Energy density of an EMwave =21εE02, whare E0 is the amplitude of the wave. Since total energy is same for both cylinders $\begin{aligned}
& \left(\frac{1}{2} \varepsilon E_1^2\right) \pi R_1^2 L_1=\left(\frac{1}{2} \varepsilon E_2^2\right) \pi R_2^2 L_2 \
& \Rightarrow E_1^2 R_1^2 L_1=E_2^2 R_2^2 L_2 \
& \text { or } E_2=\frac{E_1 R_1}{R_2} \sqrt{\frac{L_1}{L_2}}=\frac{100 \mathrm{d}}{(\mathrm{d} / 2)} \sqrt{\frac{L_1}{L 1}}=200 \mathrm{N} / \mathrm{C} \
& \qquad \quad\left[\because L_1=L_2=200 \mathrm{cm}\right]
\end{aligned}\RightarrowTheamplitudeofcorrespondingE Mwaveis200N/CorthewaveisE=200 \sin (\omega t-k x) \mathrm{NC}^{-1}$
Due to presence of an em-wave whose electric component is given by E=100 sin (ω t- kx) NC^-1, a cylinder of length 200 cm holds certain amount of…
JEE Main 2025 — Physics Electromagnetism
2025mcqmedium
Due to presence of an em-wave whose electric component is given by E=100sin(ωt−kx)NC−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
Official previous-year question
Held on 28 Jan 2025 · Verified 6 Jul 2026.
Options
- A
400sin(ωt−kx)NC−1
- B
200sin(ωt−kx)NC−1
- C
50sin(ωt−kx)NC−1
- D
25sin(ωt−kx)NC−1
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