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166
CHAPTER 19
19.59
The magnetic force is very small in comparison to the weight of the ball, so we treat the
motion as that of a freely falling body. Then, as the ball approaches the ground, it has
velocity components with magnitudes of
0
20.0 m s
xx
vv
==
()
, and
( )
22
0
2
0
2
9.80 m s
20.0 m
19.8 m s
yy
y
vva
y
=+∆
=
+
−
−
=
The velocity of the ball is perpendicular to the magnetic field and, just before it reaches
the ground, has magnitude
28.1 m s
xy
v
=+
=
. Thus, the magnitude of the magnetic
force is
6
6
sin
5.00
10
C
90.0
1.41
10
N
m
Fq
v
B
28.1 m s
0.010 0 T sin
θ
−
−
=
=×
°
19.60
We are given that the field at points on the axis of the disk varies as
3
Bkh
=
, where
k
is
a constant and

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