An electromagnetic wave with an intensity of 375 W m ⁻² propagates in a vacuum. A charged particle of charge 2 C moves with a speed of 1.5 10⁷ m s ⁻¹ along the direction of propagation of the wave. The maximum magnetic force experienced by the charge is : [Given : Speed of light c = 3 10⁸ m s ⁻¹ , Permeability of free space ₀ = 4 10⁻⁷ T m A ⁻¹ ]

Step-by-step solution

The intensity I of an electromagnetic wave is related to the electric field amplitude E₀ by the formula: I = E₀² 2 ₀c We can calculate the value of ₀c : ₀c = (4 10⁻⁷) (3 10⁸) = 120 Substituting the given intensity: 375 = E₀² 2 120 E₀² = 375 240 = 90000 E₀ = 300 V m ⁻¹ The amplitude of the magnetic field B₀ is: B₀ = E₀ c = 300 3 10⁸ = 10⁻⁶ T The maximum magnetic force F_ m on the moving charge is: F_ m = qvB₀ F_ m = (2 10⁻⁶) (1.5 10⁷) 10⁻⁶ = 3 10⁻⁵ N Answer: 3 10⁻⁵ N