Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The electric field in an electromagnetic wave is given as <img alt="\mathrm{\bar{E}=20\sin \omega \left ( t-\frac{x}{c} \right )\bar{j}NC^{-1}}" src="https://entrancecorner.oncodecogs.com

The electric field in an electromagnetic wave is given as

\mathrm{\bar{E}=20\sin \omega \left ( t-\frac{x}{c} \right )\bar{j}NC^{-1}}

where \omega and c are angular frequency and velocity of electromagnetic wave respectively. the energy contained
in a volume of 5\times 10^{-4}m^{3} will be (Given \mathrm{\varepsilon _{o}=8.85\times 10^{-12}c^{2}/Nm^{2}} )

Option: 1

88.5\times 10^{-13}\mathrm{J}


Option: 2

17.7\times 10^{-13}\mathrm{J}


Option: 3

8.85\times 10^{-13}\mathrm{J}


Option: 4

28.5\times 10^{-13}\mathrm{J}


Answers (1)

best_answer

\overrightarrow{\mathrm{E}}=20 \sin w\left[t-\frac{x}{c}\right]

E _o=20
\begin{aligned} & \text { Energy density }=\frac{1}{2} \varepsilon_0 \mathrm{E}_{\mathrm{o}}^2 \\ & =\frac{1}{2} \times 8.85 \times 10^{-12} \times 400 \\ & =200 \times 8.85 \times 10^{-12} \times 5 \times 10^{-4} \\ & =8.85 \times 10^{-12} \times 10^{-4} \times 1000 \\ & \text { Energy }=8.85 \times 10^{-13} \mathrm{~J} \end{aligned}

Posted by

Gunjita

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main Result 2025 LIVE: NTA JEE result by tomorrow at jeemain.nta.nic.in, scorecard, marks calculator
anam-khan

JEE Main final answer key 2025 delayed; aspirants heap abuse, anger upon NTA, demand another exam body
anam-khan

JEE Main results 2025 by April 19, final answer key at 2 pm: NTA confirms result date

Latest Question