The energy density of an electromagnetic wave in a vacuum is given by the relation:

$A$ 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 $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.

An electromagnetic wave is propagating in vacuum along $-\hat{j}$ direction. The magnetic field of the wave is given by $\vec{B} = (2 \times 10^{-8}) \cos [\pi \times 10^{15}(t + \frac{y}{c})] \hat{k} \text{ T}$. The electric field $\vec{E}$ of this wave is $(c = \text{speed of light})$

The electric field in $NC^{-1}$ of an electromagnetic wave is given by $E = 36 \sqrt{\pi} \sin(\omega t - kx)$. The average energy density of the electromagnetic wave due to the electric field is (Given: $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ Nm^2 C^{-2}$)

In a plane electromagnetic wave travelling in free space,the electric field component oscillates sinusoidally at a frequency of $2.0 \times 10^{10} \text{ Hz}$ and amplitude $48 \text{ V m}^{-1}$. Then the amplitude of the oscillating magnetic field is: (Speed of light in free space $= 3 \times 10^8 \text{ m s}^{-1}$)

The ratio of the electric field $(E)$ to the magnetic field $(B)$ in an electromagnetic wave gives which physical quantity?