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 velocity $\vec{v}=0.8 c \hat{j}$ ( $c$ is the speed of light in vaccum). The instantaneous force experienced by the particle is:

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

(A) $\mathrm{E}_{0} \mathrm{q}(0.8 \hat{\mathrm{i}}-\hat{\mathrm{j}}+0.4 \hat{\mathrm{k}})$

(B) $\mathrm{E}_{0} \mathrm{q}(0.4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+0.8 \hat{\mathrm{k}})$

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

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · EM Waves

The electric fields of two plane electromagnetic plane waves in vacuum are given by

E_{1} = E_{0} \hat{j} cos (ω t - kx)

and

E_{2} = E_{0} \hat{k} cos (ω t - ky)

t = 0

, a particle of charge

q

is at origin with velocity

v = 0.8 c \hat{j}

(

c

is the speed of light in vaccum). The instantaneous force experienced by the particle is:

Answer: (C)

Each wave carries a magnetic field of magnitude

\frac{E _{0}}{c}

perpendicular to its own direction of travel, so

B_{1} = \frac{E _{0}}{c} \hat{k} cos (ω t - k x)

and

B_{2} = \frac{E _{0}}{c} \hat{i} cos (ω t - k y)

. At

t = 0

and the origin every cosine is

1

, giving

E_{1} = E_{0} \hat{j}

E_{2} = E_{0} \hat{k}

B_{1} = \frac{E _{0}}{c} \hat{k}

and

B_{2} = \frac{E _{0}}{c} \hat{i}

. With

v = 0.8 c \hat{j}

, the magnetic part of the force is

q v \times (B_{1} + B_{2}) = 0.8 E_{0} q (\hat{j} \times \hat{k} + \hat{j} \times \hat{i}) = 0.8 E_{0} q (\hat{i} - \hat{k})

. Adding the electric force

q E_{0} (\hat{j} + \hat{k})

, the total is

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

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Single correct option