### Related Formula For a plane electromagnetic wave propagating in a given direction: 1. Peak electric field amplitude relates to peak magnetic field amplitude via: E_0 = B_0 cdot c$E_0 = B_0 \cdot c$ 2. The directional orientation unit vectors satisfy the cross product relation: hatE = hatB times hatc$\hat{E} = \hat{B} \times \hat{c}$ where hatc$\hat{c}$ points along the wave propagation vector direction. ### Core Logic Given the wave equation format, the phase term left(t - fraczcright)$\left(t - \frac{z}{c}\right)$ shows that propagation is along the positive z-axis : hatc = hatk$\hat{c} = \hat{k}$ The magnetic field direction unit vector is : hatB = fracsqrt32hati + frac12hatj$\hat{B} = \frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}$ Compute the electric field direction vector using the cross product relation : hatE = hatB times hatk = left(fracsqrt32hati + frac12hatjright) times hatk$\hat{E} = \hat{B} \times \hat{k} = \left(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\right) \times \hat{k}$ hatE = fracsqrt32(hati times hatk) + frac12(hatj times hatk)$\hat{E} = \frac{\sqrt{3}}{2}(\hat{i} \times \hat{k}) + \frac{1}{2}(\hat{j} \times \hat{k})$ Using unit vector properties (hati times hatk = -hatj$\hat{i} \times \hat{k} = -\hat{j}$ and hatj times hatk = hati$\hat{j} \times \hat{k} = \hat{i}$): hatE = -fracsqrt32hatj + frac12hati = frac12hati - fracsqrt32hatj quad text$\hat{E} = -\frac{\sqrt{3}}{2}\hat{j} + \frac{1}{2}\hat{i} = \frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j} \quad \text{}$ With peak amplitude E_0 = 30c$E_0 = 30c$ , the resulting vector equation is: vecE = left(frac12hati - fracsqrt32hatjright)30csinleft[omegaleft(t-fraczcright)right]$\vec{E} = \left(\frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j}\right)30c\sin\left[\omega\left(t-\frac{z}{c}\right)\right]$ ### Pattern Recognition The vectors vecE$\vec{E}$, vecB$\vec{B}$, and the propagation direction are always mutually perpendicular. Since vecE cdot vecB = 0$\vec{E} \cdot \vec{B} = 0$, you can quickly double-check your answer by verifying that the \dot product of the final vecE$\vec{E}$ and vecB$\vec{B}$ direction options equals zero. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

### Related Formula p = fracEc$p = \frac{E}{c}$ ### Core Logic Assertion (A) is false because electromagnetic waves carry both energy and finite radiation momentum (p = E/c$p = E/c$). Reason (R) is correct because the rest mass of a photon equals zero. ### Chapter Mix Class 12 Physics: Electromagnetic Waves

### Related Formula Capacitive reactance: X_C = frac1omega C$X_C = \frac{1}{\omega C}$ Conduction Current (I_textrms$I_{\text{rms}}$): I_textrms = fracV_textrmsX_C = V_textrms cdot omega C$I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} = V_{\text{rms}} \cdot \omega C$ Continuity relation: I_c = I_d$I_c = I_d$ ### Core Logic Given parameters: C = 200mathrm~pF = 200 times 10^-12mathrm~F$C = 200\mathrm{~pF} = 200 \times 10^{-12}\mathrm{~F}$, V_textrms = 230mathrm~V$V_{\text{rms}} = 230\mathrm{~V}$, omega = 300mathrm~rad/s$\omega = 300\mathrm{~rad/s}$. Calculate current: I = V_textrms cdot omega C = 230 times 300 times 200 times 10^-12$I = V_{\text{rms}} \cdot \omega C = 230 \times 300 \times 200 \times 10^{-12}$ ### Step 1: Simplify Numerical Calculation I = 230 times 60000 times 10^-12 = 13.8 times 10^-6mathrm~A = 13.8mathrm~mu A$I = 230 \times 60000 \times 10^{-12} = 13.8 \times 10^{-6}\mathrm{~A} = 13.8\mathrm{~\mu A}$ By Maxwell's electromagnetic formulation, the displacement current I_d$I_d$ within the dielectric space matches the exterior conduction current I_c$I_c$ seamlessly in magnitude. ### Pattern Recognition Factual invariant: Conduction current always equals displacement current inside a standard layout loop context (I_c = I_d$I_c = I_d$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves Class 12 Physics: Alternating Current

### Related Formula The relation between the amplitudes of electric field E$E$ and magnetic field B$B$ in an EM wave is: c = fracEB$c = \frac{E}{B}$ where: * c = 3 times 10^8text m/s$c = 3 \times 10^8\text{ m/s}$ is the speed of light. The directions are related by the vector cross product: hatE times hatB = hatv$\hat{E} \times \hat{B} = \hat{v}$ where hatv$\hat{v}$ is the direction of propagation of the EM wave. ### Core Logic Given: * Wave propagation direction, hatv = hati$\hat{v} = \hat{i}$ (along X-direction) * Electric field vector direction, hatE = hatj$\hat{E} = \hat{j}$ * Electric field magnitude, E = 9.6text V/m$E = 9.6\text{ V/m}$ ### Step 1: Calculate Magnetic Field Magnitude Using the magnitude relation: B = fracEc = frac9.63 times 10^8 = 3.2 times 10^-8text T$B = \frac{E}{c} = \frac{9.6}{3 \times 10^8} = 3.2 \times 10^{-8}\text{ T}$ ### Step 2: Determine Magnetic Field Direction Using the direction cross-product rule: hatE times hatB = hatv$\hat{E} \times \hat{B} = \hat{v}$ hatj times hatB = hati$\hat{j} \times \hat{B} = \hat{i}$ Since we know that hatj times hatk = hati$\hat{j} \times \hat{k} = \hat{i}$, the direction of the magnetic field must be hatk$\hat{k}$: hatB = hatk$\hat{B} = \hat{k}$ Combining the magnitude and direction: vecB = 3.2 times 10^-8hatktext T$\vec{B} = 3.2 \times 10^{-8}\hat{k}\text{ T}$ ### Pattern Recognition A propagation along +x$+x$ with electric field along +y$+y$ strictly mandates the magnetic field must point along +z$+z$ (+y times +z = +x$+y \times +z = +x$). This instantly eliminates Option 2 and Option 3, leaving only magnitude verification. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

### Related Formula I = frac12 epsilon_0 E_0^2 c$I = \frac{1}{2} \epsilon_0 E_0^2 c$ Where E_0$E_0$ is the amplitude of the electric field (200text V/m$200\text{ V/m}$) and c = 3 times 10^8text m/s$c = 3 \times 10^8\text{ m/s}$. ### Core Logic Substitute the constants into the equation: I = frac12 times (8.85 times 10^-12) times (200)^2 times (3 times 10^8)$I = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (200)^2 \times (3 \times 10^8)$ ### Step 1: Compute value I = frac12 times 8.85 times 10^-12 times 4 times 10^4 times 3 times 10^8$I = \frac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^4 \times 3 \times 10^8$ I = 2 times 8.85 times 3 times 10^0$I = 2 \times 8.85 \times 3 \times 10^0$ I = 53.1text W/m^2$I = 53.1\text{ W/m}^2$ ### Pattern Recognition Isolate powers of ten first (10^-12 times 10^4 times 10^8 = 10^0$10^{-12} \times 10^4 \times 10^8 = 10^0$) to streamline intermediate tracking accuracy on calculation variables. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves