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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

  1. AI = 2c U_E , B₀ = 2 ₀ U_E
  2. BI = 2c U_E , B₀ = 2 ₀ U_E
  3. CI = c U_E , B₀ = 2 ₀ U_E
  4. 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

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