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![**Transcription for Educational Website: Electromagnetic Wave Analysis**
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### Problem Statement:
Suppose a 60Hz EM wave is a sinusoidal wave propagating in the z direction with \(\mathbf{E}\) pointing in the x direction and \(E_0 = 2.00 \text{V/m}\). Write the vector expressions for the electric field and magnetic field vectors as a function of position and time.
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This problem involves understanding the propagation of an electromagnetic (EM) wave in free space. Let's break down the given information and derive the required expressions.
#### Given Information:
1. **Frequency of EM wave (\(f\))**: 60 Hz.
2. **Propagation direction**: Along the z-axis.
3. **Electric field (\(\mathbf{E}\)) direction**: Along the x-axis.
4. **Amplitude of Electric field (\(E_0\))**: 2.00 V/m.
#### Required:
1. Vector expression for the electric field (\(\mathbf{E}\)).
2. Vector expression for the magnetic field (\(\mathbf{B}\)).
### Solution:
1. **Wave Number (\(k\))** and **Angular Frequency (\(\omega\))**:
\[
k = \frac{2\pi}{\lambda}, \quad \omega = 2\pi f
\]
Here \(\lambda\) is the wavelength, and we use the fact that the speed of light \(c = f \lambda\). For an EM wave in a vacuum: \(c \approx 3 \times 10^8 \text{ m/s}\).
\[
\lambda = \frac{c}{f} = \frac{3 \times 10^8 \text{ m/s}}{60 \text{ Hz}} = 5 \times 10^6 \text{ m}
\]
\[
k = \frac{2\pi}{5 \times 10^6 \text{ m}} \approx 1.256 \times 10^{-6} \text{ m}^{-1}
\]
\[
\omega = 2\pi \times 60 \text{ Hz} \approx 377 \text{ rad/s }
\]
2. **Electric Field (\(\mathbf{E}\))**:
The electric field \(\mathbf{E}\](https://content.bartleby.com/qna-images/question/bb3b63b4-702e-4c2c-936a-1674e5e2ea60/c26fdbf2-5476-4e1e-891d-c2d4aa9adec1/a62yjrg_processed.jpeg)
Transcribed Image Text:**Transcription for Educational Website: Electromagnetic Wave Analysis**
---
### Problem Statement:
Suppose a 60Hz EM wave is a sinusoidal wave propagating in the z direction with \(\mathbf{E}\) pointing in the x direction and \(E_0 = 2.00 \text{V/m}\). Write the vector expressions for the electric field and magnetic field vectors as a function of position and time.
---
This problem involves understanding the propagation of an electromagnetic (EM) wave in free space. Let's break down the given information and derive the required expressions.
#### Given Information:
1. **Frequency of EM wave (\(f\))**: 60 Hz.
2. **Propagation direction**: Along the z-axis.
3. **Electric field (\(\mathbf{E}\)) direction**: Along the x-axis.
4. **Amplitude of Electric field (\(E_0\))**: 2.00 V/m.
#### Required:
1. Vector expression for the electric field (\(\mathbf{E}\)).
2. Vector expression for the magnetic field (\(\mathbf{B}\)).
### Solution:
1. **Wave Number (\(k\))** and **Angular Frequency (\(\omega\))**:
\[
k = \frac{2\pi}{\lambda}, \quad \omega = 2\pi f
\]
Here \(\lambda\) is the wavelength, and we use the fact that the speed of light \(c = f \lambda\). For an EM wave in a vacuum: \(c \approx 3 \times 10^8 \text{ m/s}\).
\[
\lambda = \frac{c}{f} = \frac{3 \times 10^8 \text{ m/s}}{60 \text{ Hz}} = 5 \times 10^6 \text{ m}
\]
\[
k = \frac{2\pi}{5 \times 10^6 \text{ m}} \approx 1.256 \times 10^{-6} \text{ m}^{-1}
\]
\[
\omega = 2\pi \times 60 \text{ Hz} \approx 377 \text{ rad/s }
\]
2. **Electric Field (\(\mathbf{E}\))**:
The electric field \(\mathbf{E}\
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