Protons having a kinetic energy of 5 MeV are moving in the positive x-direction and enter a magnetic field `B=0.0500 hat k T` directed out of the plane of the page and extending from `x=0 to x=1 m` as shown in Fig. 1.34.
a. Calculate the y-component of the protons' momentum as they leave the magnetic field.
b. Find the angle `phi` between the initial velocity vector of the proton beam and the velocity vector after the beam emerges from the field. Ignore relativistic effects and note that `1 eV=1.60 xx 10^(-19) J.`

Explanation

(a) `E=5MeV=5xx1.6xx10^-13J`
`1/2mv^2=5xx1.6xx10^-13`
`v=sqrt((16xx10^-13)/m)`
`r=(mv)/(eB)`
`1=r sin phi`
`implies sin phi=1/r=(eB)/(mv)`
y-component of the velocity is `v sin phi`.
Hence y-component of momentum is `mvsinphi=eB`