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Electromagnetic Waves question

2022 · 25 Jun · Shift 2 · Q58
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  5. /2022 · 25 Jun · Shift 2 · Q58

Electromagnetic Waves question

2022 · 25 Jun · Shift 2 · Q58

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
The electromagnetic waves travel in a medium at a speed of 2.0 ×\times× 108 m/s. The relative permeability of the medium is 1.0. The relative permittivity of the medium will be :
  1. A
    2.25
  2. B
    4.25
  3. C
    6.25
  4. D
    8.25
View written solutionFree

Correct answer: A

  1. Speed of electromagnetic waves in a medium

    The speed of an electromagnetic wave in a medium is v=cμrεrv = \frac{c}{\sqrt{\mu_r \varepsilon_r}}v=μr​εr​​c​ where:

    • v=2.0×108 m/sv = 2.0 \times 10^8\ \text{m/s}v=2.0×108 m/s
    • c=3.0×108 m/sc = 3.0 \times 10^8\ \text{m/s}c=3.0×108 m/s
    • μr=1.0\mu_r = 1.0μr​=1.0
    • εr=?\varepsilon_r = ?εr​=?
  2. Substitute the given values

    2.0×108=3.0×1081⋅εr2.0 \times 10^8 = \frac{3.0 \times 10^8}{\sqrt{1 \cdot \varepsilon_r}}2.0×108=1⋅εr​​3.0×108​

    Cancelling 10810^8108: 2=3εr2 = \frac{3}{\sqrt{\varepsilon_r}}2=εr​​3​

  3. Solve for εr\varepsilon_rεr​

    εr=32\sqrt{\varepsilon_r} = \frac{3}{2}εr​​=23​

    Squaring both sides: εr=(32)2=94=2.25\varepsilon_r = \left(\frac{3}{2}\right)^2 = \frac{9}{4} = 2.25εr​=(23​)2=49​=2.25

  4. Check options

    • A: 2.252.252.25 ✅
    • B: 4.254.254.25
    • C: 6.256.256.25
    • D: 8.258.258.25

Therefore, the correct answer is A.

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