NTA Abhyas JEE Main2020PhysicsElectromagnetic WavesPractice
In an electromagnetic wave, the maximum value of the electric field is 100   V  m - 1 . The average intensity is [ ε 0 = 8 . 8 × 10 - 12   C - 2   N - 1   m 2 ]
Options
- A13 . 2   W   m - 2
- B36.5 W m - 2
- C46.7 W m - 2
- D765 W m - 2
Correct answer
A. 13 . 2   W   m - 2
Step-by-step solution
I avg = 1 2 c ε 0 E 0 2 ⇒ I avg = 1 2 × 3 × 10 8 × 8 . 8 × 10 - 12 × 100 2 ⇒ I avg = 13 . 2   W   m - 2