ExamBro
ExamBro
JEE Mains · Physics · STD 12 - Electromagnetic waves
  • JEE Mains 2023
  • Medium
  • Numerical

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

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(A) \(1\)

Step-by-step Solution

Detailed explanation

\(\frac{\text { Electric energy density }}{\text { Magnetic energy density }}=\frac{\frac{1}{2} \in_0 \mathrm{E}_{\mathrm{rms}}^2}{\left(\frac{\mathrm{B}_{\mathrm{rms}}^2}{2 \mu_0}\right)}\)

\(=\left(\frac{\mathrm{E}_{\mathrm{rms}}}{\mathrm{B}_{\mathrm{rms}}}\right)^2 \cdot \mu_0 \in_0 \quad\left[\mathrm{C}=\frac{1}{\mu_0 \epsilon_0}\right]\)

\(=\frac{\mathrm{C}^2}{\mathrm{C}^2}=1\)

Same subject
Explore more questions on app