AP EAMCET · PHYSICS · Electromagnetic Waves
The electric field in \(\mathrm{NC}^{-1}\) of an electromagnetic wave is \(\mathrm{E}=36 \sqrt{\pi} \operatorname{Sin}(\omega \mathrm{t}-\mathrm{kx})\). The average energy density of the electromagnetic wave due to the electric field is \(\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right)\)
- A \(36 \times 10^{-9} \mathrm{Jm}^{-3}\)
- B \(18 \times 10^{-9} \mathrm{Jm}^{-3}\)
- C \(36 \times 10^{-7} \mathrm{Jm}^{-3}\)
- D \(18 \times 10^{-7} \mathrm{Jm}^{-3}\)
Answer & Solution
Correct Answer
(B) \(18 \times 10^{-9} \mathrm{Jm}^{-3}\)
Step-by-step Solution
Detailed explanation
The average encrgy density of the electromagnetic wave is given by \(\begin{aligned} & \mathrm{U}_{\mathrm{av}}=\frac{1}{2} \epsilon_0 \mathrm{E}^2 \\ & =\frac{1}{2} \times 8.86 \times 10^{-12} \times(36)^2 \pi=18 \times 10^{-9} \mathrm{~J} / \mathrm{m}^3 \end{aligned}\)
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