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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?

Step by step solution

01

Understanding the Problem

We need to find the intensity of the radio wave, the magnetic-field amplitude, and the total power output of the ISS transmitter based on the given electric-field amplitude and frequency. Finally, we will list any assumptions made.

02

Calculate Intensity

The intensity of an electromagnetic wave can be calculated using the formula \( I = \frac{1}{2} c \varepsilon_0 E^2 \), where \( c \) is the speed of light \( \sim 3 \times 10^8 \; \text{m/s} \), \( \varepsilon_0 \) is the permittivity of free space \( \sim 8.85 \times 10^{-12} \; \text{F/m} \), and \( E \) is the electric field amplitude. Substituting the values, we find:\[I = \frac{1}{2} \times 3 \times 10^8 \times 8.85 \times 10^{-12} \times (0.090)^2 = 1.066 \times 10^{-2} \; \text{W/m}^2\]

03

Calculate Magnetic Field Amplitude

The magnetic field amplitude \( B \) is related to electric field amplitude \( E \) by the relation \( cB = E \). Thus, \( B = \frac{E}{c} \). Using \( E = 0.090 \; \text{V/m} \) and \( c \= 3 \times 10^8 \; \text{m/s} \), we find:\[B = \frac{0.090}{3 \times 10^8} = 3 \times 10^{-10} \; \text{T}\]

04

Calculate Total Power Output

Total power \( P \) is found by multiplying the intensity \( I \) by the surface area through which the wave is passing. Assume the transmission is isotropic (uniform in all directions). The surface area at a distance \( r = 2500 \; \text{m} \) is \( 4\pi r^2 \). Thus,\[P = I \times 4\pi (2500)^2 = 1.066 \times 10^{-2} \times 4\pi \times (2500)^2 = 8.34 \times 10^3 \; \text{W}\]

05

State Assumptions

We assumed that the waves are isotropically distributed in free space without any obstruction or absorption. Also, the medium is a vacuum or behaves as one, as found in most space environments.

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Key Concepts

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

Radio Waves

Radio waves are a type of electromagnetic radiation. They have a wide range of applications, from broadcasting television and radio programs, to communication between devices and even astronomical observation.
These waves have long wavelengths and low frequencies compared to other types of electromagnetic radiation like visible light or X-rays.
In the context of this exercise, the radio waves being analyzed have a frequency of 244 MHz. This frequency is commonly used in various communication technologies. Understanding these features is key to evaluating radio transmitters' performance accurately.

• Frequency: Higher frequency waves can carry more data, but they may face more attenuation when passing through objects.
• Wavelength: Longer wavelengths (like those of radio waves) can travel longer distances and penetrate through obstacles better than shorter wavelengths.

Understanding the basic properties of radio waves helps in evaluating their performance and calculating their corresponding electromagnetic parameters with precision.

Intensity Calculation

The intensity of an electromagnetic wave indicates how much energy is transported per unit time per unit area by the wave. It's an important measure as it directly relates to the power received by a detector like the one used in the exercise.
To calculate intensity (I) for the given radio waves, the formula used is \[ I = \frac{1}{2} c \varepsilon_0 E^2 \]where:

• \( c \): speed of light (~3 x 10^8 m/s)
• \( \varepsilon_0 \): permittivity of free space (~8.85 x 10^-12 F/m)
• \( E \): electric field amplitude

Plugging in these values provides the intensity at a location 2.5 km away from the transmitter. A precise calculation gives \( I = 1.066 \times 10^{-2} \; \text{W/m}^2 \).
The intensity tells us how powerful the radio waves are by the time they reach the detector.

Magnetic Field Amplitude

The relationship between electric fields and magnetic fields is central to understanding electromagnetic waves.
For a plane electromagnetic wave in free space, the magnetic field amplitude (B) is related to the electric field amplitude (E) by the equation:\[ cB = E \]
Thus, if we have the electric field amplitude, we can calculate the magnetic field amplitude as\[ B = \frac{E}{c} \]
Using the given electric field amplitude \( E = 0.090 \; \text{V/m} \) and the speed of light \( c = 3 \times 10^8 \; \text{m/s} \), the magnetic field amplitude can be calculated to be \( B = 3 \times 10^{-10} \; \text{T} \).
This reveals the strength of the magnetic component of the electromagnetic wave at the given distance.

Power Output of Transmitter

The power output of a transmitter defines how much energy the transmitter sends out in the form of electromagnetic waves, measured typically in watts.
To calculate this, we use the previously found intensity and consider the energy distribution over a spherical surface area, assuming an isotropic transmission (equal in all directions).
The total power P is therefore calculated using the formula: \[ P = I \times 4 \pi r^2 \]
where:

• \( I \): intensity of the wave
• \( r \): distance from the source (2500 m, in this case)

Substituting the known values into the formula gives the total power output as \( P = 8.34 \times 10^3 \; \text{W} \).
This calculation allows for understanding how efficiently the transmitter is sending out waves and how potential environmental factors (like space conditions) may impact this efficiency.