JEE PYQ: Electromagnetic Waves Question 23
Question 23 - 2020 (08 Jan 2020 Shift 1)
The electric field of a plane electromagnetic wave is given by $\vec{E} = E_0 \frac{\hat{i} + \hat{j}}{\sqrt{2}} \cos(kz + \omega t)$. At $t = 0$, a positively charged particle is at the point $(x, y, z) = \left(0, 0, \frac{\pi}{k}\right)$. If its instantaneous velocity at $(t = 0)$ is $v_0 \hat{k}$, the force acting on it due to the wave is:
(1) parallel to $\frac{\hat{i} + \hat{j}}{\sqrt{2}}$
(2) zero
(3) antiparallel to $\frac{\hat{i} + \hat{j}}{\sqrt{2}}$
(4) parallel to $\hat{k}$
Show Answer
Answer: (3)
Solution
At $t = 0$, $z = \pi/k$: $\vec{E} = -\frac{E_0}{\sqrt{2}}(\hat{i} + \hat{j})$. Force due to electric field is in direction $-(\hat{i}+\hat{j})/\sqrt{2}$. Net force is antiparallel to $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$.
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