The electric fields of two plane electromagnetic plane waves in vacuum are given by
E₁⃗ = E₀ĵ cos ( ω t - kx ) and
E₂⃗ = E₀k̂cos ( ω t - ky )
At t = 0, a particle of charge q is at origin with
a velocity v⃗ = 0.8cĵ (c is the speed of light in vacuum). The instantaneous force experienced by the particle is :

Solution

Physics sir · 4 steps

Get each wave's B from its travel direction, add both at the origin, then use the full Lorentz force.

1. Rule for B. In a plane EM wave, |B⃗| = |E⃗|/c, and E⃗ × B⃗ points along the direction of travel. That fixes the direction of B⃗ for each wave.

   Wave	Travels along	E⃗ along	B⃗ along (so E⃗ × B⃗ is right)
   1: cos(ω t - kx)	+î	ĵ	k̂, since ĵ × k̂ = î
   2: cos(ω t - ky)	+ĵ	k̂	î, since k̂ × î = ĵ

2. Fields at origin. At t = 0 and the origin every cosine equals 1. So E⃗ = E₀(ĵ + k̂) and B⃗ = E₀/c(î + k̂).
3. Magnetic part. v⃗ × B⃗ = 0.8c ĵ × E₀/c(î + k̂) = 0.8E₀(î - k̂), using ĵ × î = -k̂ and ĵ × k̂ = î.

   F⃗ = qE₀[(ĵ + k̂) + (0.8 î - 0.8 k̂)] = E₀ q(0.8 î + ĵ + 0.2 k̂)

4. So. The instantaneous force is E₀ q(0.8 î + ĵ + 0.2 k̂), which is option (c).

The idea to keep. For any EM wave, get B⃗ from the rule that E⃗ × B⃗ points along the travel direction. Then use q(E⃗ + v⃗ × B⃗) with every wave included.

Answer from the NTA final answer key. Solution by PrepWiser.