Electromagnetic Waves

The charge on the capacitor plates at time $t$ is given by $q=I t$

Electric field $=\varepsilon=\frac{\sigma}{\varepsilon_0}=\frac{q}{A \varepsilon_0}=\frac{I t}{A \varepsilon_0}$

Flux through area $\frac{5 \mathrm{~A}}{6}$ is $\phi_{\varepsilon}=\left(\frac{5 \mathrm{~A}}{6}\right) \varepsilon$

$ \Rightarrow \quad \phi_{\varepsilon}=\frac{5 \mathrm{~A}}{6} \times \frac{I t}{A \varepsilon_0} \Rightarrow \phi_{\varepsilon}=\frac{5 I t}{6 \varepsilon_0} $

Now, displacement current $=I_d=\varepsilon_0 \frac{d \phi_e}{d t}$

                                         $ \begin{aligned} & I_d=\varepsilon_0 \frac{d}{d t}\left[\frac{5 I t}{6 \varepsilon_0}\right] \\ & I_d=\frac{5}{6} I \end{aligned} $

The direction of propagation of EM waves is always perpendicular to both the electric and magnetic fields. Therefore, $+v e~ x$ and $+v e~ y$ directions should be the directions of the electric and magnetic fields, respectively.

As we know that, ${E \over B} = c$

so, $B = {E \over c} = {8 \over {3 \times {{10}^8}}} = 2.66 \times {10^{ - 8}}$

When light travel in meta-material, physical characteristics remain unchanged.

$\therefore$ $v = {c \over {|n|}}$

where, | n | is the relative refractive index of meta-material.

As, $n = {c \over v} \Rightarrow |n| = {{{n_{med}}} \over {{n_{air}}}}$

$ \Rightarrow \left| n \right| = {{{\lambda _{air}}} \over {{\lambda _{med}}}} \Rightarrow {\lambda _{med}} = {{{\lambda _{air}}} \over {\left| n \right|}}$

Given, magnetic field,

B = 10 $\times$ 10^$-$8 sim(1 $\times$ 10⁷z $-$ 3.6 $\times$ 10¹⁵t) T ......... (i)

Standard equation of magnetic field,

B = B₀ sin(kz $-$ $\omega$t) ....... (ii)

On comparing Eqs. (i) and (ii), we get

B₀ = 10 $\times$ 10^$-$8 T

The average intensity of the beam,

${I_{av}} = {{B_0^2} \over {2{\mu _0}}} = {1 \over 2} \times {{{{(10 \times {{10}^{ - 8}})}^2} \times 3 \times {{10}^8}} \over {4\pi \times {{10}^{ - 7}}}}$ [$\because$ $c = 3 \times {10^8}$ ms^$-$1]

= 1.19 W/m²