JEE PYQ: Electromagnetic Waves Question 25
Question 25 - 2020 (09 Jan 2020 Shift 1)
The electric fields of two plane electromagnetic plane waves in vacuum are given by $\vec{E}_1 = E_0 \hat{j} \cos(\omega t - kx)$ and $\vec{E}_2 = E_0 \hat{k} \cos(\omega t - ky)$. At $t = 0$, a particle of charge $q$ is at origin with a velocity $\vec{v} = 0.8c\hat{j}$. The instantaneous force experienced by the particle is:
(1) $E_0 q(0.8\hat{i} - \hat{j} + 0.4\hat{k})$
(2) $E_0 q(0.4\hat{i} - 3\hat{j} + 0.8\hat{k})$
(3) $E_0 q(-0.8\hat{i} + \hat{j} + \hat{k})$
(4) $E_0 q(0.8\hat{i} + \hat{j} + 0.2\hat{k})$
Show Answer
Answer: (4)
Solution
At $t=0$, $x=y=0$: $\vec{E} = E_0(\hat{j}+\hat{k})$, $\vec{B} = \frac{E_0}{c}(\hat{k}+\hat{i})$. $\vec{F} = qE_0(\hat{j}+\hat{k}) + 0.8qE_0\hat{j}\times(\hat{k}+\hat{i}) = E_0 q(0.8\hat{i}+\hat{j}+0.2\hat{k})$.
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