Due to presence of an em-wave whose electric component is given by mathrmE = 100sin (omega t - kx)mathrmNC^-1$\mathrm{E} = 100\sin (\omega t - kx)\mathrm{NC}^{-1}$ , a cylinder of length 200~mathrmcm$200~\mathrm{cm}$ holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds the same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as

### Related Formula rho_E = frac12varepsilon_0 E_rms^2$$\rho_E = \frac{1}{2}\varepsilon_0 E_{rms}^2$$ rho_B = fracB_rms^22mu_0$$\rho_B = \frac{B_{rms}^2}{2\mu_0}$$ p_r = fracIc$$p_r = \frac{I}{c}$$ (radiation pressure for completely absorbing surface) ### Core Logic Statement I: In an EM wave, energy is stored in both the electric and magnetic fields. The energy densities are equal: U_E = U_B$U_E = U_B$ because E = cB$E = cB$ and c^2 = frac1mu_0 epsilon_0$c^2 = \frac{1}{\mu_0 \epsilon_0}$. This is a correct fact. Statement II: Electromagnetic waves carry momentum (p = U/c$p = U/c$). When they strike a surface, they transfer this momentum, creating radiation pressure. This is also a correct fact. ### Step 1: Final Evaluation Since both facts are universally true theoretical properties of EM waves, both statements are correct. ### Pattern Recognition Standard NCERT theory. Energy density is strictly symmetric 50-50 between E and B. Momentum transfer to$\to$ pressure is the definitive signature of the particle-like nature (photons) of EM waves. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

### Related Formula U_texttotal average = frac12epsilon_0 E_0^2$$U_{\text{total average}} = \frac{1}{2}\epsilon_0 E_0^2$$ ### Core Logic For an electromagnetic wave, the total average energy density is the sum of the average energy density of the electric field and the magnetic field. They are equal, so: U_textavg = U_E + U_B = 2U_E = 2 left( frac14epsilon_0 E_0^2 right) = frac12epsilon_0 E_0^2$$U_{\text{avg}} = U_E + U_B = 2U_E = 2 \left( \frac{1}{4}\epsilon_0 E_0^2 \right) = \frac{1}{2}\epsilon_0 E_0^2$$ Where E_0$E_0$ is the amplitude of the electric field. ### Step 2: Substitution Given: E_0 = 50 mathrm\, V/m$E_0 = 50 \mathrm{\, V/m}$ epsilon_0 = 8.85 times 10^-12 mathrm\, C^2/(Ncdot m^2)$\epsilon_0 = 8.85 \times 10^{-12} \mathrm{\, C^2/(N\cdot m^2)}$ U_textavg = frac12 times (8.85 times 10^-12) times (50)^2$$U_{\text{avg}} = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (50)^2$$ U_textavg = frac12 times 8.85 times 10^-12 times 2500$$U_{\text{avg}} = \frac{1}{2} \times 8.85 \times 10^{-12} \times 2500$$ U_textavg = 1.10625 times 10^-8 mathrm\, J/m^3$$U_{\text{avg}} = 1.10625 \times 10^{-8} \mathrm{\, J/m^3}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves