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 $\in_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$)

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · EM Waves

A plane electromagnetic wave propagating in x-direction is described by

E_{y} = (200 Vm^{- 1}) sin [1.5 \times 1 0^{7} t - 0.05 x]

;
The intensity of the wave is :
(Use

\in_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}

)

(A) $35.4 Wm^{- 2}$
(B) $53.1 Wm^{- 2}$ ✓ Correct
(C) $26.6 Wm^{- 2}$
(D) $106.2 Wm^{- 2}$

Answer: (B)

Solution

The time-averaged intensity of an EM wave is

I = \frac{1}{2} ε_{0} E_{0}^{2} c

With

E_{0} = 200

V/m and

c = 3 \times 1 0^{8}

m/s:

I = \frac{1}{2} \times (8.85 \times 1 0^{- 12}) \times (200)^{2} \times (3 \times 1 0^{8}) = 53.1 W/m^{2}

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Single correct option

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