/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 (I) Estimate the wavelength for ... [FREE SOLUTION] | 91Ó°ÊÓ

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(I) Estimate the wavelength for 1.9 -GHz cell phone reception.

Short Answer

Expert verified
The wavelength is approximately 0.158 meters.

Step by step solution

01

Understand the Formula

The wavelength of a wave can be calculated using the formula \( \lambda = \frac{c}{f} \), where \( \lambda \) is the wavelength, \( c \) is the speed of light in a vacuum, approximately \( 3 \times 10^8 \) meters per second, and \( f \) is the frequency of the wave.
02

Identify the Given Frequency

In the problem, the frequency for 1.9-GHz cell phone reception is given as \( 1.9 \text{ GHz} \). Convert this frequency from gigahertz to hertz by noting that \( 1 \text{ GHz} = 10^9 \text{ Hz} \). Therefore, \( 1.9 \text{ GHz} = 1.9 \times 10^9 \text{ Hz} \).
03

Apply the Formula

Substitute the known values into the wavelength formula: \( \lambda = \frac{3 \times 10^8 \text{ m/s}}{1.9 \times 10^9 \text{ Hz}} \).
04

Calculate the Wavelength

Perform the calculation: \[ \lambda = \frac{3 \times 10^8}{1.9 \times 10^9} = \frac{3}{1.9} \times 10^{-1} \]. Solve this to get \( \lambda \approx 0.158 \text{ meters} \).

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Key Concepts

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

Electromagnetic Waves
Electromagnetic waves are a fascinating part of physics that essentially represent the combined oscillations of electric and magnetic fields traveling through space. These waves can vary greatly in wavelength and frequency, ranging from very long radio waves to extremely short gamma rays. The range of all possible frequencies of electromagnetic radiation is known as the electromagnetic spectrum, which includes the visible light that we can see and other forms of radiation like X-rays and microwaves.
This concept is key when talking about electromagnetic waves in communication technologies such as cell phones, where signals like the 1.9 GHz wave discussed in the problem are used to transmit information wirelessly. Cell phone signals are a type of radio wave, positioned on the spectrum right around microwaves, which is why understanding their wavelength helps in designing and optimizing communication systems.
Speed of Light
The speed of light, denoted by the symbol \( c \), is a fundamental constant crucial in the world of physics, especially in the context of electromagnetic waves. It represents the speed at which electromagnetic waves propagate through the vacuum of space, approximately equal to \( 3 \times 10^8 \) meters per second.
Understanding this speed is important not only in the calculation of wavelengths but also in the broader scope of theories and principles, such as Einstein's theory of relativity where it plays a central role. It's the ultimate speed limit of the universe, meaning nothing can travel faster than light in vacuum. This makes it exceptionally significant when calculating wave characteristics like wavelength, as it remains unchanging across all electromagnetic waves, providing a reliable constant against which frequency can be measured.
Frequency Conversion
Frequency conversion is an important step in accurately determining the wavelength of an electromagnetic wave. It involves translating the given frequency to a standardized unit, usually hertz (Hz), which is simple and widespread in scientific calculations.
In our exercise, we were given a frequency in gigahertz (GHz), specifically \( 1.9 \, \text{GHz} \), and needed to convert it to hertz. To do so, remember that \( 1 \, \text{GHz} = 10^9 \, \text{Hz} \). Thus, \( 1.9 \, \text{GHz} \) is equal to \( 1.9 \times 10^9 \, \text{Hz} \).
This conversion ensures that when we plug frequency into the wavelength formula \( \lambda = \frac{c}{f} \), it matches the units of the speed of light, \( \text{m/s} \), allowing us to accurately calculate the wavelength. Without this step, calculations could be less accurate, leading to errors in understanding and application.

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Most popular questions from this chapter

(1) Determine the rate at which the electric field changes between the round plates of a capacitor, 6.0 \(\mathrm{cm}\) in diameter, if the plates are spaced 1.1 \(\mathrm{mm}\) apart and the voltage across them is changing at a rate of 120 \(\mathrm{V} / \mathrm{s}\) .

(II) How much energy is transported across a \(1.00 \mathrm{~cm}^{2}\) area per hour by an EM wave whose \(E\) field has an rms strength of \(32.8 \mathrm{mV} / \mathrm{m} ?\)

(III) Suppose that a circular parallel-plate capacitor has radius \(R_{0}=3.0 \mathrm{~cm}\) and plate separation \(d=5.0 \mathrm{~mm} .\) A sinusoidal potential difference \(V=V_{0} \sin (2 \pi f t)\) is applied across the plates, where \(V_{0}=150 \mathrm{~V}\) and \(f=60 \mathrm{~Hz} .(a)\) In the region between the plates, show that the magnitude of the induced magnetic field is given by \(B=B_{0}(R) \cos (2 \pi f t)\) where \(R\) is the radial distance from the capacitor's central axis. (b) Determine the expression for the amplitude \(B_{0}(R)\) of this time-dependent (sinusoidal) field when \(R \leq R_{0}\), and when \(R>R_{0}\). (c) Plot \(B_{0}(R)\) in tesla for the range \(0 \leq R \leq 10 \mathrm{~cm}\).

Suppose that a right-moving EM wave overlaps with a leftmoving EM wave so that, in a certain region of space, the total electric field in the \(y\) direction and magnetic field in the \(z\) direction are given by \(E_{y}=E_{0} \sin (k x-\omega t)+E_{0} \sin (k x+\omega t)\) and \(B_{z}=B_{0} \sin (k x-\omega t)-B_{0} \sin (k x+\omega t),(a)\) Find the mathematical expression that represents the standing electric and magnetic waves in the \(y\) and \(z\) directions, respectively. (b) Determine the Poynting vector and find the \(x\) locations at which it is zero at all times.

(II) A 15.8-mW laser puts out a narrow beam \(2.00 \mathrm{~mm}\) in diameter. What are the rms values of \(E\) and \(B\) in the beam?

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