Question

hapter 20 Problem 20.17 A 2.5-MeV (kinetic energy) proton enters a 0.31-T field, in a plane perpendicular to the field Part A What is the radius of its path? Express your answer to two significant figures and include the appropriate units. HÅ T= 0.52234 m Submit Previous Answers Request Answer * Incorrect; Try Again; 4 attempts remaining Provide Feedback

Answer 1

= 2,5 MeV 0.31T Given K.E = Kinetic energy 2.5x108 ev B = Magnetic field = Mpa Mass of Proton 1.6x10 - 27 kg qp = charge of preoton = 1.6 x 10 -19 lev = 19 1.6 x 10 Using equation for forece CF) exporcienced by a charged particle in a magnetic field (8) - VUB ma - Centrcipefol ma qUB R acceleration) F 2 bul [ > mees R = QEB U mu ER qB Kinetic energy = Ka E = mo 2x RIE عی ax RE - me mp Substituting 'i' from equation into equation 2 KE max me R qB » R = VaxkiExmp qB Selbstëfeeting -27 values in equation 6 2x 2.5 x 106 x 1.6x10 x 1.6 X10-9 1.6x 10-19 x 0.31 R=

* R = -20 3.577708 X 10 20 4.96 x 10 0.721 m R = Radius = 0.72 m