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An electromagnetic standing wave in a certain material has frequency \(1.20 \times 10^{10} \mathrm{Hz}\) and speed of propagation \(210 \times 10^{8} \mathrm{m} / \mathrm{s}\) . (a) What is the distance between a nodal plane of \(\overrightarrow{\boldsymbol{B}}\) and the closest antinodal plane of \(\overrightarrow{\boldsymbol{B}} ?\) (b) What is the distance between an antinodal plane of \(\overrightarrow{\boldsymbol{E}}\) and the closest antinodal plane of \(\overrightarrow{\boldsymbol{B}}\) ? (c) What is the distance between a nodal plane of \(\overrightarrow{\boldsymbol{E}}\) and the closest nodal plane of \(\overrightarrow{\boldsymbol{B}} ?\)

Short Answer

Expert verified
(a) 4.375 m; (b) 8.75 m; (c) 8.75 m.

Step by step solution

01

Calculate the wavelength

The wavelength \( \lambda \) of a wave is given by the formula \( \lambda = \frac{v}{f} \), where \( v \) is the speed of the wave and \( f \) is the frequency. Here, \( v = 210 \times 10^8 \ \mathrm{m/s} \) and \( f = 1.20 \times 10^{10} \ \mathrm{Hz} \). Calculate \[ \lambda = \frac{210 \times 10^8}{1.20 \times 10^{10}} = 17.5 \ \mathrm{m} \].
02

Determine distance from nodal plane to antinodal plane for \( \overrightarrow{B} \)

For a standing wave, the distance between a node and an antinode, where the wave is at maximum and minimum amplitude, is \( \frac{\lambda}{4} \). Using the wavelength from Step 1, \( \frac{17.5}{4} = 4.375 \ \mathrm{m} \).
03

Determine distance between antinodal planes of \( \overrightarrow{E} \) and \( \overrightarrow{B} \)

The antinode of \( \overrightarrow{E} \) corresponds to a node of \( \overrightarrow{B} \). Therefore, the distance between an antinode of \( \overrightarrow{E} \) and an antinode of \( \overrightarrow{B} \) is \( \frac{\lambda}{2} \). Calculate \( \frac{17.5}{2} = 8.75 \ \mathrm{m} \).
04

Determine distance between nodal planes of \( \overrightarrow{E} \) and \( \overrightarrow{B} \)

Similar to the antinodal distance, the distance between a node of \( \overrightarrow{E} \) and a node of \( \overrightarrow{B} \) is \( \frac{\lambda}{2} \). Thus, the distance is \( 8.75 \ \mathrm{m} \).

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Key Concepts

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

Understanding Standing Waves in Electromagnetic Waves
A standing wave occurs when two waves of identical frequency and amplitude move through each other in opposite directions. In electromagnetic waves, these are typically the electric (\(\overrightarrow{E}\)) and magnetic (\(\overrightarrow{B}\)) fields. As these wavelengths interact, they form stationary patterns characterized by nodes, where wave displacement is zero, and antinodes, where wave displacement is at a maximum.
The beauty of standing waves is in their consistent patterns. You can easily identify nodes and antinodes because they remain fixed in space. Understanding how these patterns form helps in visualizing waves in materials, an important analysis for fields such as optics and radio broadcasting.
Wavelength Calculation
The wavelength (\(\lambda\)) is a fundamental property of any wave, indicating the distance over which the wave's shape repeats. It can be calculated using the formula:
\[ \lambda = \frac{v}{f} \]
where \(v\) is the speed of the wave and \(f\) is the frequency.
  • In the given exercise, the speed of the wave is \(210 \times 10^8\) m/s, and the frequency is \(1.20 \times 10^{10}\) Hz. Calculating gives us a wavelength of \(17.5\) meters.

This wavelength is vital for determining the distance between features in the wave, such as the nodes and antinodes.
Wave Interference
Wave interference is when two or more waves pass through each other, and it plays a critical role in forming standing waves. This phenomenon can be constructive or destructive, depending on whether waves align in phase or out of phase, respectively.

In constructive interference, waves align perfectly, their amplitudes combined to form larger amplitudes, seen at antinodes in standing waves. In destructive interference, the waves cancel each other out, forming nodes.

This exercise particularly focuses on the distances between these formations, using the calculated wavelength to find:
  • Distance from a node to an antinode as \(\frac{\lambda}{4}\), given as \(4.375\) meters.
  • The distance between successive nodes or antinodes as \(\frac{\lambda}{2}\), calculated as \(8.75\) meters.
Role of Electric and Magnetic Fields
In electromagnetic waves, electric (\(\overrightarrow{E}\)) and magnetic (\(\overrightarrow{B}\)) fields oscillate perpendicularly to each other and to the direction of wave propagation. They come together to influence how the wave behaves.
The interaction of \(\overrightarrow{E}\) and \(\overrightarrow{B}\) fields is crucial for determining nodal and antinodal planes in standing waves. Specifically:
  • Antinodes of one field coincide with nodes of the other field, demonstrating how tightly intertwined these fields are in a standing wave.
  • The distances between nodal and antinodal planes of \(\overrightarrow{E}\) and \(\overrightarrow{B}\) fields, like those described in the exercise, further illustrate their interplay.
Understanding these concepts helps clarify how electromagnetic waves transmit information through space, impacting everything from wireless communication to modern physics theories.

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Most popular questions from this chapter

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