A plane electromagnetic wave propagating in x -direction is described by E_y=(200 Vm^-1) \sin [1.5 \times 10^7 t-0.05 x] ; The intensity of the wave is : (Use \epsilon_0=8.85 \times 10^-12 C^2 ~N^-1 ~m^-2 )

Electromagnetic Waves PYQ — JEE Main 2024 · 27 Jan · Shift 1 · Q76 with Solution

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Correct answer: B

The given electromagnetic wave is

E_y=(200\,\text{V m}^{-1})\sin\left(1.5\times 10^7 t-0.05x\right)

So the amplitude is

E_0=200\,\text{V m}^{-1}

The average intensity is

I=\frac{1}{2}c\epsilon_0 E_0^2

For an electromagnetic wave,

c=\frac{\omega}{k}

From the wave equation,

\omega=1.5\times 10^7\,\text{rad s}^{-1}, \qquad k=0.05\,\text{m}^{-1}

Thus,

c=\frac{1.5\times 10^7}{0.05}=3\times 10^8\,\text{m s}^{-1}

I=\frac{1}{2}(3\times 10^8)(8.85\times 10^{-12})(200)^2

Now,

(200)^2=40000=4\times 10^4

So,

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

Combine powers of 10:

10^8\cdot 10^{-12}\cdot 10^4=10^0=1

Hence,

I=\frac{1}{2}(3)(8.85)(4)

I=\frac{1}{2}(106.2)

I=53.1\,\text{W m}^{-2}

Therefore, the correct option is

\boxed{\text{B }\;53.1\,\text{W m}^{-2}}