Suppose a 60HZ EM wave is a sinusoidal wave propagating in the z direction with E pointing in the x direction and Eo = 2.00V/m. Write the vector expressions for the electric field and magnetic field vectors as a function of position and time.

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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}\
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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