JEE Main 2025 — Electromagnetic Waves Question with Solution

Energy density of an $E{M}_{\text{wave }}=\frac{1}{2}\epsilon E_0^2$, whare $E_0$ is the amplitude of the wave. Since total energy is same for both cylinders $\begin{aligned} & \left(\frac{1}{2} \varepsilon E_1^2\right) \pi R_1^2 L_1=\left(\frac{1}{2} \varepsilon E_2^2\right) \pi R_2^2 L_2 \\ & \Rightarrow E_1^2 R_1^2 L_1=E_2^2 R_2^2 L_2 \\ & \text { or } E_2=\frac{E_1 R_1}{R_2} \sqrt{\frac{L_1}{L_2}}=\frac{100 \mathrm{~d}}{(\mathrm{~d} / 2)} \sqrt{\frac{L_1}{L 1}}=200 \mathrm{~N} / \mathrm{C} \\ & \qquad \quad\left[\because L_1=L_2=200 \mathrm{~cm}\right] \end{aligned}$ $\Rightarrow$ The amplitude of corresponding $EM$ wave is 200 N/C or the wave is $E=200\sin (\omega t-kx){\mathrm{NC}}^{-1}$