The electric field associated with an electromagnetic wave propagating in a dielectric medium is given by
$\overrightarrow{E}=30(2\hat{x}+\hat{y})sin\left[2\pi \left(5\times {10}^{14}t-\frac{{10}^{7}z}{3}\right)\right]{\text{V m}}^{-1}$
Which of the following option(s) is/are correct?
Given: The speed of light in vacuum, c = 3 × 10⁸ms^–1

Correct Options:

1.

2. The refractive index of the medium is 2.

Step-by-Step Explanation:

1. Understanding the given Electric Field equation:
The given electric field of the electromagnetic wave is:
$\overrightarrow{E}=30(2\hat{x}+\hat{y})sin\left[2\pi \left(5\times {10}^{14}t-\frac{{10}^{7}z}{3}\right)\right]{\text{V m}}^{-1}$
This can be written in standard form $\overrightarrow{E}={\overrightarrow{E}}_{0}sin(\omega t-kz)$, where:
- Amplitude vector: ${\overrightarrow{E}}_0=60\hat{x}+30\hat{y}{\text{V m}}^{-1}$
- Angular frequency: $\omega =2\pi \times 5\times {10}^{14}{\text{rad s}}^{-1}={10}^{15}\pi {\text{rad s}}^{-1}$
- Wave propagation vector: $\overrightarrow{k}=k\hat{z}=\frac{2\pi \times {10}^7}{3}\hat{z}{\text{m}}^{-1}$ (so wave is propagating in the +z direction, $\hat{n}=\hat{z}$).

2. Finding the speed of light in the medium (v) and refractive index (n):
The phase velocity $v$ of the wave in the dielectric medium is given by:
$v=\frac{\omega }{k}=\frac{{10}^{15}\pi }{\left(\frac{2\pi \times {10}^7}{3}\right)}=\frac{3\times {10}^{15}}{2\times {10}^7}=1.5\times {10}^8{\text{m s}}^{-1}$

Now, calculate the refractive index $n$ of the medium:
$n=\frac{c}{v}=\frac{3\times {10}^8}{1.5\times {10}^8}=2$
Thus, the statement "The refractive index of the medium is 2" is correct.

3. Determining the Magnetic Field ($\overrightarrow{B}$):
The relation between magnetic field amplitude ${\overrightarrow{B}}_0$, electric field amplitude ${\overrightarrow{E}}_0$, wave propagation direction $\hat{n}=\hat{z}$, and phase velocity $v$ is given by:
${\overrightarrow{B}}_0=\frac{\hat{n}\times {\overrightarrow{E}}_0}{v}$

Substitute the values of and ${\overrightarrow{E}}_0=60\hat{x}+30\hat{y}$:
$\hat{z}\times (60\hat{x}+30\hat{y})=60(\hat{z}\times \hat{x})+30(\hat{z}\times \hat{y})=60\hat{y}-30\hat{x}$

Therefore, the magnetic field amplitude vector is:
${\overrightarrow{B}}_0=\frac{-30\hat{x}+60\hat{y}}{1.5\times {10}^8}=-2\times {10}^{-7}\hat{x}+4\times {10}^{-7}\hat{y}{\text{Wb m}}^{-2}$

The x-component of the magnetic field vector $B_x$ is given by:
$B_x=-2\times {10}^{-7}sin\left[2\pi \left(5\times {10}^{14}t-\frac{{10}^{7}z}{3}\right)\right]{\text{Wb m}}^{-2}$
Thus, the option representing $B_x$ is correct.