A radio station on the surface of the earth radiates 50 kW . If transmitter radiates equally in all directions avove the surface of the earth . Find the amplitude of electric field detected 100 km away :

Options

A. `2.45 Vm^(-1)`
B. `2.45 xx10^(-1)Vm^(-1)`
C. `2.45xx10^(-2) Vm^(-1)`
D. `2.45xx10^(-3)Vm^(-1)`

Explanation

To find the amplitude of the electric field detected 100 km away from a radio station radiating 50 kW of power equally in all directions, we can follow these steps:

### Step 1: Understand the relationship between power, intensity, and electric field
The intensity \( I \) of the electromagnetic wave can be defined as the power per unit area. For a point source radiating equally in all directions, the intensity at a distance \( r \) from the source is given by:

\[ I = \frac{P}{A} = \frac{P}{4\pi r^2} \]

where \( P \) is the power radiated and \( A \) is the surface area of a sphere with radius \( r \).

### Step 2: Calculate the intensity at 100 km
Given:
- Power \( P = 50 \, \text{kW} = 50 \times 10^3 \, \text{W} \)
- Distance \( r = 100 \, \text{km} = 100 \times 10^3 \, \text{m} \)

Substituting these values into the intensity formula:

\[ I = \frac{50 \times 10^3}{4\pi (100 \times 10^3)^2} \]

### Step 3: Calculate the area
The area \( A \) at distance \( r \) is:

\[ A = 4\pi (100 \times 10^3)^2 = 4\pi (10^4)^2 = 4\pi \times 10^{10} \, \text{m}^2 \]

### Step 4: Calculate the intensity
Now substituting the area back into the intensity formula:

\[ I = \frac{50 \times 10^3}{4\pi \times 10^{10}} \]

Calculating \( 4\pi \):

\[ 4\pi \approx 12.56 \]

Thus,

\[ I \approx \frac{50 \times 10^3}{12.56 \times 10^{10}} \approx \frac{50}{12.56} \times 10^{-7} \approx 3.98 \times 10^{-6} \, \text{W/m}^2 \]

### Step 5: Relate intensity to electric field amplitude
The intensity \( I \) can also be expressed in terms of the electric field amplitude \( E_0 \):

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

Where:
- \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \) (permittivity of free space)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)

### Step 6: Solve for \( E_0 \)
Rearranging the equation to solve for \( E_0 \):

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

Substituting the values:

\[ E_0 = \sqrt{\frac{2 \times 3.98 \times 10^{-6}}{8.85 \times 10^{-12} \times 3 \times 10^8}} \]

### Step 7: Calculate \( E_0 \)
Calculating the denominator:

\[ \epsilon_0 c = 8.85 \times 10^{-12} \times 3 \times 10^8 \approx 2.655 \times 10^{-3} \]

Now substituting back into the equation for \( E_0 \):

\[ E_0 = \sqrt{\frac{7.96 \times 10^{-6}}{2.655 \times 10^{-3}}} \approx \sqrt{3.00 \times 10^{-3}} \approx 0.05477 \, \text{V/m} \]

### Step 8: Final Result
Thus, the amplitude of the electric field detected 100 km away is approximately:

\[ E_0 \approx 0.05477 \, \text{V/m} \approx 5.48 \times 10^{-2} \, \text{V/m} \]