Question

A proton enters a region of constant magnetic field, perpendicular to the fie and after being accelerated from rest by an electric field through an electric potential difference of - 350 V. Determine the magnitude of the magnetic field, if the proton travels in a circular path with a radius of 21 cm. mt As shown in the figure below, when a charged particle enters a region of magnetic field traveling in a direction perpendicular to the field, it will travel in a circular path. If the magnitude of the magnetic field is 0.130 T, the speed of the particle is 12.0 x 106 m/s, the radius of the trajectory of the path is 109.9 cm, and the charge has a magnitude of 3.20 - 10-19 C, determine the following (a) sign of the charged particle positive o negative (b) mass of the particle (Use the conversion 1 u = 1.66 x 10-27 kg.) A beam of alpha particles is fired horizontally with a speed of 1.5 - 105 m/s into a region where there is a vertical magnetic field of magnitude 0.388 T. An alpha particle is the nucleus of a helium atom and as such, consists of two protons, two neutrons, and has a mass of 6.64 x 10-27 kg. (a) Determine the length of time it will take an alpha particle to move halfway around a complete circle. (b) Does the time found in part (a) increase, decrease, or stay the same, if the speed of the alpha particle is doubled? decrease o stay the same Tolonost na belah

Answer 1

quB =) mie 298 R a me 4B -27 5 6.64x10 x 1.5x10 2x16x10ls x 0.388 time to crm circle - R = 0.00802 1907 m 208021907 cm aplete half 28/2 UR ITX 0.008021907 1.5x 105 -7 ita 1.68x10 S b) af ola 20 Rama' и х2 qB aB 2 (mo ABA 2 2R 3 t, 2012/2 HR! T (2R) כ ہوہ 20 it at remains the same