The equation of the electric field of an electromagnetic wave propagating through free space is given by: E = √(377) sin (6.27 × 10³ t - 2.09 × 10⁻⁵ x) N/C The average power of the electromagnetic wave is ((1)/(α)) W/m². The value of α is. ( Take √((μ₀)/(εo)) = 377 in SI units) Numerical answer type. Work it out, then open the answer.

Answer: 2. Average intensity I = (E₀²)/(2)√((ε₀)/(μ₀)) = (E₀²)/(2 × 377) E₀ = √(377) ⇒ E₀² = 377 I = (377)/(754) = 1/2 W/m² So (1)/(α) = 1/2, giving α = 2.

The equation of the electric field of an electromagnetic wave… | JEE Main 2026 (23 Jan, Shift 1)

Q3JEE Main 2026 · 23 JAN · SHIFT 1Medium

Energy Density & Intensity

The equation of the electric field of an electromagnetic wave propagating through free space is given by: $E = \sqrt{377} \sin\left(6.27 \times 10^{3} t - 2.09 \times 10^{-5} x\right)\text{ N/C}$

The average power of the electromagnetic wave is $\left(\frac{1}{\alpha}\right)\text{ W/m}^2$ . The value of $\alpha$ is $\underline{\qquad}\,.\ \Big($ Take $\sqrt{\frac{\mu_0}{\varepsilon_o}} = 377$ in SI units $\Big)$

Numerical answer type. Work it out, then open the answer.

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Answer: 2

Average intensity $I = \dfrac{E_0^2}{2}\sqrt{\dfrac{\varepsilon_0}{\mu_0}} = \dfrac{E_0^2}{2 \times 377}$
$E_0 = \sqrt{377} \Rightarrow E_0^2 = 377$
$I = \dfrac{377}{754} = \dfrac{1}{2}\ \text{W/m}^2$
So $\dfrac{1}{\alpha} = \dfrac{1}{2}$ , giving $\alpha = 2$ .

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