Question

An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.40 kW. A receiving antenna 60.0 cm long is at a location 4.00 mi from the transmitter. Compute the amplitude of the emf that is induced by this signal between the ends of the receiving antenna.

Answer 1

To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.

The intensity of the wave at the receiver is

`I = (P_(avg))/(A)`

`I = (P_(avg))/(4\pi r^2)`

`I = (3.4*10^3)/(4\pi(4*1609.34)^2) \rightarrow 1mile = 1609.3m`

`I = 6.529*10^(-6)W/m^2`

The amplitude of electric field at the receiver is

`I = (E_(max)^2)/(2\mu_0 c)`

`E_(max)= √(2I\mu_0 c)`

The amplitude of induced emf by this signal between the ends of the receiving antenna is

`\epsilon_(max) = E_(max) d`

`\epsilon_(max) = √(2I \mu_0 cd)`

Here,

I = Current

`\mu_0` = Permeability at free space

c = Light speed

d = Distance

Replacing,

`\epsilon_(max) = \sqrt{2(6.529*10^(-6))(4\pi*10^(-7))(3*10^(8))(60.0*10^(-2))}`

`\epsilon_(max) = 0.05434V`

Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V