Question

A1.9-MeV (kinetic energy) proton enters a 0.29-t field, in a plane perpendicular to the field. what is the radius of its path?

Answer 1

Kinetic energy

KE=(1/2)mv²

(1.9*10^6*1.6*10^-19)=(1/2)*(1.67*10^-27)*v²

v=1.91*10⁷ m/s

magnetic force =centripetal force

qvB=mv²/r

=>r=mv/qB =(1.67*10^-27)*(1.91*10⁷)/(1.6*10^-19)*0.29

r=0.6867 m

Answer 2

For a proton
Magnetic field B=0.29T
Kinetic energy
K=1.9 MeV= 1.9*1.6*10^-19*10^6=1.6*1.9*10^-13=3.04

Answer 3

the formula is $r=\frac{mv}{qB}$

                     given $\frac{1}{2}mv^{2}=$ $\frac{1}{2}mv^{2}=1.9*1000000*1.6*10^{-19}$

                             here $9.11*10^{-31}$

$v=8.17*10^{3}$ m/s

there fore $r=\frac{9.11*10^{-31}*8.17*10^{3}}{1.6*10^{-19}*0.29}$

$r=1.6*10^{11} m$