First slide

Introduction-Electromagnetic waves

Question

The electric fields of two plane electromagnetic plane waves in vacuum are given by ${\vec{E}}_{1} = E_{0} \hat{j}$ $\cos (ω t - k x)$ and ${\vec{E}}_{2} = E_{0} \hat{k} \cos (ω t - k y)$ . At $t = 0$ , a particle of charge q is at origin with a velocity $\vec{v} = 0.8 c \hat{j}$ (c is the speed of light in vacuum). The instantaneous force experienced by the particle is :

Difficult

A

$E_{0} q (0.8 \hat{i} - \hat{j} + 0.4 \hat{k})$
B

$E_{0} q (0.8 \hat{i} + \hat{j} + 0.2 \hat{k})$
C

$E_{0} q (0.4 \hat{i} - 3 \hat{j} + 0.8 \hat{k})$
D

$E_{0} q (- 0.8 \hat{i} + \hat{j} + \hat{k})$

Solution

For $\vec{E_{1}}$ corresponding $\vec{B_{1}} = \frac{E_{0}}{C} \hat{k} \cos (ω t - k x)$
For $\vec{E_{2}}$ corresponding $\vec{B_{2}} = \frac{E_{0}}{C} \hat{i} \cos (ω t - k y)$
$\begin{array}{l} T o t a l f o r c e a c t i n g a t t = 0, x = 0 i s \\ {\vec{F}}_{n e t} = q {\vec{E}}_{n e t} + q (\vec{v} \times {\vec{B}}_{n e t}) \\ h e r e {\vec{E}}_{n e t} = {\vec{E}}_{1} + {\vec{E}}_{2} = E_{0} \hat{j} + E_{0} \hat{k} \\ a n d \\ {\vec{B}}_{n e t} = {\vec{B}}_{1} + {\vec{B}}_{2} = \frac{E_{0}}{C} \hat{k} + \frac{E_{0}}{C} \hat{i} \\ \Rightarrow \vec{F} = q (E_{0} \hat{j} + E_{0} \hat{k}) + q (0.8 c \hat{j} \times (\frac{E_{0}}{C} \hat{k} + \frac{E_{0}}{C} \hat{i})) \\ \Rightarrow \vec{F} = q E_{0} \hat{j} + q E_{0} \hat{k} + 0.8 q E_{0} \hat{i} - 0.8 q E_{0} \hat{k} \\ \Rightarrow \vec{F} = q E_{0} (0.8 \hat{i} + \hat{j} + 0.2 \hat{k}) \end{array}$

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