### 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$p = E/c$)[cite: 708, 709]. Reason (R) is correct because the rest mass of a photon equals zero. ### Chapter Mix Class 12 Physics: Electromagnetic Waves