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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Electromagnetic Waves

NTA Abhyas 2020 · Physics · Electromagnetic Waves

In an electromagnetic wave, the maximum value of the electric field...

In an electromagnetic wave, the maximum value of the electric field is 100Vm−1100Vm^{- 1}. The average intensity is [ε0=8.8×10−12C−2N−1m2{ε}_{0} = 8. 8 \times {10}^{- 12}C^{- 2}N^{- 1}m^{2}]

Options

  1. A13.2Wm−213. 2Wm^{- 2}
  2. B36.5Wm−236.5Wm^{- 2}
  3. C46.7Wm−246.7Wm^{- 2}
  4. D765Wm−2765Wm^{- 2}

Correct answer

A. 13.2Wm−213. 2Wm^{- 2}

Step-by-step solution

Iavg=12cε0(E0)2{I}_{avg} = \dfrac{1}{2}c{ε}_{0}\left({E}_{0}\right)^{2}

⇒Iavg=12×3×108×8.8×10−12×(100)2⇒ {I}_{avg} = \dfrac{1}{2} \times 3 \times {10}^{8} \times 8. 8 \times {10}^{- 12} \times \left(100\right)^{2}

⇒Iavg=13.2Wm−2⇒ {I}_{avg} = 13. 2Wm^{- 2}

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