Electric field intensity of an electromagnetic wave is represented as E= ( 100 N/C ) sin ` ( omega t - kx )` . Amount of energy stored in a cylindrical volume of cross section 10 `cm^2` and length 100 cm , along X -axis is given by `0.443 xx 10^(-2x )` Joules . what is the value of x ?

Explanation

To solve the problem, we need to find the value of \( x \) given the electric field intensity of an electromagnetic wave and the amount of energy stored in a cylindrical volume.

### Step-by-Step Solution:

1. **Identify the Electric Field Intensity**:
The electric field intensity \( E \) is given by:
\[ E = 100 \, \text{N/C} \cdot \sin(\omega t - kx) \]
Here, the peak value of the electric field \( E_0 = 100 \, \text{N/C} \).

2. **Calculate the Energy Density**:
The energy density \( u \) of an electromagnetic wave is given by the formula:
\[ u = \frac{1}{2} \epsilon_0 E^2 \]
where \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, \text{F/m} \).

Substituting \( E_0 \) into the equation:
\[ u = \frac{1}{2} \epsilon_0 (E_0)^2 = \frac{1}{2} \cdot 8.85 \times 10^{-12} \cdot (100)^2 \]
\[ u = \frac{1}{2} \cdot 8.85 \times 10^{-12} \cdot 10000 = 4.425 \times 10^{-8} \, \text{J/m}^3 \]

3. **Calculate the Volume of the Cylinder**:
The volume \( V \) of the cylindrical volume is given by:
\[ V = \text{Cross-sectional Area} \times \text{Length} \]
The cross-sectional area is \( 10 \, \text{cm}^2 = 10 \times 10^{-4} \, \text{m}^2 \) and the length is \( 100 \, \text{cm} = 1 \, \text{m} \).
\[ V = (10 \times 10^{-4} \, \text{m}^2) \times (1 \, \text{m}) = 10 \times 10^{-4} \, \text{m}^3 = 10^{-3} \, \text{m}^3 \]

4. **Calculate the Total Energy Stored**:
The total energy \( U \) stored in the cylindrical volume is:
\[ U = u \cdot V = (4.425 \times 10^{-8} \, \text{J/m}^3) \cdot (10^{-3} \, \text{m}^3) \]
\[ U = 4.425 \times 10^{-11} \, \text{J} \]

5. **Set Up the Equation for Comparison**:
We are given that the energy stored is also represented as:
\[ U = 0.443 \times 10^{-2x} \, \text{J} \]
Setting the two expressions for \( U \) equal to each other:
\[ 4.425 \times 10^{-11} = 0.443 \times 10^{-2x} \]

6. **Solve for \( x \)**:
Rearranging the equation:
\[ 10^{-2x} = \frac{4.425 \times 10^{-11}}{0.443} \]
Calculate the right side:
\[ 10^{-2x} \approx 10^{-10} \quad (\text{since } \frac{4.425}{0.443} \approx 10) \]
Taking logarithm on both sides:
\[ -2x = -10 \implies 2x = 10 \implies x = 5 \]

### Final Answer:
The value of \( x \) is \( 5 \).