CHAPTER 29—THE MAGNETIC FIELD
ActivPhysics
can help with these problems: Activities 13.4, 13.6, 13.7, 13.8
Section 29

2:—The Magnetic Force and Moving Charge
Problem
1.
(a) What is the minimum magnetic field needed to exert a 5
10
15
.4
×

N force on an electron moving at
21
10
7
.
?
×
m/s
(b) What magnetic field strength would be required if the field were at 45
°
to the electron’s velocity?
Solution
(a) From Equation 291b,
B
F e
=
= v
sin
,
θ
which is a minimum when sin
θ
=
1 (the magnetic field perpendicular to the
velocity). Thus,
B
min
( .4
) ( .
)( .
)
.
.
=
×
×
×
=
×
=



5
10
16
10
21
10
161
10
161
15
19
7
3
N
C
m/s
T
G.
=
(b) For
θ
=
°
45 ,
B
B
B
=
° =
=
min
min
sin
.
=
45
2
22 7
G.
Problem
2.
An electron moving at right angles to a 0.10T magnetic field experiences an acceleration of 6 0
10
15
.
.
×
m/s
2
(a) What
is the electron’s speed? (b) By how much does its
speed
change in
1
10
9
ns (
s
=

) ?
Solution
(a) If the magnetic force is the only one of significance acting in this problem, then
F
ma
e
B
=
=
v
sin
.
θ
Thus,
v
=
=
=
=
×
×
×
° =
×


ma eB
sin
( .
)(
) ( .
)( .
) sin
.42
θ
911
10
6
10
16
10
01
90
3
10
31
15
19
5
kg
m/s
C
T
m/s.
2
(b) Since
F
v
B
»
×
is
perpendicular to
v
,
the magnetic force on a charged particle changes its direction, but not its speed.
Problem
3.
What is the magnitude of the magnetic force on a proton moving at 2 5
10
5
.
.
m/s (a) at right angles; (b) at 30
°
;
(c) parallel to a magnetic field of 0 50
.
T?
Solution
From Equation 291b,
F
e B
F
=
=
°
=
×
×
=
×


v
sin
,
,
( .
)( .
)( .
)
.
,
θ
θ
so (a) when
C
m/s
T
N
90
16
10
2 5
10
0 5
2 0
10
19
5
14
(b)
F
F
e B
=
×
° =
×
=
° =


( .
) sin
.
,
sin
.
2 0
10
30
10
10
0
0
14
14
N
N
and (c)
v
Problem
Solution
The magnetic force is
q
v
B
v
v
v
×
=
×
=
×
(
)(
)( .
)
,
1
20
01
2
C
m/s
T
N
where
μ
μ
î
î
is a unit vector in the direction of the
velocity of the particle when it first enters the field region. (a) If
,
. ( )
,
v
v
v
=
×
=
=
î
î
b
For
0
j
the direction of the force
is
j
×
= 
î
k
(along the negative
z
axis), while (c)
.
(cos
sin
)
,
k
î
î
î
î
k
×
=
×
=
°
+
°
×
= 
j
j
(d)
v
45
45
2
=
so the force is
(
)
.

2 N
μ
k
Problem
5.
A particle carrying a
50
C
μ
charge moves with velocity
v
=
+
5 0
3 2
.
.
î
k
m/s through a uniform magnetic field
B
=
+
9
6 7
.4
.
î
j
T.
(a) What is the force on the particle? (b) Form the dot products
F
v
F
B
⋅
⋅
and
to show explicitly
that the force is perpendicular to both
v
and
B
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
2
CHAPTER 29
Solution
(a) From Equation 272,
F
v
B
k
=
×
=
+
×
+
=
×
×
+
×

q
(
)(
.
)
( .4
.
)
(
)(
.
.
.4
50
5
3 2
9
6 7
50
10
5
6 7
3 2
9
6
C
m/s
T
N
μ
î
î
k
j
j
3 2
6 7
1072
1504
1675
10
3
.
.
)
(
.
.
.
)
×
=

+
+
×

î
î
k
j
N.
(The magnitude and direction can be found from the components, if
desired.) (b) The dot products
F
v
F
B
⋅
⋅
and
are, respectively, proportional to (
.
)( )
( .
)( . )
,

+
=
1 072
5
1675
3 2
0
and
(
.
)( .4)
( .
)( . )
,

+
=
1 072
9
1504
6 7
0
since the cross product of two vectors is perpendicular to each factor. (We did not round
off the components of
F
, so that the vanishing of the dot products could be exactly confirmed.)
Problem
6.
Moving in the
x
direction, a particle carrying
1 0
.
C
μ
experiences no force. Moving with speed
v
at 30
°
to the
x
axis,
the particle experiences a magnetic force of 2 0
.
N. What magnetic force would it experience if it moved along the
y
axis with speed
v
?
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Hoston,Amahd,Bakanowski
 Force, Magnetic Field, j magnetic field

Click to edit the document details