Question

The average power output of a point source of an electromagnetic radiation is 1080 W. The maximum value of the rms value of the electric field at a distance of 3 m from the source is

• A. \( 20 \, \text{Vm}^{-1} \)
• B. \( 40 \, \text{Vm}^{-1} \)
• C. \( 60 \, \text{Vm}^{-1} \)
• D. \( 90 \, \text{Vm}^{-1} \)

Hint:

In electromagnetic waves, the relationship between power and electric field can help determine the strength of the field at a given distance from the source.

Answer 1

The Correct Option is C

The average power of an electromagnetic wave is related to the electric field \( E \) by the formula: \[ P = \frac{E_{\text{rms}}^2}{R} \] Where \( P \) is the power and \( E_{\text{rms}} \) is the rms value of the electric field. Given the power \( P = 1080 \, \text{W} \) and the distance \( r = 3 \, \text{m} \), we can use the relationship for the electric field in an electromagnetic wave to find the maximum value of the electric field at this distance. From the formula, we calculate the value of \( E_{\text{rms}} \), which corresponds to \( 60 \, \text{Vm}^{-1} \). Thus, the maximum value of the rms value of the electric field is \( \boxed{60 \, \text{Vm}^{-1}} \).