JEE MainPhysicsElectromagnetic Waves
The electric field of a plane electromagnetic wave propagating in a non-magnetic dielectric medium is given by E = 600 (2 10^6 x - 3 10¹⁴ t) j V/m. The average intensity of the electromagnetic wave is (Take ₀ = 4 10⁻⁷ T m/A)
Options
- A1500 W/m ^2
- B750 W/m ^2
- C3000 W/m ^2
- D6000 W/m ^2
Correct answer
C. 3000 W/m ^2
Step-by-step solution
From the given equation of the electric field, the wave number k = 2 10^6 rad/m and the angular frequency = 3 10¹⁴ rad/s. The speed of the electromagnetic wave in the medium is v = k = 3 10¹⁴ 2 10^6 = 1.5 10^8 m/s For a non-magnetic medium, the permeability is ₀ . The average intensity of the electromagnetic wave is given by I = E₀^2 2 ₀ v Substituting the values, we get I = (600)^2 2 4 10⁻⁷ 1.5 10^8 I = 360000 120 = 3000 W/m ^2 Answer: 3000 W/m ^2