A plane EM wave travelling along z direction is described by
`E=E_0sin (kz-omegat) hati and B=B_0 sin (kz-omegat)hatj`.
show that
(i) The average energy density of the wave is given by `u_(av)=1/4 epsilon_0 E_0^2+1/4 (B_0^2)/(mu_0)`.
(ii) The time averaged intensity of the wave is given by `I_(av)=1/2 cepsilon_0E_0^2`.

Explanation

(i) The electriomagnetic wave carry energy which is due to electric field vector and magnetic field vector. In 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 dut to electric field E is
`U_(E) =1/2 epsilon_(0)E^(2)`
The energy density due to magnetic field B is
`U_(B) =1/2 (B^(2))/(mu_(0))`
Total average energy density of electromagnetic wave
`U_(av) =U_(E) +U_(B) =(1)/(2) epsilon_(0)E^(2) +(1)/(2)(B^(2))/(mu_(0))`
Let the EM wave be propagating along z-direction . the electric field vector and magnetic field vector be represented by
`E=E_(0) sin (kz -omegat)`
`B=B_(0) sin (kz - omegat)`
The time average value of `E^(2)` over complete cycle `=(E_(0)^(2))/(2)`
and time average value of `B^(2)` over complete cycle `=(B_(0)^(2))/(2)`
`u_(av) =1/2 (epsilon_(0)E_(0)^(2))/(2)+(1)/(2) mu_(0) ((B_(0)^(2))/(2))`
`=(1^(E))/(4)epsilon_(0)E_(0)^(2) +(B_(0)^(2))/(4mu_(0))`
(ii) We know that `E_(0) =cB_(0) " and " c=(1)/(sqrt(mu_(0)epsilon_(0))`
`:. 1/4 (B_(0)^(2))/(mu_(0)) =1/4 (E_(0)^(2)//c^(2))/(mu_(0))=(E_(0)^(2))/(4mu_(0)) xx mu_(0)epsilon_(0) =1/4 epsilon_(0) E_(0)^(2)`
`:." "U_(B) =U_(E)`
`"Hence"" "u_(av)=1/4 epsilon_(0) E_(0)^(2) +1/4(B_(0)^(2))/(mu_(0))`
`=1/4 epsilon_(0) E_(0)^(2) +1/4 epsilon_(0)E_(0)^(2)`
`-1/2 epsilon_(0) E_(0)^(2) =1/2 (B_(0)^(2))/(mu_(0))`
Time average intensity of the wave
`I_(av) =U_(av)c=1/2epsilon_(0)E_(0)^(2) c=1/2 epsilon_(0) E_(0)^(2)`