JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2021
- Hard
- Numerical
The electric field in an electromagnetic wave is given by \({E}=\left(50\, {NC}^{-1}\right) \sin \omega({t}-{x} / {c})\) The energy contained in a cylinder of volume \({V}\) is \(5.5 \times 10^{-12} \, {J}\). The value of \({V}\) is \(......{cm}^{3}\) \(\left(\right.\) given \(\left.\in_{0}=8.8 \times 10^{-12} \,{C}^{2} {N}^{-1} {m}^{-2}\right)\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(500\)
Step-by-step Solution
Detailed explanation
\({E}=50 \sin \left(\omega {t}-\frac{\omega}{{c}} \cdot {x}\right)\)
Energy density \(=\frac{1}{2} \in_{0} {E}_{0}^{2}\)
Energy for volume \({V}=\frac{1}{2} \in_{0} {E}_{0}^{2} \cdot {V}=5.5 \times 10^{-12}\)
\(\frac{1}{2} 8.8 \times 10^{-12} \times 2500 {V}=5.5 \times 10^{-12}\)
\({V}=\frac{5.5 \times 2}{2500 \times 8.8}=.0005\, {m}^{3}\)
\(=.0005 \times 10^{6}\,({c} . {m})^{3}\)
\(=500({c} \, \cdot {m})^{3}\)
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