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

$I = \frac{P_{avg}}{A}$

$I = \frac{P_{avg}}{4\pi r^2}$

$I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m$

$I = 6.529*10^{-6}W/m^2$

The amplitude of electric field at the receiver is

$I = \frac{E_{max}^2}{2\mu_0 c}$

$E_{max}= \sqrt{2I\mu_0 c}$

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

$\epsilon_{max} = E_{max} d$

$\epsilon_{max} = \sqrt{2I \mu_0 cd}$

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

6 0

4 years ago

When finding the upper bound of the density, you put what number in the denominator?

lilavasa [31]

Answer: The result of "the upper bound of the density" does not go on the denominator.
So simplified, no. The answer is no.

3 0

3 years ago