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It has been proposed to place solar-power-collecting satellites in earth orbit. The power they collect would be beamed down to the earth as microwave radiation. For a microwave beam with a cross-sectional area of 36.0 \(\mathrm{m}^{2}\) and a total power of 2.80 \(\mathrm{kW}\) at the earth's surface, what is the amplitude of the electric field of the beam at the earth's surface?

Short Answer

Expert verified
The amplitude of the electric field is approximately 24.36 V/m.

Step by step solution

01

Understanding the Problem

We are given the cross-sectional area of the microwave beam as 36.0 \(\mathrm{m}^{2}\) and the total power as 2.80 \(\mathrm{kW}\). We need to find the amplitude of the electric field at the Earth's surface.
02

Power and Intensity Relationship

First, we identify the formula to relate power \(P\), intensity \(I\), and area \(A\). The intensity \(I\) is the power per unit area, given by:\[I = \frac{P}{A}\]Substitute \(P = 2800 \mathrm{W}\) and \(A = 36.0 \mathrm{m}^{2}\) into the equation.
03

Calculate the Intensity

Substituting the given values into the intensity formula:\[I = \frac{2800 \mathrm{W}}{36.0 \mathrm{m}^2} = 77.78 \mathrm{W/m}^2\]This is the intensity of the microwave beam at the Earth's surface.
04

Relate Intensity to Electric Field Amplitude

The intensity \(I\) of an electromagnetic wave is related to the amplitude of the electric field \(E_0\) by the equation:\[I = \frac{1}{2} c \varepsilon_0 E_0^2\]where \(c\) is the speed of light \(3 \times 10^8 \mathrm{m/s}\) and \(\varepsilon_0\) is the permittivity of free space \(8.85 \times 10^{-12} \mathrm{C^2/(N \cdot m^2)}\).
05

Solve for Electric Field Amplitude

Rearrange the formula to solve for \(E_0\):\[E_0 = \sqrt{\frac{2I}{c \varepsilon_0}}\]Substitute \(I = 77.78 \mathrm{W/m}^2\), \(c = 3 \times 10^8 \mathrm{m/s}\), and \(\varepsilon_0 = 8.85 \times 10^{-12} \mathrm{C^2/(N \cdot m^2)}\) into the equation to calculate \(E_0\).
06

Calculate the Result

Perform the calculation:\[E_0 = \sqrt{\frac{2 \times 77.78}{3 \times 10^8 \times 8.85 \times 10^{-12}}} \approx 24.36 \mathrm{V/m}\]Thus, the amplitude of the electric field of the beam at the Earth's surface is approximately 24.36 \(\mathrm{V/m}\).

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Key Concepts

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

Electric Field Amplitude
Electric field amplitude is a measure of the strength of the electric field at a particular point. For electromagnetic waves such as microwaves, this corresponds to the maximum electric field strength in the wave. Understanding how to calculate this is crucial, especially in applications like satellite solar power where microwaves are used to transfer energy.
To find the electric field amplitude (\(E_0\)), you need to understand its relationship with wave intensity. The intensity of an electromagnetic wave is directly related to the square of its electric field amplitude. The formula used is: \[I = \frac{1}{2} c \varepsilon_0 E_0^2\], where \(I\) is the intensity, \(c\) is the speed of light, and \(\varepsilon_0\) is the permittivity of free space. Solving for \(E_0\) involves rearranging this formula and substituting known values for these constants. This reveals the intensity directly affects how strong the electric field's amplitude is.
Electromagnetic Waves
Electromagnetic waves encompass a broad spectrum of wave types, including visible light, radio waves, and microwaves. These waves can travel through the vacuum of space and are characterized by oscillating electric and magnetic fields.
Microwaves, like those used in satellite solar power transmission, fall into a specific part of this electromagnetic spectrum. They have longer wavelengths than visible light, making them suitable for transmitting energy over long distances without significant loss. This characteristic is why they are chosen for satellite-based solar power systems, allowing satellites to send collected energy to Earth efficiently.
Power and Intensity Relationship
Understanding the power and intensity relationship is essential when assessing how energy transmission works. Power refers to the total energy distributed over time, while intensity represents power spread out over an area.
For example, with a microwave beam, its intensity (\(I\)) can be calculated by dividing the power (\(P\)) by the cross-sectional area (\(A\)) of the beam: \[I = \frac{P}{A}\]. Given a total power of 2800 \(\mathrm{W}\) and an area of 36.0 \(\mathrm{m}^2\), you can determine the beam's intensity at the Earth's surface. This calculation aids in understanding the beam's effectiveness in energy transfer.
Satellite Solar Power
Satellite solar power involves collecting solar energy in space using satellites and transmitting it to Earth. This futuristic approach circumvents many issues associated with ground-based solar power collection, like nighttime and weather interferences.
By capturing solar energy above the terrestrial layer, satellites can utilize powerful microwave beams to send energy consistently to designated receptors on Earth. The collected energy is converted into microwaves and directed to a ground-based station where it can be used or stored. This method not only maximizes energy utilization but also demonstrates the application of electromagnetic waves in future power solutions.

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Most popular questions from this chapter

A circular loop of wire can be used us a radio antenna. If a 18.0-cm-diameter antenna is located 2.50 \(\mathrm{km}\) from a 95.0 -MHz source with a total power of 55.0 \(\mathrm{kW}\) , what is the maximum emf induced in the loop? (Assume that the plane of the antenna loop is perpendicular to the direction of the radiation's magnetic field and that the source radiates uniformly in all directions.)

You are a NASA mission specialist on your first flight aboard the space shuttle. Thanks to your extensive training in pliysics, you have been assigned to evaluate the performance of a new radio transmitter on board the International Space Station (ISS). Perched on the shuttle's movable arm, you aim a sensitive detector at the ISS, which is 2.5 \(\mathrm{km}\) away. You find that the electric-field amplitude of the radio waves coming from the ISS transmitter is 0.090 \(\mathrm{V} / \mathrm{m}\) and that the frequency of the waves is 244 \(\mathrm{MHz}\) . Find the following: (a) the intensity of the radio wave at your location; (b) the magnetic-field amplitude of the wave at your location; (c) the total power output of the ISS radio transmitter. (d) What assumptions, if any, did you make in your calculations?

The electric-field amplitude near a certain radio transmitter is \(3.85 \times 10^{-3} \mathrm{V} / \mathrm{m}\) What is the amplitude of \(\overrightarrow{\boldsymbol{B}} ?\) How does this compare in magnitude with the earth's field?

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?

(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?

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