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

To solve the problem, we need to find the average energy density and the time-averaged intensity of a plane electromagnetic wave described by the electric field \( E = E_0 \sin(kz - \omega t) \hat{i} \) and the magnetic field \( B = B_0 \sin(kz - \omega t) \hat{j} \).

### Part (i): Average Energy Density

1. **Energy Density in Electric Field**:
The energy density \( u_E \) in the electric field is given by:
\[
u_E = \frac{1}{2} \epsilon_0 E^2
...