A particle with a mass of 2.0 10⁻¹⁰ g and a charge of 2.0 10⁻⁸ C is projected at a speed of 2.0 10^3 m s ⁻¹ into a region containing a uniform magnetic field of 0.10 T . The particle's velocity is perpendicular to the magnetic field. Determine the radius of the circular path described by the particle, as well as its time period.

Step-by-step solution

Given mass of the particle, m = 2.0 10⁻¹⁰ g = 2.0 10⁻¹³ kg Charge, q = 2.0 10⁻⁸ C Speed, v = 2.0 10^3 m s ⁻¹ Magnetic field, B = 0.10 T The radius of the circular path is given by: r = mv qB Substituting the given values: r = 2.0 10⁻¹³ 2.0 10^3 2.0 10⁻⁸ 0.10 r = 4.0 10⁻¹⁰ 2.0 10⁻⁹ = 0.2 m = 20 cm The time period of revolution is given by: T = 2 m qB = 2 r v Substituting the values: T = 2 3.14 0.2 2.0 10^3 T = 1.256 2.0 10^3 = 6.28 10⁻⁴ s 6.3 10⁻⁴ s