A light beam is described by E = 800 sin ω ( t - x/c ). An electron is allowed to move normal to the propagation of light beam with a speed of 3 × 10⁷ ms⁻¹. What is the maximum magnetic force exerted on the electron?

Solution

Physics sir · 4 steps

Get the peak magnetic field from B₀ = E₀/c, then the force is just evB₀.

1. Read the amplitude. The wave is E = 800 sin ω(t - x/c), so the peak electric field is E₀ = 800 V m⁻¹.
2. Peak magnetic field. In an EM wave the field amplitudes are linked by B₀ = E₀/c = 800/(3 × 10⁸) = 2.67 × 10⁻⁶ T.

   Fₘₐₓ = evB₀ = e v E₀/c

3. Put in numbers. The force is largest when v⃗ is perpendicular to B⃗. Then Fₘₐₓ = 1.6 × 10⁻¹⁹ × 3 × 10⁷ × 800/(3 × 10⁸) = 1.6 × 10⁻¹⁹ × 80 = 1.28 × 10⁻¹⁷ N.
4. So. 1.28 × 10⁻¹⁷ N is the same as 12.8 × 10⁻¹⁸ N, which is option (d).

The idea to keep. In any EM wave B₀ = E₀/c. The magnetic force on a charge is qvB, about v/c times the electric force qE, so it is much weaker unless the charge moves near light speed.

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