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JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
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
- A \(0\)
- B parallel to \(\frac{\mathrm{i}+\mathrm{j}}{\sqrt{2}}\)
- C antiparallel to \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\)
- D parallel to \(\hat{\mathrm{k}}\)
Answer & Solution
Correct Answer
(C) antiparallel to \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{E}}+\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})\) \(\overrightarrow{\mathrm{E}}=\mathrm{E}_{0}\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right) \cos \pi\)…
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