Question

If the rms value of the electric field of electromagnetic waves at a distance of 3 m from a point source is \( 3 \, \text{N C}^{-1} \), then the power of the source is

• A. 10.8 W
• B. 8.1 W
• C. 5.4 W
• D. 2.7 W

Hint:

- Intensity of EM wave: \( I = \epsilon_0 c E_{rms}^2 \). - For a point source radiating power P uniformly, intensity at distance r: \( I = \frac{P}{4\pi r^2} \). - Equate these: \( \frac{P}{4\pi r^2} = \epsilon_0 c E_{rms}^2 \). - Use \( k_e = \frac{1}{4\pi\epsilon_0} \approx 9 \times 10^9 \, \text{SI units} \). So \( 4\pi\epsilon_0 = 1/k_e \). - Speed of light \( c \approx 3 \times 10^8 \, \text{m/s} \).

Answer 1

The Correct Option is D

The intensity \(I\) of an electromagnetic wave is related to the rms value of the electric field \( E_{rms} \) by: \[ I = \frac{1}{2} \epsilon_0 c E_0^2 = \epsilon_0 c E_{rms}^2 \] where \( E_0 \) is the amplitude of the electric field (\( E_0 = \sqrt{2}E_{rms} \)), \( \epsilon_0 \) is the permittivity of free space (\( \approx 8.
854 \times 10^{-12} \, \text{F/m} \)), and \(c\) is the speed of light (\( \approx 3 \times 10^8 \, \text{m/s} \)).
A simpler form using \( \frac{1}{\mu_0 \epsilon_0} = c^2 \) and \( \frac{E_{rms}}{B_{rms}} = c \), is \( I = \frac{E_{rms}^2}{Z_0} \) where \(Z_0 = \sqrt{\mu_0/\epsilon_0} \approx 377 \Omega\).
Also \( I = \frac{E_{rms} B_{rms}}{\mu_0} = \frac{E_{rms}^2}{c\mu_0} \).
This can also be written using \(c = 1/\sqrt{\mu_0 \epsilon_0}\), so \(c\mu_0 = \sqrt{\mu_0/\epsilon_0}\).
The standard formula is \( I = \epsilon_0 c E_{rms}^2 \).
Given \( E_{rms} = 3 \, \text{N C}^{-1} \).
Distance from point source \( r = 3 \) m.
For a point source radiating uniformly in all directions, the intensity \(I\) at a distance \(r\) is related to the power \(P\) of the source by: \[ I = \frac{P}{4\pi r^2} \] Equating the two expressions for intensity: \[ \frac{P}{4\pi r^2} = \epsilon_0 c E_{rms}^2 \] \[ P = 4\pi r^2 \epsilon_0 c E_{rms}^2 \] We know that \( \frac{1}{4\pi\epsilon_0} = k_e \approx 9 \times 10^9 \, \text{N m}^2 \text{C}^{-2} \).
So, \( 4\pi\epsilon_0 = \frac{1}{k_e} = \frac{1}{9 \times 10^9} \).
\[ P = \frac{1}{9 \times 10^9} \cdot r^2 \cdot c \cdot E_{rms}^2 \] Substitute the values: \( r=3 \) m, \( c = 3 \times 10^8 \) m/s, \( E_{rms} = 3 \) N/C.
\[ P = \frac{1}{9 \times 10^9} \times (3)^2 \times (3 \times 10^8) \times (3)^2 \] \[ P = \frac{1}{9 \times 10^9} \times 9 \times (3 \times 10^8) \times 9 \] \[ P = \frac{1}{10^9} \times (3 \times 10^8) \times 9 = \frac{3 \times 9 \times 10^8}{10^9} = \frac{27}{10} = 2.
7 \, \text{W} \] This matches option (4).