Q3 / 15 · NTA Abhyas (JEE Main) 2020 · Physics · Electromagnetic Waves

In an electromagnetic wave, the maximum value of the electric field is 100 V m-1. The average intensity is [ε0=8.8×10-12 C-2 N-1 m2]

  • Correct answer
    A
    13.2 W m-2
  • B
    36.5 W m-2
  • C
    46.7 W m-2
  • D
    765 W m-2

Solution

Iavg=12cε0E02

⇒Iavg=12×3×108×8.8×10-12×1002

⇒Iavg=13.2 W m-2

Hence, the correct answer is option: 'A'.

More Electromagnetic Waves Questions