JEE MainPhysicsElectromagnetic Waves
A plane electromagnetic wave propagating in free space has an average electric energy density U_E . The total intensity I of the wave and the peak magnetic field B₀ are respectively given by ( c is the speed of light in free space, ₀ is the permeability of free space):
Options
- AI = 2c U_E , B₀ = 2 ₀ U_E
- BI = 2c U_E , B₀ = 2 ₀ U_E
- CI = c U_E , B₀ = 2 ₀ U_E
- DI = c U_E , B₀ = 2 ₀ U_E
Correct answer
A. I = 2c U_E , B₀ = 2 ₀ U_E
Step-by-step solution
In an electromagnetic wave, the average electric energy density U_E is equal to the average magnetic energy density U_B . The total average energy density of the wave is U = U_E + U_B = 2U_E . The intensity I of the electromagnetic wave is the total energy crossing per unit area per unit time, which is given by I = U c . Therefore, I = 2c U_E . The average magnetic energy density is related to the peak magnetic field B₀ by U_B = B₀^2 4 ₀ . Since U_B = U_E , we have: B₀^2 4 ₀ = U_E B₀^2 = 4 ₀ U_E B₀ = 2 ₀ U_E Answer