A plane EM wave travelling along z-direction is described byE=E0 sin(kz-ωt)i^ and B=B0 sin(kz-ωt)j^. Show that:

(i) the average energy density of the wave is given by \(U_{av}=\frac14\varepsilon_0E^2_0+\frac14\frac{B^2_0}{\mu_0}\)

(ii) the time averaged intensity of the wave is given by \(I_{av}=\frac12c\varepsilon_0E^2_0\).

Hint: The average energy density of the wave is distributed equally between the electric field and the magnetic field.
(I) Step 1: The electromagnetic wave carries energy which is due to electric field vector and magnetic field vector. In an electromagnetic wave, E and B vary from point to point and from moment to moment. Let E and B be their time averages.
The energy density due to electric field E is
UE=12ε0E2
The energy density due to magnetic field B is:
UB=1 B22 μ0
Total average energy density of electromagnetic wave:
Uav=UE=12ε0E2+12B2μ0
Step 2: Let the EM wave be propagating along z-direction. The electric field vector and magnetic field vector be represented by
E=E0sin(kz-ωt)
B=B0sin(kz-ωt)
The time average value of E2 over complete cycle=E022
The time average value of B2 over complete cycle=B022
Uav=12ε0E022+12μ0B022
=14ε0E02+B024μ0
(ii) Step 3: We know that E0=cB0 and c=1μ0ε0
    14B02μ0=14E02/c2μ0=E024μ0×μ0ε0=14ε0E02
    UB=UE
Hence,  Uav=14ε0E02+14B024μ0
   =14ε0E02+14ε0E02
  =12ε0E02=12B02μ0
Time average intensity of the wave:
Iav=UavC=12ε0E02c=12ε0E02