Question

1a) A charged particle with a charge-to-mass ratio of |q|/m = 5.7 × 10⁸ C/kg travels on a circular path that is perpendicular to a magnetic field whose magnitude is 0.55 T. How much time does it take for the particle to complete one revolution?

1b) Suppose that an ion source in a mass spectrometer produces doubly ionized gold ions (Au²⁺), each with a mass of 3.27 × 10^-25 kg. The ions are accelerated from rest through a potential difference of 1.30 kV. Then, a 0.710-T magnetic field causes the ions to follow a circular path. Determine the radius of the path.

thank you!

Answer 1

a) The time taken to complete one revolution is, 2πη T qB 27 (5.7x108)(0.55) = 2.0x10-83 =20x10° =20 ns b) The kinetic energy acquired by the ions is, 1 - mv2 = qV 2

2qV V= V m Thus the radius of the path followed by the ions is, mi qB m = 2qV qB V m 2mV VqB2 2(3.27x10-2)(1.30x109) (2)(1.60x10-1')(0.710) = 7.30 x 10-2 m = 7.30 cm