Electromagnetic Waves

JEE Main level, single correct, medium

67 of 165

Question 67 of 165

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

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Answer: (A) 2.252.25

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{\mu \epsilon }}...\left(i\right)

where μ\mu and ϵ\epsilon are absolute permeability and absolute permittivity of the medium.
Now, μ=μ0μr\mu ={\mu }_{0}{\mu }_{r} and ϵ=ϵ0ϵr\epsilon ={\epsilon }_{0}{\epsilon }_{r}
Then, we have,

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

orϵr=c2v2μr=(3×108)2(2×108)2=94=2.25or {\epsilon }_{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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