Question

What will be the radius of curvature of the path of a 3.0 KeV proton in ap perpendicular magnetic field of magnitude 0.8 T ?

Answer 1

Magnetic field B = 0.8 T

Charge on proton q = 1.6*10^-19 C

Mass of the proton m = 1.67*10^-27 Kg

From Conservation of energy

                                 (1/2)mv^2 = 3*10^3 eV

                    0.835*10^-27 v^2 = 3*10^3 * 1.6*10^-19

                                            v^2 = 5.748*10^11

                                               v = 7.581*10^5 m/s

Its clear that centipetal force = force acting on the proton

                              mv^2/r = Bqv

                                   mv/r = Bq

                                      mv = Bqr

          Radius of curvature r = mv/Bq

                                            = (1.67*10^-27*7.581*10^5)/(0.8*1.6*10^-19)

                                            = 9.89*10^-3 m or 9.89 mm.