A plane electromagnetic wave propagating in x -direction is described by E y = (200 Vm -1 ) sin [1.5 × 10 7 t -0.05 x ] The intensity of the wave is : (Use ε 0 =8.85 × 10 -12 C 2 N -1 m -2 )

Physics · Electromagnetic Waves

JEE Main 2024 — 27 January, Shift 1 — Question 43

A plane electromagnetic wave propagating in x -direction is described by $\mathrm{E}_{\mathrm{y}}=\left(200 \mathrm{Vm}^{-1}\right) \sin \left[1.5 \times 10^{7} \mathrm{t}-0.05 \mathrm{x}\right]$

The intensity of the wave is : (Use $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$ )

Option A:
$35.4 \mathrm{Wm}^{-2}$
Option B:
$53.1 \mathrm{Wm}^{-2}$
Correct
Option C:
$26.6 \mathrm{Wm}^{-2}$
Option D:
$106.2 \mathrm{Wm}^{-2}$

Answer: B

Step-by-step solution

$\mathrm{I}=\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2} \times \mathrm{c}$

$\mathrm{I}=\frac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^{4} \times 3 \times 10^{8}$

$\mathrm{I}=53.1 \mathrm{~W} / \mathrm{m}^{2}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 27 January, Shift 1
Subject: Physics
Chapter: Electromagnetic Waves
Topic: Power , Energy and Intensity of EM Waves