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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

  1. A \(1\)
  2. B \(\frac{1}{2}\)
  3. C \(2\)
  4. D \(4\)
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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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