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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Electromagnetic Waves

NTA Abhyas 2020 · Physics · Electromagnetic Waves

Electromagnetic waves travel in a medium with a speed of 2 × 10^ 8...

Electromagnetic waves travel in a medium with a speed of \text{2} \times {\text{10}^{8} \text{m}s^{- 1}. The relative permeability of the medium is 11. Find the relative permittivity.

Options

  1. A2.252.25
  2. B3.253.25
  3. C3.703.70
  4. D4.204.20

Correct answer

A. 2.252.25

Step-by-step solution

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

The speed of electromagnetic waves in a medium is given by

v=1\muε,....(i)\text{v} = \dfrac{1}{\sqrt{\text{\mu} \text{ε}}} \text{,} \text{....(i)}

where μ\mu and εε are absolute permeability and absolute permittivity of the medium.

Now, μ=μ0μr\mu = {μ}_{0} {μ}_{r} and ε=ε0εrε = {ε}_{0} {ε}_{r}

Therefore, the equation (i) becomes

v=1μ0μr×ε0εr=1μ0ε0×1μrεr=cμrεr\text{v} = \dfrac{1}{\sqrt{{μ}_{0} {μ}_{r} \times {ε}_{0} {ε}_{r}}} = \dfrac{1}{\sqrt{{μ}_{0} {ε}_{0}}} \times \dfrac{1}{\sqrt{{μ}_{r} {ε}_{r}}} = \dfrac{\text{c}}{\sqrt{{μ}_{r} {ε}_{r}}}

orεr=c2v2μr=(3×108)2(2×108)2×1=2.25\text{or} {ε}_{r} = \dfrac{c^{2}}{v^{2} {μ}_{r}} = \dfrac{{\left(3 \times {10}^{8}\right)}^{2}}{{\left(2 \times {10}^{8}\right)}^{2} \times 1} = \text{2.25}
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