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Television Broadcasting. Public television station KQED in San Francisco broadcasts a sinusoidal radio signal at a power of \(316 \mathrm{~kW}\). Assume that the wave spreads out uniformly into a hemisphere above the ground. At a home \(5.00 \mathrm{~km}\) away from the antenna, (a) what average pressure does this wave exert on a totally reflecting surface, (b) what are the amplitudes of the electric and magnetic fields of the wave, and (c) what is the average density of the energy this wave carries? (d) For the energy density in part (c), what percentage is due to the electric field and what percentage is due to the magnetic field?

Short Answer

Expert verified
(a) \( 1.34 \times 10^{-8} \text{ N/m}^2 \); (b) 38.8 V/m, 1.29 T; (c) \( 6.70 \times 10^{-9} \text{ J/m}^3 \); (d) 50% each.

Step by step solution

01

Calculate Intensity at the Distance

To find the intensity of the wave at 5.00 km from the station, we use the formula for intensity of a wave spreading over a hemisphere, given by \( I = \frac{P}{2\pi R^2} \). Here, \( P = 316 \text{ kW} = 316 \times 10^3 \text{ W} \) and \( R = 5000 \text{ m} \) (since 5.00 km = 5000 m). Substituting these into the formula gives:\[ I = \frac{316 \times 10^3}{2\pi \times (5000)^2} = 2.01 \text{ W/m}^2. \]
02

Calculate Average Pressure on Reflecting Surface

The average pressure exerted by a totally reflecting surface is given by \( p_{\text{avg}} = \frac{2I}{c} \), where \( c = 3.00 \times 10^8 \text{ m/s} \) is the speed of light. With \( I = 2.01 \text{ W/m}^2 \), the average pressure is:\[ p_{\text{avg}} = \frac{2 \times 2.01}{3.00 \times 10^8} = 1.34 \times 10^{-8} \text{ N/m}^2. \]
03

Calculate Amplitudes of Electric and Magnetic Fields

From the intensity, the electric field amplitude \( E_0 \) can be calculated using \( I = \frac{1}{2} \varepsilon_0 c E_0^2 \), where \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2 \). Solving for \( E_0 \): \[ E_0 = \sqrt{\frac{2I}{\varepsilon_0 c}} = \sqrt{\frac{2 \times 2.01}{8.85 \times 10^{-12} \times 3.00 \times 10^8}} = 38.8 \text{ V/m}. \]The magnetic field amplitude \( B_0 \) can be found using \( B_0 = \frac{E_0}{c} \):\[ B_0 = \frac{38.8}{3.00 \times 10^8} = 1.29 \times 10^{-7} \text{ T}. \]
04

Calculate Average Energy Density

The average energy density \( u \) of the wave is given by \( u = \frac{I}{c} \). With the intensity \( I = 2.01 \text{ W/m}^2 \), calculate:\[ u = \frac{2.01}{3.00 \times 10^8} = 6.70 \times 10^{-9} \text{ J/m}^3. \]
05

Determine Energy Density Contributions

The energy density is equally divided between the electric and magnetic fields in an electromagnetic wave, so each contributes 50%. Therefore, the electric field contributes \( 50\% \) of the energy density and the magnetic field also contributes \( 50\% \).

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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 fascinating aspects of physics that encompass many everyday applications, such as television broadcasting. These waves consist of oscillating electric and magnetic fields that propagate through space at the speed of light. This unique ability allows them to carry information over significant distances without the need for a physical medium. In television broadcasting, sinusoidal radio signals are a subtype of electromagnetic waves used to transmit data from a station to homes.
Understanding how these waves spread out in the environment—such as the hemispherical spreading noted in the exercise—helps in determining how the signal intensity dissipates over distance. This dispersion affects how we calculate variables such as power intensity and pressures on surfaces that the waves encounter.
Electromagnetic waves not only transmit signals but also possess inherent energy, leading to phenomena like exerting pressure on obstacles they meet. This happens because they carry momentum, and when they reflect or transmit through materials, they exert forces proportional to their intensity. Thus, electromagnetic waves are not only a medium for communication but also demonstrate the principles of momentum and force in physics.
Power Intensity
In the context of physics, power intensity is a crucial concept that describes how much power flows through a particular area. For television broadcasting, it indicates how concentrated the radio signal's power is at a given point away from the source. The intensity of a wave is defined as the power per unit area and is typically measured in watts per square meter (W/m²).
When electromagnetic waves spread from a point source, such as a broadcasting antenna, they distribute uniformly in all directions over a hypothetical surface. To find the intensity at a point 5.00 km away from the broadcasting station, as given in the exercise, one can use the formula for hemispherical spreading:
  • The formula: \( I = \frac{P}{2\pi R^2} \), where \(P\) is the power and \(R\) is the distance,
The calculations show how the power's concentration decreases as the distance from the source increases.
This concept ties into the broader understanding of how broadcast signals weaken with distance, a critical factor in managing the reach and clarity of television signals. This attenuation is crucial in practical engineering to ensure that home viewers receive clean and stable transmissions.
Energy Density
Energy density refers to the amount of energy stored in a given system per unit volume. In electromagnetic waves, it indicates how much energy is carried through a specified volume of space. This measurement is relatively small but significant in assessing the potential impact of waves within a given area.
The average energy density is computed from the intensity of the wave and the speed of light, represented by the formula:
  • The formula is: \( u = \frac{I}{c} \), where \(I\) is the intensity and \(c\) is the speed of light.
This calculation shows how the energy carried by the wave disperses through space.
Energy density is an essential parameter in physics as it quantifies the wave's ability to perform work or cause change, such as exerting pressure or moving particles. By understanding energy density, scientists and engineers can better grasp how energy is distributed in electromagnetic phenomena and utilize it in technological advancements like radio and television.
Electric and Magnetic Fields
Electric and magnetic fields are the core components of electromagnetic waves. These fields oscillate perpendicular to each other and the direction of wave propagation. The amplitudes of these fields are crucial because they describe the strength of the wave.
The electric field amplitude is tied to how much force a charged particle would experience within the field. It is calculated using the relationship between intensity and the fundamental constants of permittivity of free space and the speed of light:
  • The amplitude formula: \( E_0 = \sqrt{\frac{2I}{\varepsilon_0 c}} \)
The magnetic field amplitude follows from the electric field, given by \(B_0 = \frac{E_0}{c}\).
  • As the electric field helps impart momentum, the magnetic field complements this by ensuring the consistency and stability of the wave propelling through space.
Understanding these fields is fundamental to comprehending electromagnetic wave behavior.
The symmetry between electric and magnetic energy, each contributing 50% to the total energy density, highlights the balanced relationship within electromagnetic phenomena, vital for designing technologies reliant on wave dynamics.

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Most popular questions from this chapter

Solar Sail 2. NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, lowmass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective? Why? (b) The total power output of the sun is \(3.9 \times 10^{26} \mathrm{~W}\). How large a sail is necessary to propel a \(10,000-\mathrm{kg}\) spacecraft against the gravitational force of the sun? Express your result in square kilometers. (c) Explain why your answer to part (b) is independent of the distance from the sun.

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 harmful to the skin and is necessary for the production of vitamin D. UVB, with a wavelength between \(280 \mathrm{~nm}\) and \(320 \mathrm{~nm}\), 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 intense light source radiates uniformly in all directions. At a distance of \(5.0 \mathrm{~m}\) from the source, the radiation pressure on a perfectly absorbing surface is \(9.0 \times 10^{-6} \mathrm{~Pa}\). What is the total average power output of the source?

A space probe \(2.0 \times 10^{10} \mathrm{~m}\) from a star measures the total intensity of electromagnetic radiation from the star to be \(5.0 \times 10^{3} \mathrm{~W} / \mathrm{m}^{2}\). If the star radiates uniformly in all directions, what is its total average power output?

A cylindrical conductor with a circular cross section has a radius \(a\) and a resistivity \(\rho\) and carries a constant current \(I\). (a) What are the magnitude and direction of the electric-field vector \(\overrightarrow{\boldsymbol{E}}\) at a point just inside the wire at a distance \(a\) from the axis? (b) What are the magnitude and direction of the magneticfield vector \(\overrightarrow{\boldsymbol{B}}\) at the same point? (c) What are the magnitude and direction of the Poynting vector \(\overrightarrow{\boldsymbol{S}}\) at the same point? (The direction of \(\vec{S}\) is the direction in which electromagnetic energy flows into or out of the conductor.) (d) Use the result in part (c) to find the rate of flow of energy into the volume occupied by a length \(l\) of the conductor. (Hint: Integrate \(\vec{S}\) over the surface of this volume.) Compare your result to the rate of generation of thermal energy in the same volume. Discuss why the energy dissipated in a currentcarrying conductor, due to its resistance, can be thought of as entering through the cylindrical sides of the conductor.

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