The Electric field of a plane electromagnetic wave is given by$\overrightarrow{E}=E_0\frac{\widehat{i}+\widehat{j}}{\sqrt{2}}\cos \left(Kz+\omega t\right)$. At t=0, a positively charged particle is at the point  $\left(x,y,z\right)=\left(0,0,\frac{\pi }{K}\right)$.If its instantaneous velocity at (t=0) is  $v_0\widehat{k}$ , the force acting on it due to the wave is:

detailed solution

Correct option is B

At t=0, x =K ​Electric field =E→=E0i^+j^2cosπ=−i^+j^2E0 ∴  Force due to electric field is in direction  −i^+j^2Direction ofelectromagnetic wave is in −z direction​∴C^=−k^ is wave direction.Let B^ is the unit vector along the direction of magnetic field.​ B^=C^×E^ = −k^×−i^+j^2=−i^+j^2Now, Force due to magnetic field is in direction qv→×B→  and  v→||k^ Force due to magnetic field is F^=k^×-i^+j^2 =-i^-j^2=-i^+j^2We can see that direction of both electric and magnetic force is same.∴  Net force is antiparallel to  i^+j^2