Electromagnetic Waves

JEE Main level, single correct, medium

68 of 165

Question 68 of 165

The electric field in an electromagnetic wave is given by E=56.5sin⁡ω(t−xc)NC−1E=56.5\sin \omega \left(\dfrac{t-x}{c}\right){NC}^{-1}. Find the intensity of the wave if it is propagating along xx-axis in the free space. (Given ϵ0=8.85×10−12C2N−1m−2{ϵ}_{0}=8.85\times {10}^{-12}{C}^{2}{N}^{-1}{m}^{-2})

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Answer: (C) 4.24Wm−24.24{Wm}^{-2}

Given, E=56.5sin⁡ω(t−xc)NC−1E=56.5\sin \omega \left(\dfrac{t-x}{c}\right)N{C}^{-1}
Comparing it with the general equation of electromagnetic wave,
E=E0sin⁡(ωt−kx)E={E}_{0}\sin \left(\omega t-kx\right)
We have, E0=56.5NC−1{E}_{0}=56.5N{C}^{-1}
So, the intensity will be I=12ϵ0E02cI=\dfrac{1}{2}{\epsilon }_{0}{E}_{0}^{2}c
=12×8.85×10−12×(56.5)2×3×108=\dfrac{1}{2}\times 8.85\times {10}^{-12}\times {\left(56.5\right)}^{2}\times 3\times {10}^{8}
=4.24Wm−2=4.24W{m}^{-2}

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