The electric field associated with an electromagnetic wave propagating in a dielectric medium is given by Which of the following option(s) is(are) correct? [Given: The speed of light in vacuum, ] [2023]

Question

The electric field associated with an electromagnetic wave propagating in a dielectric medium is given by $\vec{E} = 30 (2 \hat{x} + \hat{y}) \sin [2 π (5 \times 10^{14} t - \frac{10^{7}}{3} z)] {Vm}^{- 1} .$

Which of the following option(s) is(are) correct? [Given: The speed of light in vacuum, $c = 3 \times 10^{8} {ms}^{- 1}$ ] [2023]

Smart Achievers · Accepted Answer

(1, 4)

Speed of light in the medium, $C_{medium} = \frac{ω}{k} = \frac{5 \times 10^{14}}{\frac{10^{7}}{3}}$

$= 1.5 \times 10^{8} m/s$ $∴$ $μ = \frac{C_{air}}{C_{medium}} = 2$

Also,

$C_{medium} = \frac{E}{B}$ $\Rightarrow B = \frac{E}{C_{m}} = \frac{30 \sqrt{5}}{1.5 \times 10^{8}} = 2 \sqrt{5} \times 10^{- 7}$

$B_{d i r e c t i o n}$ $= \vec{V} \times \vec{E} \equiv \hat{k} \times (2 \hat{i} + \hat{j}) = \frac{2 \hat{j} - \hat{i}}{\sqrt{5}}$

$∴$ $B_{x} = 2 \times 10^{- 7} (- \hat{i} + 2 \hat{j}) \sin [2 π (5 \times 10^{14} t - \frac{10^{7}}{3} z)]$

Solution

1 $B_{x} = - 2 \times 10^{- 7} \sin [2 π (5 \times 10^{14} t - \frac{10^{7}}{3} z)] Wb m^{- 2} .$

2 $B_{y} = 2 \times 10^{- 7} \sin [2 π (5 \times 10^{14} t - \frac{10^{7}}{3} z)] Wb m^{- 2} .$

3 The wave is polarized in the $x y$ -plane with polarization angle $30 °$ with respect to the $x$ -axis.

4 The refractive index of the medium is 2.

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Solution

Q. The electric field associated with an electromagnetic wave propagating in a dielectric medium is given by E→=30(2x^+y^)sin[2π(5×1014t-1073z)] Vm-1. Which of the following option(s) is(are) correct [Given: The speed of light in vacuum, c=3×108 ms-1] [2023]

1 Bx=-2×10-7sin[2π(5×1014t-1073z)] Wb m-2.

2 By=2×10-7sin[2π(5×1014t-1073z)] Wb m-2.

3 The wave is polarized in the xy-plane with polarization angle 30° with respect to the x-axis.

4 The refractive index of the medium is 2.

Ans. (1, 4) Speed of light in the medium, Cmedium=ωk=5×10141073 =1.5×108 m/s ∴ μ=CairCmedium=2 Also, Cmedium=EB⇒B=ECm=3051.5×108=25×10-7 Bdirection=V→×E→≡k^×(2i^+j^)=2j^-i^5 ∴ Bx=2×10-7(-i^+2j^)sin[2π(5×1014t-1073z)]

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1 $B_{x} = - 2 \times 10^{- 7} \sin [2 π (5 \times 10^{14} t - \frac{10^{7}}{3} z)] Wb m^{- 2} .$

2 $B_{y} = 2 \times 10^{- 7} \sin [2 π (5 \times 10^{14} t - \frac{10^{7}}{3} z)] Wb m^{- 2} .$

3 The wave is polarized in the $x y$ -plane with polarization angle $30 °$ with respect to the $x$ -axis.