(i) Consider the figure given below

During the propagation of electromagnetic wave a long $z$-axis, let electric field vector $(\mathbf{E})$ be along $x$-axis and magnetic field vector $\mathbf{B}$ along $y$-axis, i.e., $\mathbf{E}=E_0 \hat{\mathbf{i}}$ and $\mathbf{B}=B_0 \hat{\mathbf{j}}$.

Line integral of $E$ over the closed rectangular path 1234 in $x$ - $z$ plane of the figure

$$\begin{aligned} \oint \mathrm{E} \cdot \mathrm{dl} & =\int_1^2 \mathrm{E} \cdot \mathrm{dl}+\int_2^3 \mathrm{E} \cdot \mathrm{dl}+\int_3^4 \mathrm{E} \cdot \mathrm{dl}+\int_4^1 \mathrm{E} \cdot \mathrm{dl} \\ & =\int_1^2 \mathrm{E} \cdot \mathrm{dl} \cos 90+\int_2^3 \mathrm{E} \cdot \mathrm{dl} \cos 0+\int_3^4 \mathrm{E} \cdot \mathrm{dl} \cos 90+\int_4^1 \mathrm{E} \cdot \mathrm{dl} \cos 180 \Upsilon \\ & =E_0 h\left[\sin \left(k z_2-\omega t\right)-\sin \left(k z_1-g \omega t\right)\right] \end{aligned}$$

(ii) For evaluating $\int \mathrm{B} \cdot \mathrm{ds}$, let us consider the rectangle 1234 to be made of strips of are $d s=h d z \text { each. }$

$$\begin{aligned} & \qquad \begin{aligned} \int \mathrm{B} \cdot \mathrm{ds}=\int \mathrm{B} \cdot \mathrm{ds} \cos 0 & =\int \mathrm{B} \cdot \mathrm{ds}=\int_{z_1}^{z_2} B_0 \sin (k z-\omega t) h d z \\ & =\frac{-B_0 h}{k}\left[\cos \left(k z_2-\omega t\right)-\cos \left(k z_1-\omega t\right)\right] \end{aligned} \end{aligned}$$

(iii) Given, $\oint \mathrm{E} \cdot \mathrm{dl}=\frac{-d \phi_B}{d t}=-\frac{d}{d t} \oint \mathrm{~B} \cdot \mathrm{ds}$ Putting the values from Eqs. (i) and (ii), we get $$ \begin{aligned} & E_0 h\left[\sin \left(k z_2-\omega t\right)-\sin \left(k z_1-\omega t\right)\right] \\ & =\frac{-d}{d t}\left[\frac{B_y h}{k}\left\{\cos \left(k z_2-\omega t\right)-\cos \left(k z_1-\omega t\right)\right]\right. \\ & =\frac{B_y h}{k} \omega\left[\sin \left(k z_2-\omega t\right)-\sin \left(k z_1-\omega t\right)\right] \\ \Rightarrow \quad E_0 & =\frac{B_0 \omega}{k}=B_y c \left(\because \frac{\omega}{k}=c\right)\\ \Rightarrow \quad \frac{E_0}{B_0} & =c \end{aligned}$$

(iv) For evaluating $\boldsymbol{\rho B} \cdot \mathrm{dl}$, let us consider a loop 1234 in $y-z$ plane as shown in figure given below

$$\begin{aligned} \oint \mathbf{B} \cdot \mathbf{d l} & =\int_1^2 \mathbf{B} \cdot \mathbf{d l}+\int_2^3 \mathbf{B} \cdot \mathbf{d l}+\int_3^4 \mathbf{B} \cdot \mathbf{d l}+\int_4^1 \mathbf{B} \cdot \mathbf{d l} \\ & =\int_1^2 \mathbf{B} \cdot \mathbf{d l} \cos 0+\int_2^3 \mathbf{B} \cdot \mathbf{d l} \cos 90 \Upsilon+\int_3^4 \mathbf{B} \cdot \mathbf{d l} \cos 180 \Upsilon_{+}+\int_4^1 \mathbf{B} \cdot \mathbf{d l} \cos 90 \Upsilon \\ & =B_0 h\left[\sin \left(k z_1-\omega t\right)-\sin \left(k z_2-\omega t\right)\right.\quad\text{.... (iii)} \end{aligned}$$

Now to evaluate $\phi_E=\int \mathbf{E} \cdot \mathrm{ds}$, let us consider the rectangle 1234 to be made of strips of area $h d_2$ each.

$$ \begin{aligned} & \phi_E=\int \mathrm{E} \cdot \mathrm{ds}=\int E d s \cos 0=\int E d s=\int_{z_1}^{z_2} E_0 \sin \left(k z_1-\omega t\right) h d z \\ & =-\frac{E_0 h}{k}\left[\cos \left(k z_2-\omega t\right)-\cos \left(k z_1-\omega t\right)\right] \\ & \therefore \quad \frac{d \phi_E}{d t}=\frac{E_0 h \omega}{k}\left[\sin \left(k z_1-\omega t\right)-\sin \left(k z_2-\omega t\right)\right] \quad\text{.... (iv)}\\ & \text { In } \quad \oint \mathrm{B} \cdot \mathrm{dl}=\alpha_0\left(I+\frac{\varepsilon_0 d \phi_E}{d t}\right), I=\text { conduction current } \\ & =0 \text { in vacuum } \\ & \therefore \quad \quad \quad \oint\mathrm{B} \cdot \mathrm{dl}=\alpha_0 \varepsilon \frac{d \phi_E}{d t} \end{aligned}$$

Using relations obtained in Eqs. (iii) and (iv) and simplifying, we get

$$\begin{aligned} B_0 & =E_0 \frac{\omega \propto_0 \varepsilon_0}{k} \\ \Rightarrow\quad \frac{E_0}{B_0} \frac{\omega}{k} & =\frac{1}{\propto_0 \varepsilon_0} \\ \text{But}\quad\frac{E_0}{B_0} & =c \text { and } \omega=c k \\ \Rightarrow\quad c \cdot c & =\frac{1}{\alpha_0 \varepsilon_0}, \text { therefore } c=\frac{1}{\sqrt{\alpha_0 \varepsilon_0}} \end{aligned}$$

(i) The electromagnetic 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 due to electric field $E$ is

$$u_E=\frac{1}{2} \varepsilon_0 E^2$$

The energy density due to magnetic field $B$ is

$$u_B=\frac{1}{2} \frac{B^2}{\alpha_0}$$

Total average energy density of electromagnetic wave

$$u_{a v}=u_E+u_B=\frac{1}{2} \varepsilon_0 E^2+\frac{1}{2} \frac{B^2}{\alpha_0}$$

Let the EM wave be propagating along $z$-direction. The electric field vector and magnetic field vector be represented by

$$\begin{aligned} & E=E_0 \sin (k z-\omega t) \\ & B=B_0 \sin (k z-\omega t) \end{aligned}$$

The time average value of $E^2$ over complete cycle $=\frac{E_0^2}{2}$

and time average value of $B^2$ over complete cycle $=\frac{B_0^2}{2}$

$$\begin{aligned} u_{\mathrm{av}} & =\frac{1}{2} \frac{\varepsilon_0 E_0^2}{2}+\frac{1}{2} \alpha_0\left(\frac{B_0^2}{2}\right) \\ & =\frac{1}{4} \varepsilon_0 E_0^2+\frac{B_0^2}{4 \alpha_0} \end{aligned}$$

$$\begin{aligned} &\text { (ii) We know that } E_0=c B_0 \text { and } c=\frac{1}{\sqrt{\alpha_0 \varepsilon_0}}\\ &\begin{aligned} & \therefore \quad \frac{1}{4} \frac{B_0^2}{\alpha_0}=\frac{1}{4} \frac{E_0^2 / C^2}{\alpha_0}=\frac{E_0^2}{4 \propto_0} \times \propto_0 \varepsilon_0=\frac{1}{4} \varepsilon_0 E_0^2 \\ & \therefore \quad u_B=u_E \end{aligned} \end{aligned}$$

$$\begin{aligned} \text{Hence,}\quad U_{\mathrm{av}} & =\frac{1}{4} \varepsilon_0 E_0^2+\frac{1}{4} \frac{B_0^2}{\alpha_0} \\ & =\frac{1}{4} \varepsilon_0 E_0^2+\frac{1}{4} \varepsilon_0 E_0^2 \\ & =\frac{1}{2} \varepsilon_0 E_0^2=\frac{1}{2} \frac{B_0^2}{\alpha_0} \end{aligned}$$

$$\begin{aligned} &\text { Time average intensity of the wave }\\ &I_{\mathrm{av}}=U_{\mathrm{av}} C=\frac{1}{2} \varepsilon_0 E_0^2 C=\frac{1}{2} \varepsilon_0 E_0^2 \end{aligned}$$