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In a certain experiment, a radio transmitter emits sinusoidal electromagnetic waves of frequency 110.0\(\mathrm { MHz }\) 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
(a) 1.364 meters between nodal planes. (b) The cavity is 10.912 meters long for the eighth harmonic.

Step by step solution

01

Understanding the Frequency and Wavelength Relationship

Since the frequency \(f\) is given as 110.0 MHz, we first need to convert this frequency to hertz: \(110.0 \times 10^6\) Hz. The speed of light \(c\) in a vacuum is approximately \(3.00 \times 10^8\) m/s. To find the wavelength \(\lambda\), use the formula \(\lambda = \frac{c}{f}\). Substitute the values: \(\lambda = \frac{3.00 \times 10^8}{110.0 \times 10^6}\) meters.
02

Calculating the Wavelength and Nodal Distance

Using the formula from Step 1, calculate the wavelength: \(\lambda = \frac{3.00 \times 10^8}{110.0 \times 10^6} = 2.727\) meters. The distance between nodal planes for a standing wave is half the wavelength, so the nodal distance is \(\frac{\lambda}{2} = \frac{2.727}{2} = 1.364\) meters.
03

Understanding Harmonics and Length of the Cavity

In a standing wave pattern described by the \(n\)-th harmonic, the length \(L\) of the cavity supports \(n\) half-wavelengths, i.e., \(L = n \times \frac{\lambda}{2}\). For the eighth harmonic \(n = 8\), substitute the value of \(\lambda\) from Step 2: \(L = 8 \times \frac{2.727}{2}\).
04

Calculating the Length of the Cavity

Perform the calculation from Step 3: \(L = 8 \times \frac{2.727}{2} = 8 \times 1.364 = 10.912\) meters. Therefore, the length of the cavity for the eighth harmonic is 10.912 meters.

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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 one of the fundamental types of waves that occur in nature. These waves are composed of oscillating electric and magnetic fields that travel together through space. They do not require a medium to propagate, meaning they can travel through a vacuum, like outer space. Some common examples include light, radio, and microwaves.

In the context of the problem, the mentioned radio waves are electromagnetic waves. They have a frequency of 110 MHz. The frequency denotes how often the wave oscillates in one second. Higher frequencies mean more oscillations and typically shorter wavelengths, as seen with waves like visible light, compared to radio waves.
  • These waves are typically described by their frequency and wavelength.
  • The speed of electromagnetic waves in a vacuum is approximately 3.00 x \(10^8\) m/s.
  • They can be reflected, refracted, and form standing waves in confined spaces like cavities.
Understanding these principles is crucial to grasping how standing wave patterns form in experiments and nature.
Harmonics
Harmonics play a key role in the study of waves, especially in standing wave systems. When describing waves in a bounded environment, like our narrow cavity from the exercise, harmonics come into play.

A harmonic is essentially a mode of vibration, where the system supports certain frequencies that are whole-number multiples of a fundamental frequency. In standing wave patterns, these harmonics are counted by the number of nodes, or nodal planes, present in the system.
  • A nodal plane is where the wave has zero amplitude.
  • The first harmonic is the fundamental, the lowest frequency, while other harmonics are higher frequencies.
  • For an nth harmonic, the system accommodates "n" half-wavelengths in its defined length.
In the exercise, the standing wave is stated to be in its eighth harmonic. This means the length of the cavity equals eight half-wavelengths of the radio wave, highlighting the direct relationship between harmonics and the physical dimensions they manifest.
Wavelength
Wavelength is a fundamental property of waves, describing the distance over which the wave's shape repeats. It is usually represented by the Greek letter \(\lambda\) (lambda).

In electromagnetic waves, such as the radio waves in the exercise, the wavelength is inversely proportional to its frequency, as per the equation \(\lambda = \frac{c}{f}\), where \(c\) is the speed of light. For radio waves with a frequency of 110 MHz, the calculated wavelength is approximately 2.727 meters.
  • Wavelength describes the distance between crest to crest or trough to trough.
  • Shorter wavelengths correlate with higher frequencies.
  • In standing waves, the distance between consecutive nodal planes is half the wavelength.
This relationship is pivotal for determining physical dimensions and characteristics in wave-based experiments and applications like the one described.
Nodal Planes
Nodal planes are key features in the phenomena of standing waves. When two identical waves, such as the ones bouncing between reflectors in a cavity, interfere, they form a pattern of alternating nodes and antinodes.

Nodes, or nodal lines when referring to two-dimensional patterns, are places where destructive interference occurs. Here, the waves cancel each other out completely, resulting in zero amplitude.
  • Nodal planes occur every half-wavelength apart.
  • They serve as stable regions within a standing wave where no wave activity occurs.
  • The presence of these nodal planes helps in identifying the harmonic modes in bounded systems.
Knowing the distance between nodal planes, like in the exercise, is useful to determine standing wave parameters such as the length of cavities or pipes in acoustical and electromagnetic applications.

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Most popular questions from this chapter

A plane sinusoidal electromagnetic wave in air has a wavelength of 3.84\(\mathrm { cm }\) and an \(\vec { \boldsymbol { E } }\) -field amplitude of 1.35\(\mathrm { V } / \mathrm { m }\) . (a) What is the frequency? (b) What is the \(\vec { \boldsymbol { B } }\) -field amplitude? (c) What is the intensity? (d) What average force does this radiation exert on a totally absorbing surface with area 0.240\(\mathrm { m } ^ { 2 }\) perpendicular to the direction of propagation?

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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, 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 \(\vec { E } ( x , t ) = E _ { y } ( x , t ) \hat { J }\) propagating in the \(+ x\) -direction within a conductor is $$\frac { \partial ^ { 2 } E _ { y } ( x , t ) } { \partial x ^ { 2 } } = \frac { \mu } { \rho } \frac { \partial E _ { y } ( x , t ) } { \partial t }\( where \)\mu$$ is the permeability of the conductor and \(\rho\) is its resistivity. (a) A solution to this wave equation is $$E _ { y } ( x , t ) = E _ { \max } e ^ { - k _ { c } x } \cos \left( k _ { C } x - \omega t \right)$$ where \(k _ { \mathrm { C } } = \sqrt { \omega \mu / 2 \rho } .\) 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 \(i ^ { 2 } R\) heating within the conductor, raising its temperature. Where does the energy to do this come from? \(( \mathrm { c } )\) Show that the electric-field amplitude decreases by a factor of 1\(/ e\) in a distance \(1 / k _ { \mathrm { C } } = \sqrt { 2 p / \omega \mu } ,\) and calculate this distance for a radio wave with frequency \(f = 1.0 \mathrm { MHz }\) in copper (resistivity \(1.72 \times 10 ^ { - 8 } \Omega \cdot\) 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 can- not penetrate through copper or other metals, and why radio reception is poor inside a metal structure.

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