/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 A future space station in orbit ... [FREE SOLUTION] | 91Ó°ÊÓ

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A future space station in orbit about the earth is being powered by an electromagnetic beam from the earth. The beam has a cross-sectional area of \(135 \mathrm{m}^{2}\) and transmits an average power of \(1.20 \times 10^{4} \mathrm{W} .\) What are the rms values of the (a) electric and (b) magnetic fields?

Short Answer

Expert verified
The rms electric field is 182.5 V/m, and the rms magnetic field is 6.08 x 10^-7 T.

Step by step solution

01

Understand the Given Data

We are given:- Cross-sectional area of the beam, \(A = 135 \, \text{m}^2\)- Power transmitted by the beam, \(P = 1.20 \times 10^4 \, \text{W}\)
02

Calculate the Intensity of the Beam

The intensity \(I\) of the beam is the power per unit area. Therefore, we have:\[I = \frac{P}{A} = \frac{1.20 \times 10^4 \, \text{W}}{135 \, \text{m}^2} = 88.89 \, \text{W/m}^2\]
03

Relate Intensity to Electric Field

The intensity \(I\) relates to the rms electric field \(E_{\text{rms}}\) by the formula:\[I = \frac{1}{2} c \varepsilon_0 E_{\text{rms}}^2\]where \(c\) is the speed of light \(3 \times 10^8 \, \text{m/s}\) and \(\varepsilon_0\) is the permittivity of free space \(8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2\).
04

Solve for Electric Field

Rearrange the formula from Step 3 to solve for \(E_{\text{rms}}\):\[E_{\text{rms}} = \sqrt{\frac{2I}{c \varepsilon_0}} = \sqrt{\frac{2 \times 88.89}{3 \times 10^8 \times 8.85 \times 10^{-12}}}\]Calculate it to find \(E_{\text{rms}} = 182.5 \, \text{V/m}\).
05

Relate Electric Field to Magnetic Field

The rms magnetic field \(B_{\text{rms}}\) relates to the rms electric field \(E_{\text{rms}}\) through the equation:\[B_{\text{rms}} = \frac{E_{\text{rms}}}{c}\]Substitute the known values to find \(B_{\text{rms}}\).
06

Solve for Magnetic Field

Substitute the value from Step 4 into the equation from Step 5:\[B_{\text{rms}} = \frac{182.5}{3 \times 10^8} = 6.08 \times 10^{-7} \, \text{T}\]
07

Concluding Step: Review and Validate

Make sure the units are consistent and the computations logically follow the steps. We get the electric field \(E_{\text{rms}} = 182.5 \, \text{V/m}\) and the magnetic field \(B_{\text{rms}} = 6.08 \times 10^{-7} \, \text{T}\) as the final answers.

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Key Concepts

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

Electric Field
The electric field is a fundamental concept in electromagnetics, representing the force per unit charge experienced by a stationary charge. In the context of electromagnetic waves, the electric field oscillates perpendicular to the direction of wave propagation. This oscillation is essential for the wave's energy transmission.

In this exercise, we calculate the root mean square (rms) value of the electric field for an electromagnetic beam. The rms value is a kind of average that helps in understanding the effective strength of the oscillating field. It is derived from the intensity of the wave, using the formula:

\[E_{\text{rms}} = \sqrt{\frac{2I}{c \varepsilon_0}}\]

Here,
  • \(I\) is the intensity of the wave, given as the power per unit area.
  • \(c\) is the speed of light, approximately \(3 \times 10^8 \text{m/s}\).
  • \(\varepsilon_0\) is the permittivity of free space, valued at \(8.85 \times 10^{-12} \text{C}^2/\text{N} \cdot \text{m}^2\).
Through this equation, we transform the power details of the beam into an understanding of the electric field's amplitude, enhancing our grasp of electromagnetic phenomena.
Magnetic Field
The magnetic field forms an inseparable pair with the electric field in electromagnetic waves. For any wave propagating through space, the magnetic field oscillates perpendicularly to both the electric field and the direction of wave propagation. This interaction between the electric and magnetic fields sustains the wave's momentum and energy.

In electromagnetic waves, the rms value of the magnetic field is directly related to the rms value of the electric field through the speed of light \(c\). This relationship is expressed by:

\[B_{\text{rms}} = \frac{E_{\text{rms}}}{c}\]

This formula shows us:
  • How the magnetic field's magnitude is derived from the electric field.
  • It is inverse-proportional to the speed of light, emphasizing the intertwined nature of light speed and electromagnetic interactions.
  • Allows us to understand the balance and proportionality between the two fields.
By understanding this proportional relationship, we gain insight into how electromagnetic waves transmit energy and interact with the environment.
Intensity of Electromagnetic Waves
Intensity is a measure of the power per unit area carried by a wave. It specifies how much energy is transmitted or distributed over a particular area, providing insight into the wave's strength and impact.

In this exercise, the intensity \(I\) of our electromagnetic beam was calculated using:

\[I = \frac{P}{A}\]

Where:
  • \(P\) is the power of the beam, representing the total energy delivered per second.
  • \(A\) is the cross-sectional area of the beam, indicating how widely the energy is spread.
This calculation shows how energy is distributed across space, allowing us to infer the impact of the beam at various distances. Understanding wave intensity is crucial for multiple applications, such as telecommunications and energy technologies, as it directly influences how waves interact with physical materials and systems.

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Most popular questions from this chapter

Obtain the wavelengths in vacuum for (a) blue light whose frequency is \(6.34 \times 10^{14} \mathrm{Hz},\) and (b) orange light whose frequency is \(4.95 \times 10^{14} \mathrm{Hz}\) Express your answers in nanometers ( \(1 \mathrm{nm}=10^{-9} \mathrm{m}\) ).

A heat lamp emits infrared radiation whose rms electric field is \(E_{\mathrm{rm}}=2800 \mathrm{N} / \mathrm{C}\) (a) What is the average intensity of the radiation? (b) The radiation is focused on a person's leg over a circular area of radius \(4.0 \mathrm{cm} .\) What is the average power delivered to the leg? (c) The portion of the leg being irradiated has a mass of \(0.28 \mathrm{kg}\) and a specific heat capacity of \(3500 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) How long does it take to raise its temperature by \(2.0 \mathrm{C}^{\circ}\) ? Assume that there is no other heat transfer into or out of the portion of the leg being heated.

A distant galaxy is simultaneously rotating and receding from the earth. As the drawing shows, the galactic center is receding from the earth at a relative speed of \(u_{\mathrm{G}}=1.6 \times 10^{6} \mathrm{m} / \mathrm{s} .\) Relative to the center, the tangential speed is \(v_{\mathrm{T}}=0.4 \times 10^{6} \mathrm{m} / \mathrm{s}\) for locations \(A\) and \(B,\) which are equidistant from the center. When the frequencies of the light coming from regions \(A\) and \(B\) are measured on earth, they are not the same and each is different from the emitted frequency of \(6.200 \times 10^{14} \mathrm{Hz}\). Find the measured frequency for the light from (a) region \(A\) and (b) region \(B\).

The magnitude of the electric field of an electromagnetic wave increases from 315 to \(945 \mathrm{N} / \mathrm{C}\). (a) Determine the wave intensities for the two values of the electric field. (b) What is the magnitude of the magnetic field associated with each electric field? (c) Determine the wave intensity for each value of the magnetic field.

The drawing shows light incident on a polarizer whose transmission axis is parallel to the \(z\) axis. The polarizer is rotated clockwise through an angle \(\alpha .\) The average intensity of the incident light is \(7.0 \mathrm{W} / \mathrm{m}^{2}\). Determine the average intensity of the transmitted light for each of the six cases shown in the table.

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