The electric field of a plane electromagnetic wave is given by E^{\to}=E_0{i^{\^}+j^{\^}}/\sqrt2 \cos (kz+\omegat) 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 \upsilon_0k^{\^}, the force acting on it due to the wave is : [7 Jan 2020 Shift 2]

Correct Answer is : antiparallel to ^i+^j√2

parallel to {i^{\^}+j^{\^}}/\sqrt2

antiparallel to {i^{\^}+j^{\^}}/\sqrt2

JEE Main 7 Jan 2020 Shift 2 Solved Paper

Section: Physics

Question : 5 of 75

Marks: +1, -0

The electric field of a plane electromagnetic wave is given by

→

=E0

√2

‌cos(kz+ωt) At t=0, a positively charged particle is at the point (x,y,z)=(0,0,

). If its instantaneous velocity at (t=0) is υ0

, the force acting on it due to the wave is :

[7 Jan 2020 Shift 2]

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