JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
In a plane \(EM\) wave, the electric field oscillates sinusoidally at a frequency of \(5 \times 10^{10} \mathrm{~Hz}\) and an amplitude of \(50 \mathrm{Vm}^{-1}\). The total average energy density of the electromagnetic field of the wave is _______. [Use \(\varepsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2\) ]
- A \(1.106 \times 10^{-8} \mathrm{Jm}^{-3}\)
- B \(4.425 \times 10^{-8} \mathrm{Jm}^{-3}\)
- C \(2.212 \times 10^{-8} \mathrm{Jm}^{-3}\)
- D \(2.212 \times 10^{-10} \mathrm{Jm}^{-3}\)
Answer & Solution
Correct Answer
(A) \(1.106 \times 10^{-8} \mathrm{Jm}^{-3}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{U}_{\mathrm{E}}=\frac{1}{2} \epsilon_0 \mathrm{E}^2\) \(\mathrm{U}_{\mathrm{E}}=\frac{1}{2} \times 8.85 \times 10^{-12} \times(50)^2\) \(=1.106 \times 10^{-8} \mathrm{~J} / \mathrm{m}^3\)
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