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271
M
AGNETIC
F
IELD AND
M
AGNETIC
F
ORCES
27.1.
IDENTIFY
and
SET UP:
Apply Eq.(27.2) to calculate
.
F
!
Use the cross products of unit vectors from Section 1.10.
EXECUTE:
( ) ( )
44
&&
4.19 10 m/s
3.85 10 m/s
=+
×
+−
×
vi
j
!
(a)
()
&
1.40 T
=
B
i
!
( )
( ) ( )
84
4
1.24 10 C 1.40 T
q
−
⎡
⎤
=×=
− ×
×
×
−
×
×
⎣
⎦
F
vB
ii
j
i
!!
!
&& &
0,
×=
×=−
j
ik
( )
( )( ) ( )
4
6.68 10 N
−
−
=−
×
−
×
− =−
×
F
kk
!
EVALUATE:
The directions of
and
!
!
are shown in Figure 27.1a.
The righthand rule gives that
×
!
!
is directed
out of the paper (+
z
direction). The charge is
negative so
F
!
is opposite to
;
×
!
!
Figure 27.1a
F
!
is in the

z
−
direction. This agrees with the direction calculated with unit vectors.
(b) EXECUTE:
&
1.40 T
=
B
k
!
( )
( ) ( )
4
q
−
⎡
⎤
+ ×
×−
×
×
⎣
⎦
F
j
k
!
&&&
,
− ×=
jjki
( )
( ) ( ) ( )
4
4
&
7.27 10 N
−−
−
−
⎡
⎤
×
− +
×
=
×
+
×
⎣
⎦
Fjii
j
!
EVALUATE:
The directions of
and
!
!
are shown in Figure 27.1b.
The direction of
F
!
is opposite to
×
!
!
since
q
is negative. The direction of
F
!
computed
from the righthand rule agrees qualitatively
with the direction calculated with unit vectors.
Figure 27.1b
27.2.
IDENTIFY:
The net force must be zero, so the magnetic and gravity forces must be equal in magnitude and
opposite in direction.
SET UP:
The gravity
force is downward so the force from the magnetic field must be upward. The charge±s
velocity and the forces are shown in Figure 27.2. Since the charge is negative, the magnetic force is opposite to the
righthand rule direction. The minimum magnetic field is when the field is perpendicular to
v
!
. The force is also
perpendicular to
B
!
, so
B
!
is either eastward or westward.
EXECUTE:
If
B
!
is eastward, the righthand rule direction is into the page and
B
F
!
is out of the page, as required.
Therefore,
B
!
is eastward.
sin
mg
q vB
φ
=
.
90
=
°
and
32
48
(0.195 10 kg)(9.80 m/s )
1.91 T
(4.00 10 m/s)(2.50 10 C)
mg
B
vq
−
−
×
==
=
××
.
27
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View Full Document272
Chapter 27
EVALUATE:
The magnetic field could also have a component along the northsouth direction, that would not
contribute to the force, but then the field wouldn&t have minimum magnitude.
Figure 27.2
27.3.
IDENTIFY:
The force
F
!
on the particle is in the direction of the deflection of the particle. Apply the righthand
rule to the directions of
v
!
and
B
!
. See if your thumb is in the direction of
F
!
, or opposite to that direction. Use
sin
Fq
v
B
φ
=
with
90
=
°
to calculate
F
.
SET UP:
The directions of
v
!
,
B
!
and
F
!
are shown in Figure 27.3.
EXECUTE: (a)
When you apply the righthand rule to
v
!
and
B
!
, your thumb points east.
F
!
is in this direction,
so the charge is positive.
(b)
63
sin
(8.50 10 C)(4.75 10 m/s)(1.25 T)sin90
0.0505 N
v
B
−
==
×
×
=
°
EVALUATE:
If the particle had negative charge and
v
!
and
B
!
are unchanged, the particle would be deflected
toward the west.
Figure 27.3
27.4.
IDENTIFY:
Apply Newton&s second law, with the force being the magnetic force.
SET UP:
±± ±
×−
j i= k
EXECUTE:
mq
×
F
=a
=vB
!!
gives
q
m
×
=
vB
a
!
!
!
and
84
3
±±
(1.22 10 C)(3.0 10 m/s)(1.63 T) (
)
±
(0.330 m/s ) .
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This note was uploaded on 03/14/2008 for the course MSE 250 taught by Professor Jabbour during the Spring '08 term at University of Arizona Tucson.
 Spring '08
 Jabbour
 Materials Science And Engineering

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