/*! 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 16 A sinusoidal electromagnetic wav... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area \(0.500 \mathrm{~m}^{2}\). At the window, the electric field of the wave has rms value \(0.0400 \mathrm{~V} / \mathrm{m}\). How much energy does this wave carry through the window during a 30.0 s commercial?

Short Answer

Expert verified
To solve this problem, we need to find the power per unit area, then multiply that by the area and time to find the total energy that the wave carries through the window. In calculation, the final answer will be the total energy.

Step by step solution

01

Numerical Information Extraction

Identify all relevant quantities given in the exercise: area (A) = 0.500 m², electric field rms value (E) = 0.0400 V/m, and time (t) = 30.0 s. The speed of light (c) is always, approximately, \(3.0 \times 10^{8}\) m/s and the permittivity of free space (εₒ) is approximately \(8.85 \times 10^{-12}\) C²/N.m². These are constants of nature that are always the same, and are typically given in physics problems.
02

Calculate the Power Density

To find the power (the amount of energy it gives off per second), we first have to calculate the power density (P/A). The formula for the power carried per unit area (S) in an electromagnetic wave is given by \(S = 0.5 \times c \times \varepsilon_{0} \times E^{2}\). By substituting the given values in, we get \(S = 0.5 \times 3.0 \times 10^{8} \textrm{m/s} \times 8.85 \times 10^{-12} \textrm{C}^{2}/\textrm{N} .\textrm{m}^{2} \times (0.0400 \textrm{V/m})^{2}\).
03

Calculate the Total Power

To find the total power we need to multiply the power density by the total area. The total power (P) can be found by multiplying the power density by the area of the window: \(P = S \times A\). Substitute the values found in the previous steps to find the answer.
04

Calculate the Energy

Next, the total energy (E) can now be found just by multiplying that power by the total time, according to the formula \(E = P \times t\). Substitute the values found in the previous steps to find the answer.

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 91Ó°ÊÓ!

Key Concepts

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

Power Density
The concept of power density is critical when studying electromagnetic waves, especially in applications like broadcasting. Power density refers to the amount of power passing through a certain area perpendicular to the wave's direction. In mathematical terms, power density is denoted as 'S' and can be calculated using the equation:
\[ S = \frac{P}{A} \]
where 'S' is the power density, 'P' is the power, and 'A' is the area. However, for electromagnetic waves, the formula for power density is more specific due to their intrinsic properties and is given by:
\[ S = \frac{1}{2} c \varepsilon_{0} E^{2} \]
Here, \( c \) represents the speed of light, \( \varepsilon_{0} \) is the permittivity of free space, and \( E \) is the rms value of the electric field. The factor of 1/2 appears in the equation to account for the average power over the wave's cycle, as electromagnetic waves oscillate sinusoidally.
RMS Value of Electric Field
The term 'rms' stands for root mean square, and it provides a way of expressing an average value of an oscillating quantity, such as an electric field which varies over time. In the case of electromagnetic waves, the electric field undergoes sinusoidal oscillations, resulting in oscillating energy and power. The rms value is particularly useful because it enables calculation of the power that would be equivalent if the electric field maintained a constant value rather than oscillating.
The rms value of an electric field (E) is calculated using the formula:
\[ E_{rms} = \frac{E_{max}}{\sqrt{2}} \]
where \( E_{max} \) is the maximum value of the electric field. When we deal with power and energy calculations of electromagnetic waves, as in the given exercise, the rms value of the electric field is used rather than the maximum to account for its average effect over time.
Electromagnetic Wave Equation
The electromagnetic wave equation is a core component in understanding how electromagnetic waves propagate through space. Derived from Maxwell's equations, it describes the relationship between electric and magnetic fields that comprise the electromagnetic wave. The basic wave equation is a second-order partial differential equation that can be expressed for the electric field E or magnetic field B of the wave.
For a sinusoidal wave, which is common in many applications such as radio broadcasting, the electric field component can be written as:
\[ E(x,t) = E_{max} \cos(2\pi(ft - \frac{x}{\lambda})) \]
Here, \( E(x,t) \) represents the electric field at a point 'x' at a time 't', \( E_{max} \) is the maximum electric field, 'f' is the frequency of the wave, '\lambda' is the wavelength, and '\( 2\pi(ft - \frac{x}{\lambda}) \)' is the phase of the wave. This equation embodies the wave-like nature of electromagnetic radiation and allows us to predict how the fields within the wave will behave as they travel 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 source of sinusoidal electromagnetic waves radiates uniformly in all directions. At a distance of \(10.0 \mathrm{~m}\) from this source, the amplitude of the electric field is measured to be \(3.50 \mathrm{~N} / \mathrm{C}\). What is the electric-field amplitude \(20.0 \mathrm{~cm}\) from the source?

A sinusoidal electromagnetic wave emitted by a mobile 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. Calculate (a) the frequency of the wave; (b) the magnetic-field amplitude; (c) the intensity of the wave.

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 \(\mathrm{WCCO}\) 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.

Interplanetary space contains many small particles referred to as interplanetary dust. Radiation pressure from the sun sets a lower limit on the size of such dust particles. To see the origin of this limit, consider a spherical dust particle of radius \(R\) and mass density \(\rho\). (a) Write an expression for the gravitational force exerted on this particle by the sun (mass \(M\) ) when the particle is a distance \(r\) from the sun. (b) Let \(L\) represent the luminosity of the sun, equal to the rate at which it emits energy in electromagnetic radiation. Find the force exerted on the (totally absorbing) particle due to solar radiation pressure, remembering that the intensity of the sun's radiation also depends on the distance \(r\). The relevant area is the cross-sectional area of the particle, not the total surface area of the particle. As part of your answer, explain why this is so. (c) The mass density of a typical interplanetary dust particle is about \(3000 \mathrm{~kg} / \mathrm{m}^{3} .\) Find the particle radius \(R\) such that the gravitational and radiation forces acting on the particle are equal in magnitude. The luminosity of the sun is \(3.9 \times 10^{26} \mathrm{~W}\). Does your answer depend on the distance of the particle from the sun? Why or why not? (d) Explain why dust particles with a radius less than that found in part (c) are unlikely to be found in the solar system. [Hint: Construct the ratio of the two force expressions found in parts (a) and (b).]

A monochromatic light source with power output \(60.0 \mathrm{~W}\) radiates light of wavelength \(700 \mathrm{nm}\) uniformly in all directions. Calculate \(E_{\max }\) and \(B_{\max }\) for the \(700 \mathrm{nm}\) light at a distance of \(5.00 \mathrm{~m}\) from the source.

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.