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The currents in household wiring and power lines alternate at a frequency of \(60.0 \mathrm{Hz}\). (a) What is the wavelength of the EM waves emitted by the wiring? (b) Compare this wavelength with Earth's radius. (c) In what part of the EM spectrum are these waves?

Short Answer

Expert verified
(a) The wavelength is \( 5 \times 10^6 \text{ m} \). (b) It is slightly smaller than Earth's radius. (c) These waves are in the ELF range of the EM spectrum.

Step by step solution

01

Understanding the Relationship between Frequency and Wavelength

The speed of electromagnetic waves in a vacuum, which is the speed of light, is approximately \( c = 3 \times 10^8 \text{ m/s} \). The relationship between speed \( c \), frequency \( f \), and wavelength \( \lambda \) of a wave is given by the equation: \[ c = f \times \lambda \]Given that the frequency \( f = 60.0 \text{ Hz} \), we can rearrange the equation to find the wavelength:\[ \lambda = \frac{c}{f} \]
02

Calculating the Wavelength

Using the equation from the previous step, we substitute the given values:\[ \lambda = \frac{3 \times 10^8 \text{ m/s}}{60.0 \text{ Hz}} \]\[ \lambda = 5 \times 10^6 \text{ m} \]Thus, the wavelength of the EM waves is \( 5 \times 10^6 \text{ meters} \).
03

Comparing the Wavelength to Earth's Radius

Earth's average radius is approximately \( 6.371 \times 10^6 \text{ m} \). We compare this with the wavelength calculated:\( 5 \times 10^6 \text{ m} \) versus \( 6.371 \times 10^6 \text{ m} \).The wavelength of the EM waves is slightly smaller than Earth's radius.
04

Identifying the Spectrum Part

The calculated wavelength \( 5 \times 10^6 \text{ m} \) lies in the extremely low frequency (ELF) range within the electromagnetic spectrum. Radio waves, particularly those with low frequencies, can have very long wavelengths, such as this one.

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Key Concepts

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

Wavelength Calculation
In physics, understanding how to calculate the wavelength of electromagnetic (EM) waves is crucial. The wavelength (4") is determined by dividing the speed of light (3 98 ext{ m/s}") by the frequency (6 ext{ Hz}") of the wave. This calculation helps us find out how long the wave is from one peak to the next. Wavelength calculation is an important step in analyzing EM waves. By knowing the frequency, such as the typical household 60 Hz, we can use the formula:\[ \lambda = \frac{c}{f} \] Wavelength helps us understand how different EM waves interact with the environment and affect us.Using this equation, for a household frequency, the wavelength is \( 5 \times 10^6 \text{ meters} \), illustrating the vast stretch these waves encompass.
Frequency and Wavelength Relationship
The frequency and wavelength of electromagnetic waves are interconnected through the speed of light. Frequency refers to how often the wave oscillates, while wavelength is the distance between consecutive peaks of the wave. Given the speed of light (3 98 ext{ m/s}") is constant in a vacuum, changes in one will inversely affect the other.The equation connecting frequency (6 ext{ Hz}") and wavelength (d") is:\[ c = f \times \lambda \]If frequency increases, wavelength decreases, and vice versa. This relationship helps us determine characteristics of electromagnetic waves, from radio to gamma rays, across diverse applications.Understanding this relationship is key in fields such as telecommunications and astronomy.
Electromagnetic Spectrum
The electromagnetic spectrum is the range of all types of EM radiation. EM waves vary from very long radio waves to extremely short gamma rays. The spectrum includes multiple regions such as radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. Each type of wave has unique properties:
  • Radio waves typically have the longest wavelengths.
  • Visible light has wavelengths that the human eye can detect.
  • Gamma rays possess very high energy and short wavelengths.
The placement of a wave within the spectrum is determined by its frequency and wavelength. This allows us to categorize and study different waves for numerous scientific and practical uses.
Extremely Low Frequency (ELF)
The extremely low frequency (ELF) range lies at one end of the electromagnetic spectrum. ELF waves have frequencies from 3 to 30 Hz and correspondingly long wavelengths, stretching even thousands of kilometers. These waves are produced by natural phenomena like the planet's atmosphere and man-made sources such as power lines and submarine communications. Despite their long wavelengths and low frequency, ELF waves can penetrate deep into water and the Earth's surface, making them valuable for certain kinds of communication.
Speed of Light
The speed of light is a fundamental constant of nature, approximately \( 3 \times 10^8 \text{ meters per second}\) in a vacuum. It's crucial in connecting frequency and wavelength of electromagnetic waves. Since light speed is constant, it governs how variations in frequency and wavelength interact. For example:
  • If frequency increases, wavelength must decrease to maintain the light's speed.
  • These principles are applied in technologies involving optics, wireless communications, and more.
Understanding the speed of light helps correctly navigate the relationship between wavelength and frequency.

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Most popular questions from this chapter

A magnetic dipole antenna is used to detect an electromagnetic wave. The antenna is a coil of 50 turns with radius \(5.0 \mathrm{cm} .\) The EM wave has frequency \(870 \mathrm{kHz}\) electric field amplitude $0.50 \mathrm{V} / \mathrm{m},\( and magnetic field amplitude \)1.7 \times 10^{-9} \mathrm{T} .$ (a) For best results, should the axis of the coil be aligned with the electric field of the wave, or with the magnetic field, or with the direction of propagation of the wave? (b) Assuming it is aligned correctly, what is the amplitude of the induced emf in the coil? (Since the wavelength of this wave is much larger than \(5.0 \mathrm{cm},\) it can be assumed that at any instant the fields are uniform within the coil.) (c) What is the amplitude of the emf induced in an electric dipole antenna of length \(5.0 \mathrm{cm}\) aligned with the electric field of the wave?
How long does it take sunlight to travel from the Sun to Earth?
Calculate the frequency of an EM wave with a wavelength the size of (a) the thickness of a piece of paper \((60 \mu \mathrm{m}),(\mathrm{b})\) a \(91-\mathrm{m}\) -long soccer field, (c) the diameter of Earth, (d) the distance from Earth to the Sun.
Energy carried by an EM wave coming through the air can be used to light a bulb that is not connected to a battery or plugged into an electric outlet. Suppose a receiving antenna is attached to a bulb and the bulb is found to dissipate a maximum power of 1.05 W when the antenna is aligned with the electric field coming from a distant source. The wavelength of the source is large compared to the antenna length. When the antenna is rotated so it makes an angle of \(20.0^{\circ}\) with the incoming electric field, what is the power dissipated by the bulb?
The magnetic field in a radio wave traveling through air has amplitude $2.5 \times 10^{-11} \mathrm{T}\( and frequency \)3.0 \mathrm{MHz}$. (a) Find the amplitude and frequency of the electric field. (b) The wave is traveling in the \(-y\) -direction. At \(y=0\) and \(t=0\), the magnetic field is $1.5 \times 10^{-11} \mathrm{T}\( in the \)+z$ -direction. What are the magnitude and direction of the electric field at \(y=0\) and \(t=0 ?\)
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