JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
For the plane electromagnetic wave given by \(\mathrm{E}=\mathrm{E}_0 \sin (\omega \mathrm{t}-\mathrm{kx})\) and \(\mathrm{B}=\mathrm{B}_0 \sin (\omega \mathrm{t}-\mathrm{kx})\), the ratio of average electric energy density to average magnetic energy density is
- A \(1\)
- B \(\frac{1}{2}\)
- C \(2\)
- D \(4\)
Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
\(\frac{\text { Electric energy density }}{\text { Magnetic energy density }}=\frac{\frac{1}{2} \in_0 \mathrm{E}_{\mathrm{rms}}^2}{\left(\frac{\mathrm{B}_{\mathrm{rms}}^2}{2 \mu_0}\right)}\)…
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