MAGNETISM: FORCE AND FIELD
26
E
XERCISES
Section 26.2 Magnetic Force and Field
17.
I
NTERPRET
This problem is about the magnetic force exerted on a moving electron.
D
EVELOP
The magnetic force on a charge
q
moving with velocity
v
G
is given by Equation 26.1:
.
B
F
qv
B
=
×
G
G
G
The
magnitude of
B
F
G
is




 
sin
B
B
F
F
qv
B
q vB
θ
=
=
×
=
G
G
G
E
VALUATE
(a)
The magnetic field is a minimum when
sin
1
θ
=
(the magnetic field perpendicular to the velocity).
Thus,
15
3
min
19
7
5.4
10
N
1.61
10
T
16.1 G


(1.6
10
C)(2.1
10
m/s)
B
F
B
q v
−
−
−
×
=
=
=
×
=
×
×
(b)
For
45 ,
θ
=
°
the magnetic field is
15
min
19
7
5.4
10
N
2
22.7 G


sin
(1.6
10
C)(2.1
10
m/s)sin 45
B
F
B
B
q v
θ
−
−
×
=
=
=
=
×
×
°
A
SSESS
The magnetic force on the electron is very tiny. The magnetic field required to produce this force can be
compared to the Earth’s magnetic field, which is about 1 G.
18.
(a)
If the magnetic force is the only one of significance acting in this problem, then
sin .
F
ma
ev B
θ
=
=
Thus,
31
15
2
19
5
v
/
sin
(9.11
10
kg)(6
10
m/s
)/(1.6
10
C)(0.1 T)sin 90
3.42
10
m/s.
ma eB
θ
−
−
=
=
×
×
×
° =
×
(b)
Since
F
v
B
×
G
G
G
z
is perpendicular to
the magnetic force on a charged particle changes its direction, but not its speed.
,
v
G
19.
I
NTERPRET
In this problem we are asked to find the magnetic force exerted on a moving proton.
D
EVELOP
The magnetic force on a charge
q
moving with velocity
v
G
is given by Equation 26.1:
.
B
F
qv
B
=
×
G
G
G
The magnitude of
B
F
G
is




 
sin
B
B
F
F
qv
B
q vB
θ
=
=
×
=
G
G
G
The charge of the proton is
19
1.6
10
C.
q
−
=
×
E
VALUATE
(a)
When
90 ,
θ
=
°
the magnitude of the magnetic force is
19
5
14
sin 90
(1.6
10
C)(2.5
10
m/s)(0.5 T)
2.0
10
N
B
F
qvB
−
−
=
° =
×
×
=
×
(b)
When
30 ,
θ
=
°
the force is
19
5
14
sin 30
(1.6
10
C)(2.5
10
m/s)(0.5 T)sin30
1.0
10
N
B
F
qvB
−
−
=
° =
×
×
° =
×
(c)
When
0 ,
θ
=
°
the force is
sin 0
0.
B
F
qvB
=
° =
)
A
SSESS
The magnetic force is a maximum
,max
(
 
B
F
q vB
=
when
90
θ
=
°
and a minimum
when
,min
(
0
B
F
=
)
0 .
θ
=
°
20.
An electron, moving perpendicularly to the Earth’s magnetic field near the surface, experiences a maximum
magnetic force of
19
4
3
19
31
v
2
/
(1.6
10
C)(10
T)
2(10
eV)(1.6
10
J/eV)/(9.11
10
kg)
e B
eB
K m
−
−
−
−
=
×
×
×
.
.
16
3
10
N.
−
×
The weight of an electron near the Earth’s surface is
31
2
30
(9.11
10
kg)(9.8 m/s
)
9
10
N,
mg
−
−
=
×
×
.
a factor of nearly
times smaller.
14
3 10
−
×
21.
I
NTERPRET
This problem is about the speed of a given charge if it is to pass through the velocity selector
undeflected.
26.1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document