- JEE Mains 2020
- Hard
- MCQ
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
Answer & Solution
(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
Its corresponding magnetic field will be
\(\overrightarrow{\mathrm{B}}_{1}=\frac{\mathrm{E}_{0}}{\mathrm{c}} \hat{\mathrm{k}} \cos (\omega \mathrm{t}-\mathrm{kx})\)
\(\overrightarrow{\mathrm{E}}_{2}=\mathrm{E}_{0} \hat{\mathrm{k}} \cos (\omega \mathrm{t}-\mathrm{ky})\)
\(\overrightarrow{\mathrm{B}}_{2}=\frac{\mathrm{E}_{0}}{\mathrm{c}} \hat{\mathrm{i}} \cos (\omega \mathrm{t}-\mathrm{ky})\)
Net force on charge particle
\(=\mathrm{q} \overrightarrow{\mathrm{E}}_{1}+\mathrm{q} \overrightarrow{\mathrm{E}}_{2}+\mathrm{q} \overrightarrow{\mathrm{v}} \times \mathrm{B}_{1}+\mathrm{q} \overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}_{2}\)
\(=\mathrm{q} \mathrm{E}_{0} \hat{\mathrm{j}}+\mathrm{q} \mathrm{E}_{0} \hat{\mathrm{k}}+\mathrm{q}(0.8 \mathrm{c} \hat{\mathrm{j}}) \times\left(\frac{\mathrm{E}_{0}}{\mathrm{c}} \hat{\mathrm{k}}\right)+\mathrm{q}(0.8 \mathrm{c} \hat{\mathrm{j}}) \times\left(\frac{\mathrm{E}_{0}}{\mathrm{c}} \hat{\mathrm{i}}\right)\)
\(=\mathrm{q} \mathrm{E}_{0} \hat{\mathrm{j}}+\mathrm{q} \mathrm{E}_{0} \hat{\mathrm{k}}+0.8 \mathrm{q} \mathrm{E}_{0} \hat{\mathrm{i}}-0.8 \mathrm{q} \mathrm{E}_{0} \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{F}}=\mathrm{q} \mathrm{E}_{0}[0.8 \hat{\mathrm{i}}+1 \hat{\mathrm{j}}+0.2 \hat{\mathrm{k}}]\)
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