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Chapter 24: Problem 6

A flat coil of wire is used with an \(L C\) -tuned circuit as a receiving antenna. The coil has a radius of \(0.25 \mathrm{m}\) and consists of 450 turns. The transmitted radio wave has a frequency of \(1.2 \mathrm{MHz}\). The magnetic field of the wave is parallel to the normal to the coil and has a maximum value of \(2.0 \times 10^{-13} \mathrm{T}\). Using Faraday's law of electromagnetic induction and the fact that the magnetic field changes from zero to its maximum value in one-quarter of a wave period, find the magnitude of the average emf induced in the antenna during this time.

Short Answer

Expert verified
The magnitude of the average induced emf is approximately \(8.49 \times 10^{-5} \text{ V}\).

Step by step solution

01

Calculate Frequency in Hertz

The frequency of the radio wave is given as \(1.2 \text{ MHz}\), which means it needs to be converted to Hertz. \[1.2 \text{ MHz} = 1.2 \times 10^6 \text{ Hz}\]
02

Determine the Wave Period

The period \(T\) of the wave is the reciprocal of its frequency. \[T = \frac{1}{1.2 \times 10^6 \text{ Hz}} \approx 8.33 \times 10^{-7} \text{ s}\]
03

Compute Time for Magnetic Field Change

The magnetic field changes from zero to its maximum value in one-quarter of a wave period.\[\text{Time} = \frac{T}{4} = \frac{8.33 \times 10^{-7} \text{ s}}{4} \approx 2.08 \times 10^{-7} \text{ s}\]
04

Calculate Coil's Cross-sectional Area

The area \( A \) of the coil is given by the formula for the area of a circle. With a radius \( r = 0.25 \text{ m} \), the area is:\[A = \pi r^2 = \pi (0.25)^2 \approx 0.196 \text{ m}^2\]
05

Use Faraday's Law of Induction

Faraday's law states that the emf \( \varepsilon \) induced is the change in magnetic flux \( \Delta \Phi \) over time \( \Delta t \):\[\varepsilon = -N \frac{\Delta \Phi}{\Delta t}\]Where \( N \) is the number of turns (450), and \( \Delta \Phi = B_{\text{max}} \cdot A \) is the change in magnetic flux. With \( B_{\text{max}} = 2.0 \times 10^{-13} \text{ T} \):\[\Delta \Phi = 2.0 \times 10^{-13} \times 0.196 \approx 3.92 \times 10^{-14} \text{ T m}^2\]Now calculate the emf:\[\varepsilon = -450 \frac{3.92 \times 10^{-14}}{2.08 \times 10^{-7}} \approx -8.49 \times 10^{-5} \text{ V}\]
06

Present the Magnitude of Induced EMF

Since we are asked for the magnitude, disregard the negative sign which indicates direction. Thus, the magnitude of the average induced emf is:\[\varepsilon \approx 8.49 \times 10^{-5} \text{ V}\]

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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 a type of wave that can travel through the vacuum of outer space. They are formed when an electric field couples with a magnetic field. These two fields oscillate perpendicular to each other and perpendicular to the direction of energy propagation. Electromagnetic waves include a variety of waves from radio waves to gamma rays, all of which travel at the speed of light in a vacuum. Understanding electromagnetic waves is vital because they form the basis for wireless communication, including radio waves, which are forms of low-frequency electromagnetic waves used in broadcasting and communication systems. Knowing the properties of these waves helps in designing and understanding antennas and receivers that detect and translate these waves into usable signals.
Radio Frequency
Radio frequency (RF) refers to the range of electromagnetic frequencies between 3 kHz and 300 GHz. RF is associated with radio wave transmission and is used in various applications such as broadcasting, megahertz-based communication devices, and radar. At the heart of radio frequency lies the fundamental principle that these waves can induce an electromagnetic signal in a receiving antenna, such as the one described in the exercise. These frequencies are used because they can travel long distances and penetrate materials, making them ideal for communication. In the exercise, the radio frequency was given in megahertz (MHz), specifically 1.2 MHz, illustrating how these frequencies are practical in everyday technologies. The speed and oscillation of these frequencies facilitate the broadcasting and reception of signals over vast distances.
Magnetic Flux
Magnetic flux, symbolized as \( \Phi \), represents the total magnetic field passing through a specified area. Imagine it as the number of magnetic field lines crossing through a loop or surface. It is calculated using the formula: \( \Phi = B \cdot A \cdot \cos(\theta) \), where \( B \) is the magnetic field strength, \( A \) is the area the field lines pass through, and \( \theta \) is the angle between the field lines and the normal to the surface. Magnetic flux is a pivotal concept in Faraday's Law of Electromagnetic Induction, which states that a change in magnetic flux through a coil induces an electromotive force (emf) in the coil. In the exercise problem, the change in magnetic flux was calculated and used to determine the induced emf in the coil. The concept of changing magnetic flux is central to understanding how electrical energy can be generated from a moving magnetic field, a key principle behind electric generators and transformers.

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Most popular questions from this chapter

Unpolarized light whose intensity is \(1.10 \mathrm{W} / \mathrm{m}^{2}\) is incident on the polarizer in Figure \(24.20 .\) (a) What is the intensity of the light leaving the polarizer? (b) If the analyzer is set at an angle of \(\theta=75^{\circ}\) with respect to the polarizer, what is the intensity of the light that reaches the photocell?

The light beam in the figure passes through a polarizer whose transmission axis makes an angle \(\phi\) with the vertical. The beam is partially polarized and partially unpolarized, and the average intensity \(\bar{S}_{0}\) of the incident light is the sum of the average intensity \(\bar{S}_{0, \text { polar }}\) of the polarized light and the average intensity \(S_{0, \text { unpolar }}\) of the unpolarized light; \(\bar{S}_{0}=S_{0, \text { polar }}+S_{0, \text { unpolar }}\) The intensity \(\bar{S}\) of the transmitted light is also the sum of two parts: \(\bar{S}=\bar{S}_{\text {pobar }}+\bar{S}_{\text {unpola }}\). As the polarizer is rotated clockwise, the intensity of the transmitted light has a minimum value of \(\bar{S}=2.0 \mathrm{W} / \mathrm{m}^{2}\) when \(\phi=20.0^{\circ}\) and has a maximum value of \(\bar{S}=8.0 \mathrm{W} / \mathrm{m}^{2}\) when the angle is \(\phi=\phi_{\max } .\) Concepts: (i) How is \(\bar{S}_{\text {unpolar }}\) related to \(\bar{S}_{0, \text { unpolur }} ?\) (ii) How is \(\bar{S}_{\text {polur }}\) related to \(\bar{S}_{0, \text { polar }}\) ? (iii) The minimum transmitted intensity is \(2.0 \mathrm{W} / \mathrm{m}^{2} .\) Why isn't it \(0 \mathrm{W} / \mathrm{m}^{2}\) ? Calculations: (a) What is the intensity \(\bar{S}_{0, \text { unpolar }}\) of the incident light that is unpolarized? (b) What is the intensity \(\bar{S}_{0, \text { polur }}\) of the incident light that is polarized?

Multiple-Concept Example 5 discusses the principles used in this problem. A neodymium-glass laser emits short pulses of high-intensity electromagnetic waves. The electric field of such a wave has an rms value of \(E_{\mathrm{rms}}=2.0 \times 10^{9} \mathrm{N} / \mathrm{C} .\) Find the average power of each pulse that passes through a \(1.6 \times 10^{-5}-\mathrm{m}^{2}\) surface that is perpendicular to the laser beam.

(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is \(3.85 \times 10^{8} \mathrm{m}\). Find the time it took for his voice to reach the earth via radio waves. (b) Someday a person will walk on Mars, which is \(5.6 \times 10^{10} \mathrm{m}\) from the earth at the point of closest approach. Determine the minimum time that will be required for a message from Mars to reach the earth via radio waves.

In a dentist's office an X-ray of a tooth is taken using X-rays that have a frequency of \(6.05 \times 10^{18} \mathrm{Hz} .\) What is the wavelength in vacuum of these X-rays?
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