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

We are given an EM wave with electric field $$E = 100\sin(\omega t - kx)$$ NC$$^{-1}$$ inside a cylinder of length 200 cm. A second cylinder has the same length but half the diameter, and must hold the same amount of EM energy.

The average energy density of an EM wave is proportional to $$E_0^2$$:
$$u = \frac{1}{2}\epsilon_0 E_0^2$$
Total energy stored in a cylinder of volume $$V$$ is: $$U = u \times V = \frac{1}{2}\epsilon_0 E_0^2 \times V$$

For a cylinder with diameter $$d$$ and length $$L$$: $$V = \frac{\pi d^2}{4} \cdot L$$. The second cylinder has half the diameter, so its cross-sectional area is $$(1/2)^2 = 1/4$$ of the first, giving $$V_2 = \frac{V_1}{4}$$.

Equating the energies, $$\frac{1}{2}\epsilon_0 E_1^2 V_1 = \frac{1}{2}\epsilon_0 E_2^2 V_2$$ implies $$E_1^2 V_1 = E_2^2 \cdot \frac{V_1}{4}$$, so $$E_2^2 = 4E_1^2$$ and hence $$E_2 = 2E_1 = 2 \times 100 = 200 \text{ NC}^{-1}$$.

The modified electric field is $$E = 200\sin(\omega t - kx)$$ NC$$^{-1}$$.

The correct answer is Option B) $$200\sin(\omega t - kx)$$ NC$$^{-1}$$