A plane electromagnetic wave of frequency $50 \mathrm{MHz}$ travels in free space. If the average energy…

A plane electromagnetic wave of frequency $50 \mathrm{MHz}$ travels in free space. If the average energy densities in the electric field and magnetic field are $\mathrm{K}_{\mathrm{E}}$ and $\mathrm{K}_{\mathrm{B}}$ respectively, then the correct option in the following is
  1. $\mathrm{K}_{\mathrm{E}}=\mathrm{K}_{\mathrm{B}}$
  2. $\mathrm{K}_{\mathrm{E}}=\mathrm{K}_{\mathrm{B}}=0$
  3. $\mathrm{K}_{\mathrm{E}}>\mathrm{K}_{\mathrm{B}}$
  4. $\mathrm{K}_{\mathrm{E}} < \mathrm{K}_{\mathrm{B}}$

Solution

$\mathrm{K}_{\mathrm{E}}=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2=\frac{1}{2} \varepsilon_0 \mathrm{~B}_0^2 \mathrm{C}^2$ $=\frac{1}{2} \varepsilon_0 B_0^2 \times \frac{1}{\mu_0 \varepsilon_0}=\frac{B_0^2}{2 \mu_0}=K_B$

Asked in: AP EAMCET 2022 (05 Jul Shift 2)

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