/*! 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 13 Radio station WCCO in Minneapoli... [FREE SOLUTION] | 91影视

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Radio station WCCO in Minneapolis broadcasts at a frequency of \(830 \mathrm{kHz}\). At a point some distance from the transmitter, the magnetic- field amplitude of the electromagnetic wave from WCCO is \(4.82 \times 10^{-11} \mathrm{~T}\). Calculate (a) the wavelength; (b) the wave number; (c) the angular frequency; (d) the electric-field amplitude.

Short Answer

Expert verified
The wavelength is approximately 361.45 m, the wave number is 0.0174 m鈦宦, the angular frequency is 5.215 脳 10鈦 rad/s, and the electric-field amplitude is 0.0145 V/m.

Step by step solution

01

Calculate the Wavelength

The wavelength of a wave is given by the formula \( \lambda = \frac{c}{f} \), where \( c \) is the speed of light \( (3 \times 10^8 \text{ m/s}) \) and \( f \) is the frequency. Here, the frequency \( f \) is \( 830 \times 10^3 \text{ Hz} \) (since \( 1 \text{ kHz} = 1000 \text{ Hz} \)). Substituting the given values, \( \lambda = \frac{3 \times 10^8 \text{ m/s}}{830 \times 10^3 \text{ Hz}} \approx 361.45 \text{ m} \).
02

Calculate the Wave Number

The wave number \( k \) is given by \( k = \frac{2\pi}{\lambda} \). From Step 1, we found the wavelength \( \lambda \approx 361.45 \text{ m} \). Substituting, \( k = \frac{2\pi}{361.45} \approx 0.0174 \text{ m}^{-1} \).
03

Calculate the Angular Frequency

The angular frequency \( \omega \) is given by \( \omega = 2\pi f \). With \( f = 830 \times 10^3 \text{ Hz} \), we substitute to find \( \omega = 2\pi \times 830 \times 10^3 \approx 5.215 \times 10^6 \text{ rad/s} \).
04

Calculate the Electric-Field Amplitude

The amplitude of the electric field \( E_0 \) can be calculated using the relation \( E_0 = cB_0 \), where \( B_0 \) is the magnetic field amplitude. Given \( B_0 = 4.82 \times 10^{-11} \text{ T} \), \( E_0 = 3 \times 10^8 \text{ m/s} \times 4.82 \times 10^{-11} \text{ T} \approx 0.0145 \text{ V/m} \).

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Key Concepts

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

Wavelength Calculation
Understanding the concept of wavelength is essential when dealing with electromagnetic waves. Wavelength, denoted by \( \lambda \), represents the distance over which a wave's shape repeats. In simple terms, it's the distance between consecutive crests or troughs in a wave. The formula to calculate wavelength is \( \lambda = \frac{c}{f} \), where \( c \) is the speed of light, approximately \( 3 \times 10^8 \text{ m/s} \), and \( f \) is the frequency of the wave.

For example, if you have a radio station broadcasting at a frequency of \( 830 \text{ kHz} \), you convert this to Hz by multiplying by 1000, giving \( 830 \times 10^3 \text{ Hz} \). Substituting into the equation, you get \( \lambda = \frac{3 \times 10^8 \text{ m/s}}{830 \times 10^3 \text{ Hz}} \approx 361.45 \text{ m} \). This tells us that each wave produced by the radio station is about 361.45 meters long.
Wave Number
The wave number is a concept tied to how waves are quantified in terms of their length. It tells us the number of wave cycles in a unit distance and is denoted by \( k \). Mathematically, it is related to the wavelength through the formula \( k = \frac{2\pi}{\lambda} \). The wave number is expressed in units of \( \text{m}^{-1} \), which communicates the intensity and spatial frequency of the wave.

In the context of our earlier problem, where the wavelength of a radio wave was calculated as approximately \( 361.45 \text{ m} \), the wave number is \( k = \frac{2\pi}{361.45} \approx 0.0174 \text{ m}^{-1} \). This value implies how densely packed the wave cycles are in a meter. The larger the wave number, the shorter the wavelength and more cycles per meter.
Angular Frequency
Angular frequency is a key concept in wave physics. Represented as \( \omega \), it describes how fast an object moves through its wave cycles. This concept is similar to frequency but takes into account the full cycle of the wave in radians. The formula used to find angular frequency is \( \omega = 2\pi f \), where \( f \) is the frequency in Hertz.

When considering a radio station with a frequency of \( 830 \times 10^3 \text{ Hz} \), the angular frequency is calculated as \( \omega = 2\pi \times 830 \times 10^3 \approx 5.215 \times 10^6 \text{ rad/s} \). This tells us how quickly the electromagnetic wave oscillates as it travels; the higher this number, the faster the wave cycles through its oscillations.
Electric Field Amplitude
The electric field amplitude in an electromagnetic wave is a measure of the wave鈥檚 strength or intensity. The electric field amplitude \( E_0 \) is related to the magnetic field amplitude \( B_0 \) in this manner: \( E_0 = cB_0 \), where \( c \) is again the speed of light.

Given a magnetic field amplitude \( B_0 = 4.82 \times 10^{-11} \text{ T} \), we find the electric field amplitude using the formula \( E_0 = 3 \times 10^8 \text{ m/s} \times 4.82 \times 10^{-11} \text{ T} \approx 0.0145 \text{ V/m} \). This means that the intensity of the radio station鈥檚 electromagnetic wave has an electric field amplitude of \( 0.0145 \text{ V/m} \).
  • This electric field amplitude determines how strongly the wave can influence charges it encounters.
  • Higher electric field amplitudes mean stronger waves.
By understanding this value, engineers and scientists can better design and position antennas for optimal signal reception.

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Most popular questions from this chapter

An electromagnetic wave with frequency \(5.70 \times 10^{14} \mathrm{~Hz}\) propagates with a speed of \(2.17 \times 10^{8} \mathrm{~m} / \mathrm{s}\) in a certain piece of glass. Find (a) the wavelength of the wave in the glass; (b) the wavelength of a wave of the same frequency propagating in air; (c) the index of refraction \(n\) of the glass for an electromagnetic wave with this frequency; (d) the dielectric constant for glass at this frequency, assuming that the relative permeability is unity.

(a) How much time does it take light to travel from the moon to the earth, a distance of \(384,000 \mathrm{~km}\) ? (b) Light from the star Sirius takes \(8.61\) years to reach the earth. What is the distance from earth to Sirius in kilometers?

The electric field of a sinusoidal electromagnetic wave obeys the equation \(E=(375 \mathrm{~V} / \mathrm{m}) \cos \left[\left(1.99 \times 10^{7} \mathrm{rad} / \mathrm{m}\right) x+\right.\) \(\left.\left(5.97 \times 10^{15} \mathrm{rad} / \mathrm{s}\right) t\right] .\) (a) What are the amplitudes of the electric and magnetic fields of this wave? (b) What are the frequency, wavelength, and period of the wave? Is this light visible to humans? (c) What is the speed of the wave?

A sinusoidal electromagnetic wave emitted by a cellular phone has a wavelength of \(35.4 \mathrm{~cm}\) and an electric-field amplitude of \(5.40 \times 10^{-2} \mathrm{~V} / \mathrm{m}\) at a distance of \(250 \mathrm{~m}\) from the phone. \(\mathrm{Cal}-\) culate (a) the frequency of the wave; (b) the magnetic-field amplitude; (c) the intensity of the wave.

A source of sinusoidal electromagnetic waves radiates uniformly in all directions. At \(10.0 \mathrm{~m}\) from this source, the amplitude of the electric field is measured to be \(1.50 \mathrm{~N} / \mathrm{C}\). What is the electric-field amplitude at a distance of \(20.0 \mathrm{~cm}\) from the source?

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