/*! 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

91影视

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 \(4.82 \times 10 ^ { - 111 } \mathrm { T.Calculate }\) (a) the wavelength; (b) the wave number; (c) the angular frequency; (d) the electric-field amplitude.

Short Answer

Expert verified
(a) 361.45 m; (b) 0.01738 m鈦宦; (c) 5.216 脳 10鈦 rad/s; (d) 14.46 V/m.

Step by step solution

01

Determine the wavelength

The speed of light, denoted as \( c \), is a constant \( 3.00 \times 10^8 \, \text{m/s} \). To find the wavelength \( \lambda \), we use the formula: \[ \lambda = \frac{c}{f} \] where \( f \) is the frequency. Given \( f = 830\, \text{kHz} = 830 \times 10^3 \, \text{Hz} \), thus, \( \lambda = \frac{3.00 \times 10^8}{830 \times 10^3} = 361.45 \, \text{m} \).
02

Calculate the wave number

The wave number \( k \) is given by the formula: \[ k = \frac{2\pi}{\lambda} \] We have already calculated \( \lambda = 361.45 \, \text{m} \). Plugging it into the formula gives \( k = \frac{2\pi}{361.45} = 0.01738 \, \text{m}^{-1} \).
03

Calculate the angular frequency

The angular frequency \( \omega \) is given by the formula: \[ \omega = 2\pi f \] where \( f = 830 \times 10^3 \, \text{Hz} \). Therefore, \( \omega = 2\pi \times 830 \times 10^3 = 5.216 \times 10^6 \, \text{rad/s} \).
04

Calculate the electric-field amplitude

The amplitude of the electric field \( E_0 \) is related to the magnetic field amplitude \( B_0 \) by the equation: \[ E_0 = c B_0 \] Given \( B_0 = 4.82 \times 10^{-11} \, \text{T} \), we find \( E_0 = 3.00 \times 10^8 \times 4.82 \times 10^{-11} = 14.46 \, \text{V/m} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Wavelength Calculation
The wavelength of an electromagnetic wave, like those broadcast by radio stations, represents the distance over which the wave's shape repeats. It is denoted by the Greek letter lambda (\( \lambda \)). To find this for a wave, we use the formula:\[ \lambda = \frac{c}{f} \]where \( c \) is the speed of light (approximately \( 3.00 \times 10^8 \) meters per second) and \( f \) is the wave's frequency. For the radio station WCCO broadcasting at \( 830 \) kHz (or \( 830 \times 10^3 \) Hz), inserting these values into the formula gives us:\[ \lambda = \frac{3.00 \times 10^8}{830 \times 10^3} = 361.45 \, \text{m} \]Thus, the wavelength is about \( 361.45 \) meters. This means that each full cycle of the wave stretches over \( 361.45 \) meters in space.
Wave Number
The wave number provides us with quantifiable insight into the number of wave cycles within a given length. It is defined as the number of wavelengths per unit distance, typically expressed in radians per meter (\( \text{m}^{-1} \)).To calculate the wave number \( k \), we use the formula:\[ k = \frac{2\pi}{\lambda} \]Using the wavelength we found earlier, \( 361.45 \, \text{m} \), the wave number is calculated as follows:\[ k = \frac{2\pi}{361.45} = 0.01738 \, \text{m}^{-1} \]This value signifies that approximately \( 0.01738 \) cycles fit into each meter of this electromagnetic wave.
Angular Frequency
Angular frequency is a pivotal concept that aids in understanding how quickly the wave oscillates in terms of angles, specifically radians. Represented by \( \omega \), angular frequency connects the frequency of a wave to a rotational perspective:\[ \omega = 2\pi f \]Using the frequency \( f = 830 \times 10^3 \, \text{Hz} \) for the radio station's broadcast, the angular frequency can be computed as:\[ \omega = 2\pi \times 830 \times 10^3 = 5.216 \times 10^6 \, \text{rad/s} \]This result, \( 5.216 \times 10^6 \, \text{rad/s} \), elucidates how many radians the wave travels over in one second, which is invaluable in applications involving wave interference and modulation.
Electric Field Amplitude
The electric field amplitude (\( E_0 \)) of an electromagnetic wave indicates the maximum strength of the electric field component of the wave. This measurement is in volts per meter (\( \text{V/m} \)) and is crucial for understanding the wave's energy and intensity.The relationship between the magnetic field amplitude (\( B_0 \)) and the electric field amplitude is described by the equation:\[ E_0 = c B_0 \]where \( B_0 \) is given as \( 4.82 \times 10^{-11} \, \text{T} \) (tesla) for this particular wave. Plugging this value into the equation produces:\[ E_0 = 3.00 \times 10^8 \times 4.82 \times 10^{-11} = 14.46 \, \text{V/m} \]Therefore, the electric field amplitude of this electromagnetic wave from station WCCO is \( 14.46 \, \text{V/m} \), a value signifying the peak electric field our wave achieves as it propagates through space.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A small helium-neon laser emits red visible light with a power of 4.60\(\mathrm { mW }\) in a beam that has a diameter of 2.50\(\mathrm { mm }\) . (a) What are the amplitudes of the electric and magnetic fields of the light? (b) What are the average energy densities associated with the electric field are the magnetic field? (c) What is the total energy contained in a \(1.00 - \mathrm { m }\) length of the beam?

In a certain experiment, a radio transmitter emits sinusoidal electromagnetic waves of frequency 110.0\(\mathrm { MHz }\) in opposite directions inside a narrow cavity with reflectors at both ends, causing a standing-wave pattern to occur. (a) How far apart are the nodal planes of the magnetic field? (b) If the standing-wave pattern is determined to be in its eighth harmonic, how long is the cavity?

You want to support a sheet of fireproof paper horizon tally, using only a vertical upward beam of light spread uniformly over the sheet. There is no other light on this paper. The sheet measures 22.0\(\mathrm { cm }\) by 28.0\(\mathrm { cm }\) and has a mass of 1.50\(\mathrm { g }\) . (a) If the paper is black and hence absorbs all the light that hits it, what must be the intensity of the light beam? (b) For the light in part (a), what are the amplitudes of its electric and magnetic fields? (c) If the paper is white and hence reflects all the light that hits it, what intensity of light beam is needed to support it? (d) To see if it is physically reasonable to expect to support a sheet of paper this way, calculate the intensity in a typical 0.500 -mW laser beam that is 1.00\(\mathrm { mm }\) in diameter, and compare this value with your answer in part (a).

An intense light source radiates uniformly in all directions. At a distance of 5.0\(\mathrm { m }\) from the source, the radiation pressure on a perfectly absorbing surface is \(9.0 \times 10 ^ { - 6 }\) Pa. What is the total average power output of the source?

A sinusoidal electromagnetic wave is propagating in vacuum in the \(+ z\) -direction. If at a particular instant and at a certain point in space the electric field is in the \(+ x - x -\) -direction and has mag- nitude \(4.00 \mathrm { V } / \mathrm { m } ,\) what are the magnitude and direction of the magnetic field of the wave at this same point in space and instant in time?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.