A portion is bent into a circular path of radius 2 m be a ma

A portion is bent into a circular path of radius 2 m be a magnetic field of 0.5 T. What is the momentum of the proton. What is its kinetic energy. What is its dimensionless speed factor (beta)

Solution

magnetic force provides the centipetal force,

if r be the radius,

a.) qvb=mv²/r,

mv=qbr

monentum=.5*2*1.6*10^-19=8.32*10^-19

^{B. kinetic enerygy}

^{q*v*B*r=m*v*v}

^{calculating v,}

^v=q*r*b/m,

v=8.32*10²⁷/1.6*10^19,

v=5.2*10⁸

so kinetic energy,1/2*m*v²

1/2*1.6*10^-27*27.04*10¹⁶

21.632*10^-11..

c.speed factor

qb/2*3.14*1.6*10^-27

=.08*10⁸