A plane electromagnetic wave-varies sinusoidally at 900 MHz as it travels along the positive X Direction. The peak value of the electric field is `2mv//m` and is directed`pm`direction. The average power per unit area perpendicular to the direction of propagation is

Options

A. `2W//m^(2)`
B. `10W//m^(2)`
C. `2xx10^(-9)W//m^(2)`
D. `5.31xx10^(-9)W//m^(2)`

Answer

To find the average power per unit area (intensity) of the given electromagnetic wave, we can use the formula for the average power per unit area (intensity) of an electromagnetic wave:

\[ I = \frac{1}{2} \epsilon_0 E_0^2 c \]

Where:
- \( I \) is the average power per unit area (intensity).
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{F/m} \).
- \( E_0 \) is the peak value of the electric field.
- \( c \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \, \text{m/s} \).

### Step-by-step solution:

1. **Identify the given values:**
- Frequency \( f = 900 \, \text{MHz} = 900 \times 10^6 \, \text{Hz} \)
- Peak electric field \( E_0 = 2 \, \text{mV/m} = 2 \times 10^{-3} \, \text{V/m} \)

2. **Calculate the speed of light \( c \):**
- The speed of light \( c \) is a constant value, \( c = 3 \times 10^8 \, \text{m/s} \).

3. **Substitute the values into the intensity formula:**
\[ I = \frac{1}{2} \epsilon_0 E_0^2 c \]
\[ I = \frac{1}{2} \times (8.85 \times 10^{-12} \, \text{F/m}) \times (2 \times 10^{-3} \, \text{V/m})^2 \times (3 \times 10^8 \, \text{m/s}) \]

4. **Calculate \( E_0^2 \):**
\[ E_0^2 = (2 \times 10^{-3})^2 = 4 \times 10^{-6} \, \text{V}^2/\text{m}^2 \]

5. **Plug in the values:**
\[ I = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (4 \times 10^{-6}) \times (3 \times 10^8) \]

6. **Calculate the product:**
\[ I = \frac{1}{2} \times 8.85 \times 4 \times 3 \times 10^{-12} \times 10^{-6} \times 10^8 \]
\[ = \frac{1}{2} \times 106.2 \times 10^{-10} \]
\[ = 53.1 \times 10^{-10} \, \text{W/m}^2 \]
\[ = 5.31 \times 10^{-9} \, \text{W/m}^2 \]

7. **Final result:**
The average power per unit area (intensity) is:
\[ I = 5.31 \times 10^{-9} \, \text{W/m}^2 \]

### Conclusion:
The average power per unit area perpendicular to the direction of propagation is \( 5.31 \times 10^{-9} \, \text{W/m}^2 \).