### 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

### Related Formula According to the Maxwell-Ampere law, displacement current I_d$I_d$ is defined as: I_d = epsilon_0 fracmathrmdPhi_Emathrmdt$I_d = \epsilon_0 \frac{\mathrm{d}\Phi_E}{\mathrm{d}t}$ ### Core Logic Since I_d$I_d$ represents a physical current, its dimensions must match those of standard conduction electric current ([A]$[A]$ or [I]$[I]$). ### Step 1: Dimensional Analysis Let us verify using fundamental dimensions: - Permittivity of free space [epsilon_0] = [M^-1 L^-3 T^4 A^2]$[\epsilon_0] = [M^{-1} L^{-3} T^4 A^2]$ - Electric flux [Phi_E] = [M L^3 T^-3 A^-1]$[\Phi_E] = [M L^3 T^{-3} A^{-1}]$ - Time [t] = [T]$[t] = [T]$ left[ epsilon_0 fracmathrmdPhi_Emathrmdt right] = [M^-1 L^-3 T^4 A^2] times frac[M L^3 T^-3 A^-1][T] = [A]$\left[ \epsilon_0 \frac{\mathrm{d}\Phi_E}{\mathrm{d}t} \right] = [M^{-1} L^{-3} T^4 A^2] \times \frac{[M L^3 T^{-3} A^{-1}]}{[T]} = [A]$ This confirms the quantity is dimensionally equivalent to Electric current. ### Pattern Recognition Sees: epsilon_0 fracmathrmdPhi_Emathrmdt$\epsilon_0 \frac{\mathrm{d}\Phi_E}{\mathrm{d}t}$. Shortcut: Instantly identify this as Maxwell's expression for displacement current. Displacement current has the exact same dimensions as regular conduction current. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

### Related Formula hatc = hatE times hatB$\hat{c} = \hat{E} \times \hat{B}$ where, hatc$\hat{c}$ = unit vector in the direction of wave propagation hatE$\hat{E}$ = unit vector of the electric field hatB$\hat{B}$ = unit vector of the magnetic field ### Core Logic Given that the wave propagates along the +x$+x$ direction: hatc = hati$\hat{c} = \hat{i}$ Let us test the combination E_y$E_y$ and B_z$B_z$, which implies hatE = hatj$\hat{E} = \hat{j}$ and hatB = hatk$\hat{B} = \hat{k}$: hatE times hatB = hatj times hatk = hati$\hat{E} \times \hat{B} = \hat{j} \times \hat{k} = \hat{i}$

This cross product yields exactly the direction of propagation +x$+x$. Therefore, E_y$E_y$ and B_z$B_z$ are appropriate components representing the cross-orthogonal vectors of the EM wave. ### Pattern Recognition Always remember the cyclic relation hati rightarrow hatj rightarrow hatk$\hat{i} \rightarrow \hat{j} \rightarrow \hat{k}$. Since propagation is hati$\hat{i}$, our cross product must be hatj times hatk$\hat{j} \times \hat{k}$. Thus electric field along y$y$ matches magnetic field along z$z$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

### Related Formula I_d = I_c = C fracdVdt$I_d = I_c = C \frac{dV}{dt}$ C = fracepsilon_0 Ad = fracepsilon_0 pi r^2d$C = \frac{\epsilon_0 A}{d} = \frac{\epsilon_0 \pi r^2}{d}$ ### Core Logic Combining the current expression with capacitance relations yields: I = left(fracepsilon_0 pi r^2dright) fracdVdt$I = \left(\frac{\epsilon_0 \pi r^2}{d}\right) \frac{dV}{dt}$ Isolating plate separation distance d$d$: d = fracepsilon_0 pi r^2I cdot left(fracdVdtright)$d = \frac{\epsilon_0 \pi r^2}{I} \cdot \left(\frac{dV}{dt}\right)$ Substitute the parameters specified by the problem layout: - r = 10mathrm~cm = 0.1mathrm~m$r = 10\mathrm{~cm} = 0.1\mathrm{~m}$ - epsilon_0 = 9 times 10^-12$\epsilon_0 = 9 \times 10^{-12}$ - pi = 22/7$\pi = 22/7$ - I = 0.15mathrm~A$I = 0.15\mathrm{~A}$ - fracdVdt = 7 times 10^8mathrm~V/s$\frac{dV}{dt} = 7 \times 10^{8}\mathrm{~V/s}$ d = frac(9 times 10^-12) times left(frac227right) times (0.1)^20.15 times (7 times 10^8)$d = \frac{(9 \times 10^{-12}) \times \left(\frac{22}{7}\right) \times (0.1)^2}{0.15} \times (7 \times 10^8)$ d = frac9 times 10^-12 times 22 times 0.01 times 10^80.15 = frac198 times 10^-40.15$d = \frac{9 \times 10^{-12} \times 22 \times 0.01 \times 10^8}{0.15} = \frac{198 \times 10^{-4}}{0.15}$ d = 1320 times 10^-6mathrm~m = 1320mumathrmm$d = 1320 \times 10^{-6}\mathrm{~m} = 1320\mu\mathrm{m}$ Thus, the integer value for the distance is 1320. ### Pattern Recognition The charging current matches displacement current identically. Treat the system as a standard linear differential capacitor setup using I = C fracdVdt$I = C \frac{dV}{dt}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves Class 12 Physics: Electrostatic Potential and Capacitance

### 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