chap27 - 27.1 a F b F qv B 1.24 10 8 C 3.85 104 m s(1.40 T...

27.1: a) )
ˆ ˆ
T)( )(1.40 s m 10 3.85 C)( 10 24 . 1 ( 4 8 i j B v F        



q
.
ˆ
N) 10 68 . 6 ( 4 k F     

b) B v F  

q
ˆ
)( s m 10 19 . 4 ( )
ˆ ˆ
)( s m 10 3.85 T)[( C)(1.40 10 24 . 1 ( 4 4 8 i k j F           
.
ˆ
N) 10 27 . 7 (
ˆ
N) 10 68 . 6 ( 4 4 j i F       
27.2: Need a force from the magnetic field to balance the downward gravitational force.
Its magnitude is:
T. 91 . 1
) s m 10 C)(4.00 10 50 . 2 (
) s m kg)(9.80 10 95 . 1 (
4 8
2
4

 

   


qv
mg
B mg qvB
The right-hand rule requires the magnetic field to be to the east, since the velocity
is northward, the charge is negative, and the force is upwards.
27.3: By the right-hand rule, the charge is positive.
27.4:
m
q
q m
B v
a B v a F

    
.
ˆ
) s m 330 . 0 (
kg 10 81 . 1
)
ˆ ˆ
T)( )(1.63 s m 10 C)(3.0 10 22 . 1 (
2
3
4 8
k
i j
a  

  
 



27.5: See figure on next page. Let ,
0
qvB F  then:
0
F F
a
 in the k
ˆ
 direction
0
F F
b
 in the j
ˆ
 direction
, 0 
c
F since B and velocity are parallel
o 45 sin
0
F F
d
 in the j
ˆ
 direction
0
F F
e
 in the )
ˆ ˆ
( k j   direction
27.6: a) The smallest possible acceleration is zero, when the motion is parallel to the
magnetic field. The greatest acceleration is when the velocity and magnetic field are at
right angles:
. s m 10 25 . 3
kg) 10 (9.11
T) 10 )(7.4 s m 10 C)(2.50 10 6 . 1 (
2
16
31
2 6 19
 

  
 

 
m
qvB
a
b) If . 5 . 14 25 . 0 sin
sin
) s m 10 25 . 3 (
4
1
2
16 
        

m
qvB
a
27.7:
 60 sin ) T 10 C)(3.5 10 (1.6
N 10 60 . 4
sin
sin
3 19 -
15


 

   


B q
F
v B v q F
. s m 10 49 . 9 6  

27.8: a) )].
ˆ
( )
ˆ
( [ )]
ˆ ˆ
( )
ˆ ˆ
( )
ˆ ˆ
( [ i j k k k j k i B v F
y x z z y x z
v v qB v v v qB q           
Set this equal to the given value of F to obtain:
s m 106
T) 1.25 C)( 10 5.60 (
N) 10 40 . 7 (
9
7
 
   






z
y
x
qB
F
v
. s m 6 . 48
T) 1.25 C)( 10 5.60 (
N) 10 40 . 3 (
9
7
 
  
 
 


z
x
y
qB
F
v
b) The value of
z
v is indeterminate.
c) . 90 ; 0    

     
y
z
x
x
z
y
z z y y x x
F
qB
F
F
qB
F
F v F v F v F v
27.9: s m 10 80 . 3
ˆ
, 3      
y y
v with v q j v B v F
, 0 N, 10 60 . 7 3     
y x
F F and
N 10 20 . 5
3 
  
z
F
z y y z z y x
B qv B v B v q F    ) (
T 0.256 )] s m 10 3.80 C)( 10 ([7.80 N) 10 60 . 7 ( 3 6 3          
y x z
qv F B
, 0 ) (   
z x x z y
B v B v q F which is consistent with F as given in the problem. No
force component along the direction of the velocity.
x y x y y x z
B qv B v B v q F     ) (
T 175 . 0    
y z x
qv F B
b)
y
B is not determined. No force due to this component of B along ; v measurement
of the force tells us nothing about .
y
B
c)
        

N) 10 7.60 T)( 175 . 0 (
3
z z y y x x
F B F B F B F B
N)
3
10 5.20 T)( 0.256 (

  
B F B ; 0   and F are perpendicular (angle is
) 90
o

27.10: a) The total flux must be zero, so the flux through the remaining surfaces must be
120 . 0  Wb.