Q70 / 75 · JEE Main 2026 (23 January Shift 1) · Physics
The equation of the electric field of an electromagnetic wave propagating through free space is given by : $E=\sqrt{377} \sin \left(6.27 \times 10^{3} t-2.09 \times 10^{-5} x\right) \mathrm{N} / \mathrm{C}$
The average power of the electromagnetic wave is $\left(\frac{1}{\alpha}\right) \mathrm{W} / \mathrm{m}^{2}$. The value of $\alpha$ is $\_\_\_\_$
(Take $\sqrt{\frac{\mu_{0}}{\varepsilon_{o}}}=377$ in SI units)
The average power of the electromagnetic wave is $\left(\frac{1}{\alpha}\right) \mathrm{W} / \mathrm{m}^{2}$. The value of $\alpha$ is $\_\_\_\_$
(Take $\sqrt{\frac{\mu_{0}}{\varepsilon_{o}}}=377$ in SI units)
Correct answer: 2
Solution
From the electric field equation, amplitude:
$E_0 = \sqrt{377}$ N/C
Average intensity of EM wave:
$I_{avg} = \frac{E_0^2}{2Z_0}$ where $Z_0 = \sqrt{\mu_0/\varepsilon_0} = 377$ Ω
$I_{avg} = \frac{(\sqrt{377})^2}{2 \times 377} = \frac{377}{2 \times 377} = \frac{1}{2}$ W/m²
Given $I_{avg} = \frac{1}{\alpha}$ W/m², we have $\frac{1}{\alpha} = \frac{1}{2}$
Therefore, $\alpha = 2$
$E_0 = \sqrt{377}$ N/C
Average intensity of EM wave:
$I_{avg} = \frac{E_0^2}{2Z_0}$ where $Z_0 = \sqrt{\mu_0/\varepsilon_0} = 377$ Ω
$I_{avg} = \frac{(\sqrt{377})^2}{2 \times 377} = \frac{377}{2 \times 377} = \frac{1}{2}$ W/m²
Given $I_{avg} = \frac{1}{\alpha}$ W/m², we have $\frac{1}{\alpha} = \frac{1}{2}$
Therefore, $\alpha = 2$
Hence, the correct answer is 2.