The electric field of an electromagnetic wave is given by `E = (50 N C ^(-1)) sin omega (t - x//c).`
The energy contained in a cylinder of cross section `10 cm ^(2) and ` length l /10 along the x-axis is `5.5 xx 10 ^(-12)J.` The vlaue of l is

Options

A. 3
B. 9
C. 4
D. 5

Explanation

To solve the problem, we need to find the value of \( l \) given the energy contained in a cylinder with a specified electric field. Here’s a step-by-step solution:

### Step 1: Identify the given parameters
- Electric field \( E_0 = 50 \, \text{N/C} \)
- Cross-sectional area \( A = 10 \, \text{cm}^2 = 10 \times 10^{-4} \, \text{m}^2 = 10^{-3} \, \text{m}^2 \)
- Energy \( U = 5.5 \times 10^{-12} \, \text{J} \)
- Length of the cylinder \( L = \frac{l}{10} \)

### Step 2: Calculate the average energy density
The average energy density \( u \) of an electromagnetic wave is given by:
\[ u = \frac{1}{2} \epsilon_0 E_0^2 \]
where \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \) (permittivity of free space).

### Step 3: Substitute the values into the energy density formula
Substituting the values into the formula:
\[ u = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (50)^2 \]
Calculating \( 50^2 = 2500 \):
\[ u = \frac{1}{2} \times (8.85 \times 10^{-12}) \times 2500 \]
\[ u = \frac{1}{2} \times 2.2125 \times 10^{-8} = 1.10625 \times 10^{-8} \, \text{J/m}^3 \]

### Step 4: Relate energy to volume
The total energy \( U \) contained in the cylinder can be expressed as:
\[ U = u \times V \]
where \( V \) is the volume of the cylinder given by \( V = A \times L \).

### Step 5: Substitute the volume into the energy equation
Substituting \( V = A \times \frac{l}{10} \):
\[ U = u \times A \times \frac{l}{10} \]
Substituting the known values:
\[ 5.5 \times 10^{-12} = (1.10625 \times 10^{-8}) \times (10^{-3}) \times \frac{l}{10} \]

### Step 6: Solve for \( l \)
Rearranging the equation to solve for \( l \):
\[ 5.5 \times 10^{-12} = 1.10625 \times 10^{-11} \times \frac{l}{10} \]
Multiplying both sides by 10:
\[ 5.5 \times 10^{-11} = 1.10625 \times 10^{-11} l \]
Dividing both sides by \( 1.10625 \times 10^{-11} \):
\[ l = \frac{5.5 \times 10^{-11}}{1.10625 \times 10^{-11}} \approx 4.97 \, \text{m} \]

### Step 7: Final answer
Rounding to the nearest integer, we find:
\[ l \approx 5 \, \text{m} \]

### Conclusion
The value of \( l \) is approximately \( 5 \, \text{m} \).

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