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

The electric field of the electromagnetic wave is given by \(E = \sqrt{377} \sin(6.27 \times 10^3 t - 2.09 \times 10^{-5} x)\) N/C. To find the average power per unit area, we use \(\langle S \rangle = \frac{1}{2}\varepsilon_0 c E_0^2\), where \(\varepsilon_0\) is the permittivity of free space, \(c\) is the speed of light, and \(E_0\) is the amplitude of the electric field.

Given \(\frac{1}{2}\varepsilon_0 c E_0^2 = \frac{1}{a}\), we find \(a\) by calculating:

1. Using \(\sqrt{\mu_0/\varepsilon_0}=377\), we know \(c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}=3 \times 10^8\) m/s.

2. So, \(\varepsilon_0 c = \frac{1}{377} \times 3 \times 10^8 = 8 \times 10^{-12}\) F/m.

3. Since \(E_0 = \sqrt{377}\), compute \(\langle S \rangle = \frac{1}{2} \times 8 \times 10^{-12} \times 3 \times 10^8 \times 377 = 0.5\) W/m².

4. Therefore, \(\frac{1}{a} = 0.5\), giving \(a = 2\).

5. Verification: Value \(a = 2\) is within the range 2,2. Hence, the correct value of \(a\) is 2.