The electric field part of an electromagnetic wave in a medium is represented by
`E_(x)=0,`
`E_(y)=2.5(N)/(C) cos[(2pixx10^(6)(rad)/(m))t-(pixx10^(-2)(rad)/(s))x]`
`E_(z)=0`.
The wave is

Options

A. Moving along –x direction with frequency `10^(6)` Hz and wavelength 200m
B. Moving along y direction with frequency `2pi xx10^(6)` Hz and wavelength 200m
C. Moving along x direction with frequency `10^(6)` Hz and wavelength 100m
D. Moving along x direction with frequency `10^(6)` Hz and wavelength 200m

Explanation

To solve the problem, we need to analyze the given electric field of the electromagnetic wave and determine its properties, including the direction of propagation, frequency, and wavelength.

### Step-by-Step Solution:

1. **Identify the Electric Field Components**:
The electric field components are given as:
- \( E_x = 0 \)
- \( E_y = 2.5 \, \text{N/C} \cos\left[(2\pi \times 10^6 \, \text{rad/m})t - \left(\pi \times 10^{-2} \, \text{rad/s}\right)x\right] \)
- \( E_z = 0 \)

Here, the electric field is only in the y-direction.

2. **Determine the Direction of Propagation**:
Since the electric field is in the y-direction and the wave is propagating in the x-direction (as indicated by the argument of the cosine function), we conclude that the wave is moving along the positive x-axis.

3. **Extract the Wavenumber \( k \)**:
The wavenumber \( k \) is given by the coefficient of \( x \) in the cosine argument:
\[ k = \pi \times 10^{-2} \, \text{rad/m} \]

4. **Calculate the Wavelength \( \lambda \)**:
The relationship between wavenumber \( k \) and wavelength \( \lambda \) is given by:
\[ k = \frac{2\pi}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{\pi \times 10^{-2}} = 200 \, \text{m} \]

5. **Extract the Angular Frequency \( \omega \)**:
The angular frequency \( \omega \) is given by the coefficient of \( t \) in the cosine argument:
\[ \omega = 2\pi \times 10^6 \, \text{rad/s} \]

6. **Calculate the Frequency \( f \)**:
The relationship between angular frequency \( \omega \) and frequency \( f \) is:
\[ \omega = 2\pi f \]
Rearranging gives:
\[ f = \frac{\omega}{2\pi} = \frac{2\pi \times 10^6}{2\pi} = 10^6 \, \text{Hz} \]

7. **Final Conclusion**:
The wave is propagating along the positive x-direction with a frequency of \( 10^6 \, \text{Hz} \) and a wavelength of \( 200 \, \text{m} \).

### Summary:
- **Direction of Propagation**: Positive x-direction
- **Frequency**: \( 10^6 \, \text{Hz} \)
- **Wavelength**: \( 200 \, \text{m} \)