Verified Solution • 2025–26 Session moderate

QJEEM22Phe22e: JEE Main 2022 Physics - Class 12 JEEM22Phe22e Solution: The electric field in an electromagnetic wave is given by E =…

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

(Given : ε\varepsilon 0 = 8.85 ×\times 10 −- 12C2N −- 1m −- 2)

Multiple Choice Options

  • A 5.65 Wm −- 2
  • B 4.24 Wm −- 2
  • C 1.9 ×\times 10 −- 7 Wm −- 2
  • D 56.5 Wm −- 2

Step-by-Step Verified Solution

I=12ε0E02cI = {1 \over 2}{\varepsilon _0}E_0^2c

=128.5×10−12×(56.5)2×3×108= {1 \over 2} 8.5 \times {10^{ - 12}} \times {(56.5)^2} \times 3 \times {10^8}

=4.24= 4.24 W/m2

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