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Electromagnetic Waves question

2023 · Shift 2 · Q41
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Electromagnetic Waves question

2023 · Shift 2 · Q41

JEE AdvancedPhysicsElectromagnetic WavesMultiple correct+4 / −2
The electric field associated with an electromagnetic wave propagating in a dielectric medium is given by E⃗=30(2x^+y^)sin⁡[2π(5×1014t−1073z)]Vm−1\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}E=30(2x^+y^​)sin[2π(5×1014t−3107​z)]Vm−1. Which of the following option(s) is(are) correct? [Given: The speed of light in vacuum, c=3×108 m s−1c=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}c=3×108 m s−1 ]
  1. A
    Bx=−2×10−7sin⁡[2π(5×1014t−1073z)]Wbm−2B_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}Bx​=−2×10−7sin[2π(5×1014t−3107​z)]Wbm−2.
  2. B
    By=2×10−7sin⁡[2π(5×1014t−1073z)]Wbm−2B_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}By​=2×10−7sin[2π(5×1014t−3107​z)]Wbm−2.
  3. C
    The wave is polarized in the xyx yxy-plane with polarization angle 30∘30^{\circ}30∘ with respect to the xxx-axis.
  4. D
    The refractive index of the medium is 2.
View written solutionFree

Correct answer: A, D

  1. Given electric field
E⃗=30(2x^+y^)sin⁡[2π(5×1014t−1073z)] V m−1\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]\,\text{V m}^{-1}E=30(2x^+y^​)sin[2π(5×1014t−3107​z)]V m−1

So,

Ex=60sin⁡ϕ,Ey=30sin⁡ϕE_x = 60\sin\phi, \qquad E_y = 30\sin\phiEx​=60sinϕ,Ey​=30sinϕ

where

ϕ=2π(5×1014t−1073z)\phi = 2\pi\left(5\times 10^{14}t-\frac{10^7}{3}z\right)ϕ=2π(5×1014t−3107​z)

The wave propagates along the +z+z+z direction because the phase is of the form ωt−kz\omega t-kzωt−kz.

  1. Find angular frequency and wave number

Comparing with

sin⁡(ωt−kz)\sin(\omega t-kz)sin(ωt−kz)

we get

ω=2π(5×1014),k=2π(1073)\omega = 2\pi(5\times 10^{14}), \qquad k = 2\pi\left(\frac{10^7}{3}\right)ω=2π(5×1014),k=2π(3107​)

Hence wave speed in the medium is

v=ωk=2π(5×1014)2π(107/3)=5×1014⋅3107=1.5×108 m s−1v = \frac{\omega}{k} = \frac{2\pi(5\times 10^{14})}{2\pi(10^7/3)} = \frac{5\times 10^{14}\cdot 3}{10^7} = 1.5\times 10^8\,\text{m s}^{-1}v=kω​=2π(107/3)2π(5×1014)​=1075×1014⋅3​=1.5×108m s−1

Therefore refractive index,

n=cv=3×1081.5×108=2n = \frac{c}{v} = \frac{3\times 10^8}{1.5\times 10^8} = 2n=vc​=1.5×1083×108​=2

So Option D is correct.

  1. Relation between E⃗\vec EE and B⃗\vec BB

For an electromagnetic wave,

B⃗=1v k^×E⃗\vec B = \frac{1}{v}\,\hat k\times \vec EB=v1​k^×E

Here k^=z^\hat k = \hat zk^=z^, so

B⃗=1v z^×E⃗\vec B = \frac{1}{v}\,\hat z\times \vec EB=v1​z^×E

Now,

E⃗=(60x^+30y^)sin⁡ϕ\vec E = (60\hat x+30\hat y)\sin\phiE=(60x^+30y^​)sinϕ

Thus,

z^×E⃗=60(z^×x^)sin⁡ϕ+30(z^×y^)sin⁡ϕ\hat z\times \vec E = 60(\hat z\times \hat x)\sin\phi + 30(\hat z\times \hat y)\sin\phiz^×E=60(z^×x^)sinϕ+30(z^×y^​)sinϕ

Using

z^×x^=y^,z^×y^=−x^\hat z\times \hat x = \hat y, \qquad \hat z\times \hat y = -\hat xz^×x^=y^​,z^×y^​=−x^

we get

z^×E⃗=(60y^−30x^)sin⁡ϕ\hat z\times \vec E = (60\hat y-30\hat x)\sin\phiz^×E=(60y^​−30x^)sinϕ

Therefore,

B⃗=11.5×108(60y^−30x^)sin⁡ϕ\vec B = \frac{1}{1.5\times 10^8}(60\hat y-30\hat x)\sin\phiB=1.5×1081​(60y^​−30x^)sinϕ

So,

Bx=−301.5×108sin⁡ϕ=−2×10−7sin⁡ϕ TB_x = -\frac{30}{1.5\times 10^8}\sin\phi = -2\times 10^{-7}\sin\phi\,\text{T}Bx​=−1.5×10830​sinϕ=−2×10−7sinϕT By=601.5×108sin⁡ϕ=4×10−7sin⁡ϕ TB_y = \frac{60}{1.5\times 10^8}\sin\phi = 4\times 10^{-7}\sin\phi\,\text{T}By​=1.5×10860​sinϕ=4×10−7sinϕT

Hence:

  • Option A: correct
  • Option B: incorrect
  1. Polarization

The electric field has fixed direction along

2x^+y^2\hat x+\hat y2x^+y^​

Hence it is linearly polarized in the xyxyxy-plane.

The angle θ\thetaθ with the xxx-axis is

tan⁡θ=EyEx=3060=12\tan\theta = \frac{E_y}{E_x} = \frac{30}{60} = \frac{1}{2}tanθ=Ex​Ey​​=6030​=21​

So,

θ=tan⁡−1(12)≈26.6∘\theta = \tan^{-1}\left(\frac{1}{2}\right) \approx 26.6^\circθ=tan−1(21​)≈26.6∘

This is not 30∘30^\circ30∘.

So Option C is incorrect.

  1. Final evaluation of options
  • A: Correct
  • B: Incorrect
  • C: Incorrect
  • D: Correct

Thus the correct options are:

A,D\boxed{A, D}A,D​
  1. Comparison with stored answer

Stored correct answer: A, D

Derived answer: A, D

They match.

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