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NEETPhysicsElectromagnetic Waves

If the total average energy density of an electromagnetic wave propagating in vacuum is U , then the amplitude of the oscillating magnetic field is: (where ₀ is the permeability of free space)

Options

  1. A₀ U
  2. B2 ₀ U
  3. C4 ₀ U
  4. DU 2 ₀

Correct answer

B. 2 ₀ U

Step-by-step solution

The total average energy density U of an electromagnetic wave is the sum of the average electric energy density ( u_E ) and the average magnetic energy density ( u_B ). Since u_E = u_B , the total average energy density is U = 2u_B . The average energy density of the magnetic field is given by u_B = B_ rms ^2 2 ₀ . Since B_ rms = B₀ 2 , we have u_B = B₀^2 4 ₀ . Therefore, the total average energy density is: U = 2 ( B₀^2 4 ₀ ) = B₀^2 2 ₀ Rearranging for the amplitude of the magnetic field B₀ gives: B₀^2 = 2 ₀ U B₀

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