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Electromagnetic Waves question

2021 · 1 Sep · Shift 2 · Q64
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  5. /2021 · 1 Sep · Shift 2 · Q64

Electromagnetic Waves question

2021 · 1 Sep · Shift 2 · Q64

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
Electric field of plane electromagnetic wave propagating through a non-magnetic medium is given by E = 20cos(2 ×\times× 1010 t −-− 200x) V/m. The dielectric constant of the medium is equal to : (Take μ\muμ r = 1)
  1. A
    9
  2. B
    2
  3. C
    13{1 \over 3}31​
  4. D
    3
View written solutionFree

Correct answer: A

  1. Write the given wave in standard form

    For a plane electromagnetic wave, E=E0cos⁡(ωt−kx)E = E_0 \cos(\omega t - kx)E=E0​cos(ωt−kx)

    Comparing with the given expression E=20cos⁡(2×1010t−200x) V/mE = 20\cos(2 \times 10^{10} t - 200x)\ \text{V/m}E=20cos(2×1010t−200x) V/m

    we get: ω=2×1010 rad/s,k=200 rad/m\omega = 2 \times 10^{10}\ \text{rad/s}, \qquad k = 200\ \text{rad/m}ω=2×1010 rad/s,k=200 rad/m

  2. Find the wave speed in the medium

    The phase velocity is v=ωkv = \frac{\omega}{k}v=kω​

    So, v=2×1010200=108 m/sv = \frac{2 \times 10^{10}}{200} = 10^8\ \text{m/s}v=2002×1010​=108 m/s

  3. Use the relation between speed and dielectric constant

    In a medium, v=1μεv = \frac{1}{\sqrt{\mu \varepsilon}}v=με​1​

    Since the medium is non-magnetic, μr=1\mu_r = 1μr​=1

    Also, v=cμrεr=cεrv = \frac{c}{\sqrt{\mu_r \varepsilon_r}} = \frac{c}{\sqrt{\varepsilon_r}}v=μr​εr​​c​=εr​​c​

    Therefore, εr=cv\sqrt{\varepsilon_r} = \frac{c}{v}εr​​=vc​

    Taking c=3×108 m/sc = 3 \times 10^8\ \text{m/s}c=3×108 m/s

    we get εr=3×108108=3\sqrt{\varepsilon_r} = \frac{3 \times 10^8}{10^8} = 3εr​​=1083×108​=3

    Hence, εr=9\varepsilon_r = 9εr​=9

  4. Match with the options

    The dielectric constant of the medium is: 9\boxed{9}9​

    So the correct option is A.

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