Electromagnetic Waves

JEE Main level, single correct, medium

42 of 165

Question 42 of 165

A plane electromagnetic wave propagating in xx-direction is described by
Ey=(200Vm−1)sin⁡[1.5×107t−0.05x];{E}_{y}=\left(200V{m}^{-1}\right)\sin \left[1.5\times {10}^{7}t-0.05x\right];The intensity of the wave is :
(Use 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: (B) 53.1Wm−253.1W{m}^{-2}

The magnitude of the electric field component of the electromagnetic wave is given by
E0=200Vm−1{E}_{0}=200V{m}^{-1}
The formula to calculate the intensity of the electromagnetic wave is given by
I=12ϵ0cE02...(1)I=\dfrac{1}{2}{\epsilon }_{0}c{E}_{0}^{2}...\left(1\right)
From equation (1), it follows that
I=12×8.85×10−12×3×108×(200)2Wm−2=53.1Wm−2\begin{matrix}I & = & \dfrac{1}{2}\times 8.85\times {10}^{-12}\times 3\times {10}^{8}\times {\left(200\right)}^{2}W{m}^{-2} \\ & = & 53.1W{m}^{-2}\end{matrix}

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