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A plane electromagnetic wave travels in a medium of relative permeability \mathrm{1.61} and relative permittivity \mathrm{6.44}.If magnitude of magnetic intensity is \mathrm{4.5 \times 10^{-2} \mathrm{Am}^{-1}}at a point, what will be the approximate magnitude of electric field intensity at that point?

(Given : Permeability of free space \mathrm{\mu_{0}=4 \pi \times 10^{-7} \mathrm{NA}^{-2}}, speed of light in vacuum \mathrm{c=3 \times 10^{8} \mathrm{~ms}^{-1}})

Option: 1

\mathrm{16.96\, \mathrm{Vm}^{-1}}


Option: 2

\mathrm{2.25 \times 10^{-2}\, \mathrm{Vm}^{-1}}


Option: 3

\mathrm{8.48 \, Vm^{-1}}


Option: 4

\mathrm{6.75 \times 10^{6} \: \mathrm{Vm}^{-1}}


Answers (1)

\mathrm{\mu _{r}= 1.61}
\mathrm{\xi _{r}= 6.44}
We know that ,

\mathrm{B= \mu\, H= \mu _{0}\,\mu _{r}\, H }
\mathrm{\frac{E}{B} = V= \frac{1}{\sqrt{\mu \, \xi }}= \frac{C}{\sqrt{\mu _{r}\, \xi _{r}}}

\mathrm{E= \frac{3\times 10^{8}\times 1.61\times \left ( 4\pi\times 10^{-7} \right )\left ( 4.5\times 10^{-2} \right )}{\sqrt{1.61\times 6.44}}}
\mathrm{E= 8.478\left ( \frac{V}{m} \right )}
The correct option is (3)


 

Posted by

Kshitij

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