Due to presence of an em-wave whose electric component is given by \(\mathrm{E}=100 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{NC}^{-1}\), 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

Energy density of an \(E M_{\text {wave }}=\frac{1}{2} \varepsilon E_0^2\), 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 \(E M\) wave is 200 N/C
or the wave is \(E=200 \sin (\omega t-k x) \mathrm{NC}^{-1}\)