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JEE Main 2022 (25 Jun, Shift 1) · Energy Density & Intensity

Energy Density & Intensity question from JEE Main 2022 (25 Jun, Shift 1)

Q15JEE Main 2022 · 25 JUN · SHIFT 1Easy

Energy Density & Intensity

The electric field in an electromagnetic wave is given by E = 56.5 sin ω\omega(t −- x/c) NC−1{}^{-1}. Find the intensity of the wave if it is propagating along x-axis in the free space.

(Given: ε\varepsilon0 = 8.85 ×\times 10−-12C2N−1 m−-1\,\text{m}-2)

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Answer: (b)

  1. I=12ε0cE02I = \tfrac{1}{2}\varepsilon_0 c E_0^2 with E0=56.5E_0 = 56.5 N/C
  2. I=12×8.85×10−12×3×108×56.52=1.3275×10−3×3192I = \tfrac{1}{2} \times 8.85 \times 10^{-12} \times 3 \times 10^{8} \times 56.5^2 = 1.3275 \times 10^{-3} \times 3192
  3. I≈4.24I \approx 4.24 W m−2^{-2}
  4. So the answer is (b) 4.24 W m−2^{-2}.
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