Question 18: EM waves

Question

The electric field associated with an electromagnetic wave propagating in a dielectric medium is given by $\vec{E}=30(2 \hat{x}+\hat{y}) \sin \left[2 \pi\left(5 	imes 10^{14} t-\frac{10^7}{3} zight)ight] \mathrm{Vm}^{-1}$. Which of the following option(s) is(are) correct?
[Given: The speed of light in vacuum, $c=3 	imes 10^8 \mathrm{~m} \mathrm{~s}^{-1}$ ]

Accepted Answer

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 $$ V = \frac{3 	imes 5 	imes 10^{14}}{10^7} = 1.5 	imes 10^8 	ext{ m/s} $$.
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, $$ \mu = \frac{3 	imes 10^8}{1.5 	imes 10^8} = 2 $$.
- Hence, the refractive index $$ \mu $$ is 2.
Therefore, the option (D) is the answer.
Given, $\vec{E}=30(2 \hat{x}+\hat{y}) \sin \left[2 \pi\left(5 	imes 10^{14} t-\frac{10^7}{3} zight)ight] \mathrm{Vm}^{-1}$
Now magnitude of
$$
\begin{aligned}
B & =\frac{E}{v} \\
\Rightarrow B & =\frac{30 \sqrt{5}}{1.5 	imes 10^8}
\end{aligned}
$$
$\begin{aligned} & 	an \alpha=\frac{1}{2} \\ & \sin \alpha=\frac{1}{\sqrt{5}}\end{aligned}$
The direction of $\vec{B}$ is along the direction of $\vec{v} 	imes \vec{E}$
i.e. $\hat{v} 	imes \hat{E}=\hat{B}$
Now $ B_x=-B \sin \alpha$
$$
\Rightarrow B_x=-\frac{30 \sqrt{5}}{1.5 	imes 10^8} 	imes \frac{1}{\sqrt{5}}=-2 	imes 10^7
$$
Therefore, option (A) is the answer.

Answer

$B_x=-2 	imes 10^{-7} \sin \left[2 \pi\left(5 	imes 10^{14} t-\frac{10^7}{3} zight)ight] \mathrm{Wb} \mathrm{m}^{-2}$.

Answer

$B_y=2 	imes 10^{-7} \sin \left[2 \pi\left(5 	imes 10^{14} t-\frac{10^7}{3} zight)ight] \mathrm{Wb} \mathrm{m}^{-2}$.

Answer

The wave is polarized in the $x y$-plane with polarization angle $30^{\circ}$ with respect to the $x$-axis.

Answer

The refractive index of the medium is 2.

EM waves

Choose the correct answer: