🛠️ JEE🧲 Physics

Due to presence of an em-wave whose electric component is given by \(E=100\sin (\omega t-kx)N{C}^{-1}\), a cylinder of l…

Q1 FREE PREVIEW

Due to presence of an em-wave whose electric component is given by \(E=100\sin (\omega t-kx)N{C}^{-1}\), a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as

a

\begin{equation}
25 \sin (\omega t-k x) N C^{-1}
\end{equation}

b

\begin{equation}
50 \sin (\omega t-k x) N C^{-1}
\end{equation}

c

\begin{equation}
400 \sin (\omega t-k x) N C^{-1}
\end{equation}

d

\begin{equation}
200 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{NC}^{-1}
\end{equation}

✓ Correct answer: d)

\begin{equation}
200 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{NC}^{-1}
\end{equation}

Explanation

\(\text{ Average energy density }=\frac{1}{2}{\epsilon }_{0}{E}_{0}^{2}\\ \text{ Average energy }\left({E}_{AV}\right)=\frac{1}{2}{\epsilon }_{0}{E}_{0}^{2}AL\\ =\frac{1}{2}{\epsilon }_{0}{E}_{0}^{2}\pi {\left(\frac{D}{2}\right)}^{2}L\\ =\frac{\pi {\epsilon }_{0}{E}_{0}^{2}{D}^{2}L}{8}\\ As,{E}_{1}={E}_{2}\\ \frac{\pi {\epsilon }_{0}{E}_{01}^{2}{D}^{2}L}{8}=\frac{\pi {\epsilon }_{0}{E}_{02}^{2}{\left(\frac{D}{2}\right)}^{2}L}{8}\\ \Rightarrow {E}_{01}^{2}=\frac{{E}_{02}^{2}}{4}\\ {E}_{02}=2{E}_{01}=200\)

Practice more JEE Physics PYQs

See every question on Electromagnetic Waves, or browse the full JEE question bank.

See all questions on Electromagnetic Waves →