1. Energy stored in an electromagnetic wave

For an electromagnetic wave, the energy density is $u = \frac{1}{2}\varepsilon_0 E^2 + \frac{1}{2\mu_0}B^2.$ For a plane wave, electric and magnetic energy densities are equal, so $u = \varepsilon_0 E^2.$ Thus, the energy stored in a volume $V$ is proportional to $U \propto E^2 V.$

2. Volume of the cylinder

Volume of a cylinder is $V = \pi r^2 l.$ Both cylinders have the same length, but the second cylinder has half the diameter, hence also half the radius.

So if the first radius is $r$, the second radius is $r/2$. Therefore, $V_2 = \pi \left(\frac{r}{2}\right)^2 l = \frac{1}{4}\pi r^2 l = \frac{V_1}{4}.$

3. Equal energy condition

Given both cylinders hold the same amount of electromagnetic energy, $U_1 = U_2.$ Using $U \propto E^2 V$, $E_1^2 V_1 = E_2^2 V_2.$ Since $V_2 = \frac{V_1}{4},$ we get $E_1^2 V_1 = E_2^2 \frac{V_1}{4}.$ Cancel $V_1$: $E_1^2 = \frac{E_2^2}{4}$ $E_2^2 = 4E_1^2$ $E_2 = 2E_1.$

4. Substitute the given field amplitude

The original electric field is $E_1 = 100\sin(\omega t-kx)\,\text{N C}^{-1}.$ So the new amplitude must be twice this: $E_2 = 200\sin(\omega t-kx)\,\text{N C}^{-1}.$

5. Check options

• A: $50\sin(\omega t-kx)$ ❌
• B: $400\sin(\omega t-kx)$ ❌
• C: $200\sin(\omega t-kx)$ ✅
• D: $25\sin(\omega t-kx)$ ❌

Therefore, the correct option is C.