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

The electric field of a plane electromagnetic wave is given by vecE(t)

1.	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
Answer» zero antiparallel to `(hati+hatj)/SQRT2` PARALLEL to `(hati+hatj)/sqrt2` parallel to `hatk` Solution :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`.

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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

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