- Home
- Standard 12
- Physics
The electric field of a plane electromagnetic wave is given by
$\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}} \cos (\mathrm{kz}+\omega \mathrm{t})$ At $\mathrm{t}=0,$ a positively charged particle is at the point $(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(0,0, \frac{\pi}{\mathrm{k}}\right) .$ If its instantaneous velocity at $(t=0)$ is $v_{0} \hat{\mathrm{k}},$ the force acting on it due to the wave is
$0$
parallel to $\frac{\mathrm{i}+\mathrm{j}}{\sqrt{2}}$
antiparallel to $\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}$
parallel to $\hat{\mathrm{k}}$
Solution
$\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{E}}+\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})$
$\overrightarrow{\mathrm{E}}=\mathrm{E}_{0}\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right) \cos \pi$
$=-\mathrm{E}_{0} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}$
as $\quad \overrightarrow{E} \times \overrightarrow{B}=\overrightarrow{\mathrm{c}}$
$+\mathrm{E}_{0}\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right) \times \overrightarrow{\mathrm{B}}=\mathrm{c} \hat{\mathrm{k}}$
$\Rightarrow \overrightarrow{\mathrm{B}}=-\left(\frac{\hat{\mathrm{i}}-\hat{\mathrm{j}}}{\sqrt{2}}\right) \frac{\mathrm{E}_{0}}{\mathrm{c}}$
$\overline{\mathrm{F}}=\mathrm{q}\left(-\mathrm{E}_{0} \frac{(\hat{\mathrm{i}}+\hat{\mathrm{j}})}{\sqrt{2}}-\frac{\mathrm{v}_{0} \hat{\mathrm{k}}}{\mathrm{c}} \times(\hat{\mathrm{i}}-\hat{\mathrm{j}}) \mathrm{E}_{\mathrm{o}}\right)$
since $\frac{v_{0}}{c}<<1$
$\Rightarrow \mathrm{F}$ is antiparallel to $\frac{\mathrm{i}+\mathrm{j}}{\sqrt{2}}$
