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164
CHAPTER 19
19.54
For the rail to move at constant velocity, the net force acting on it must be zero. Thus, the
magnitude of the magnetic force must equal that of the friction force giving
( )
k
BIL
mg
µ
=
, or
( )
( )( )
( )
()
( )
2
2
0.100
0.200 kg
9.80 m s
3.92
10
T
10.0 A
0.500 m
mg
IL
−
==
=
×
k
B
19.55
The magnetic force acting on each type particle supplies the centripetal acceleration for
that particle. Thus,
2
qvB
mv
r
=
or
rm
v
q
B
=
.
After completing one half of the circular paths, the two types of particle are separated by
the difference in the diameters of the two paths. Therefore,
21
2
1
5
27
19
2
2
2
.00 10 ms
23.4
20.0
10
kg
1.60
10
C
0.200 T
2.13
10
m
2.13 cm
v
dr
r
mm
qB
−
−
−
∆=
−
=
−
×
=−

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