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 solve the problem of finding the average power per unit area of a plane electromagnetic wave, we can follow these steps:

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

### Step 2: Calculate the speed of light
The speed of light \( c \) is a constant:
\[ c = 3 \times 10^8 \, \text{m/s} \]

### Step 3: Calculate the peak value of the magnetic field \( B_0 \)
Using the relationship between the electric field and magnetic field in an electromagnetic wave:
\[ B_0 = \frac{E_0}{c} \]
Substituting the values:
\[ B_0 = \frac{2 \times 10^{-3}}{3 \times 10^8} \]
\[ B_0 = \frac{2}{3} \times 10^{-11} \, \text{T} \]

### Step 4: Calculate the energy density \( u \) of the electromagnetic wave
The energy density \( u \) is given by the sum of the electric and magnetic energy densities:
\[ u = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0} \]
Since \( \frac{E^2}{c^2} = B^2 \), we can simplify this to:
\[ u = \epsilon_0 E^2 \]
Where \( \epsilon_0 \) (the permittivity of free space) is approximately:
\[ \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \]
Now substituting the values:
\[ u = \epsilon_0 E_0^2 \]
\[ u = 8.85 \times 10^{-12} \times (2 \times 10^{-3})^2 \]
\[ u = 8.85 \times 10^{-12} \times 4 \times 10^{-6} \]
\[ u = 35.4 \times 10^{-18} \, \text{J/m}^3 \]

### Step 5: Calculate the average power per unit area \( S \)
The average power per unit area \( S \) is given by:
\[ S = u \cdot c \]
Substituting the values:
\[ S = 35.4 \times 10^{-18} \times 3 \times 10^8 \]
\[ S = 106.2 \times 10^{-10} \]
\[ S = 1.062 \times 10^{-9} \, \text{W/m}^2 \]

### Final Answer
The average power per unit area perpendicular to the direction of propagation is:
\[ S \approx 1.06 \times 10^{-9} \, \text{W/m}^2 \]

---