AP EAMCET · PHYSICS · Electromagnetic Waves
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
- A \(\mathrm{K}_{\mathrm{E}}=\mathrm{K}_{\mathrm{B}}\)
- B \(\mathrm{K}_{\mathrm{E}}=\mathrm{K}_{\mathrm{B}}=0\)
- C \(\mathrm{K}_{\mathrm{E}}>\mathrm{K}_{\mathrm{B}}\)
- D \(\mathrm{K}_{\mathrm{E}} < \mathrm{K}_{\mathrm{B}}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{K}_{\mathrm{E}}=\mathrm{K}_{\mathrm{B}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{K}_{\mathrm{E}}=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2=\frac{1}{2} \varepsilon_0 \mathrm{~B}_0^2 \mathrm{C}^2\) \(=\frac{1}{2} \varepsilon_0 B_0^2 \times \frac{1}{\mu_0 \varepsilon_0}=\frac{B_0^2}{2 \mu_0}=K_B\)
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