A beam of light travelling along X-axis is described by the electric field E_y=900 sin ω(t-x / c). The ratio of electric force to magnetic force on a charge q moving along Y-axis with a speed of 3 × 10⁷ ms⁻¹ will be :
(Given speed of light =3 × 10⁸ ms⁻¹)

Solution

Physics sir · 4 steps

In an EM wave E = cB, so the electric to magnetic force ratio is just c over the charge's speed.

1. Find the field directions. The wave moves along x and E points along y. B must be perpendicular to both, so it points along z. Its size is B = E/c.
2. Write both forces. The electric force is F_E = qE. The charge moves along y and B is along z, so they are at 90° and the magnetic force has its full value, F_B = qvB.

   F_E/F_B = qE/qvB = E/(v (E/c)) = c/v

3. Put in numbers. c/v = (3 × 10⁸)/(3 × 10⁷) = 10. The amplitude 900 cancels, so it does not affect the ratio.
4. So. The electric force is 10 times the magnetic force, and the ratio is 10 : 1.

The idea to keep. On a charge in an EM wave, F_E/F_B = c/v when v is perpendicular to B. A slow charge feels mostly the electric force; the two forces become equal only as v approaches c.

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