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A monochromatic light source emits \(100 \mathrm{W}\) of electromagnetic power uniformly in all directions. (a) Calculate the average electric-field energy density \(1.00 \mathrm{m}\) from the source. (b) Calculate the average magnetic- field energy density at the same distance from the source. (c) Find the wave intensity at this location.

Short Answer

Expert verified
After performing the above calculations, we find that the wave intensity at a distance of 1m from the light source is \(I = \frac{100}{4\pi}\) W/m². The average electric-field energy density \(u_E\) and average magnetic-field energy density \(u_B\) at that location will both be equal to \(I/c\) joules/m³. Interpretation of values and units can help assess the accuracy of the result.

Step by step solution

01

Establish Relevant Equations

Start by defining the relevant equations needed to solve the problem: \[Wave\:Intensity = \frac{Power}{Surface\:Area}\] This equation shows that the wave intensity is the power divided by the surface area over which it is distributed. Since the light is emitted uniformly in all directions, this area is the surface of a sphere of radius 1m, and the equation is \[Wave\:Intensity = \frac{Power}{4\pi r^2}\] where \(r\) represents the distance from the light source. There are also two formulas associated with magnetic-field energy density and electric-field energy density: \[u_E = \frac{1}{2}\varepsilon_0 E^2,\] \[u_B = \frac{1}{2u_0}B^2, \] where \(u_0\) and \(E\) are the permeability and permittivity of free space respectively, \(B\) is the magnetic field and \(E\) is the electric field. Given that \(u_E = u_B\) for electromagnetic waves in free space.
02

Computing Wave Intensity

The given parameters are the power \(P = 100W\) and the distance \(r = 1.00m\). Plug these values into the wave intensity formula: \[Wave Intensity = \frac{100}{4\pi(1.00)^2} \], this will give the wave intensity value.
03

Calculating Electric-field Energy Density

Knowing the wave intensity and the speed of light in free space, we can use the equation connecting wave intensity and electric field energy density. Considering that the propagation of the electromagnetic wave happens in free space, we take \( \varepsilon_0\) to be \(8.85×10^{-12}\), and \( v = c = 3.00 × 10^8 m/s\) that is the speed of light in free space. Plugging these into the equation we have: \[ u_E = \frac{Wave\:Intensity}{c}\]. This will give the electric-field energy density value.
04

Obtaining Magnetic-field Energy Density

Given that the electric and magnetic field energy densities are equal in free space, we conclude that \(u_B = u_E\). This will give the magnetic-field energy density.

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Key Concepts

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

Electric-field Energy Density
The electric-field energy density is a way to understand how much energy is stored in the electric field component of an electromagnetic wave. It is denoted by \( u_E \) and is calculated using the formula:
  • \( u_E = \frac{1}{2}\varepsilon_0 E^2 \)
where \( \varepsilon_0 \) is the permittivity of free space, and \( E \) is the electric field strength. In simpler terms, it describes how much energy is held in the electric field part of the wave, per unit volume.
For electromagnetic waves in free space, the formula simplifies further because of the known relationship between wave intensity and energy density. The wave intensity \( I \) relates to the electric field energy density through:
  • \( u_E = \frac{I}{c} \)
where \( c \) is the speed of light in free space. This expression highlights that a higher wave intensity or a stronger wave increases the energy density in the electric field.
Understanding electric-field energy density helps in comprehending how energy is transferred in space through electromagnetic waves.
Magnetic-field Energy Density
Magnetic-field energy density, similar to electric-field energy density, quantifies the energy contained within the magnetic component of an electromagnetic wave. This energy density is represented by \( u_B \) and is determined using the formula:
  • \( u_B = \frac{1}{2\mu_0}B^2 \)
where \( \mu_0 \) is the permeability of free space, and \( B \) is the magnetic field strength. Essentially, \( u_B \) tells us how much energy is held in the magnetic field portion per unit volume.
In a situation where electromagnetic waves travel through free space, the energy density in the magnetic field equals that in the electric field due to their intrinsic relationship:
  • \( u_B = u_E \)
This equality arises because electromagnetic waves possess an equal partition of energy between their electric and magnetic components. This concept underscores the harmonious and interconnected nature of electromagnetic wave dynamics.
Grasping magnetic-field energy density helps in visualizing the distribution of energy between the components of an electromagnetic field.
Wave Intensity
Wave intensity is a measure of how much power an electromagnetic wave transmits through a particular area. It is a crucial concept that summarizes the energy conveyance capacity of the wave and is represented by \( I \). The intensity is calculated using the equation:
  • \( I = \frac{Power}{Surface Area} \)
For a wave emitting uniformly in all directions, forming a sphere around the source, the intensity further becomes:
  • \( I = \frac{Power}{4\pi r^2} \)
where \( r \) is the distance from the wave source. This demonstrates that intensity decreases with the square of the distance from the source, a representation of how the energy spreads over a larger area.
For practical calculations, the power is the energy output, and \( 4\pi r^2 \) is the surface area of the sphere created by the wave's reach. Hence, wave intensity communicates how concentrated the wave's power is at a particular distance.
Understanding wave intensity provides insights into how strong or weak a wave's influence, in terms of energy transfer, can be at different distances from the source.

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Most popular questions from this chapter

One goal of the Russian space program is to illuminate dark northern cities with sunlight reflected to Earth from a \(200-\mathrm{m}\) diameter mirrored surface in orbit. Several smaller prototypes have already been constructed and put into orbit. (a) Assume that sunlight with intensity \(1340 \mathrm{W} / \mathrm{m}^{2}\) falls on the mirror nearly perpendicularly and that the atmosphere of the Earth allows \(74.6 \%\) of the energy of sunlight to pass through it in clear weather. What is the power received by a city when the space mirror is reflecting light to it? (b) The plan is for the reflected sunlight to cover a circle of diameter \(8.00 \mathrm{km} .\) What is the intensity of light (the average magnitude of the Poynting vector) received by the city? (c) This intensity is what percentage of the vertical component of sunlight at Saint Petersburg in January, when the sun reaches an angle of \(7.00^{\circ}\) above the horizon at noon?

An astronaut, stranded in space \(10.0 \mathrm{m}\) from his spacecraft and at rest relative to it, has a mass (including equipment) of 110 kg. Because he has a \(100-W\) light source that forms a directed beam, he considers using the beam as a photon rocket to propel himself continuously toward the spacecraft. (a) Calculate how long it takes him to reach the spacecraft by this method. (b) What If? Suppose, instead, that he decides to throw the light source away in a direction opposite the spacecraft. If the mass of the light source is \(3.00 \mathrm{kg}\) and, after being thrown, it moves at \(12.0 \mathrm{m} / \mathrm{s}\) relative to the recoiling astronaut, how long does it take for the astronaut to reach the spacecraft?

The intensity of solar radiation at the top of the Earth's atmosphere is \(1340 \mathrm{W} / \mathrm{m}^{2} .\) Assuming that \(60 \%\) of the incoming solar energy reaches the Earth's surface and assuming that you absorb \(50 \%\) of the incident energy, make an order-of-magnitude estimate of the amount of solar energy you absorb in a \(60-\) min sunbath.

In the absence of cable input or a satellite dish, a television set can use a dipole-receiving antenna for VHF channels and a loop antenna for UHF channels (Fig. Q34.12). The UHF antenna produces an emf from the changing magnetic flux through the loop. The TV station broadcasts a signal with a frequency \(f\), and the signal has an electric-field amplitude \(E_{\max }\) and a magnetic-field amplitude \(B_{\max }\) at the location of the receiving antenna. (a) Using Faraday's law, derive an expression for the amplitude of the emf that appears in a single-turn circular loop antenna with a radius \(r,\) which is small compared with the wavelength of the wave. (b) If the electric field in the signal points vertically, what orientation of the loop gives the best reception?

An electromagnetic wave in vacuum has an electric field amplitude of \(220 \mathrm{V} / \mathrm{m} .\) Calculate the amplitude of the corresponding magnetic field.

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