Electromagnetic Waves

JEE Main level, numerical answer, medium

13 of 165

Question 13 of 165

The equation of the electric field of an electromagnetic wave propagating through free space is given by : E=377sin⁡(6.27×103t−2.09×10−5x)N/CE=\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 (1α)W/m2\left(\frac{1}{\alpha}\right) \mathrm{W} / \mathrm{m}^{2}. The value of α\alpha is ____\_\_\_\_
(Take μ0εo=377\sqrt{\frac{\mu_{0}}{\varepsilon_{o}}}=377 in SI units)

Show answer and solution

Answer: 2

From the electric field equation, amplitude:
E0=377E_0 = \sqrt{377} N/C
Average intensity of EM wave:
Iavg=E022Z0I_{avg} = \frac{E_0^2}{2Z_0} where Z0=μ0/ε0=377Z_0 = \sqrt{\mu_0/\varepsilon_0} = 377 Ω
Iavg=(377)22×377=3772×377=12I_{avg} = \frac{(\sqrt{377})^2}{2 \times 377} = \frac{377}{2 \times 377} = \frac{1}{2} W/m²
Given Iavg=1αI_{avg} = \frac{1}{\alpha} W/m², we have 1α=12\frac{1}{\alpha} = \frac{1}{2}
Therefore, α=2\alpha = 2

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Electromagnetic Waves (Electromagnetic Waves and Maxwell's equations), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.