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Q.

The Electric field of a plane electromagnetic wave is given byE→=E0i^+j^2cosKz+ωt. At t=0, a positively charged particle is at the point  x,y,z=0,0,πK.If its instantaneous velocity at (t=0) is  v0k^ , the force acting on it due to the wave is:

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a

zero

b

Antiparallel to  i^+j^2

c

Parallel to  i^+j^2

d

Parallel to  k^

answer is B.

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Detailed Solution

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
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