The electric field of a plane electromagnetic wave is given by `vecE(t)=E_(0)(hati+hatj)/sqrt2cos(omegat+kz)`. At t = 0, a positively charged particle is at the point `(x,y,z)=(0,0,pi/k)`. If its instantaneous velocity at t = 0 is `v_(0)hatk`, the force acting on it due to the wave is

Options

A. zero
B. antiparallel to `(hati+hatj)/sqrt2`
C. parallel to `(hati+hatj)/sqrt2`
D. parallel to `hatk`

Answer

B

Explanation

Direction of force due to electric field = `-(hati+hatj)/sqrt2`
Because, at t = 0, `vecE=(-(hati+hatj))/sqrt2E_(0)`
`rArr` Direction of force `q(vecvxxvecB)` due to magnetic field is parallel to `vecE" as "vecv||hatk`
`therefore` Resultant force is anti-parallel to `((hati+hatj))/sqrt2`.