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