QJEEAdv23Phdce5: JEE Advanced 2023 Physics - Class 12 JEEAdv23Phdce5 Solution: The electric field associated with an electromagnetic wave…

Question

Verified Solution • 2025–26 Session moderate

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 \times 10^{14} t-\frac{10^7}{3} z\right)\right] \mathrm{Vm}^{-1}

. Which of the following option(s) is(are) correct?

[Given: The speed of light in vacuum,

c=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}

]

A $B_x=-2 \times 10^{-7} \sin \left[2 \pi\left(5 \times 10^{14} t-\frac{10^7}{3} z\right)\right] \mathrm{Wb} \mathrm{m}^{-2}$ .
B $B_y=2 \times 10^{-7} \sin \left[2 \pi\left(5 \times 10^{14} t-\frac{10^7}{3} z\right)\right] \mathrm{Wb} \mathrm{m}^{-2}$ .
C The wave is polarized in the $x y$ -plane with polarization angle $30^{\circ}$ with respect to the $x$ -axis.
D The refractive index of the medium is 2.

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

\begin{aligned} &amp; \tan \alpha=\frac{1}{2} \\\\ &amp; \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.

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