The electric field associated with an electromagnetic wave propagating in a dielectric medium is given by 𝐸⃗ = 30(2𝑥̂ + 𝑦̂)sin [2𝜋 (5 × 10 14𝑡 − 107 3 𝑧)] V m−1 . Which of the following option(s) is(are) correct? [Given: The speed of light in vacuum, 𝑐 = 3 × 108 m s −1 ] (A) 𝐵𝑥 = −2 × 10 −7 sin [2𝜋 (5 × 10 14 𝑡 − 107 3 𝑧)] Wb m−2 . (B) 𝐵𝑦 = 2 × 10 −7 sin [2𝜋 (5 × 10 14 𝑡 − 107 3 𝑧)] Wb m−2 . (C) The wave is polarized in the 𝑥𝑦-plane with polarization angle 30° with respect to the 𝑥-axis. (D) The refractive index of the medium is 2.

To determine which of the given options are correct, we need to analyze the provided electric field and use the relationships between the electric field, magnetic field, and the properties of the electromagnetic wave in a dielectric medium.

Given:

• Electric field:

  $$\mathbf{E}=30(2\widehat{x}+\widehat{y})\sin \left[2\pi (5\times 10^{14}t-\frac{10^7}{3}z)\right]\text{V/m}$$

• Speed of light in vacuum: $c=3\times 10^8\text{m/s}$

Step 1: Determine the wave vector and angular frequency

The general form of the electric field for a plane wave is:

$$\mathbf{E}={\mathbf{E}}_0\sin (\mathbf{k}\cdot \mathbf{r}-\omega t)$$

Comparing with the given electric field:

$$\omega =2\pi \times 5\times 10^{14}\text{rad/s}$$$$k=\frac{2\pi \times 10^7}{3}{\text{m}}^{-1}$$

Step 2: Determine the magnetic field

The magnetic field $\mathbf{B}$ is related to the electric field $\mathbf{E}$ by:

$$\mathbf{B}=\frac{1}{c}(\widehat{k}\times \mathbf{E})$$

Here, $\widehat{k}$ is the unit vector in the direction of propagation, which is the $z$-direction. Thus:

$$\widehat{k}=\widehat{z}$$

The electric field components are:

$$E_x=60\sin \left[2\pi (5\times 10^{14}t-\frac{10^7}{3}z)\right]$$$$E_y=30\sin \left[2\pi (5\times 10^{14}t-\frac{10^7}{3}z)\right]$$

Using the cross product:

$$\mathbf{B}=\frac{1}{c}(\widehat{z}\times \mathbf{E})=\frac{1}{c}(-E_y\widehat{x}+E_x\widehat{y})$$

Substituting the values:

$$B_x=-\frac{E_y}{c}=-\frac{30}{3\times 10^8}\sin \left[2\pi (5\times 10^{14}t-\frac{10^7}{3}z)\right]=-10^{-7}\sin \left[2\pi (5\times 10^{14}t-\frac{10^7}{3}z)\right]$$$$B_y=\frac{E_x}{c}=\frac{60}{3\times 10^8}\sin \left[2\pi (5\times 10^{14}t-\frac{10^7}{3}z)\right]=2\times 10^{-7}\sin \left[2\pi (5\times 10^{14}t-\frac{10^7}{3}z)\right]$$

Step 3: Check the options

• Option (A):

  $$B_x=-2\times 10^{-7}\sin \left[2\pi (5\times 10^{14}t-\frac{10^7}{3}z)\right]$$

  This is incorrect because our calculation shows $B_x=-10^{-7}\sin (\dots )$.

• Option (B):

  $$B_y=2\times 10^{-7}\sin \left[2\pi (5\times 10^{14}t-\frac{10^7}{3}z)\right]$$

  This is correct as it matches our calculation.

• Option (C): The wave is polarized in the $xy$-plane with a polarization angle of $30^{\circ }$ with respect to the $x$-axis.
  The electric field components are $E_x=60$ and $E_y=30$, so the polarization angle $\theta$ is:

  $$\theta ={\tan }^{-1}\left(\frac{E_y}{E_x}\right)={\tan }^{-1}\left(\frac{30}{60}\right)={\tan }^{-1}\left(\frac{1}{2}\right)\approx 26.565^{\circ }$$

  This is not exactly $30^{\circ }$, so this option is incorrect.

• Option (D): The refractive index of the medium is 2.
  The refractive index $n$ is given by:

  $$n=\frac{c}{v}$$

  where $v$ is the phase velocity of the wave in the medium. The phase velocity $v$ is:

  $$v=\frac{\omega }{k}=\frac{2\pi \times 5\times 10^{14}}{2\pi \times \frac{10^7}{3}}=\frac{5\times 10^{14}}{\frac{10^7}{3}}=1.5\times 10^8\text{m/s}$$

  Thus:

  $$n=\frac{c}{v}=\frac{3\times 10^8}{1.5\times 10^8}=2$$

  This option is correct.

Final Answer:

The correct options are (B) and (D).