A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5 $\times$ 10$-$2 Am$-$1 at a point, what will be the approximate magnitude of electric field intensity at that point?
(Given : Permeability of free space $\mu$0 = 4$\pi$ $\times$ 10$-$7 NA$-$2, speed of light in vacuum c = 3 $\times$ 108 ms$-$1)
Solution
<p>H = 4.5 $\times$ 10<sup>$-$2</sup></p>
<p>So B = $\mu$<sub>0</sub>$\mu$H</p>
<p>Thus $E = {c \over n}B$ (where n $\Rightarrow$ refractive index)</p>
<p>So $$E = {{3 \times {{10}^8} \times 4\pi \times {{10}^{ - 7}} \times 1.61 \times 4.5 \times {{10}^{ - 2}}} \over {\sqrt {1.61 \times 6.44} }}$$</p>
<p>$E = 8.48$</p>
About this question
Subject: Physics · Chapter: Electromagnetic Waves · Topic: Properties of EM Waves
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