/*! 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 12 A radio broadcasts at a waveleng... [FREE SOLUTION] | 91Ó°ÊÓ

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A radio broadcasts at a wavelength of \(3.06 \mathrm{~m}\). Near the broadcast tower the electric-field amplitude is \(2700 \mathrm{~N} / \mathrm{C}\). Calculate (a) the frequency, (b) the wave number, and (c) the magnetic-field amplitude of the electromagnetic wave.

Short Answer

Expert verified
(a) Frequency: \(9.8039 \times 10^7\,\mathrm{Hz}\); (b) Wave number: \(2.052 \,\mathrm{m}^{-1}\); (c) Magnetic-field amplitude: \(9 \times 10^{-6} \, \mathrm{T}\).

Step by step solution

01

Calculate the Frequency

To find the frequency of the wave, use the formula for the speed of light: \[ c = \lambda \cdot f \] where \( c = 3 \times 10^8 \mathrm{~m/s} \), "\( \lambda = 3.06 \mathrm{~m} \)" is the wavelength, and \( f \) is the frequency. Solve for \( f \):\[ f = \frac{c}{\lambda} = \frac{3 \times 10^8}{3.06} \approx 9.8039 \times 10^7 \mathrm{~Hz} \]
02

Calculate the Wave Number

The wave number \( k \) is given by the formula:\[ k = \frac{2\pi}{\lambda} \]Substitute "\( \lambda = 3.06 \mathrm{~m} \)" into the equation:\[ k = \frac{2\pi}{3.06} \approx 2.052 \mathrm{~m}^{-1} \]
03

Calculate the Magnetic-field Amplitude

To determine the magnetic-field amplitude \( B_0 \), use the relationship between the electric field \( E_0 \) and the magnetic field in an electromagnetic wave:\[ B_0 = \frac{E_0}{c} \]Given \( E_0 = 2700 \mathrm{~N/C} \), solve for \( B_0 \):\[ B_0 = \frac{2700}{3 \times 10^8} \approx 9 \times 10^{-6} \mathrm{~T} \]

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Key Concepts

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

Radio Frequency
Electromagnetic waves, like those used for radio broadcasts, display a variety of frequencies. Frequency, denoted as \( f \), represents the number of wave cycles that pass a fixed point in one second. It is measured in hertz (Hz). When dealing with radio waves, we're often discussing frequencies that fall into the range of 3 kHz to 300 GHz. The higher the frequency, the higher the energy of the wave.
For an electromagnetic wave, the frequency is calculated using the formula: \( c = \lambda \cdot f \). Here, \( c \) is the speed of light, approximately \( 3 \times 10^8 \) m/s. The wavelength \( \lambda \) is the distance over which the wave's shape repeats. For example, in our exercise, the wavelength is given as \( 3.06 \) m. Solving for \( f \), we divide the speed of light by the wavelength:
\[ f = \frac{3 \times 10^8}{3.06} \] This results in a frequency of approximately \( 98.039 \) MHz, which is typical for FM radio stations. Understanding frequency helps us comprehend how energy and information travel across distances without wires.
Wave Number
The wave number is another crucial concept in the context of electromagnetic waves, representing the spatial frequency of a wave. It tells us how many wave cycles exist per unit distance and is denoted by \( k \). The wave number is expressed in reciprocal meters (m\(^{-1}\)).
Mathematically, wave number \( k \) is defined as:
\[ k = \frac{2\pi}{\lambda} \] The factor of \( 2\pi \) arises because a complete wave cycle corresponds to a full rotation of a circle, or \( 360^\circ \). In the original exercise, substituting the given wavelength \( \lambda = 3.06 \) m:
\[ k = \frac{2\pi}{3.06} \] This yields a wave number of approximately \( 2.052 \) m\(^{-1}\).
The wave number is vital in the study of wave phenomena, as it relates physical space to the wave's behavior, much like frequency relates time for oscillations. It's an indicator of how "compact" the waves are; more compact waves have a higher wave number, indicating more cycles in a given length.
Magnetic Field Amplitude
In electromagnetic waves, the electric field and magnetic field produce each other. The magnetic field amplitude \( B_0 \) describes the maximum strength of the magnetic part of the wave. It is essential for understanding the wave's ability to exert force on a magnetic or charged particle and is measured in teslas (T).
The relationship between electric field amplitude \( E_0 \) and magnetic field amplitude in electromagnetic waves is given by:\[ B_0 = \frac{E_0}{c} \]where \( c \) is the speed of light.
From our exercise, with \( E_0 = 2700 \) N/C and substituting \( c = 3 \times 10^8 \) m/s:\[ B_0 = \frac{2700}{3 \times 10^8} \approx 9 \times 10^{-6} \] TThis calculation shows us that even when electric fields can be quite strong, the magnetic field component is typically much weaker in terms of tesla, verifying the interdependence and coexistent nature of electric and magnetic fields in waves of this kind.
Understanding the relation between these fields is crucial for many applications, from the design of antennas to the analysis of radio signal propagation.

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