Q27 (2026)
A plane electromagnetic wave is moving in free space with velocity $c = 3 \times 10^{8} \text{ m/s}$ and its electric field is given as $\vec{E} = 54 \sin (\text{kz} - \omega t) \hat{j} \, \text{V/m}$ , where $\hat{j}$ is the unit vector along y-axis. The magnetic field vector $\vec{B}$ of the wave is :
- $-1.8 \times 10^{-7} \sin (\mathrm{kz} - \omega t) \hat{\mathrm{i}} \text{ T}$
- $1.4 \times 10^{-7} \sin (\mathrm{kz} - \omega t) \hat{\mathrm{k}} \text{ T}$
- $1.4 \times 10^{-7} \sin (\mathrm{kz} - \omega t) \hat{\mathrm{i}} \text{ T}$
- $+1.8 \times 10^{-7} \sin (\mathrm{kz} - \omega t) \hat{\mathrm{i}} \text{ T}$
### Related Formula
$$B_0 = \frac{E_0}{c}$$
$$\hat{c} = \hat{E} \times \hat{B}$$
Where $\hat{c}$ is the direction of wave propagation.
### Core Logic
From the argument of sine $(kz - \omega t)$, the wave is propagating along the positive z-axis. Thus, $\hat{c} = \hat{k}$.
The electric field oscillates along the y-axis, so $\hat{E} = \hat{j}$.
### Step 1: Find Magnetic Field Direction
Using the cross-product relation for directions:
$$\hat{B} = \hat{c} \times \hat{E} = \hat{k} \times \hat{j} = -\hat{i}$$
### Step 2: Find Magnetic Field Amplitude
$$B_0 = \frac{E_0}{c} = \frac{54}{3 \times 10^8} = 18 \times 10^{-8} = 1.8 \times 10^{-7} \text{ T}$$
### Step 3: Assemble the Final Vector
$$\vec{B} = 1.8 \times 10^{-7} \sin (kz - \omega t) (-\hat{i}) = -1.8 \times 10^{-7} \sin (kz - \omega t) \hat{i} \text{ T}$$
### Pattern Recognition
Always use $\hat{E} \times \hat{B} = \hat{c}$. Given $\hat{c} = \hat{k}$ and $\hat{E} = \hat{j}$, $\hat{B}$ must be $-\hat{i}$ to satisfy $\hat{j} \times (-\hat{i}) = \hat{k}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Waves
