The electric field of a plane electromagnetic wave is given by
E⃗ = E₀(î + ĵ)/√2 cos ( kz + ω t )
At t = 0, a positively charged particle is at the point (x, y, z) = ( 0,0,π/k ).
If its instantaneous velocity at (t = 0) is v₀k̂ , the force acting on it due to the wave is :

Solution

Physics sir · 4 steps

Put in z = π/k first: the cosine is -1, so the field points backwards, and the magnetic force does too.

1. Field at the charge. At t = 0 and z = π/k, the phase is kz + ω t = π, and cos π = -1. So E⃗ = -E₀ (î + ĵ)/√2, and the electric force qE⃗ is antiparallel to (î + ĵ)/√2.
2. Find B. The phase kz + ω t means the wave travels along -k̂. So B⃗ = 1/c(-k̂) × E⃗. Since -k̂ × (î + ĵ) = î - ĵ, at this point B⃗ = -E₀/c (î - ĵ)/√2.
3. Magnetic force. qv⃗ × B⃗ = q v₀ k̂ × (-E₀/c (î - ĵ)/√2). Using k̂ × î = ĵ and k̂ × ĵ = -î, this is -(q v₀ E₀)/c (î + ĵ)/√2, in the same direction as the electric force.

   F⃗ = qE⃗ + qv⃗ × B⃗ = -qE₀(1 + v₀/c)(î + ĵ)/√2

4. So. Both parts point along -(î + ĵ)/√2, so the net force is antiparallel to (î + ĵ)/√2. The answer is (c).

The idea to keep. For a charge in an EM wave, first find the sign of the cosine at that point. The magnetic force qv⃗ × B⃗ is only v/c times qE, so it cannot cancel the electric force for v < c.

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