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`.

Options

A. 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))`
B. The time averaged intensity of the wave is given by `I_(av)=(1)/(2)c epsilon_(0)E_(0)^(2)`
C. Both (a) and (b)
D. None of these

Explanation

To solve the problem step by step, we will address both parts (i) and (ii) of the question regarding the average energy density and the time-averaged intensity of a plane electromagnetic wave.

### Part (i): Average Energy Density

1. **Understanding Energy Density**:
The energy density \( u \) of an electromagnetic wave is the sum of the energy densities due to the electric field \( E \) and the magnetic field \( B \). The energy density for the electric field is given by:
\[ u_E = \frac{1}{2} \epsilon_0 E^2 \]
and for the magnetic field:
\[ u_B = \frac{1}{2} \frac{B^2}{\mu_0} \]
Therefore, the total energy density \( u \) is:
\[ u = u_E + u_B = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0} \]

2. **Substituting the Electric and Magnetic Fields**:
Given the electric field \( E = E_0 \sin(kz - \omega t) \) and the magnetic field \( B = B_0 \sin(kz - \omega t) \), we can find the average values of \( E^2 \) and \( B^2 \) over one complete cycle.

3. **Calculating Average Values**:
The average value of \( E^2 \) over one cycle is:
\[ \langle E^2 \rangle = \frac{E_0^2}{2} \]
Similarly, for the magnetic field:
\[ \langle B^2 \rangle = \frac{B_0^2}{2} \]

4. **Substituting Average Values into Energy Density**:
Now substituting these average values into the total energy density formula:
\[ u_{av} = \frac{1}{2} \epsilon_0 \langle E^2 \rangle + \frac{1}{2} \frac{\langle B^2 \rangle}{\mu_0} \]
This gives:
\[ u_{av} = \frac{1}{2} \epsilon_0 \left(\frac{E_0^2}{2}\right) + \frac{1}{2} \frac{\left(\frac{B_0^2}{2}\right)}{\mu_0} \]
Simplifying this results in:
\[ u_{av} = \frac{1}{4} \epsilon_0 E_0^2 + \frac{1}{4} \frac{B_0^2}{\mu_0} \]

5. **Final Expression**:
Thus, we have shown that:
\[ u_{av} = \frac{1}{4} \epsilon_0 E_0^2 + \frac{1}{4} \frac{B_0^2}{\mu_0} \]

### Part (ii): Time-Averaged Intensity

1. **Understanding Intensity**:
The intensity \( I \) of an electromagnetic wave is defined as the power per unit area, and it can be expressed in terms of the energy density and the speed of light \( c \):
\[ I = u \cdot c \]

2. **Using Average Energy Density**:
From part (i), we have the average energy density \( u_{av} \). Now, substituting this into the intensity formula:
\[ I_{av} = u_{av} \cdot c \]

3. **Substituting for \( u_{av} \)**:
We can substitute our expression for \( u_{av} \):
\[ I_{av} = \left(\frac{1}{4} \epsilon_0 E_0^2 + \frac{1}{4} \frac{B_0^2}{\mu_0}\right) c \]

4. **Relating Electric and Magnetic Fields**:
We know that the speed of light \( c \) is related to the electric and magnetic fields by:
\[ c = \frac{E_0}{B_0} \]
Therefore, we can express \( B_0 \) in terms of \( E_0 \):
\[ B_0 = \frac{E_0}{c} \]

5. **Substituting \( B_0 \)**:
Now substituting \( B_0 \) into the intensity equation:
\[ I_{av} = \left(\frac{1}{4} \epsilon_0 E_0^2 + \frac{1}{4} \frac{\left(\frac{E_0}{c}\right)^2}{\mu_0}\right) c \]

6. **Simplifying**:
Since \( \frac{1}{\mu_0} = \epsilon_0 c^2 \), we can simplify further:
\[ I_{av} = \frac{1}{4} \epsilon_0 E_0^2 c + \frac{1}{4} \frac{E_0^2}{\mu_0 c} c \]
This leads to:
\[ I_{av} = \frac{1}{2} c \epsilon_0 E_0^2 \]

7. **Final Expression**:
Thus, we have shown that:
\[ I_{av} = \frac{1}{2} c \epsilon_0 E_0^2 \]

### Summary of Results
- (i) The average energy density 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 is given by:
\[ I_{av} = \frac{1}{2} c \epsilon_0 E_0^2 \]