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8.Electromagnetic waves
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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
$4$
(JEE MAIN-2023)
Solution
$\frac{\text { Electric energy density }}{\text { Magnetic energy density }}=\frac{\frac{1}{2} \in_0 \mathrm{E}_{\mathrm{rms}}^2}{\left(\frac{\mathrm{B}_{\mathrm{rms}}^2}{2 \mu_0}\right)}$
$=\left(\frac{\mathrm{E}_{\mathrm{rms}}}{\mathrm{B}_{\mathrm{rms}}}\right)^2 \cdot \mu_0 \in_0 \quad\left[\mathrm{C}=\frac{1}{\mu_0 \epsilon_0}\right]$
$=\frac{\mathrm{C}^2}{\mathrm{C}^2}=1$
Standard 12
Physics
