StreakPeaked· Practice

ExamsJEE MainPhysics

The magnetic field of a plane electromagnetic wave is given by: B = B₀î [cos(kz-ω t)] + B₁ĵcos(kz+ω t) where B₀ = 3×10⁻⁵ T and B₁ = 2×10⁻⁶ T. The rms value of the force experienced by a stationary charge Q = 10⁻⁴ C at z=0 is closest to:

  1. 0.6 N
  2. 0.1 N
  3. 0.9 N
  4. 3×10⁻² N

Correct answer: 0.6 N

Solution

A charge at rest feels no magnetic force, only F = qE with E = cB. E0 = c*B0 = 9000 V/m and E1 = c*B1 = 600 V/m are perpendicular, so Erms = sqrt((E0^2+E1^2)/2) ~ 6378 V/m and F = qErms = 1e-4*6378 ~ 0.6 N.

Related JEE Main Physics questions

⚔️ Practice JEE Main Physics free + battle 1v1 →