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Testing a Space Radio Transmitter. You are a NASA mission specialist on your first flight aboard the space shuttle. Thanks to your extensive training in physics, 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\(\mathrm { MH } z\) . 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

Calculate the Intensity of the Radio Wave (Part a)

To find the intensity \( I \) of a radio wave, we use the formula:\[I = \frac{1}{2} c \varepsilon_0 E^2\]where \( c \) is the speed of light \( (3 \times 10^8 \, \text{m/s}) \), \( \varepsilon_0 \) is the permittivity of free space \( (8.85 \times 10^{-12} \, \text{F/m}) \), and \( E \) is the electric-field amplitude \( (0.090 \, \text{V/m}) \).Substituting the values:\[I = \frac{1}{2} \times 3 \times 10^8 \, \text{m/s} \times 8.85 \times 10^{-12} \, \text{F/m} \times (0.090 \, \text{V/m})^2\]After calculation:\[I \approx 1.07 \times 10^{-5} \, \text{W/m}^2\]

02

Calculate the Magnetic-Field Amplitude (Part b)

The relationship between the electric field \( E \) and the magnetic field \( B \) in an electromagnetic wave is given by:\[B = \frac{E}{c}\]where \( E = 0.090 \, \text{V/m} \) and \( c = 3 \times 10^8 \, \text{m/s} \).Substitute the values:\[B = \frac{0.090 \, \text{V/m}}{3 \times 10^8 \, \text{m/s}}\]After calculation:\[B \approx 3.0 \times 10^{-10} \, \text{T}\]

03

Calculate the Total Power Output (Part c)

The total power output \( P \) can be found if we assume isotropic radiation, allowing us to use the formula:\[P = I \cdot 4 \pi r^2\]where \( I = 1.07 \times 10^{-5} \, \text{W/m}^2 \) and the distance \( r = 2.5 \, \text{km} = 2500 \, \text{m} \).Substitute the values:\[P = 1.07 \times 10^{-5} \, \text{W/m}^2 \times 4 \times \pi \times (2500 \, \text{m})^2\]After calculation:\[P \approx 8.41 \, \text{kW}\]

04

Identify Assumptions Made (Part d)

In the calculations above, the key assumption is that the radio transmitter emits radiation isotropically, meaning it radiates energy uniformly in all directions. Additionally, free space conditions are assumed, with no absorbing or reflecting materials that could alter the wave attributes.

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Key Concepts

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

Electromagnetic Waves

Electromagnetic waves are waves that propagate through space and carry energy. They are composed of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of the wave's travel. These waves can travel through a vacuum, requiring no medium.

Some common types of electromagnetic waves include visible light, radio waves, and X-rays. The speed of electromagnetic waves in a vacuum is approximately **300,000 kilometers per second**, denoted by the constant **c**. This is the fastest speed at which information or energy can travel.

The ability of electromagnetic waves to carry information makes them crucial in communication technologies, such as radio and television broadcasting. In the context of the ISS radio transmitter, these electromagnetic waves are the medium by which data is sent back to Earth.

Electric Field Amplitude

The electric field of an electromagnetic wave has a certain strength, which is referred to as its amplitude. Specifically, for radio waves, this is the maximum strength of the electric field oscillating at any point. Amplitude is measured in volts per meter (**V/m**).

In our example, the electric field amplitude detected was **0.090 V/m**. The greater the amplitude, the more energy the wave carries.

• The electric field amplitude provides a direct measure of the wave's power at a particular location.
• The intensity, which is the power per unit area, can be derived using this amplitude.

By calculating the intensity using the electric field amplitude, and the known constants of speed of light and permittivity of free space, we determine the energy density of the electromagnetic wave.

Magnetic Field Amplitude

Alongside the electric field in an electromagnetic wave, there exists a magnetic field, and its strength is described by the magnetic amplitude. The electric field amplitude and magnetic field amplitude are related through the wave's speed; in vacuum, this relationship is governed by:

• The formula: \( B = \frac{E}{c} \)
• Where **E** is the electric field strength and **c** is the speed of light.

In our problem, using the formula with the given electric field amplitude of **0.090 V/m**, we calculate the magnetic field amplitude as approximately **3.0 x 10鈦宦光伆 Tesla (T)**. This highlights the proportional relationship between both components of the wave.

The magnetic field is weaker than the electric field but it maintains the same oscillation pattern and directions.

Permittivity of Free Space

The permittivity of free space, denoted as \( \varepsilon_0 \), is a fundamental physical constant that influences how electric fields interact in a vacuum. It has a standard value of approximately **8.85 x 10鈦宦孤� Farads per meter (F/m)**.

• Permittivity defines the ability of a material, or vacuum in this case, to permit electric field lines and affect how electric forces are transmitted through space.
• It plays a crucial role in calculating the intensity and energy density of electromagnetic waves.

In the ISS example, \( \varepsilon_0 \) is used in the formula for calculating wave intensity to account for the medium through which the wave propagates. Understanding permittivity helps in conceptualizing how electromagnetic waves transfer energy through free space.

Isotropic Radiation

Isotropic radiation refers to radiation that is emitted uniformly in all directions. This is an often-used assumption in physics to simplify calculations, assuming a source radiates equally across the full sphere around it.

For the ISS radio transmitter, assuming isotropic radiation allows us to calculate total power output by considering energy dispersal in a sphere of area **4蟺r虏**.

• Such simplifications are crucial for handling large distances and complex systems.
• It provides a baseline model to analyze wave transport without local variations in emission.

However, in real-world applications, perfect isotropy is rare due to interference, obstacles, or focused broadcasts, which necessitate adjustments to this model.