Discussion
The electric field of an electromagnetic wave is \(\vec{E} = i\sqrt{2}\sin(kz - \omega t)\) V/m. The average flow energy per unit area per unit time, due to this wave is :
Answer: B
The average energy flow is \(27 \times 10^{-4}\) W/m\(^{2}\).
The time-averaged intensity of a plane wave with amplitude \(E_{0}\) is
\[ \begin{aligned} \langle S\rangle &= \tfrac{1}{2}c\varepsilon_{0}E_{0}^{2} \\ &= \tfrac{1}{2}(3 \times 10^{8})(8.85 \times 10^{-12})(\sqrt{2})^{2} \\ &= 2.66 \times 10^{-3}\ \text{W/m}^{2} \end{aligned} \]
which is about \(27 \times 10^{-4}\) W/m\(^{2}\). Equivalently, \(E_{0}^{2}/(2Z_{0}) = 2/754\). The other options differ by powers of ten: they correspond to errors in the exponent of \(\varepsilon_{0}\) or c, not to a different physical result.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development