Electromagnetic Waves

1. Calculation of the Speed of Light in a Medium:

- The speed of light in a medium, denoted as $ V $, is given by the formula $ V = \frac{W}{k} $.

- Substituting values, we get $ V = \frac{3 \times 5 \times 10^{14}}{10^7} = 1.5 \times 10^8 \text{ m/s} $.

2. Determination of the Refractive Index:

- The refractive index, $ \mu $, is the ratio of the speed of light in vacuum, $ C $, to the speed of light in the medium, $ V $.

- Substituting values, $ \mu = \frac{3 \times 10^8}{1.5 \times 10^8} = 2 $.

- Hence, the refractive index $ \mu $ is 2.

Therefore, the option (D) is the answer.

Given, $\vec{E}=30(2 \hat{x}+\hat{y}) \sin \left[2 \pi\left(5 \times 10^{14} t-\frac{10^7}{3} z\right)\right] \mathrm{Vm}^{-1}$

Now magnitude of

$ \begin{aligned} B & =\frac{E}{v} \\\\ \Rightarrow B & =\frac{30 \sqrt{5}}{1.5 \times 10^8} \end{aligned} $



$\begin{aligned} & \tan \alpha=\frac{1}{2} \\\\ & \sin \alpha=\frac{1}{\sqrt{5}}\end{aligned}$

The direction of $\vec{B}$ is along the direction of $\vec{v} \times \vec{E}$

i.e. $\hat{v} \times \hat{E}=\hat{B}$

Now $ B_x=-B \sin \alpha$

$ \Rightarrow B_x=-\frac{30 \sqrt{5}}{1.5 \times 10^8} \times \frac{1}{\sqrt{5}}=-2 \times 10^7 $

Therefore, option (A) is the answer.

The speed of light in vacuum is given by, $c = {1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$ and The impedance (resistance) of free space is defined a $R = \sqrt {{{{\mu _0}} \over {{\varepsilon _0}}}} $. Using this check the dimensional correctness of equations.

(a) ${\mu _0}{I^2} = {\varepsilon _0}{V^2}$

${{{\mu _0}} \over {{\varepsilon _0}}} = {{{V^2}} \over {{I^2}}} = {R^2}$

$\therefore$ ${R^2} = {R^2}$ which is dimensionally correct.

(b) ${\varepsilon _0}I = {\mu _0}V \Rightarrow {{{\varepsilon _0}} \over {{\mu _0}}} = {V \over I} \Rightarrow {1 \over {{R^2}}} = R$

which is dimensionally incorrect.

(c) $I = {\varepsilon _0}cV \Rightarrow {I \over V} = {\varepsilon _0}C = {{{\varepsilon _0}} \over {\sqrt {{\varepsilon _0}{\mu _0}} }}$ = $\sqrt {{{{\varepsilon _0}} \over {{\mu _0}}}} $ = $ {1 \over R}$

${1 \over R} = \sqrt {{{{\varepsilon _0}} \over {{\mu _0}}}} \Rightarrow {1 \over R} = {1 \over R}$

which is dimensionally correct.

(d) ${\mu _0}cI = {\varepsilon _0}V \Rightarrow {{{\mu _0}c} \over {{\varepsilon _0}}} = {V \over I}$

${{{\mu _0}} \over {{\varepsilon _0}}}{1 \over {\sqrt {{\varepsilon _0}{\mu _0}} }} = R \Rightarrow {1 \over {{\varepsilon _0}}}\sqrt {{{{\mu _0}} \over {{\varepsilon _0}}}} = R$

${R \over {{\varepsilon _0}}} = R$, which is dimensionally incorrect.