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154
CHAPTER 19
19.28
Since the path is circular, the particle moves perpendicular to the magnetic field, and the
magnetic force supplies the centripetal acceleration. Hence,
2
v
mq
v
r
=
B
, or
mv
qr
=
B
. But
the momentum is given by
()
2
p
mv
m KE
==
, and the kinetic energy of this proton is
19
61
1.60
10
J
10.0
10 eV
1.60
10
J
1 eV
KE
−
−
×
=×
2
. We then have
(
)
27
12
12
19
10
2 1.67
10
kg
1.60
10
J
2
7.88
10
T
1.60
10
C
5.80
10
m
mKE
B
qr
−−
−
−
××
=
×
19.29
For the particle to pass through with no deflection, the net force acting on it must be
zero. Thus, the magnetic force and the electric force must be in opposite directions and
have equal magnitudes. This gives
, or
which reduces to
m
FF
=
e
qvB
qE
=
vE
B
=
19.30
The speed of the particles emerging from the velocity selector is
B
=
(see Problem
29). In the deflection chamber, the magnetic force supplies the centripetal acceleration,

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