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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 0.0200 \(\mathrm{V} / \mathrm{m}\) . How much energy docs this wave carry through the window during a 30.0 -s commercial?

Short Answer

Expert verified
The wave carries approximately 4.47 脳 10鈦烩伕 Joules of energy through the window during 30 seconds.

Step by step solution

01

Understanding the RMS Electric Field

The root mean square (RMS) value of the electric field is given as 0.0200 V/m. The RMS value is related to the peak electric field (E_0) by the formula:\[ E_0 = E_{rms} \times \sqrt{2} \]Plugging in the given RMS value:\[ E_0 = 0.0200 \, \text{V/m} \times \sqrt{2} = 0.0200 \, \text{V/m} \times 1.414 = 0.0283 \, \text{V/m} \].
02

Calculate the Intensity of the Wave

The intensity of an electromagnetic wave is given by the formula:\[ I = \frac{c \varepsilon_0 E_0^2}{2} \]where:- \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \).- \( \varepsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2 \).Plugging in the values:\[ I = \frac{(3 \times 10^8 \times 8.85 \times 10^{-12} \times 0.0283^2}{2} \]\[ I = \frac{(3 \times 10^8) \times (8.85 \times 10^{-12}) \times 0.0008}{2} = \frac{5.956 \times 10^{-9}}{2} \, \approx 2.978 \times 10^{-9} \, \text{W/m}^2 \].
03

Determine Energy Passing Through the Window

Energy passing through an area is the product of intensity, area, and time. Thus, the energy \( E \) is given by:\[ E = I \times A \times t \]Here, \( A = 0.500 \, \text{m}^2 \) and \( t = 30.0 \, \text{s} \). Substitute the values:\[ E = (2.978 \times 10^{-9} \, \text{W/m}^2) \times (0.500 \, \text{m}^2) \times (30.0 \, \text{s}) \]\[ E = 4.467 \times 10^{-8} \] Joules.

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Key Concepts

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

RMS Electric Field and Its Significance
The RMS Electric Field, or the root mean square electric field, is an important measure in electromagnetic waves. It gives us the average value of the electric field within the wave over time. This is crucial because electromagnetic waves, like those from radio stations, have oscillating fields. Instead of dealing with these constantly changing values, we use RMS to simplify calculations.

The relation between RMS and the peak electric field can be quantified with the formula:\[ E_0 = E_{rms} \times \sqrt{2} \]This formula allows us to determine the maximum (or peak) amplitude of the electric field from its RMS value. For instance, if the RMS value is 0.0200 V/m, the peak electric field can be calculated as 0.0283 V/m after multiplying by \(\sqrt{2} 鈮 1.414\). This conversion is valuable for further calculations, such as wave intensity.

In practice, understanding and calculating the RMS electric field gives us a more manageable way to comprehend and work with electromagnetic waves. It helps in evaluating their effects and potential applications.
Understanding Wave Intensity
Wave Intensity is a key concept when exploring electromagnetic waves, representing the amount of energy transmitted per unit area per unit time. It is directly proportional to the square of the electric field's peak value. As the intensity gives insight into how powerful a wave is, it is very relevant in practical scenarios such as broadcasting.

The formula for intensity \(I\) of an electromagnetic wave is:\[ I = \frac{c \varepsilon_0 E_0^2}{2} \]Where:
  • \( c \) is the speed of light, approximately \(3 \times 10^8 \, \text{m/s}\).
  • \( \varepsilon_0 \) is the permittivity of free space, approximately \(8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2\).
By substituting these known constants and the calculated peak electric field, the intensity of the wave can be determined. For the given example, this results in an intensity of about \(2.978 \times 10^{-9} \, \text{W/m}^2\). This tells us how much energy from the wave hits a square meter every second.

Understanding wave intensity is essential for assessing how much power an electromagnetic wave delivers, which plays a vital role in fields like communications and power transmission.
Energy Transfer Through an Area
The concept of Energy Transfer in electromagnetic waves involves calculating how much energy is transferred through a given area over a certain time period. This is crucial in applications where predicting or measuring energy delivery is important, such as in solar panels or communication devices.

One can calculate the energy \(E\) passing through an area using the formula:\[ E = I \times A \times t \]Where:
  • \(I\) is the intensity of the wave, measured in watts per square meter (W/m虏).
  • \(A\) is the area, in this case, 0.500 m虏.
  • \(t\) is the time, which in this example is 30.0 seconds.
In the provided exercise, substituting these values gives an energy transfer of approximately \(4.467 \times 10^{-8}\) Joules. This represents the total energy that has passed through the window during the commercial.

By understanding and using these equations, one can effectively calculate how electromagnetic waves, such as those used in radio transmissions, can deliver energy across distances. This knowledge applies to escalating technology areas, from energy harvesting to data communication.

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Most popular questions from this chapter

There are two categories of ultraviolet light. Ultraviolet A (UVA) has a wavelength ranging from 320 \(\mathrm{nm}\) to 400 \(\mathrm{nm}\) . It is not so harmful to the skin and is necessary for the production of vitamin D. UVB, with a wavelength between 280 \(\mathrm{nm}\) and 320 \(\mathrm{mm}\) , is much more dangerous because it causes skin cancer. (a) Find the frequency ranges of UVA and UVB. (b) What are the ranges of the wave numbers for UVA and UVB?

An electromagnetic standing wave in air of frequency 750 MHz is set up between two conducting planes 80.0 \(\mathrm{cm}\) apart. At which positions between the planes could a point charge be placed at rest so that it would remain at rest? Explain.

The microwaves in a certain microwave oven have a wavelength of \(12.2 \mathrm{cm} .\) (a) How wide must this oven be so that it will contain five antinodal planes of the electric field along its width in the standing wave pattern? (b) What is the frequency of these microwaves? (c) Suppose a manufacturing error occurred and the oven was made 5.0 \(\mathrm{cm}\) longer than specificd in part (a). In this case, what would have to be the frequency of the microwaves for there still to be five antinodal planes of the electric field along the width of the oven?

The electron in a hydrogen atom can be considered to be in a circular orbit with a radius of 0.0529 \(\mathrm{nm}\) and a kinetic energy of 13.6 \(\mathrm{eV}\) . If the electron behaved classically, how much energy would it radiate per second (see Challenge Problem 32.57\() ?\) What does this tell you about the use of classical physics in describing the atom?

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 electricfield amplitude at a distance of 20.0 \(\mathrm{cm}\) from the source?

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