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
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Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
\(\frac{\text { Electric energy density }}{\text { Magnetic energy density }}=\frac{\frac{1}{2} \in_0 \mathrm{E}_{\mathrm{rms}}^2}{\left(\frac{\mathrm{B}_{\mathrm{rms}}^2}{2 \mu_0}\right)}\)
\(=\left(\frac{\mathrm{E}_{\mathrm{rms}}}{\mathrm{B}_{\mathrm{rms}}}\right)^2 \cdot \mu_0 \in_0 \quad\left[\mathrm{C}=\frac{1}{\mu_0 \epsilon_0}\right]\)
\(=\frac{\mathrm{C}^2}{\mathrm{C}^2}=1\)
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