Question

The equation of the electric field of an electromagnetic wave propagating through free space is given by: \(E = \sqrt{377} \sin(6.27 \times 10^3 t - 2.09 \times 10^{-5} x)\) N/C. The average power of the electromagnetic wave is (1/a) W/m². The value of a is_______. (Take \(\sqrt{\mu_0/\varepsilon_0} = 377\) in SI units)}

Hint:

Intensity is the time-averaged Poynting vector. For a sinusoidal wave, the average is always half the peak value ($\frac{1}{2} E_0 H_0$).

Answer 1

Correct Answer: 2

Step 1: Understanding the Concept:
The average power per unit area of an electromagnetic wave is its average intensity (\(I\)). It is related to the peak electric field amplitude (\(E_0\)) and the impedance of free space (\(\eta = \sqrt{\mu_0/\varepsilon_0}\)).
Step 2: Key Formula or Approach:
1. \(I = \frac{1}{2} c \varepsilon_0 E_0^2\).
2. Or \(I = \frac{E_0^2}{2\eta}\).
Step 3: Detailed Explanation:
From the equation: \(E_0 = \sqrt{377}\) N/C.
Impedance of free space \(\eta = 377 \Omega\).
Average power (Intensity): \[ I = \frac{E_0^2}{2\eta} \] \[ I = \frac{(\sqrt{377})^2}{2 \times 377} \] \[ I = \frac{377}{754} = \frac{1}{2} \text{ W/m}^2 \] Comparing with \(1/a\): \[ \frac{1}{a} = \frac{1}{2} \implies a = 2 \]
Step 4: Final Answer:
The value of a is 2.