Match the List-I with List-II

List-I	List-II
A. Radio-wave	I. is produced by Magnetron value
B. Micro-wave	II. Due to change in the vibrational modes of atoms
C. Infrared-wave	III. Due to inner shell electrons moving from higher energy level to lower energy level
D. X-ray	IV. Due to rapid acceleration of electrons

Choose the correct answer from the options given below:

### 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 v = fracomegak, quad n = fraccv = sqrtmu_r epsilon_r$$v = \frac{\omega}{k}, \quad n = \frac{c}{v} = \sqrt{\mu_r \epsilon_r}$$ ### Core Logic Wave velocity in medium: v = fracomegak = frac4.5 times 10^143 times 10^6 = 1.5 times 10^8 text m/s$$v = \frac{\omega}{k} = \frac{4.5 \times 10^{14}}{3 \times 10^6} = 1.5 \times 10^8 \text{ m/s}$$ Refractive index: n = frac3 times 10^81.5 times 10^8 = 2$$n = \frac{3 \times 10^8}{1.5 \times 10^8} = 2$$ For non-magnetic medium (mu_r = 1$\mu_r = 1$): n = sqrtepsilon_r implies 2 = sqrtepsilon_r implies epsilon_r = 4$$n = \sqrt{\epsilon_r} \implies 2 = \sqrt{\epsilon_r} \implies \epsilon_r = 4$$ ### Pattern Recognition Sees: EM wave equation in dielectric medium. Shortcut: Extract phase velocity from wave equation coefficients, find refractive index and dielectric constant. Check: Numerical answer is 4. ✓ ### 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

### Core Logic Let's systematically identify each integral equation with its corresponding physical law: (A) ointvecEcdotvecdl=-fracdphi_Bdt$\oint\vec{E}\cdot\vec{dl}=-\frac{d\phi_{B}}{dt}$ corresponds to the line integral of the electric field around a closed loop being equal to the negative rate of change of magnetic flux, which is Faraday's Law of Electromagnetic Induction (II). (B) oint vecBcdotvecdl=mu_0left(I+epsilon_0fracdphi_Edtright)$\oint \vec{B}\cdot\vec{dl}=\mu_{0}\left(I+\epsilon_{0}\frac{d\phi_{E}}{dt}\right)$ defines the Ampere-Maxwell Law (III), incorporating the displacement current. (C) ointvecEcdotvecda=fracQ_textencepsilon_0$\oint\vec{E}\cdot\vec{da}=\frac{Q_{\text{enc}}}{\epsilon_{0}}$ relates the electric flux through a closed surface to the enclosed charge, mapping to Gauss's Law of Electrostatics (IV). (D) ointvecBcdotvecdl=mu_0I$\oint\vec{B}\cdot\vec{dl}=\mu_{0}I$ is the original Ampere's Circuital Law (I) without Maxwell's correction. ### Step 1: Final Conclusion The correct matches are A-II, B-III, C-IV, D-I. ### Pattern Recognition Sees: Surface integral of E mapped to volume integral of charge → Gauss's Law. Circulation of E mapped to time derivative of B flux → Faraday's Law. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves