$A$ plane $EM$ wave travelling along $z-$ direction is described by $\vec E = E_0 \sin(kz - \omega t)\hat i$ and $\vec B = B_0 \sin(kz - \omega t)\hat j$. Show that:
$(i)$ The average energy density of the wave is given by $U_{av} = \frac{1}{4} \epsilon_0 E_0^2 + \frac{1}{4} \frac{B_0^2}{\mu_0}$.
$(ii)$ The time-averaged intensity of the wave is given by $I_{av} = \frac{1}{2} c \epsilon_0 E_0^2$.

(N/A) $(i)$ The energy density associated with an electric field $E$ is $u_E = \frac{1}{2} \epsilon_0 E^2$ and with a magnetic field $B$ is $u_B = \frac{1}{2} \frac{B^2}{\mu_0}$.
The total instantaneous energy density is $u = u_E + u_B = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0}$.
For a plane wave,$E = E_0 \sin(kz - \omega t)$ and $B = B_0 \sin(kz - \omega t)$.
The time average of $\sin^2(kz - \omega t)$ over one cycle is $\frac{1}{2}$.
Thus,$\langle E^2 \rangle = \frac{E_0^2}{2}$ and $\langle B^2 \rangle = \frac{B_0^2}{2}$.
Substituting these into the average energy density expression:
$U_{av} = \langle u_E \rangle + \langle u_B \rangle = \frac{1}{2} \epsilon_0 \left( \frac{E_0^2}{2} \right) + \frac{1}{2 \mu_0} \left( \frac{B_0^2}{2} \right) = \frac{1}{4} \epsilon_0 E_0^2 + \frac{1}{4} \frac{B_0^2}{\mu_0}$.
$(ii)$ Since $E_0 = c B_0$,we have $B_0 = \frac{E_0}{c}$. Also,$c^2 = \frac{1}{\mu_0 \epsilon_0}$,so $\frac{1}{\mu_0} = c^2 \epsilon_0$.
Substituting $B_0$ and $\frac{1}{\mu_0}$ into the energy density formula:
$U_{av} = \frac{1}{4} \epsilon_0 E_0^2 + \frac{1}{4} (c^2 \epsilon_0) \left( \frac{E_0}{c} \right)^2 = \frac{1}{4} \epsilon_0 E_0^2 + \frac{1}{4} \epsilon_0 E_0^2 = \frac{1}{2} \epsilon_0 E_0^2$.
The intensity $I_{av}$ is the energy crossing unit area per unit time,given by $I_{av} = U_{av} \cdot c$.
Therefore,$I_{av} = \left( \frac{1}{2} \epsilon_0 E_0^2 \right) c = \frac{1}{2} c \epsilon_0 E_0^2$.