In an EM wave, `E=50 sin (omega t-k x)`. If `mu=4 mu_(0)` and `varepsilon=varepsilon_(0)`, then average power per unit area (in `W/m^2` ) is

Answer

To solve the problem, we need to find the average power per unit area (intensity) of the electromagnetic wave given by the equation \( E = 50 \sin(\omega t - kx) \). The average intensity \( I \) of an electromagnetic wave can be calculated using the formula:

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

where:
- \( I \) is the average intensity (power per unit area),
- \( \epsilon_0 \) is the permittivity of free space,
- \( c \) is the speed of light in vacuum,
- \( E_0 \) is the amplitude of the electric field.

### Step 1: Identify the amplitude of the electric field
From the given equation \( E = 50 \sin(\omega t - kx) \), we can see that the amplitude \( E_0 \) is 50 V/m.

### Step 2: Calculate the speed of light \( c \)
The speed of light \( c \) in a medium is given by the formula:

\[ c = \frac{1}{\sqrt{\mu \varepsilon}} \]

Given that \( \mu = 4 \mu_0 \) and \( \varepsilon = \varepsilon_0 \), we can substitute these values into the equation:

\[ c = \frac{1}{\sqrt{4 \mu_0 \varepsilon_0}} = \frac{1}{2} \cdot \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = \frac{c_0}{2} \]

where \( c_0 \) is the speed of light in vacuum.

### Step 3: Substitute values into the intensity formula
Now we can substitute \( E_0 = 50 \, \text{V/m} \) and \( c = \frac{c_0}{2} \) into the intensity formula:

\[ I = \frac{1}{2} \epsilon_0 \left(\frac{c_0}{2}\right) E_0^2 \]

### Step 4: Simplify the expression
Substituting \( c_0 = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \):

\[ I = \frac{1}{2} \epsilon_0 \left(\frac{1}{2\sqrt{\mu_0 \varepsilon_0}}\right) E_0^2 \]

### Step 5: Calculate the average intensity
Now we can plug in the values:

1. \( E_0 = 50 \, \text{V/m} \)
2. \( \epsilon_0 \) and \( \mu_0 \) are constants, but we can use their values in the calculation.

Using \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) and \( \epsilon_0 = \frac{1}{\mu_0 c_0^2} \):

\[ I = \frac{1}{2} \cdot \epsilon_0 \cdot \frac{c_0}{2} \cdot (50)^2 \]

Calculating this gives us the average power per unit area.

### Final Calculation
After substituting the values and simplifying, we find:

\[ I = \frac{1}{2} \cdot \frac{1}{4\pi \times 10^{-7}} \cdot \frac{3 \times 10^8}{2} \cdot (50)^2 \]

Calculating this will yield the final answer for average power per unit area in \( W/m^2 \).