Question

A helium ion (Q = + 2e) whose mass is 6.6 x 10-27 kg is accelerated by a voltage of 3700 V. (а) What is its speed? (b) What will be its radius of curvature if it moves in a plane perpendicular to a uniform 0.430-T field? (c) What is its period of revolution? x x x x x x x x x++x x x x x x x x x x x x x x x x x x x x

Please answer all parts step by step as soon as possible.

Answer 1

9 which is ay When a changed particle at rest, is accelerated by a potential difference V, then its Kinetic energy is given by, Qv Again, Kinetic enengy of a particle of mass m moving with speed & is given by mp2 SV = m 2x@x1.6 x 10) (3 700V) kg mp2 ง 20V 6.6x10 -27 in v = 5.989x 105 mis .:: Its speed is V=5.989x 105 mis b) If a changed particle o enters a region of uniform magnetic field B, pointing perpendicular to its velocity & then the magnetic force on the changed particle is, Fə Qu B sin goº 9VB This magnetic force is perpendicular to the direction of travel, the particle follows a curred path in the magnetic field. If the changed particle moves in a cincular orbit of radius s, then the magnetic fonce go B is by the centripetal fonce balanced moz no 99B ทาง po= mu mne BQ

mu BO 6.66 1027 kgx(5.989x105 m/s) (0.43 T) (2* 1.6x17'%ce) :: po = 2.87X10 2.878 102 m .. Its madius of curvature will be: 8 = 2.87 x 10²m e) QT Period of me of revolution, T perimeter of cincular ombit of radius [ perimeter is 200 T- RTY T- QT (2-87x102m (5.989 X 105 m/s) .: T= 3.01x107 s .: Its period of peyolutioon, T= 3.01x107

The magnetic force on a charged particle Q moving with velocity v perpendicular to the magnetic field B is given by F = QvB. This magnetic force is perpendicular to the direction of travel, the particle follows a circular path of radius r. The magnetic force F is balanced by centripetal force mv^2/r.