/*! 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 49 A circular loop of wire can be u... [FREE SOLUTION] | 91影视

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Chapter 32: Problem 49

A circular loop of wire can be used us a radio antenna. If a 18.0-cm-diameter antenna is located 2.50 \(\mathrm{km}\) from a 95.0 -MHz source with a total power of 55.0 \(\mathrm{kW}\) , what is the maximum emf induced in the loop? (Assume that the plane of the antenna loop is perpendicular to the direction of the radiation's magnetic field and that the source radiates uniformly in all directions.)

Short Answer

Expert verified
The maximum induced emf in the loop is approximately 11.6 mV.

Step by step solution

01

Calculate the Intensity of the Electromagnetic Wave

The intensity \( I \) can be calculated using the formula \( I = \frac{P}{A} \), where \( P \) is the total power and \( A \) is the area of a sphere with radius equal to the distance from the source, given by \( 2.50 \, \mathrm{km} = 2500 \, \mathrm{m} \). Thus, the area \( A = 4 \pi r^2 \). Substituting the values we get: \[ A = 4 \pi (2500)^2 \, \mathrm{m}^2 \approx 7.85 \times 10^7 \, \mathrm{m}^2 \] \[ I = \frac{55.0 \times 10^3 \, \mathrm{W}}{7.85 \times 10^7 \, \mathrm{m}^2} \approx 0.7 \, \mathrm{W/m}^2 \]
02

Determine the Electric Field Amplitude

The electric field amplitude \( E_0 \) can be related to the intensity \( I \) using the formula \( I = \frac{c \varepsilon_0}{2} E_0^2 \). Rewriting this, we have:\[ E_0 = \sqrt{\frac{2I}{c\varepsilon_0}} \] Substitute in the known constants, speed of light \( c = 3 \times 10^8 \, \mathrm{m/s} \), and permittivity \( \varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C}^2/\mathrm{N} \, \mathrm{m}^2 \), and the intensity calculated earlier, \( I = 0.7 \, \mathrm{W/m}^2 \):\[ E_0 = \sqrt{\frac{2 \times 0.7}{3 \times 10^8 \times 8.85 \times 10^{-12}}} \approx 22.94 \, \mathrm{V/m} \]
03

Calculate the Magnetic Field Amplitude

Using the relationship between electric and magnetic fields in electromagnetic waves, \( B_0 = \frac{E_0}{c} \):\[ B_0 = \frac{22.94}{3 \times 10^8} \approx 7.65 \times 10^{-8} \, \mathrm{T} \]
04

Determine the Maximum Induced EMF in the Loop

The maximum emf \( \mathcal{E} \) induced in the loop can be found using Faraday's law of induction: \( \mathcal{E} = NAB \omega \), where \( N \) is the number of turns (assume 1), \( A \) is the area of the loop, \( B \) is the magnetic field amplitude, and \( \omega = 2\pi f \) is the angular frequency. Given, diameter = 18.0 cm, so radius \( r = 0.09 \, \mathrm{m} \), area of the loop \( A = \pi r^2 \):\[ A = \pi (0.09)^2 \approx 0.0254 \, \mathrm{m}^2 \] Frequency \( f = 95.0 \times 10^6 \, \mathrm{Hz} \), so \( \omega = 2\pi \times 95.0 \times 10^6 \approx 5.97 \times 10^8 \, \mathrm{rad/s} \). Finally, compute \( \mathcal{E} \):\[ \mathcal{E} = 1 \times 0.0254 \times 7.65 \times 10^{-8} \times 5.97 \times 10^8 \approx 1.16 \times 10^{-2} \, \mathrm{V} \]
05

Conclusion

The maximum induced emf in the antenna loop is approximately 11.6 millivolts.

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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 travel through space carrying energy through electromagnetic fields. These waves are composed of oscillating electric and magnetic fields that are perpendicular to one another and the direction of travel.
Electromagnetic waves include a wide range of frequencies, from radio waves, which have long wavelengths, to gamma rays, which have very short wavelengths. In this exercise, the focus is on radio waves.
Radio waves are used in applications such as broadcasting, satellites, and radar. They are a type of electromagnetic radiation that can be described by their frequency (given in the problem as 95.0 MHz) and their speed (the speed of light). By calculating the electromagnetic wave intensity, we can understand how much power is being transferred through a certain area from the transmitting antenna to the receiving antenna, such as the loop mentioned in the exercise.
Faraday's Law of Induction
Faraday's Law of Induction is a fundamental principle that explains how a changing magnetic field can create an electromotive force (emf) or voltage in a conductor. This is the main concept used in determining the emf induced in the antenna loop.
According to Faraday's Law, the induced emf is given by the change in magnetic flux through the loop. The magnetic flux is related to the magnetic field strength and the area it penetrates. In simpler terms, it shows how electricity can be generated from a moving magnetic field.
  • This principle is widely used in electric generators and transformers.
  • It also underlies the operation of many electrical devices and is an essential part of understanding electricity and magnetism.
  • In the given problem, Faraday's Law helps calculate the emf based on the loop area, magnetic field amplitude, and the wave frequency.
Electric Field Amplitude
The electric field amplitude, denoted as \( E_0 \), is a measure of the strength of the electric field in an electromagnetic wave. It represents the maximum value of the electric field's oscillation strength.
In the context of the provided exercise, the electric field amplitude is calculated using the intensity of the electromagnetic wave. The formula \( I = \frac{c \varepsilon_0}{2} E_0^2 \) relates intensity, the speed of light \( c \), and the permittivity of free space \( \varepsilon_0 \). By rearranging this equation, we can solve for \( E_0 \).
Understanding electric field amplitude is crucial, as it influences the energy carried by the wave and its interactions with objects such as the antenna loop. The higher the amplitude, the stronger the electric field, and the greater its potential to induce currents or forces in conductive materials.
  • This concept is vital for various applications, including radio communication and electromagnetic field analysis.
Magnetic Field Amplitude
The magnetic field amplitude, denoted as \( B_0 \), is another vital component of electromagnetic waves, indicating the maximum strength of the magnetic field in the wave.
In our problem, the relationship \( B_0 = \frac{E_0}{c} \) ties together the electric field amplitude and the magnetic field amplitude. This equation highlights how both fields are interlinked, with each affecting the other in a propagating wave.
A thorough understanding of magnetic field amplitude is important, especially when calculating effects like induced emf in loops or coils, as seen with our radio antenna exercise.
  • Magnetic field amplitude impacts how much and how effectively an electromagnetic wave can interact with conductive materials.
  • It's a valuable measure in applications such as magnetometry and electromagnetic compatibility testing.

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Most popular questions from this chapter

Interplanetary space contains many small particles referred to as interplanetary dust. Radiation pressure from the sun sets a lower limit on the size of such dust particles. To see the origin of this limit, consider a spherical dust particle of radius \(R\) and mass density \(\rho\) (a) Write an expression for the gravitational force exerted on this particle by the sun (mass \(M )\) when the particle is a distance \(r\) from the sun. (b) Let \(L\) represent the luminosity of the sun, equal to the rate at which it emits energy in electromagnetic radiation. Find the force exerted on the (totally absorbing) particle due to solar radiation pressure, remembering that the intensity of the sun's radiation also depends on the distance \(r .\) The relevant area is the cross-sectional area of the particle, \(n o t\) the total surface area of the particle. As part of your answer, explain why this is so. (c) The mass density of a typical interplanetary dust particle is about 3000 \(\mathrm{kg} / \mathrm{m}^{3}\) . Find the particle radius \(R\) such that the gravitational and radiation forces acting on the particle are equal in magnitude. The luminosity of the sun is \(3.9 \times 10^{26} \mathrm{W}\) . Does your answer depend on the distance of the particle from the sun? Why or why not? (d) Explain why dust particles with a radius less than that found in part (c) are unlikely to be found in the solar system. [Hint: Construct the ratio of the two force expressions found in parts (a) and (b).]

A simusoidal electromagnetic wave emitted by a cellular phone has a wavelength of 35.4 \(\mathrm{cm}\) and an electric-field amplitude of \(5.40 \times 10^{-2} \mathrm{V} / \mathrm{m}\) at a distance of 250 \(\mathrm{m}\) from the antenna. Calculate (a) the frequency of the wave; (b) the magnetic-field amplitude; (c) the intensity of the wave.

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 so harmful to the skin and is necessary for the production of vitamin D. UVB, with a wavelength between 280 \(\mathrm{nm}\) and 320 \(\mathrm{mm}\) , 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?

In the \(25-\) fil Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of intensity 2500 \(\mathrm{W} / \mathrm{m}^{2}\) at the floor of the facility. (This simulates the intensity of sunlight near the planet Venus.) Find the average radiation pressure (in pascals and in atmospheres) on (a) a totally absorbing section of the floor and (b) a totally reflecting section of the floor. (c) Find the average momentum density (momentum per unit volume) in the light at the floor.
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