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Question

Protons having a kinetic energy of 4.50 MeV are moving in the positive x-direction and enter a magnetic field of 0.0510 T in the z-direction, out of the plane of the page, and extending from x = 0 to x = 1.00 m as in Figure P19.65.

Figure P19.65
http://www.webassign.net/sf5/p19_67.gif

(a) Calculate the y-component of the protons' momentum as they leave the magnetic field.
kgm/s
(b) Find the angle between the initial velocity vector of the proton beam and the velocity vector after the beam emerges from the field. (Hint: Neglect relativistic effects and note that 1 eV = 1.60 10-19 J.)
°

Answer 1

This is like a projectile problem

convert eV to joules and then find v using KE formula
KE=1/2mv^2
4.5E6*1.6E-19=720E-15 J

m-proton=1.7E-27 kg

720E-15=1/2*1.7E-27*v^2 ==>v=sqrt(720E-15*2/1.7E-27)
=29.1E6 m/s.

compared to c, this is ~9% c so we can solve in a non-relativistic way

the force exerted on this particle F=qv x B so

F=-qvB in the y direction =>
a=-qvB/m=-1.6E-19*29.1E6*0.051/1.7E-27
=-139.7E12 m/s^2

This traces out a parabola

y=-ay/2vx^2 * x^2
we have to first figure out the exit point versus the vertical

x=1 m therefore y=-139.7E12/(2*29.1E6^2)=-82.5E-3 m
or below the vertical

the slope of the parabola at this point is
dy/dx=-ay/v^2*x=-139.7E12/29.1E6^2*1=-0.165
arctan(-0.165)=-9.4 degrees.

the total velocity of the exit is sqrt(vx^2+vy^2)

vx=29.1E6
from vx and x=1m we can determine how long the proton
stays in the field
the proton spends 34.4E-9 seconds in the field
so
vy=-139.7E12 m/s^2*34.4E-9=4.81E6
m*vy=4.81E6*1.7E-27=8.2E-21 this should be the y component of momentum.