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Physics · 2026

JEE Main · 23 January 2026, Shift 1 · Q48

The equation of the electric field of an electromagnetic wave propagating through free space is given by: E=√ 377 sin (6.27 × 10^3 t-2.09 × 10^-5 x)…

The equation of the electric field of an electromagnetic wave propagating through free space is given by $\displaystyle : E=\sqrt{377} \sin \left(6.27 \times 10^3 t-2.09 \times 10^{-5} x\right) \mathrm{N} / \mathrm{C}$The average power of the electromagnetic wave is $\displaystyle \left(\frac{1}{\alpha}\right) \mathrm{W} / \mathrm{m}^2$. The value of $\displaystyle \alpha$ is $\displaystyle \_\_\_\_$ $$\left(\text { Take } \sqrt{\frac{\mu_0}{\varepsilon_o}}=377 \text { in SI units }\right) $$
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JEE Main 2026 Physics question, with the answer from NTA’s final answer key. Where our answers come from.