A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5×10-2Am-1 at a point, what will be the approximate magnitude of electric field intensity at that point ?(Given : permeability of free space μ0=4π×10-7 NA-2, speed of light in vacuum c =3×108 ms-1 )

μr=1.61 ϵr=6.44 B=4.5×10−2 E=? C=1μ0ϵ0V=1μ∈ CV=μrϵr=1.61×6.44 EB=V=3×1081.61×6.44=9.32×107m/s E=4.5×10−2×9.32×107 =4.2×106

A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5×10-2Am-1 at a point, what will be the approximate magnitude of electric field intensity at that point ?(Given : permeability of free space μ0=4π×10-7 NA-2, speed of light in vacuum c =3×108 ms-1 )

μr=1.61 ϵr=6.44 B=4.5×10−2 E=? C=1μ0ϵ0V=1μ∈ CV=μrϵr=1.61×6.44 EB=V=3×1081.61×6.44=9.32×107m/s E=4.5×10−2×9.32×107 =4.2×106

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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 $μ_{0} = 4 π \times 10^{- 7}$ ${NA}^{- 2}$ , speed of light in vacuum c $= 3 \times 10^{8} {ms}^{- 1}$ )

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$2.25 \times 10^{- 2} {Vm}^{- 1}$

$16.96 {Vm}^{- 1}$

$8.48 {Vm}^{- 1}$

$6.75 \times 10^{6} {Vm}^{- 1}$

answer is C.

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Detailed Solution

$\begin{aligned} μ_{r} = 1.61 ϵ_{r} = 6.44 \\ B = 4.5 \times 10^{- 2} \\ E = ? \\ C = \frac{1}{\sqrt{μ_{0} ϵ_{0}}} V = \frac{1}{\sqrt{μ \in}} \\ \frac{C}{V} = \sqrt{μ_{r} ϵ_{r}} = \sqrt{1.61 \times 6.44} \\ \frac{E}{B} = V = \frac{3 \times 10^{8}}{\sqrt{1.61 \times 6.44}} = 9.32 \times 10^{7} m / s \\ E = 4.5 \times 10^{- 2} \times 9.32 \times 10^{7} \\ = 4.2 \times 10^{6} \end{aligned}$