170
CHAPTER 19
19.65
Note: We solve part (b) before part (a) for this problem.
(b) Since the magnetic force supplies the centripetal acceleration for this particle,
2
qvB
mv
r
=
or the radius of the path is
rm
v
q
B
=
. The speed of the particle may be
written as
()
2
vK
E
=
m
, so the radius becomes
( )( )
( )
27
6
19
-19
2 1.67
10
kg
5.00
10 eV
1.60
10
J eV
2
1.60
10
C
0.050 0 T
6.46 m
mKE
r
qB
−−
××
×
==
×
=
Consider the circular path shown at the right and
observe that the desired angle is
11
1.00 m
1.00 m
sin
sin
8.90
6.46 m
r
α
===
°
±
±
±
±
±²³³´µ
±²
±
²
(a) The constant speed of the particle is
2
E
=
m
, so
the vertical component of the momentum as the particle leaves the field is

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