### Current Attempt in Progress**Problem Statement:**In a certain UHF radio wave, the shortest distance between positions where the electric and magnetic fields are zero is 0.31 m. Determine the frequency of this radio wave.**Diagram Explanation:**The diagram depicts a segment of a UHF radio wave in which both the electric field (E) and the magnetic field (B) are plotted as sinusoidal waves. The horizontal axis represents the position along the wave, while the vertical axis represents the magnitude of the fields.- **E** (red sinusoidal wave): The electric field.- **B** (blue sinusoidal wave): The magnetic field. The distance, \( d \), is labeled on the graph, representing the shortest distance between points where both fields are zero (i.e., the nodes of the wave).### Calculate the Frequency:The shortest distance between the points where the fields are zero corresponds to half the wavelength (\(\lambda/2\)):\[ d = \frac{\lambda}{2} \]Given:\[ d = 0.31 \, \text{m} \]Thus:\[ \lambda = 2d = 2 \times 0.31 \, \text{m} = 0.62 \, \text{m} \]The speed of light (\(c\)) in a vacuum is approximately:\[ c = 3 \times 10^8 \, \text{m/s} \]Frequency (\( f \)) is related to wavelength (\(\lambda\)) and speed (\(c\)) by the equation:\[ f = \frac{c}{\lambda} \]Substitute the given values:\[ f = \frac{3 \times 10^8 \, \text{m/s}}{0.62 \, \text{m}} \]\[ f ≈ 4.84 \times 10^8 \, \text{Hz} \]Thus, the frequency of the radio wave is approximately:\[ 4.84 \times 10^8 \, \text{Hz} \]**Input Fields:**- **Number**: [input field]- **Units**: [dropdown selection]**Hint**: Use the relationship between wavelength and frequency provided by the speed of light.---This transcription is designed to educate users on how to determine the frequency of a radio wave given certain parameters,

Answered: Image /qna-images/answer/c6c7fde2-60a1-436b-840d-819f6f244169.jpg

Answered: In a certain UHF radio wave, the…

In a certain UHF radio wave, the shortest distance between positions where the electric and magnetic fields are zero is 0.31 m. Determine the frequency of this radio wave.

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter22: Magnetic Forces And Magnetic Fields

Section: Chapter Questions

Problem 7OQ

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### Current Attempt in Progress

**Problem Statement:**
In a certain UHF radio wave, the shortest distance between positions where the electric and magnetic fields are zero is 0.31 m. Determine the frequency of this radio wave.

**Diagram Explanation:**
The diagram depicts a segment of a UHF radio wave in which both the electric field (E) and the magnetic field (B) are plotted as sinusoidal waves. The horizontal axis represents the position along the wave, while the vertical axis represents the magnitude of the fields.

- **E** (red sinusoidal wave): The electric field.
- **B** (blue sinusoidal wave): The magnetic field.

The distance, \( d \), is labeled on the graph, representing the shortest distance between points where both fields are zero (i.e., the nodes of the wave).

### Calculate the Frequency:

The shortest distance between the points where the fields are zero corresponds to half the wavelength (\(\lambda/2\)):

\[ d = \frac{\lambda}{2} \]

Given:
\[ d = 0.31 \, \text{m} \]

Thus:
\[ \lambda = 2d = 2 \times 0.31 \, \text{m} = 0.62 \, \text{m} \]

The speed of light (\(c\)) in a vacuum is approximately:

\[ c = 3 \times 10^8 \, \text{m/s} \]

Frequency (\( f \)) is related to wavelength (\(\lambda\)) and speed (\(c\)) by the equation:

\[ f = \frac{c}{\lambda} \]

Substitute the given values:

\[ f = \frac{3 \times 10^8 \, \text{m/s}}{0.62 \, \text{m}} \]

\[ f ≈ 4.84 \times 10^8 \, \text{Hz} \]

Thus, the frequency of the radio wave is approximately:

\[ 4.84 \times 10^8 \, \text{Hz} \]

**Input Fields:**

- **Number**: [input field]
- **Units**: [dropdown selection]

**Hint**: Use the relationship between wavelength and frequency provided by the speed of light.

---

This transcription is designed to educate users on how to determine the frequency of a radio wave given certain parameters,

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