For the plane electromagnetic wave given by E=E0sin(ωt–kx) and B=B0sin(ωt−kx), the ratio of average electric energy density to average magnetic energy density is
Question
For the plane electromagnetic wave given by E=E0sin(ωt–kx) and B=B0sin(ωt−kx), the ratio of average electric energy density to average magnetic energy density is
Solution
The average electric energy density (U_E) is given by U_E = 1/2 ε_0 E^2, where ε_0 is the permittivity of free space and E is the electric field amplitude.
The average magnetic energy density (U_B) is given by U_B = 1/2 B^2/μ_0, where μ_0 is the permeability of free space and B is the magnetic field amplitude.
The ratio of the average electric energy density to the average magnetic energy density is therefore:
U_E / U_B = (1/2 ε_0 E^2) / (1/2 B^2/μ_0) = ε_0 μ_0 (E/B)^2
But we know that the speed of light c in a vacuum is given by 1/sqrt(ε_0 μ_0), so ε_0 μ_0 = 1/c^2.
Therefore, the ratio U_E / U_B = (E/B)^2 / c^2.
However, for an electromagnetic wave in a vacuum, the magnitudes of the electric and magnetic fields are related by E = cB.
Therefore, the ratio U_E / U_B = 1, meaning the average electric energy density is equal to the average magnetic energy density for an electromagnetic wave in a vacuum.
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