The magnetic field vector of an electromagnetic wave is given by B = B₀(î + ĵ)/√2 cos (kz - ω t); where î,ĵ represents unit vector along x and y-axis respectively. At t = 0s, two electric charges q₁ of 4π coulomb and q₂ of 2π coulomb located at ( 0,0,π/k ) and ( 0,0,3π/k ), respectively, have the same velocity of 0.5 c î, (where c is the velocity of light). The ratio of the force acting on charge q₁ to q₂ is :-

Solution

Physics sir · 4 steps

Check the phase at both charges first: if the fields match, only the charges decide the force ratio.

1. Lorentz force. Each charge feels F⃗ = q(E⃗ + v⃗ × B⃗). Both charges have the same velocity 0.5c î, so if E⃗ and B⃗ are also the same at both points, the forces differ only through q.
2. Phase at each charge. At t = 0 the phase is kz. For q₁ it is k · π/k = π, and for q₂ it is k · 3π/k = 3π.

   cos π = cos 3π = -1

3. Same fields. So B⃗ = -B₀(î + ĵ)/√2 at both charges. The electric field E⃗ = cB₀(î - ĵ)/√2 cos(kz - ω t) has the same phase, so it is also identical at both points. The bracket E⃗ + v⃗ × B⃗ is therefore the same vector for both charges.
4. So. F₁/F₂ = q₁/q₂ = 4π/2π = 2, so the ratio is 2 : 1, option (c).

The idea to keep. Points one full wavelength apart (Δ z = 2π/k) have identical E and B at every instant. With the same velocity, the force then scales only with the charge.

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