A proton is moving with a speed of 3 10⁶ m s ⁻¹ along the direction of propagation of an electromagnetic wave in a vacuum. If the maximum magnetic force experienced by the proton is 4.8 10⁻²⁰ N , the amplitude of the electric field of the electromagnetic wave is : [Given : Charge of proton = 1.6 10⁻¹⁹ C , Speed of light c = 3 10⁸ m s ⁻¹ ]

Step-by-step solution

The maximum magnetic force on a moving charge is given by F_ m = qvB₀ , where B₀ is the amplitude of the magnetic field. Substituting the given values: 4.8 10⁻²⁰ = (1.6 10⁻¹⁹) (3 10⁶) B₀ B₀ = 4.8 10⁻²⁰ 4.8 10⁻¹³ = 10⁻⁷ T The relationship between the amplitudes of the electric and magnetic fields in an electromagnetic wave is E₀ = cB₀ . E₀ = (3 10⁸) 10⁻⁷ = 30 V m ⁻¹ Answer: 30 V m ⁻¹