$A$ plane $EM$ wave travelling in vacuum along $z$-direction is given by $\vec E = E_0 \sin(kz - \omega t) \hat i$ and $\vec B = B_0 \sin(kz - \omega t) \hat j$.
$(i)$ Evaluate $\int \vec E \cdot d\vec l$ over the rectangular loop $1234$ shown in the figure.
$(ii)$ Evaluate $\int \vec B \cdot d\vec s$ over the surface bounded by loop $1234$.
$(iii)$ Use $\int \vec E \cdot d\vec l = -\frac{d\phi_E}{dt}$ to prove $\frac{E_0}{B_0} = c$.
$(iv)$ By using a similar process and the equation $\int \vec B \cdot d\vec l = \mu_0 I + \mu_0 \epsilon_0 \frac{d\phi_E}{dt}$,prove that $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.

(N/A) $(i)$ The line integral $\oint \vec E \cdot d\vec l$ over the loop $1234$ is $\int_1^2 \vec E \cdot d\vec l + \int_2^3 \vec E \cdot d\vec l + \int_3^4 \vec E \cdot d\vec l + \int_4^1 \vec E \cdot d\vec l$. Since $\vec E$ is along $\hat i$ and the segments $1-2$ and $3-4$ are along $\hat z$,the dot product is zero. For segments $2-3$ and $4-1$,$\vec E$ is parallel/anti-parallel to $d\vec l$. Thus,$\oint \vec E \cdot d\vec l = E(z_2, t)h - E(z_1, t)h = h E_0 [\sin(kz_2 - \omega t) - \sin(kz_1 - \omega t)]$.
$(ii)$ The magnetic flux $\phi_B = \int \vec B \cdot d\vec s$. With $d\vec s = h dz \hat j$,$\phi_B = \int_{z_1}^{z_2} B_0 \sin(kz - \omega t) h dz = \frac{B_0 h}{k} [\cos(kz_1 - \omega t) - \cos(kz_2 - \omega t)]$.
$(iii)$ Using Faraday's Law $\oint \vec E \cdot d\vec l = -\frac{d\phi_B}{dt}$,we differentiate the flux with respect to $t$ and equate it to the line integral. Comparing the amplitudes,we get $E_0 = c B_0$,hence $\frac{E_0}{B_0} = c$.
$(iv)$ Using Ampere-Maxwell Law $\oint \vec B \cdot d\vec l = \mu_0 \epsilon_0 \frac{d\phi_E}{dt}$,where $\phi_E = \int E dA$. Following the same derivation steps as Faraday's law,we obtain $B_0 = \mu_0 \epsilon_0 c E_0$. Substituting $E_0 = c B_0$,we get $B_0 = \mu_0 \epsilon_0 c^2 B_0$,which simplifies to $c^2 = \frac{1}{\mu_0 \epsilon_0}$ or $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.