The electric vector vibration of an electromagnetic wave is given by `E = (50NC^(-1)) sin omega(t-(x)/(c))`, The intensity of the wave is

Options

A. `2.3 Wm^(-2)`
B. `4.3 Wm^(-2)`
C. `3.3 Wm^(-2)`
D. `1.8 Wm^(-2)`

Explanation

To find the intensity of the electromagnetic wave given by the electric vector vibration \( E = (50 \, \text{N/C}) \sin(\omega(t - \frac{x}{c})) \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Maximum Electric Field (E₀)**:
From the given equation, we can identify the maximum electric field \( E_0 \):
\[ E_0 = 50 \, \text{N/C} \]

2. **Use the Intensity Formula**:
The intensity \( I \) of an electromagnetic wave can be calculated using the formula:
\[ I = \frac{1}{2} \epsilon_0 E_0^2 c \]
where:
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{F/m} \)
- \( c \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \, \text{m/s} \)

3. **Substitute the Values**:
Substitute \( E_0 \), \( \epsilon_0 \), and \( c \) into the intensity formula:
\[ I = \frac{1}{2} \times (8.85 \times 10^{-12} \, \text{F/m}) \times (50 \, \text{N/C})^2 \times (3 \times 10^8 \, \text{m/s}) \]

4. **Calculate \( E_0^2 \)**:
First, calculate \( E_0^2 \):
\[ E_0^2 = (50)^2 = 2500 \, \text{(N/C)}^2 \]

5. **Calculate the Intensity**:
Now plug this value into the intensity formula:
\[ I = \frac{1}{2} \times (8.85 \times 10^{-12}) \times 2500 \times (3 \times 10^8) \]
\[ I = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (7500 \times 10^8) \]
\[ I = \frac{1}{2} \times (66.375 \times 10^{-4}) \, \text{W/m}^2 \]
\[ I = 3.31875 \, \text{W/m}^2 \approx 3.3 \, \text{W/m}^2 \]

6. **Final Result**:
Thus, the intensity of the wave is:
\[ I \approx 3.3 \, \text{W/m}^2 \]

### Conclusion:
The intensity of the electromagnetic wave is approximately \( 3.3 \, \text{W/m}^2 \).