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Electromagnetic Waves question

2024 · 6 Apr · Shift 1 · Q64
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  5. /2024 · 6 Apr · Shift 1 · Q64

Electromagnetic Waves question

2024 · 6 Apr · Shift 1 · Q64

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
Electromagnetic waves travel in a medium with speed of 1.5×108 m s−11.5 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}1.5×108 m s−1. The relative permeability of the medium is 2.0. The relative permittivity will be:
  1. A
    4
  2. B
    1
  3. C
    2
  4. D
    5
View written solutionFree

Correct answer: C

  1. Use the speed of electromagnetic waves in a medium

    The speed of electromagnetic waves in a medium is v=1μεv = \frac{1}{\sqrt{\mu\varepsilon}}v=με​1​

    In terms of relative permeability μr\mu_rμr​ and relative permittivity εr\varepsilon_rεr​, v=cμrεrv = \frac{c}{\sqrt{\mu_r\varepsilon_r}}v=μr​εr​​c​

    where c=3×108 m s−1c = 3 \times 10^8\ \text{m s}^{-1}c=3×108 m s−1.

  2. Substitute the given values

    Given: v=1.5×108 m s−1,μr=2v = 1.5 \times 10^8\ \text{m s}^{-1}, \quad \mu_r = 2v=1.5×108 m s−1,μr​=2

    So, 1.5×108=3×1082εr1.5 \times 10^8 = \frac{3 \times 10^8}{\sqrt{2\varepsilon_r}}1.5×108=2εr​​3×108​

  3. Solve for εr\varepsilon_rεr​

    Cancel 10810^8108 from both sides: 1.5=32εr1.5 = \frac{3}{\sqrt{2\varepsilon_r}}1.5=2εr​​3​

    Rearranging, 2εr=31.5=2\sqrt{2\varepsilon_r} = \frac{3}{1.5} = 22εr​​=1.53​=2

    Squaring both sides, 2εr=42\varepsilon_r = 42εr​=4

    Hence, εr=2\varepsilon_r = 2εr​=2

  4. Match with the options

    εr=2\varepsilon_r = 2εr​=2

    So the correct option is C.

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