The electric field in a plane electromagnetic wave is given by → E = 200 [(0 . 5 × 10 ^ 3 m ^ - 1 )x - (1 . 5 × 10 ^ 11 rad s ^ - 1 )t]V m ^ - 1 ^ j . If this wave falls normally on a perfectly reflecting surface having

Correct answer: 0 E _ 0 = 200 I = (1)/(2) ε _ 0 E _ 0 ^ 2 c Radiation pressure P = (2l)/(c) = ((2)/(c))((1)/(2) ε _ 0 E _ 0 ^ 2 C) = ε _ 0 E _ 0 ^ 2 P = (8 . 85 × 10 ^ - 12 ) × 200 ^ 2 = 8 . 85 × 10 ^ - 8 × 4 = 354 10 ^ 9 So, k = 354

JEE Main 202126 AugustMorning Shift

Exam: JEE Main
Year: 2021
Date: 26 August
Shift: Morning Shift
Subject: Physics
Chapter: Electromagnetic Waves

JEE Main 2021 · 26 August · Morning Shift · Physics · Electromagnetic Waves

The electric field in a plane electromagnetic wave is given by → E...

The electric field in a plane electromagnetic wave is given by

\overset{\to}{E} = 200\cos \left[\left(0 . 5 \times {10}^{3}{m}^{- 1}\right)x - \left(1 . 5 \times {10}^{11}rad{s}^{- 1}\right)t\right]V{m}^{- 1}\overset{^}{j}

.
If this wave falls normally on a perfectly reflecting surface having an area of

100{cm}^{2}

. If the radiation pressure exerted by the E.M. wave on the surface during a

10min

exposure is

\dfrac{k}{{10}^{9}}N{m}^{- 2}

. Find the value of

k

Correct answer

Step-by-step solution

${E}_{0} = 200$

$I = \dfrac{1}{2}{ε}_{0}{E}_{0}^{2}c$
Radiation pressure
$P = \dfrac{2l}{c} = \left(\dfrac{2}{c}\right)\left(\dfrac{1}{2}{ε}_{0}{E}_{0}^{2}C\right) = {ε}_{0}{E}_{0}^{2}P = \left(8 . 85 \times {10}^{- 12}\right) \times {200}^{2} = 8 . 85 \times {10}^{- 8} \times 4 = \dfrac{354}{{10}^{9}}So, k = 354$

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