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JEE Main 202225 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
25 June
Shift
Evening Shift
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2022 · 25 June · Evening Shift · Physics · Electromagnetic Waves

The electromagnetic waves travel in a medium at a speed of 2 . 0 ×...

The electromagnetic waves travel in a medium at a speed of 2.0×108ms−12 . 0 \times {10}^{8}m{s}^{- 1}. The relative permeability of the medium is 1.01 . 0. The relative permittivity of the medium will be

Options

  1. A2.252 . 25
  2. B4.254 . 25
  3. C6.256 . 25
  4. D8.258 . 25

Correct answer

A. 2.252 . 25

Step-by-step solution

Here, v=2×108ms−1v = 2 \times {10}^{8}\text{m}{\text{s}}^{- 1} and μr=1{\mu}_{r} = 1

The speed of electromagnetic waves in a medium is given by v=1με...(i)v = \dfrac{1}{\sqrt{\muε}} . .. \left(i\right)

where μ\mu and εε are absolute permeability and absolute permittivity of the medium.
Now, μ=μ0μr\mu = {\mu}_{0}{\mu}_{r} and ε=ε0εrε = {ε}_{0}{ε}_{r}
Then, we have,

v=1μ0μr×ε0εr=1μ0ε0×1μrεr=cμrεrv = \dfrac{1}{\sqrt{{\mu}_{0}{\mu}_{r} \times {ε}_{0}{ε}_{r}}} = \dfrac{1}{\sqrt{{\mu}_{0}{ε}_{0}}} \times \dfrac{1}{\sqrt{{\mu}_{r}{ε}_{r}}} = \dfrac{c}{\sqrt{{\mu}_{r}{ε}_{r}}}

orεr=c2v2μr=(3×108)2(2×108)2=94=2.25or{ε}_{r} = \dfrac{{c}^{2}}{{v}^{2}{\mu}_{r}} = \dfrac{{\left(3 \times {10}^{8}\right)}^{2}}{{\left(2 \times {10}^{8}\right)}^{2}} = \dfrac{9}{4} = 2 . 25
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