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 × 10-2;Am -1 at a point, what will be the approximate magnitude of electric field intensity at that point? (Given: permeability of free space μ0=4 π × 10-7;NA -2, speed of light in vacuum c =3 × 108;ms -1)

(B) 2.25 \times 10^{-2}\; Vm ^{-1}

(D) 6.75 \times 10^{6} \;Vm ^{-1}

JEE Mains · Physics · STD 12 - Electromagnetic waves

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 \(c =3 \times 10^{8} \;ms ^{-1}\) )

Pick an option to check your answer.

Verified Solution

Correct Answer

(C) \(8.48\; Vm ^{-1}\)

Detailed explanation

\(\mu_{ I }=1.61 \quad \epsilon_{ I }=6.44\)

\(B =4.5 \times 10^{-2}\)

\(E =\) \(?\)

\(C =\frac{1}{\sqrt{\mu_{0} \epsilon_{0}}} V =\frac{1}{\sqrt{\mu \epsilon}}\)

\(\frac{ C }{ V }=\sqrt{\mu_{ r } \epsilon_{ r }}=\sqrt{1.61 \times 6.44}\)

\(\frac{E}{B}=V=\frac{3 \times 10^{8}}{\sqrt{1.61 \times 6.44}}=9.32 \times 10^{7} m / s\)

\(E =4.5 \times 10^{-2} \times 9.32 \times 10^{7}\)

\(=4.2 \times 10^{6}\)

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