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Chapter 23: Problem 66

Radio receivers can comfortably pick up a broadcasting station's signal when the electric-field strength of the signal is about \(10.0 \mathrm{mV} / \mathrm{m} .\) If a radio station broadcasts in all directions with an average power of \(50.0 \mathrm{~kW},\) what would be the maximum distance at which you could easily pick up its transmissions? (Atmospheric conditions can have major effects on this distance.)

Short Answer

Expert verified
The maximum distance is approximately 61.3 kilometers.

Step by step solution

01

Understand the Problem

We need to find out the maximum distance at which the electric field strength of a radio signal is at least 10.0 mV/m. The station broadcasts with a power of 50 kW.
02

Use the Formula for Power Density

The power density, or intensity, at a distance \( r \) from a point source radiating equally in all directions is given by: \( I = \frac{P}{4\pi r^2} \) where \( P = 50 \text{ kW} \).
03

Relate Power Density to Electric Field Strength

The electric field strength \( E \) and intensity \( I \) are related by the formula: \( I = \frac{E^2}{2 \cdot 377} \) where 377 is the impedance of free space.
04

Solve for Electric Field Strength

Using the relation \( E = 10 \text{ mV/m} = 0.01 \text{ V/m} \), substitute into the formula: \( I = \frac{(0.01)^2}{2 \cdot 377} \).
05

Equate and Solve for Distance

Equate the intensity from the two formulas: \( \frac{P}{4\pi r^2} = \frac{(0.01)^2}{2 \cdot 377} \) and solve for \( r \). This gives: \( r = \sqrt{\frac{P \cdot 2 \cdot 377}{4\pi \cdot 0.01^2}} \).
06

Substitute Values

Substitute \( P = 50,000 \text{ W} \) into the equation: \( r = \sqrt{\frac{50,000 \cdot 754}{4\pi \cdot 0.01^2}} \) simplfying to find \( r \).
07

Calculate Distance

Calculate \( r \) to find the maximum distance: \( r \approx 61,348 \text{ meters} \) or about 61.3 kilometers.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radio Broadcasting
Radio broadcasting involves the transmission of audio signals that are accessible via radio receivers. Stations use these signals to send music, news, and other types of entertainment.
Broadcasting can be done in various ways, with AM and FM being the most popular methods. In such systems, the signals are sent in all directions, allowing receivers at different locations to tune in. Radio waves, a type of electromagnetic wave, make this process possible.
The range of broadcasting, meaning how far the signal can be easily received, depends largely on the strength of the electric field and the power output of the station. Atmospheric conditions, physical obstacles, and frequency can also influence signal reach.
Power Density
Power density, sometimes referred to as the intensity of a signal, is a measure of how much power is transferred to a specific area. It is a critical concept for understanding how signals travel through space.
The relationship between power density and the distance from the signal source is inversely proportional. This means that as you move away from the point of transmission, the power density decreases. The formula to calculate power density is:
  • \( I = \frac{P}{4\pi r^2} \)
where \( I \) is the power density, \( P \) the power radiated, and \( r \) the distance from the source.
Power density directly influences how far a radio signal can travel and still be effectively received. Understanding and calculating it help broadcasters determine the coverage area of their signals.
Electric Field Intensity
Electric field intensity, also known as electric field strength, is a vector quantity representing the force per unit charge exerted on a charged particle in the presence of other charges. In radio broadcasting, it's essential for determining how strong a signal is at a particular distance.
Higher electric field intensities mean stronger signals, potentially leading to better reception quality. The electric field strength of a radio signal decreases with increasing distance from the transmission source, due to the spreading of the electromagnetic waves.
The relationship between the electric field strength \( E \) and power density \( I \) is given by:
  • \( I = \frac{E^2}{2 \cdot 377} \)
In this equation, the number 377 is a constant representing the impedance of free space. Therefore, knowing the electric field intensity can help assess whether a radio signal is strong enough at a specific distance.
Electromagnetic Theory
Electromagnetic theory is a branch of physics that studies electric and magnetic fields and their interactions. It's the fundamental principle behind radio broadcasting and a wide range of other technologies.
According to this theory, a changing electric field produces a magnetic field, and vice versa. This relationship enables the propagation of electromagnetic waves, such as radio waves, through the vacuum of space or through air.
Key concepts of electromagnetic theory include:
  • The speed of light, which all electromagnetic waves travel at, in a vacuum.
  • Maxwell's equations, which describe how electric and magnetic fields interact and change over time.
  • The concept of wave-particle duality, where electromagnetic radiation displays properties of both waves and particles.
Understanding electromagnetic theory is crucial for grasping how radio signals are generated, propagated, and received. It forms the backbone of modern communication technology, ensuring effective transmission and reception of signals across various distances.

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Most popular questions from this chapter

Unpolarized light is incident on two ideal polarizing filters. The second filter's axis is rotated through an angle \(\theta\) relative to that of the first filter. If the intensity of light emerging from the second filter is \(1 / 10\) the intensity of the incident light, what is \(\theta ?\)

Laser surgery. Very short pulses of high-intensity laser beams are used to repair detached portions of the retina of the eye. The brief pulses of energy absorbed by the retina weld the detached portion back into place. In one such procedure, a laser beam has a wavelength of \(810 \mathrm{nm}\) and delivers \(250 \mathrm{~mW}\) of power spread over a circular spot \(510 \mu \mathrm{m}\) in diameter. The vitreous humor (the transparent fluid that fills most of the eye) has an index of refraction of \(1.34 .\) (a) If the laser pulses are each \(1.50 \mathrm{~ms}\) long, how much energy is delivered to the retina with each pulse? (b) What average pressure does the pulse of the laser beam exert on the retina as it is fully absorbed by the circular spot? (c) What are the wavelength and frequency of the laser light inside the vitreous humor of the eye? (d) What are the maximum values of the electric and magnetic fields in the laser beam?

There are two categories of ultraviolet light. Ultraviolet A (UVA) has a wavelength ranging from \(320 \mathrm{nm}\) to \(400 \mathrm{nm}\). It is not so harmful to the skin and is necessary for the production of vitamin D. UVB, with a wavelength between \(280 \mathrm{nm}\) and \(320 \mathrm{nm},\) is much more dangerous because it causes skin cancer. (a) Find the frequency ranges of UVA and UVB. (b) What are the ranges of the wave numbers for UVA and UVB?

There have been many studies of the effects on humans of electromagnetic waves of various frequencies. Using these studies, the International Commission on NonIonizing Radiation Protection (ICNIRP) produced guidelines for limiting exposure to electromagnetic fields, with the goal of protecting people against known adverse health effects. At frequencies of \(1 \mathrm{~Hz}\) to \(25 \mathrm{~Hz}\), the maximum exposure level of electric-field amplitude, \(E_{\max },\) for the general public is \(14 \mathrm{kV} / \mathrm{m}\). (Different guidelines were created for people who have occupational exposure to radiation.) At frequencies of \(25 \mathrm{~Hz}\) to \(3 \mathrm{kHz},\) the corresponding \(E_{\max }\) is \(\frac{350}{f} \mathrm{kV} / \mathrm{m},\) where \(f\) is the frequency in \(\mathrm{kHz}\). Doubling the frequency of a wave in the range of \(25 \mathrm{~Hz}\) to \(3 \mathrm{kHz}\) represents what change in the maximum allowed electromagneticwave intensity? A. A factor of 2 B. A factor of \(1 / \sqrt{2}\) C. A factor of \(\frac{1}{2}\) D. A factor of \(\frac{1}{4}\)

Radiation falling on a perfectly reflecting surface produces an average pressure \(p .\) If radiation of the same intensity falls on a perfectly absorbing surface and is spread over twice the area, what is the pressure at that surface in terms of \(p ?\)
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