JEE Mains · Physics · STD 12 - 8. 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 \(c =3 \times 10^{8} \;ms ^{-1}\) )
- A \(16.96\; Vm ^{-1}\)
- B \(2.25 \times 10^{-2}\; Vm ^{-1}\)
- C \(8.48\; Vm ^{-1}\)
- D \(6.75 \times 10^{6} \;Vm ^{-1}\)
Answer & Solution
Correct Answer
(C) \(8.48\; Vm ^{-1}\)
Step-by-step Solution
Detailed explanation
\(\mu_{ I }=1.61 \quad \epsilon_{ I }=6.44\) \(B =4.5 \times 10^{-2}\) \(E =\) \(?\) \(C =\frac{1}{\sqrt{\mu_{0} \epsilon_{0}}} V =\frac{1}{\sqrt{\mu \epsilon}}\) \(\frac{ C }{ V }=\sqrt{\mu_{ r } \epsilon_{ r }}=\sqrt{1.61 \times 6.44}\)…
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