JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
The electric fields of two plane electromagnetic plane waves in vacuum are given by \(\overrightarrow{\mathrm{E}}_{1}=\mathrm{E}_{0} \hat{\mathrm{j}} \cos (\omega \mathrm{t}-\mathrm{kx})\) and \(\overrightarrow{\mathrm{E}}_{2}=\mathrm{E}_{0} \hat{\mathrm{k}} \cos (\omega \mathrm{t}-\mathrm{ky})\) At \(t=0,\) a particle of charge \(q\) is at origin with a velocity \(\overrightarrow{\mathrm{v}}=0.8 \mathrm{c} \hat{\mathrm{j}}\) (\(c\) is the speed of light in vacuum). The instantaneous force experienced by the particle is
- A \(\mathrm{E}_{0} \mathrm{q}(-0.8 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)
- B \(\mathrm{E}_{0} \mathrm{q}(0.8 \hat{\mathrm{i}}-\hat{\mathrm{j}}+0.4 \hat{\mathrm{k}})\)
- C \(\mathrm{E}_{0} \mathrm{q}(0.8 \hat{\mathrm{i}}+\hat{\mathrm{j}}+0.2 \hat{\mathrm{k}})\)
- D \(\mathrm{E}_{0} \mathrm{q}(0.4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+0.8 \hat{\mathrm{k}})\)
Answer & Solution
Correct Answer
(C) \(\mathrm{E}_{0} \mathrm{q}(0.8 \hat{\mathrm{i}}+\hat{\mathrm{j}}+0.2 \hat{\mathrm{k}})\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathrm{E}}_{1}=\mathrm{E}_{0} \mathrm{j} \cos (\omega \mathrm{t}-\mathrm{kx})\) Its corresponding magnetic field will be \(\overrightarrow{\mathrm{B}}_{1}=\frac{\mathrm{E}_{0}}{\mathrm{c}} \hat{\mathrm{k}} \cos (\omega \mathrm{t}-\mathrm{kx})\)…
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