Electromagnetic Waves

Given, wave is propagating along $z$-axis.

As we know that, electric field $\mathbf{E}$, magnetic field $\mathbf{B}$ and propagation $\mathbf{k}$ all are mutually perpendicular to each other.

So, $\mathbf{E}=E_0 \hat{\mathbf{i}}$ and $\mathbf{B}=B_0 \hat{\mathbf{j}}$

Given, equation of magnetic field, $B=400 \mu \mathrm{T} \sin \left[4 \times 10^{15}\left(t-\frac{x}{c}\right)\right]$

$\because$ Average energy, $E_{\mathrm{av}}=\frac{1}{2} \varepsilon_0 E_0^2=\frac{1}{2} \frac{B_0^2}{\mu_0}$ where, $\varepsilon_0$ is free space permittivity and $\mu_0$ is free space permeability.

$\begin{aligned} \therefore \quad E_{\mathrm{av}} & =\frac{1}{2} \frac{B_0^2}{\mu_0}=\frac{1}{2} \times \frac{\left(400 \times 10^{-6}\right)^2}{4 \pi \times 10^{-7}}=\frac{0.2}{\pi} \\ & =63.66 \times 10^{-3} \mathrm{~J} / \mathrm{m}^3 \end{aligned}$

Given, maximum amplitude of modulated signal,

$V_{\max }=15 \mathrm{~V}=V_m+V_c$ ..... (i)

Minimum amplitude of modulated signal,

$V_{\min }=5 \mathrm{~V}=V_m-V_c$ .... (ii)

where, $V_c, V_m$ are carries wave and modulated wave voltage.

Now, from Eqs. (i) and (ii), we get

$20=2 V_m \Rightarrow V_m=10 \mathrm{~V} $

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|}}$

Wavelength range of X-ray is $10^{-8} \mathrm{~m}$ to $10^{-12} \mathrm{~m}$. Hence, option (a) is correct.

Average intensity, $I_{\text {avg }}=\frac{P_{\text {avg }}}{A}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

where, $P_{\text {avg }}=$ average power generated by source

$ =4 \% \text { of } 100 \mathrm{~W}=\frac{4}{100} \times 100 \mathrm{~W}=4 \mathrm{~W} $

$A=$ area to which energy is transmitted

$ =4 \pi r^2=4 \pi \times(2)^2=4 \pi \times 4=16 \pi \mathrm{~m}^2 $

On putting the values into Eq. (i), we have

$ \begin{equation*} I_{\mathrm{avg}}=\frac{4}{16 \pi} \Rightarrow I_{\mathrm{avg}}=\frac{1}{4 \pi} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{equation*} $

Also average intensity,

$ \begin{equation*} I_{\mathrm{avg}}=\varepsilon_0 E_{\mathrm{rms}}^2 c \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{equation*} $

where, $\varepsilon_0=$ permittivity of free space and $E_{\text {rms }}=$ root mean square (rms) value of electric field.

$ \begin{aligned} c & =\text { speed of light in vacuum } \\ & =3 \times 10^8 \mathrm{~m} / \mathrm{s} \end{aligned} $

On putting the values into Eq. (iii), we get

$ \begin{equation*} I_{\mathrm{avg}}=\varepsilon_0 \times E_{\mathrm{rms}}^2 \times 3 \times 10^8 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv) \end{equation*} $

On equating both the Eqs. (ii) and (iv), we get

$ \frac{1}{4 \pi}=\varepsilon_0 \times E_{\mathrm{rms}}^2 \times 3 \times 10^8 $

$ \begin{aligned} & \Rightarrow \frac{1}{4 \pi \varepsilon_0 \times 3 \times 10^8}=E_{\mathrm{rms}}^2 \\ & E_{\mathrm{rms}}^2= \frac{1}{4 \pi \varepsilon_0} \times \frac{1}{3 \times 10^8} \\ &= 9 \times 10^9 \times \frac{1}{3 \times 10^8} \\ &\left(\because \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2\right) \\ &= 3 \times 10=\sqrt{30} \mathrm{~V} / \mathrm{m} \end{aligned} $

Electric field between plates of a capacitor is

$ E=V / d $

But $V=\frac{Q}{C}$ and $C=\frac{\varepsilon_0 A}{d} \Rightarrow Q=A E \varepsilon_0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \therefore \quad E=\frac{Q d}{\varepsilon_0 A d}=\frac{Q}{A \varepsilon_0} $

Displacement currrent is

$ \begin{aligned} I_d & \left.=\frac{d Q}{d t}=\frac{d}{d t}\left(E \cdot A \varepsilon_0\right) \quad \text { [from Eq. (i)] }\right .\\ & =\varepsilon_0 A\left(\frac{\Delta E}{\Delta t}\right)=\frac{4 \pi \varepsilon_0 A}{4 \pi} \times\left(\frac{\Delta E}{\Delta t}\right) \\ & =\frac{4 \pi \varepsilon_0 \cdot \pi r^2}{4 \pi} \cdot\left(\frac{\Delta E}{\Delta t}\right) \\ \Rightarrow I_d & =4 \pi \varepsilon_0\left(\frac{r^2}{4}\right)\left(\frac{\Delta E}{\Delta t}\right) \end{aligned} $

Substituting values given, we have

$ \begin{aligned} I_d & =\frac{1}{9 \times 10^9}\left(\frac{\left(10 \times 10^{-2}\right)^2}{4}\right) \times 3.6 \times 10^{12} \\ & =1 \mathrm{~A} \quad\left[\text { Since, } \frac{\Delta E}{\Delta t}=3.6 \times 10^{12} \mathrm{~V} / \mathrm{ms}\right] \end{aligned} $

Intensity of parallel beam of light is given as

$ I=\frac{1}{2} \varepsilon_0 E_0^2 c \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

where, $E_0=$ amplitude of electric field.

Here, $\quad I=\frac{15}{\pi} \frac{\mathrm{~W}}{\mathrm{~m}^2}, \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \frac{\mathrm{Nm}^2}{\mathrm{C}^2}$

$ c=3 \times 10^8 \mathrm{~ms}^{-1} $

From Eq. (i), we get

$ \begin{aligned} & E_0^2=\frac{2 \times I}{\varepsilon_0 \times c}=\frac{2 \times 4 \pi \times I}{4 \pi \varepsilon_0 \times c} \\ & \quad=\left(\frac{2 \times 4 \pi \times \frac{15}{\pi} \times 9 \times 10^9}{3 \times 10^8}\right) \\ & \Rightarrow \quad E_0^2=3600 \Rightarrow E_0=60 \mathrm{~N} / \mathrm{C} \end{aligned} $

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