/m^2`

The intensity of the radio signal is given by

`I=(P)/(A)`

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

`A=2\pi r^2 = 2\pi (8700 m)^2=4.76\cdot 10^8 m^2`

Therefore, the intensity of the signal is

`I=(50000 W)/(4.76\cdot 10^8 m^2)=1.05\cdot 10^(-4) W/m^2`

(c) 0.281 V/m

The intensity of an electromagnetic wave can be written as

`I=(1)/(2)c\epsilon_0 E^2`

where

c is the speed of light

`\epsilon_0` is the vacuum permittivity

E is the amplitude of the electric field

Re-arranging the equation, we get

`E=\sqrt{(2I)/(c \epsilon_0)}`

And substituting

`I=1.05\cdot 10^(-4) W/m^2`

we find

`E=\sqrt{(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`