JEE Main20209 Jan 2020Morning ShiftPhysicsElectromagnetic WavesActual
The electric fields of two plane electromagnetic plane waves in vacuum are given by E 1 → = E 0 j ^ cos ⁡ ω t - k x and E 2 → = E 0 k ^ cos ⁡ ω t - k y At t = 0 , a particle of charge q is at origin with a velocity v → = 08 c j ^ ( c is the speed of light in vaccum). The instantaneous force experienced by the particle is:
Options
- AE 0 q 0.8 i ^ - j ^ + 0.4 k ^
- BE 0 q 0.4 i ^ - 3 j ^ + 0.8 k ^
- CE 0 q - 0.8 i ^ + j ^ + k ^
- DE 0 q 0.8 i ^ + j ^ + 0.2 k ^
Correct answer
D. E 0 q 0.8 i ^ + j ^ + 0.2 k ^
Step-by-step solution
Magnetic field vectors associated with this electromagnetic wave are given by B → 1 = E 0 c k ^ cos k x - ω t & B → 2 = E 0 c i ^ cos k y - ω t F → = q E → + q V → × B → = q E → 1 + E → 2 + q V → × B → 1 + B → 2 By putting the value of E → 1 , E → 2 , B → 1 & B → 2 The net Lorentz force on the charged particles is F → = q E 0 0.8 cos k x - ω t i ^ + c o s k y - ω t j ^ + 0.2 c o s k y - ω t k ^ At t = 0 and at x = y = 0 F → = q E 0 0.8 i ^ + j ^ + 0.2 k ^