In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 times 10^10 mathrm~Hz$5 \times 10^{10} \mathrm{~Hz}$ and an amplitude of 50 mathrm~Vm^-1$50 \mathrm{~Vm}^{-1}$. The total average energy density of the electromagnetic field of the wave is : [Use varepsilon_0 = 8.85 times 10^-12 \, textC^2 / textNm^2$\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 / \text{Nm}^2$ ]

### Related Formula textEnergy Density = frac12 epsilon_0 mathrmE^2$$\text{Energy Density} = \frac{1}{2} \epsilon_0 \mathrm{E}^2$$ textTotal Energy = textEnergy Density times textVolume$$\text{Total Energy} = \text{Energy Density} \times \text{Volume}$$ ### Core Logic Since both cylinders hold equal amounts of electromagnetic energy: left(textEnergyright)_1 = left(textEnergyright)_2$$\left(\text{Energy}\right)_1 = \left(\text{Energy}\right)_2$$ frac12 epsilon_0 mathrmE_1^2 cdot c pi mathrmR_1^2 times mathrmL_1 = frac12 epsilon_0 mathrmE_2^2 cdot c pi mathrmR_2^2 times mathrmL_2$$\frac{1}{2} \epsilon_0 \mathrm{E}_1^2 \cdot c \pi \mathrm{R}_1^2 \times \mathrm{L}_1 = \frac{1}{2} \epsilon_0 \mathrm{E}_2^2 \cdot c \pi \mathrm{R}_2^2 \times \mathrm{L}_2$$ Since the lengths are identical (mathrmL_1 = mathrmL_2$\mathrm{L}_1 = \mathrm{L}_2$), this simplifies to: mathrmE_1^2 mathrmR_1^2 = mathrmE_2^2 mathrmR_2^2 implies mathrmE_1 mathrmR_1 = mathrmE_2 mathrmR_2$$\mathrm{E}_1^2 \mathrm{R}_1^2 = \mathrm{E}_2^2 \mathrm{R}_2^2 \implies \mathrm{E}_1 \mathrm{R}_1 = \mathrm{E}_2 \mathrm{R}_2$$ Given the second cylinder has half the diameter (and radius) of the first (mathrmR_2 = fracmathrmR_12$\mathrm{R}_2 = \frac{\mathrm{R}_1}{2}$): 100 times mathrmR_1 = mathrmE_2 times fracmathrmR_12$$100 \times \mathrm{R}_1 = \mathrm{E}_2 \times \frac{\mathrm{R}_1}{2}$$ mathrmE_2 = 200 mathrmN/C$$\mathrm{E}_2 = 200 \mathrm{N/C}$$ ### Step 1: Final Equation Match The wave equation adjusts its amplitude factor to 200sin (omega t - kx)mathrmNC^-1$200\sin (\omega t - kx)\mathrm{NC}^{-1}$, which matches option (2). ### Pattern Recognition When energy is constant and volume scales down inversely by a factor of 4 (due to mathrmR^2$\mathrm{R}^2$), the electric field strength must increase by a factor of sqrt4 = 2$\sqrt{4} = 2$ to maintain balance. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

### Related Formula For a perfectly reflecting surface, the radiation pressure P_textrad$P_{\text{rad}}$ is given by: P_textrad = frac2Ic$$P_{\text{rad}} = \frac{2I}{c}$$ where, I$I$ = intensity of the light source, c$c$ = speed of light approx 3 times 10^8mathrm~m/s$\approx 3 \times 10^8\mathrm{~m/s}$. ### Core Logic Let's first calculate the intensity I$I$ of the point source at a distance r = 2mathrm~m$r = 2\mathrm{~m}$: I = fractextPowertextArea = fracP4pi r^2$$I = \frac{\text{Power}}{\text{Area}} = \frac{P}{4\pi r^2}$$ Substitute the given values (P = 450mathrm~W$P = 450\mathrm{~W}$ and r = 2mathrm~m$r = 2\mathrm{~m}$): I = frac4504pi times 2^2 = frac45016pimathrm~W/m^2$$I = \frac{450}{4\pi \times 2^2} = \frac{450}{16\pi}\mathrm{~W/m}^2$$ ### Step 1: Calculating Radiation Pressure Now, substitute I$I$ into the radiation pressure formula: P_textrad = frac2 times left(frac45016piright)3 times 10^8 = frac90016pi times 3 times 10^8$$P_{\text{rad}} = \frac{2 \times \left(\frac{450}{16\pi}\right)}{3 \times 10^8} = \frac{900}{16\pi \times 3 \times 10^8}$$ P_textrad = frac30016pi times 10^8 = frac754pi times 10^8mathrm~N/m^2$$P_{\text{rad}} = \frac{300}{16\pi \times 10^8} = \frac{75}{4\pi \times 10^8}\mathrm{~N/m}^2$$ Using pi approx 3.1416$\pi \approx 3.1416$: P_textrad = frac754 times 3.1416 times 10^8 = frac7512.566 times 10^-8$$P_{\text{rad}} = \frac{75}{4 \times 3.1416 \times 10^8} = \frac{75}{12.566} \times 10^{-8}$$ P_textrad approx 5.968 times 10^-8mathrm~Pascals approx 6 times 10^-8mathrm~Pascals$$P_{\text{rad}} \approx 5.968 \times 10^{-8}\mathrm{~Pascals} \approx 6 \times 10^{-8}\mathrm{~Pascals}$$ ### Pattern Recognition Remember: Perfectly absorbing surface implies P = I/c$\implies P = I/c$. Perfectly reflecting surface implies P = 2I/c$\implies P = 2I/c$. Always pay close attention to the surface's properties mentioned in the prompt! ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

### Related Formula v = frac1sqrtmu epsilon = frac1sqrtmu_0mu_r cdot epsilon_0epsilon_r = fraccsqrtmu_r epsilon_r$$v = \frac{1}{\sqrt{\mu \epsilon}} = \frac{1}{\sqrt{\mu_0\mu_r \cdot \epsilon_0\epsilon_r}} = \frac{c}{\sqrt{\mu_r \epsilon_r}}$$ ### Core Logic Given parameters: mu_r = frac10pi$$\mu_r = \frac{10}{\pi}$$ epsilon_r = frac10.0885$$\epsilon_r = \frac{1}{0.0885}$$ Let's substitute these into the refractive index radical term sqrtmu_r epsilon_r$\sqrt{\mu_r \epsilon_r}$: mu_r epsilon_r = frac10pi times frac10.0885$$\mu_r \epsilon_r = \frac{10}{\pi} \times \frac{1}{0.0885}$$ ### Step 1: Simplify Numerical Ratio Using standard approximations pi approx 3.14$\pi \approx 3.14$: mu_r epsilon_r = frac103.1415 times 0.0885 approx frac100.278 approx 36$$\mu_r \epsilon_r = \frac{10}{3.1415 \times 0.0885} \approx \frac{10}{0.278} \approx 36$$ Taking the square root: sqrtmu_r epsilon_r = sqrt36 = 6$$\sqrt{\mu_r \epsilon_r} = \sqrt{36} = 6$$ Therefore, v = fracc6 implies c = 6v$v = \frac{c}{6} \implies c = 6v$. Velocity in a vacuum is exactly **6** times faster. ### Pattern Recognition The expression sqrtmu_r epsilon_r$\sqrt{\mu_r \epsilon_r}$ is identical to the definition of refractive index n$n$. Simplifying indices down to a perfect square (36 implies 6$36 \implies 6$) clarifies the ratio immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

### Related Formula I = frac12epsilon_0 E_0^2 c$$I = \frac{1}{2}\epsilon_0 E_0^2 c$$ [cite: 670] ### Core Logic Rearranging the equation for intensity I$I$ to express the peak electric field amplitude E_0$E_0$: [cite: 670] E_0^2 = frac2Iepsilon_0 c$$E_0^2 = \frac{2I}{\epsilon_0 c}$$ E_0 = sqrtfrac2Iepsilon_0 c$$E_0 = \sqrt{\frac{2I}{\epsilon_0 c}}$$ [cite: 670] Thus, the given quantity is simply the magnitude of the peak electric field E_0$E_0$[cite: 672]. The standard SI unit of an electric field is Newtons per Coulomb (mathrmNcdot C^-1$\mathrm{N\cdot C^{-1}}$) or Volts per meter (mathrmVcdot m^-1$\mathrm{V\cdot m^{-1}}$)[cite: 673]. Matching with the given structural choices, mathrmNC^-1$\mathrm{NC}^{-1}$ is the correct unit[cite: 31, 669]. ### Pattern Recognition Recognize the standard configuration for energy flux density (intensity) I = u_textavgc$I = u_{\text{avg}}c$[cite: 670]. Identifying that sqrtfrac2Iepsilon_0 c$\sqrt{\frac{2I}{\epsilon_0 c}}$ resolves to the electric field dimension directly yields the solution unit mathrmNC^-1$\mathrm{NC}^{-1}$ or mathrmV/m$\mathrm{V/m}$[cite: 670, 672, 673]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

### Related Formula L = fracmu_0 N^2 Al implies mu_0 propto L$$L = \frac{\mu_0 N^2 A}{l} \implies \mu_0 \propto L$$ [cite: 675] C = fracepsilon_0 Ad implies epsilon_0 propto C$$C = \frac{\epsilon_0 A}{d} \implies \epsilon_0 \propto C$$ [cite: 677] ### Core Logic From the basic formulas of inductance and capacitance, we can note the proportional parameters: [cite: 675, 677] fracmu_0epsilon_0 propto fracLC$$\frac{\mu_0}{\epsilon_0} \propto \frac{L}{C}$$ [cite: 678] We know that the time constant for an LR$LR$ circuit is tau = fracLR$\tau = \frac{L}{R}$ and for a RC$RC$ circuit is tau = RC$\tau = RC$[cite: 679]. Equating these time dimensions: [cite: 679] fracLR = RC implies fracLC = R^2$$\frac{L}{R} = RC \implies \frac{L}{C} = R^2$$ [cite: 679] Taking the square root or matching parameters from the text solution layout yields the characteristic dimension of resistance[cite: 679]. ### Pattern Recognition The quantity sqrtfracmu_0epsilon_0$\sqrt{\frac{\mu_0}{\epsilon_0}}$ represents the intrinsic impedance of free space, which has the value approx 377\ Omega$\approx 377\ \Omega$[cite: 679]. Hence, its square matches the dimension of resistance squared, which maps to Resistance in the choice sets[cite: 38, 674]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves