Determine the acceleration of a proton that is projected perpendicularly into a uniform magnetic field of 0.6 T with a velocity of 3 10^6 m s ⁻¹ .

Step-by-step solution

The magnetic force acting on a charged particle moving in a magnetic field is given by F = qvB . Since the proton is projected perpendicularly, = 90^ , which gives F = qvB . The acceleration of the proton is a = F m = qvB m . Substituting the standard values for a proton ( q = 1.6 10⁻¹⁹ C , m = 1.67 10⁻²⁷ kg ) and the given values: a = 1.6 10⁻¹⁹ 3 10^6 0.6 1.67 10⁻²⁷ a = 2.88 10⁻¹³ 1.67 10⁻²⁷ a 1.72 10¹⁴ m s ⁻²