The electric field of a plane electromagnetic wave, travelling in an unknown non-magnetic medium is given by, E_y=20sin(3times 10^6x-4.5times 10^14t)$E_{y}=20\sin(3\times 10^{6}x-4.5\times 10^{14}t)$ V/m (where x, t and other values have S.I. units). The dielectric constant of the medium is \_\_\_\_. (speed of light in free space is 3 times 10^8 mathrm~m/s$3 \times 10^{8} \mathrm{~m/s}$)

### Related Formula B_0 = fracE_0c$$B_0 = \frac{E_0}{c}$$ hatc = hatE times hatB$$\hat{c} = \hat{E} \times \hat{B}$$ ### Core Logic The phase of the wave is (0.6 times 10^3 x - 1.8 times 10^11 t)$(0.6 \times 10^3 x - 1.8 \times 10^{11} t)$. This indicates the wave propagates in the +x$+x$ direction, so hatc = hati$\hat{c} = \hat{i}$. The electric field oscillates along the y$y$-axis, so hatE = hatj$\hat{E} = \hat{j}$. From hatB = hatc times hatE$\hat{B} = \hat{c} \times \hat{E}$, we have hatB = hati times hatj = hatk$\hat{B} = \hat{i} \times \hat{j} = \hat{k}$. So, the magnetic field is along the z$z$-axis (B_z$B_z$). ### Step 1: Calculate Amplitude of B Wave speed v = c = fracomegak = frac1.8 times 10^110.6 times 10^3 = 3 times 10^8text m/s$v = c = \frac{\omega}{k} = \frac{1.8 \times 10^{11}}{0.6 \times 10^3} = 3 \times 10^8\text{ m/s}$. The amplitude of the magnetic field is: B_0 = fracE_0c = frac693 times 10^8 = 23 times 10^-8 = 2.3 times 10^-7text T$$B_0 = \frac{E_0}{c} = \frac{69}{3 \times 10^8} = 23 \times 10^{-8} = 2.3 \times 10^{-7}\text{ T}$$ The phase remains exactly the same as the electric field: B_z = 2.3 times 10^-7 sin(0.6 times 10^3 x - 1.8 times 10^11 t)$$B_z = 2.3 \times 10^{-7} \sin(0.6 \times 10^3 x - 1.8 \times 10^{11} t)$$ ### Pattern Recognition B_0 = E_0/c$B_0 = E_0/c$ gives the magnitude. The vector identity hatB = hatv times hatE$\hat{B} = \hat{v} \times \hat{E}$ gives the direction. Phase part never changes sign or terms between E and B equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

### Related Formula J_c = sigma E$J_c = \sigma E$ J_d = epsilon_0 fracpartial Epartial t$$J_d = \epsilon_0 \frac{\partial E}{\partial t}$$ ### Core Logic Let the electric field of the wave be E = E_0 sin(omega t - kx)$E = E_0 \sin(\omega t - kx)$. The conduction current density is: J_c = sigma E_0 sin(omega t - kx)$$J_c = \sigma E_0 \sin(\omega t - kx)$$ The maximum conduction current density is: (J_c)_textmax = sigma E_0 quad text--- (i)$$(J_c)_{\text{max}} = \sigma E_0 \quad \text{--- (i)}$$ The displacement current density is: J_d = frac1A left( epsilon_0 fracpartial (EA)partial t right) = epsilon_0 fracpartial Epartial t$$J_d = \frac{1}{A} \left( \epsilon_0 \frac{\partial (EA)}{\partial t} \right) = \epsilon_0 \frac{\partial E}{\partial t}$$ J_d = epsilon_0 E_0 omega cos(omega t - kx)$$J_d = \epsilon_0 E_0 \omega \cos(\omega t - kx)$$ The maximum displacement current density is: (J_d)_textmax = epsilon_0 E_0 omega quad text--- (ii)$$(J_d)_{\text{max}} = \epsilon_0 E_0 \omega \quad \text{--- (ii)}$$ ### Step 1: Taking the Ratio Dividing (i) by (ii): textRatio = frac(J_c)_textmax(J_d)_textmax = fracsigma E_0epsilon_0 omega E_0 = fracsigmaepsilon_0 omega$$\text{Ratio} = \frac{(J_c)_{\text{max}}}{(J_d)_{\text{max}}} = \frac{\sigma E_0}{\epsilon_0 \omega E_0} = \frac{\sigma}{\epsilon_0 \omega}$$ ### Step 2: Substitution and Calculation We know f = 100 text MHz = 10^8 text Hz$f = 100 \text{ MHz} = 10^8 \text{ Hz}$, so omega = 2pi f = 2pi times 10^8 text rad/s$\omega = 2\pi f = 2\pi \times 10^8 \text{ rad/s}$. sigma = 10 text mho/m$\sigma = 10 \text{ mho/m}$. Also, frac14piepsilon_0 = 9 times 10^9 implies frac1epsilon_0 = 4pi times 9 times 10^9$\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \implies \frac{1}{\epsilon_0} = 4\pi \times 9 \times 10^9$. textRatio = frac10 times 4pi times 9 times 10^92pi times 10^8$$\text{Ratio} = \frac{10 \times 4\pi \times 9 \times 10^9}{2\pi \times 10^8}$$ textRatio = frac360pi times 10^92pi times 10^8 = frac36002 = 1800$$\text{Ratio} = \frac{360\pi \times 10^9}{2\pi \times 10^8} = \frac{3600}{2} = 1800$$ ### Step 3: Final Conclusion The required ratio is 1800. ### Pattern Recognition The ratio of conduction to displacement current density in any medium is universally sigma / (omega epsilon_0)$\sigma / (\omega \epsilon_0)$. This dictates whether a medium behaves as a good conductor or a dielectric at a given frequency. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

### Related Formula I = frac12 epsilon_0 E_0^2 c$$I = \frac{1}{2} \epsilon_0 E_0^2 c$$ B_0 = fracE_0c = frac1c sqrtfrac2Iepsilon_0 c$$B_0 = \frac{E_0}{c} = \frac{1}{c} \sqrt{\frac{2I}{\epsilon_0 c}}$$ ### Core Logic Expressing electric field amplitude E_0$E_0$ in terms of intensity I$I$: E_0 = sqrtfrac2Iepsilon_0 c$$E_0 = \sqrt{\frac{2I}{\epsilon_0 c}}$$ Relating magnetic field amplitude B_0$B_0$ to E_0$E_0$ via B_0 = fracE_0c$B_0 = \frac{E_0}{c}$: B_0 = frac1csqrtfrac2Iepsilon_0 c = frac13 times 10^8 sqrtfrac2 times 4.0 times 10^148.85 times 10^-12 times 3 times 10^8$$B_0 = \frac{1}{c}\sqrt{\frac{2I}{\epsilon_0 c}} = \frac{1}{3 \times 10^8} \sqrt{\frac{2 \times 4.0 \times 10^{14}}{8.85 \times 10^{-12} \times 3 \times 10^8}}$$ Simplifying terms under the radical: B_0 = frac13 times 10^8 sqrtfrac8 times 10^142.655 times 10^-3 = frac13 times 10^8 sqrt3.013 times 10^17 = frac103 sqrtfrac88.85 times 3 approx 1.83 mathrm~T$$B_0 = \frac{1}{3 \times 10^8} \sqrt{\frac{8 \times 10^{14}}{2.655 \times 10^{-3}}} = \frac{1}{3 \times 10^8} \sqrt{3.013 \times 10^{17}} = \frac{10}{3} \sqrt{\frac{8}{8.85 \times 3}} \approx 1.83 \mathrm{~T}$$ ### Step 1: Final Conclusion The amplitude of the magnetic field is 1.83 mathrm~T$1.83 \mathrm{~T}$. ### Pattern Recognition EM Wave Intensity: I = frac12 c fracB_0^2mu_0 = frac12 epsilon_0 E_0^2 c$I = \frac{1}{2} c \frac{B_0^2}{\mu_0} = \frac{1}{2} \epsilon_0 E_0^2 c$. Direct sub: B_0 = sqrtfrac2 mu_0 Ic$B_0 = \sqrt{\frac{2 \mu_0 I}{c}}$ or B_0 = frac1c sqrtfrac2Iepsilon_0 c approx 1.83mathrm~T$B_0 = \frac{1}{c} \sqrt{\frac{2I}{\epsilon_0 c}} \approx 1.83\mathrm{~T}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

### Related Formula Maxwell's relation for the speed of light (c$c$) in vacuum: c = frac1sqrtmu_0 varepsilon_0$$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$$ ### Core Logic Square both sides of the speed of light formula: c^2 = frac1mu_0 varepsilon_0$$c^2 = \frac{1}{\mu_0 \varepsilon_0}$$ Since c$c$ represents the speed of light (velocity), its dimensional formula is: [c] = [L T^-1]$$[c] = [L T^{-1}]$$ Therefore, the dimensions of c^2$c^2$ are: [c^2] = [L T^-1]^2 = [L^2 T^-2]$$[c^2] = [L T^{-1}]^2 = [L^2 T^{-2}]$$ ### Step 1: Match options Express the dimensions in terms of the given variable ratios: [c^2] = fracL^2T^2$$[c^2] = \frac{L^2}{T^2}$$ This perfectly matches Option (2). ### Pattern Recognition Sees: Dimensions of vacuum permittivity-permeability inverse product. Trap: Trying to find the separate dimensions of both mu_0$\mu_0$ and varepsilon_0$\varepsilon_0$, then conducting manual division. This is extremely slow and prone to algebraic error. Shortcut: Directly identify the term as c^2$c^2$. The velocity squared has units of mathrmm^2mathrms^-2$\mathrm{m}^2\mathrm{s}^{-2}$, giving dimensions of mathrmL^2mathrmT^-2$\mathrm{L}^2\mathrm{T}^{-2}$ immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves Class 11 Physics: Units and Measurements