An electron is constrained to move along the y-axis with a speed of 0.1 c (c is the speed of light) in the presence of electromagnetic wave, whose electric field is
E⃗ = 30ĵ sin ( 1.5 × 10⁷t - 5 × 10⁻²x ) V/m.
The maximum magnetic force experienced by the electron will be :
(given c = 3 × 10⁸ ms –1 and electron charge = 1.6 × 10 –19 C)

Solution

Physics sir · 4 steps

Get B₀ from E₀/c, check that v⃗ is perpendicular to B⃗, then F = qvB₀.

1. Read the wave. The phase is ω t - kx, so the wave travels along x. The electric field is along y, so the magnetic field must be along z. Check: ω/k = (1.5 × 10⁷)/(5 × 10⁻²) = 3 × 10⁸ m s⁻¹ = c.
2. Find the peak B. In an electromagnetic wave, B₀ = E₀/c = 30/(3 × 10⁸) = 10⁻⁷ T.
3. Angle check. The electron moves along y and B⃗ points along z, so they are at 90° and sin 90° = 1. The force is largest when B reaches its peak value B₀.

   Fₘₐₓ = evB₀ = (1.6 × 10⁻¹⁹)(0.1 × 3 × 10⁸)(10⁻⁷)

4. So. Fₘₐₓ = 1.6 × 10⁻¹⁹ × 3 × 10⁷ × 10⁻⁷ = 4.8 × 10⁻¹⁹ N, which is option (a).

The idea to keep. For the magnetic part of an EM wave, always take B₀ = E₀/c. Then B⃗ is perpendicular to both E⃗ and the direction of travel, and that tells you the angle in qvB sin θ.

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