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In a certain experiment, a radio transmitter emits sinusoidal electromagnetic waves of frequency 110.0 MHz in opposite directions inside a narrow cavity with reflectors at both ends, causing a standing-wave pattern to occur. (a) How far apart are the nodal planes of the magnetic field? (b) If the standing- wave pattern is determined to be in its eighth harmonic, how long is the cavity?

Short Answer

Expert verified
The nodal planes are \(1.36\,\text{m}\) apart, and the cavity is \(10.9\,\text{m}\) long.

Step by step solution

01

Understand the Problem

We need to calculate two things: the distance between the nodal planes for magnetic waves in a standing wave pattern, and the length of the cavity given the standing wave pattern is in its eighth harmonic.
02

Calculate the Wavelength

The frequency of the waves is given as 110.0 MHz. We start by calculating the wavelength using the formula \( \lambda = \frac{c}{f} \), where \( c = 3 \times 10^8 \) m/s is the speed of light, and \( f = 110.0 \times 10^6 \) Hz.
03

Calculate the Distance Between Nodal Planes

For standing waves, the distance between nodal planes (nodes) is half the wavelength. So, calculate \( \frac{\lambda}{2} \) using the wavelength obtained in Step 2.
04

Identify Harmonic Relationship

For the eighth harmonic in a standing wave, the length of the cavity \( L \) is \( \frac{8\lambda}{2} \), as each harmonic adds a node.
05

Calculate the Cavity Length

Using the relation found in Step 4, substitute the wavelength from Step 2 to find the length of the cavity as \( L = 8 \times \frac{\lambda}{2} \).

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Key Concepts

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

Standing Wave
Standing waves are fascinating patterns that form when waves travel in opposite directions within a confined space, such as a cavity with reflective ends. These waves appear to be stationary or "standing," rather than moving through the medium. This phenomenon occurs due to the interference of incoming and reflected waves.

Standing waves have specific points called nodes and antinodes. Nodes are points where the amplitude is always zero, while antinodes are points with maximum amplitude. In an electromagnetic standing wave, nodes can refer to magnetic or electric fields, depending on the context of the problem. Knowing where these nodes lie helps in determining properties such as the length of the medium hosting the standing wave pattern.

Standing waves are a key concept in understanding how wave patterns interact within bounded surfaces, making them essential in fields like acoustics and electromagnetism.
Frequency
Frequency is the measure of how many wave cycles pass a given point in space per unit time. It is denoted by the symbol \( f \) and is measured in hertz (Hz). In the context of electromagnetic waves, the frequency determines the energy and the type of wave, such as radio waves and microwaves.

Higher frequencies mean more cycles in a given time frame. This concept is fundamental in understanding how waves like the sinusoidal electromagnetic waves in our problem, which have a frequency of 110.0 MHz, behave.

The relationship between frequency \( f \) and wavelength \( \lambda \) is given by the equation \( \lambda = \frac{c}{f} \), where \( c \) is the speed of light, a constant \( 3 \times 10^8 \) m/s. This equation is vital for determining one property if the other is known, linking frequency to other wave characteristics.
Harmonics
Harmonics are integral multiples of the fundamental frequency or the simplest mode of vibration in a standing wave. Each harmonic corresponds to a unique standing wave pattern, characterized by a series of nodes and antinodes. The first harmonic is the fundamental frequency itself, with the simplest standing wave pattern, while subsequent harmonics have more nodes and higher frequencies.

The eighth harmonic, as in our exercise, means that the frequency of the standing wave is eight times that of the fundamental frequency. - Each harmonic corresponds to a specific standing wave pattern. - More nodes and antinodes appear as the harmonic order increases. - Higher harmonics reflect higher frequencies and complex waveforms.

Understanding harmonics is crucial for analyzing musical instruments, the transmission of sound waves, and specific vibrations in confined spaces, helping to determine characteristics like the cavity length for different harmonics.
Wavelength Calculation
Calculating wavelength is fundamental when analyzing wave motion. It involves using the relationship between the speed of the wave, its frequency, and wavelength itself. Wavelength (\( \lambda \)) is calculated using the formula \( \lambda = \frac{c}{f} \).

In our example, with a frequency of 110.0 MHz: 1. Convert MHz to Hz: \( f = 110.0 \times 10^6 \) Hz. 2. Use the speed of light \( c = 3 \times 10^8 \) m/s in the formula. 3. Calculate \( \lambda \) by substituting in the values: \( \lambda = \frac{3 \times 10^8}{110.0 \times 10^6} \) meters.

This computation provides the wavelength of electromagnetic waves in our scenario. Breaking down these calculations helps ascertain essential wave properties, such as node placement or harmonics within standing waves. Understanding these techniques support broader applications in fields like telecommunications and physics.

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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, \(not\) 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 kg/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\(^{26}\) 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).]

An electromagnetic wave with frequency 5.70 \(\times\) 10\(^{14}\) Hz propagates with a speed of 2.17 \(\times\) 10\(^8\) m/s in a certain piece of glass. Find (a) the wavelength of the wave in the glass; (b) the wavelength of a wave of the same frequency propagating in air; (c) the index of refraction \(n\) of the glass for an electromagnetic wave with this frequency; (d) the dielectric constant for glass at this frequency, assuming that the relative permeability is unity.

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 electricfield vector \(\vec{E}\) 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 \(\vec{B}\) at the same point? (c) What are the magnitude and direction of the Poynting vector \(\vec{S}\) 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 current-carrying conductor, due to its resistance, can be thought of as entering through the cylindrical sides of the conductor.

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

Electromagnetic waves propagate much differently in conductors than they do in dielectrics or in vacuum. If the resistivity of the conductor is sufficiently low (that is, if it is a sufficiently good conductor), the oscillating electric field of the wave gives rise to an oscillating conduction current that is much larger than the displacement current. In this case, the wave equation for an electric field \(\vec{E} (x, t) = E_y(x, t)\hat{\jmath}\) en propagating in the +\(x\)-direction within a conductor is $${\partial^2E_y(x, t)\over \partial x^2} = {\mu \over \rho} {\partial Ey(x, t)\over \partial t}$$ where \(\mu\) is the permeability of the conductor and \(\rho\) is its resistivity. (a) A solution to this wave equation is \(E_y(x, t) = E_{max} e^{-k_C x} cos(k_Cx - \omega t)\), where \(k_C = \sqrt{(\omega \mu/2\rho}\). Verify this by substituting E_y(x, t) into the above wave equation. (b) The exponential term shows that the electric field decreases in amplitude as it propagates. Explain why this happens. (\(Hint\): The field does work to move charges within the conductor. The current of these moving charges causes \(i^2R\) heating within the conductor, raising its temperature. Where does the energy to do this come from?) (c) Show that the electric-field amplitude decreases by a factor of 1/\(e\) in a distance \(1/k_C = \sqrt{2\rho/\omega\mu}\), and calculate this distance for a radio wave with frequency \(f\) = 1.0 MHz in copper (resistivity 1.72 \(\times\) 10\(^{-8 } \Omega \bullet m\); permeability \(\mu = \mu_0\)). Since this distance is so short, electromagnetic waves of this frequency can hardly propagate at all into copper. Instead, they are reflected at the surface of the metal. This is why radio waves cannot penetrate through copper or other metals, and why radio reception is poor inside a metal structure.

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