The electric field of a plane electromagnetic wave is given by $$\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}} \cos (\mathrm{kz}+\omega \mathrm{t})$$ 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:

(D) antiparallel to $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

(B) parallel to $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · EM Waves

The electric field of a plane electromagnetic wave is given by

E = E_{0} \frac{i ^ + j ^}{2} cos (kz + ω t)

t = 0

, a positively charged particle is at the point

(x, y, z) = (0, 0, \frac{π}{k})

. If its instantaneous velocity at

(t = 0)

v_{0} \hat{k}

, the force acting on it due to the wave is:

Answer: (D)

t = 0

and

z = \frac{π}{k}

the phase

k z + ω t

equals

π

, so

cos π = - 1

and the field is

E = - E_{0} \frac{i ^ + j ^}{2}

. The electric force on the positive charge,

q E

, then points opposite to

\frac{i ^ + j ^}{2}

, that is antiparallel to it. The magnetic force stays perpendicular to the velocity

v_{0} \hat{k}

and does not change this conclusion about the direction along

\hat{i} + \hat{j}

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option