1.

Electric field in space is given by vectorE(t) = E0{(i + j)}/{√2} cos (ωt + Kz). A positively charged particle at (0, 0, π/K) is given velocity v0k at t = 0. Direction of force acting on particle is(1) f = 0(2) Antiparallel to {(i + j)}/{√2}(3) Parallel to {(i + j)}/{√2}(4) k

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Answer is (2)

Force due to electric field is in direction - {i + j}/{√2}

because at t = 0, E = - {i + j}/{√2}E0

Force due to magnetic field is in direction q(vectorv x vectorB) and vectorv || k

∴ it is parallel to vectorE

∴ net force is antiparallel to {i + j}/{√2}



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