TS EAMCET · Physics · Electromagnetic Waves
The electric field for an electromangetic wave in free space is \(\mathbf{E}=\mathbf{i} 30 \cos \left(k z-5 \times 10^8 t\right)\), where magnitude of \(E\) is in \(\mathrm{V} / \mathrm{m}\). The magnitude of wave vector, \(k\) is (velocity of em wave in free space \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\) )
- A \(0.46 \mathrm{rad} \mathrm{m}^{-1}\)
- B \(3 \mathrm{rad} \mathrm{m}{ }^{-1}\)
- C \(1.66 \mathrm{rad} \mathrm{m}^{-1}\)
- D \(0.83 \mathrm{rad} \mathrm{m} \mathrm{m}^{-1}\)
Answer & Solution
Correct Answer
(C) \(1.66 \mathrm{rad} \mathrm{m}^{-1}\)
Step-by-step Solution
Detailed explanation
The given \(\mathbf{E}=\mathbf{i} 30 \cos \left(k z-5 \times 10^8 t\right)\) We know that, \(\mathbf{E}=\mathrm{iV} \cos (k z-\omega t)\) Comparing the both equations, \(\omega=5 \times 10^8 \mathrm{rad} / \mathrm{s}\) But we know that also…
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