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\(\bullet$$\bullet\) 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 rea value 0.0200 \(\mathrm{V} / \mathrm{m} .\) How much energy does this wave carry through the window during a 30.0 s commercial?

Short Answer

Expert verified
Calculate intensity, find power, then find total energy: about 1.59 \(\times\) 10\(^{-10}\) J.

Step by step solution

01

Understanding the Problem

We're tasked to find the energy carried by a sinusoidal electromagnetic wave through a specified window area during a given time. The wave's electric field strength is known.
02

Find the Intensity of the Wave

The intensity of an electromagnetic wave, which is the power per unit area, can be found using the formula \[ I = \frac{1}{2} c \varepsilon_0 E^2 \]where \( c \) is the speed of light in vacuum \((3 \times 10^8 \mathrm{\ m/s})\), \( \varepsilon_0 \) is the permittivity of free space \((8.85 \times 10^{-12} \mathrm{\ F/m})\), and \( E \) is the electric field (0.0200 V/m).
03

Calculate Intensity Value

Substitute the known values into the formula:\[ I = \frac{1}{2} \times (3 \times 10^8) \times (8.85 \times 10^{-12}) \times (0.0200)^2 \] Calculate the intensity \( I \).
04

Compute the Power Passing Through

The power passing through the window is the product of intensity and area. Use the formula: \[ P = I \times A \] where \( A = 0.500 \ \mathrm{m^2} \).
05

Determine the Total Energy Transmitted

To calculate the total energy, use the formula\[ E = P \times t \]where \( t = 30.0 \ \mathrm{s} \) is the duration of the wave's passage through the window.
06

Complete the Calculation

Substitute the power value from Step 4 and \( t = 30.0 \mathrm{\ s} \) to find \( E \).
07

Solution Verification

Ensure that each step's calculation is correct, focusing on units used throughout the equations.

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Key Concepts

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

Electric Field Intensity
Electric field intensity, often referred to simply as the electric field, is a fundamental concept in electromagnetism. It describes the force per unit charge that a charged object would experience in an electric field. In our exercise, the electric field intensity ( E ) is given as 0.0200 V/m, representing the strength of the electric wave at the window.
This strength of the field is crucial for calculating other properties of the electromagnetic wave, like intensity and power. The electric field acts as a driving force, influencing the flow of energy through space, which helps us determine how much power is transmitted through an area, like the window.
Understanding the electric field intensity provides a foundation for exploring how waves interact with different media, and helps in calculating energy-related parameters in wave energy problems.
Permittivity of Free Space
Permittivity of free space, denoted as \( \varepsilon_0 \), is a constant that describes how electric fields interact in a vacuum. It is fundamental in equations involving electromagnetic waves and helps in determining the capacity of a vacuum to permit electric field lines.
The value of permittivity of free space is approximately \( 8.85 \times 10^{-12} \, \mathrm{F/m} \) (farads per meter). This value is used in formulas to calculate electromagnetic wave properties such as intensity. Without permittivity, it would be difficult to gauge how effectively a wave can propagate through a vacuum or air.
The permittivity of free space is introduced into equations for its essential role in bridging between electric fields and magnetic fields in vacuum, enabling accurate calculations involving wave intensity and energy transmission.
Power Calculation
The power of an electromagnetic wave indicates how much energy it can deliver. To calculate power, we use the formula \( P = I \times A \), where \( I \)is the intensity of the wave and \( A \)is the area it passes through.
Intensity \( I \)is calculated by the formula \( I = \frac{1}{2} c \varepsilon_0 E^2 \). In this case, \( c \) is the speed of light, \( \varepsilon_0 \) is the permittivity of free space, and \( E \) is the electric field intensity. Substituting these values allows us to compute the intensity before determining power.
Finally, multiply this intensity by the window area (0.500 m²) to find the power passing through. This step links how an electromagnetic wave’s inherent properties like electric field intensity, influence how much power can be harnessed or observed in a given area.
Energy Transmission
Energy transmission involves calculating how much energy an electromagnetic wave carries over time through a specific area. This is crucial as it tells us the total amount of energy transferred which is applied practically in various technologies.
To find the energy transmitted, we use the formula \( E = P \times t \), where \( P \)is the power passing through and \( t \)is the time duration (30 seconds in our problem). This approach is especially useful in scenarios like transmitting radio waves or signals where knowing the exact amount of energy passing through is essential.
By understanding how energy is transmitted, engineers and scientists can optimize systems that rely on wave propagation, ensuring efficient energy use and adequate signal strength where necessary.

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Most popular questions from this chapter

\(\bullet$$\bullet\) You (height of your eyes above the water, 1.75 \(\mathrm{m}\) ) are standing 2.00 \(\mathrm{m}\) from the edge of a 2.50 -m-deep swimming pool. You notice that you can barely see your cell phone, which went missing a few minutes before, on the bottom of the pool. How far from the side of the pool is your cell phone?

\(\bullet\) The refractive index of a certain glass is \(1.66 .\) For what angle of incidence is light that is reflected from the surface of this glass completely polarized if the glass is immersed in (a) air or (b) water?

\(\bullet$$\bullet \mathrm{A}\) sinusoidal electromagnetic wave of frequency \(6.10 \times 10^{14} \mathrm{Hz}\) travels in vacuum in the \(+x\) -direction. The magnetic field is parallel to the \(y\) -axis and has amplitude \(5.80 \times 10^{-4} \mathrm{T} .\) (a) Find the magnitude and direction of the electric field. (b) Write the wave functions for the electric and magnetic fields in the form of Equations \((23.3).\)

\(\bullet$$\bullet\) A 1.55 -m-tall fisherman stands at the edge of a lake, being watched by a suspicious trout who is 3.50 \(\mathrm{m}\) from the fisherman in the horizontal direction and 45.0 \(\mathrm{cm}\) below the surface of the water. At what angle from the vertical does the fish see the top of the fisherman's head?

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