Fundamentals of Physics
Ha
lliday & Resnic
東海大學物理系
98
學年度第
2
學期
第
1
頁，共
12
頁
系級：□
0401
物理組
□
0409
應用物理組
學號：
姓名：
.
【
CH2
8
】
Magnetic Fields
3, 5, 9, 11, 14, 15, 17, 21, 27, 29, 31, 35, 39, 43, 47, 49, 53, 55, 59,63, 6
5
【
Problem
2
8

3
】
A proton traveling at
with respect to the dire
ction
of a magnetic field of strength 2.60 mT
experiences a magnetic
force of
. Calculate (a) the proton’s speed and
(b) its kinetic
energy in electron

volts.
<
解
>
：
(a) Eq. 28

3 leads to
(b) The kinetic energy of
the proton is
,
which is equivalent to
【
Problem
28

5
】
An electron moves through a uniform magnetic field
given by
.
At a particular
instant, the electron
as velocity
and the magnetic
force acting on it is
. Find
.
<
解
>
：
Using Eq. 28

2 and Eq. 3

30, we obtain
where we use the fact that
. Since the force (at
the instant considered) is
where
, then we are led to the condition
Substituting
,
and
, we obtain
【
Problem
28

9
】
An electron has an initial velocity of
and a constant acceleration of
in a
region in which uniform electric and magnetic fields are present.
If
,
find the ele
ctric field
.
<
解
>
：
We apply
to solve for
:
Fundamentals of Physics
Ha
lliday & Resnic
東海大學物理系
98
學年度第
2
學期
第
2
頁，共
12
頁
【
Problem
28

11
】
An ion source is producing
ions, which have charge
and ma
ss
. The ions
are accelerated by a
potential difference of 10 kV and pass horizontally into a region
in which there
is a uniform vertical magnetic field of magnitude
. Calculate the strength of the smal
lest
electric field, to
be set up over the same region, that will allow the
ions to pass
through
undeflected.
<
解
>
：
Since the total force given by
vanishes, the electric field
must be
perpendicular to both the particle velocity
and the magnetic field
. The magnetic field
is perpendicular to t
he velocity, so
has magnitude
vB
and the magnitude of the
electric field is given by
. Since the particle has charge
e
and is accelerated through
a potential difference
,
and
.
Thus,
【
Problem
28

14
】
A metal strip 6.50 cm long,
0.850 cm wide, and 0.760 mm
thick moves with constant velocity
through a uniform
magnetic field
directed
perpendicular to the strip,
as shown in Fig.
28

37.
A potential
difference of
is measured
between points x and y
across the strip.
Calculate the
speed
.
（圖
28

37
）
<
解
>
：
For a free charge
q
inside the metal strip with velocity
we have
. We set
this force equal to zero and use the relation between (uniform) electric field and potential
difference. Thus,
Fundamentals of Physics
Ha
lliday & Resnic
東海大學物理系
98
學年度第
2
學期
第
3
頁，共
12
頁
【
Problem
28

15
】
In Fig. 28

38, a conducting rectangular solid of dimensions
,
, and
moves at
constant velocity
through a uniform magnetic
field
.
What
are
the resulting (a) electric field
within the solid, in unit

ve
ctor
notation, and (b) potential
difference across the solid?
（圖
28

38
）
<
解
>
：
(a) We seek the electrostatic field established by the separation of charges (brought on by the
magnetic force). With Eq. 28

10, we define the magnitude of the electric field as
.
Its direction may be inferred from Figure
28

8; its direction is opposite to that defined by
. In summary,
which insures that
vanishes.
(b) Eq. 28

9 yields
.
【
Problem
28

17
】
An electron of kinetic energy
1.20 keV circles in a plane
perpendicular to a uniform magnetic field.
The orbit radius is
25.0 cm. Find (a) the electron’s speed, (b) the magnetic field
magnitude, (c) the
circling frequency, and (d) the period of the
motion.
<
解
>
：
(a) From
we get
(b) From
we get
(c) The “orbital” frequency is
Fundamentals of Physics
Ha
lliday & Resnic
東海大學物理系
98
學年度第
2
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(d)
【
Problem
28

21
】
(a) Find the frequency of revolution of an electron
with
an energy of 100 eV in a uniform magnetic
field of magnitude
. (b) Calculate the radius of the path of this electron if
its velocity is
perpendicular to the magnetic field.
<
解
>
：
(a) The frequency of revolution is
(b) Using Eq. 28

16, we obtain
【
Problem
28

27
】
An electron follows a helical path in a uniform magnetic
field of magnitude 0.300 T. The pitch of
the path is
, and the magnitude of the magnetic f
orce on the electron
is
.
What is
the electron’s speed?
<
解
>
：
Reference to Fig. 28

11 is very useful for interpreting this problem. The distance traveled
parallel to
is
usin
g Eq. 28

17. Thus,
using the values given in this problem. Also, since the magnetic force is
, then we
find
. The speed is therefore
.
Fundamentals of Physics
Ha
lliday & Resnic
東海大學物理系
98
學年度第
2
學期
第
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頁，共
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（圖
28

1
1
）
【
Problem
28

29
】
A positron with kinetic energy 2.00 keV is projected
into a uniform magnetic field of magnitude
0.100 T, with its
velocity vector making an angle of 89.0° with
. Find (a) the
period, (b) the pitch
p, and (c)
the radius r of its helical path.
<
解
>
：
(a) If
v
is the speed of the positron then
v
sin
is the component of its velocity in the plane
that is perpendicular to the magnetic field. Here
is the angle between the velocity and
the field (89°). Newton’s secon
d law yields
eBv
sin
=
m
e
(
v
sin
)
2
/
r
, where
r
is the
radius of the orbit. Thus
r
= (
m
e
v
/
eB
) sin
. The period is given by
The equation for
r
is substituted to obtain the second expression for
T
.
(b) The pitch is the distance travel
ed along the line of the magnetic field in a time interval of
one period. Thus
p = vT
cos
. We use the kinetic energy to find the speed:
means
Thus,
(c) The orbit radius is
【
Problem
28

31
】
A certain commercial mass spectrometer (see Sample
Problem 28

3) is used to separate uranium
ions of mass
and charge
from related species. The
ions are
Fundamentals of Physics
Ha
lliday & Resnic
東海大學物理系
98
學年度第
2
學期
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accelerated through a po
tential difference of 100 kV
and then pass into a uniform magnetic field,
where they are
bent in a path of radius 1.00 m. After traveling through 180°
and passing through a
slit of width 1.00 mm and height
1.00 cm, they are collected in a cup. (a) What is
the magnitude
of
the (perpendicular) magnetic field in the separator? If the
machine is used to separate out 100 mg of
material per hour,
calculate (b) the current of the desired ions in the machine
and (c) the thermal
energy produced in the cup in 1.00 h.
<
解
>
：
(a) We solve for
B
from
m = B
2
qx
2
/8
V
(see Sample Problem 28

3):
We evaluate this expression using
x
= 2.00 m:
(b) Let
N
be the number of ions that are separated by the machine per unit time. The current
is
i
= qN
and the mass that is separated per unit time is
M = mN
, where
m
is the mass of
a single ion.
M
has the value
Since
N = M
/
m
we have
(c) Each ion deposits energy
qV
in the cup, so the energy deposited in time
t
is given by
For
t
= 1.0 h,
To obtain the second expression,
i
/
q
is substituted for
N
.
【
Problem
28

35
】
Estimate the total path length traveled by a deuteron
in the cyclotron of Sample Problem 28

5
during the
(entire)
acceleration process. Assume that the accelerating potential
between the dees is
80 kV.
<
解
>
：
We approximate the total distance by the number of revolutions times the circumference of
the orbit corresponding to the average energy. This should be a
good approximation since
the deuteron receives the same energy each revolution and its period does not depend on its
energy. The deuteron accelerates twice in each cycle, and each time it receives an energy of
qV
= 80
10
3
eV. Since its final energy is 16
.6 MeV, the number of revolutions it makes is
Its average energy during the accelerating process is 8.3 MeV. The radius of the orbit is
given by
r = mv
/
qB
, where
v
is the deuteron’s speed. Since this is given by
,
the
radius is
Fundamentals of Physics
Ha
lliday & Resnic
東海大學物理系
98
學年度第
2
學期
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7
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For the average energy
The total distance traveled is about
n
2
r
= (104)(2
)(0.375) = 2.4
10
2
m.
【
Problem
28

39
】
A 13.0 g wire of length
is suspended by a
pair of flexible leads in a uniform magnetic
field of magnitude
0.440 T (Fig. 28

44). What are
the (a) magnitude and (b) direction
(left or right)
of the current
required to remove the tension
in the supporting leads?
（圖
28

44
）
<
解
>
：
(a)
The magnetic force on the wire must be upward and have a magnitude equal to the
gravitational force
mg
on the wire. Since the field and the current a
re perpendicular to
each other the magnitude of the magnetic force is given by
F
B
= iLB
, where
L
is the
length of the wire. Thus,
(b)
Applying the right

hand rule reveals that the current must be from left to right.
【
Problem
28

43
】
A wire 50.0 cm long carries a 0.500 A current in the
positive direction of an x axis through a
magnetic field
In unit

vector notation, what is the
magnetic force on the
wire?
<
解
>
：
The magnetic force on the wire is
【
Problem
28

47
】
Figure 28

47 shows a rectangular 20

turn coil of wire, of
dimensions 10 cm by 5.0 cm. It carries a
current of 0.10 A and
is hinged along one long side. It is mounted in the xy plane, at
angle
Fundamentals of Physics
Ha
lliday & Resnic
東海大學物理系
98
學年度第
2
學期
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頁，共
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to the dir
ection of
a uniform magnetic field of
magnitude 0.50 T. In unit

vector
notation,
what is the torque acting
on the coil about the hinge
line?
（圖
28

47
）
<
解
>
：
The applied field has two components:
and
Considering ea
ch
straight

segment of the rectangular coil, we note that Eq. 28

26 produces a non

zero force
only for the component of
which is perpendicular to that segment; we also note that
the equation is effectively multiplied by
N
= 20 due to
the fact that this is a 20

turn coil.
Since we wish to compute the torque about the hinge line, we can ignore the force acting
on the straight

segment of the coil which lies along the
y
axis (forces acting at the axis of
rotation produce no torque about th
at axis). The top and bottom straight

segments
experience forces due to Eq. 28

26 (caused by the
B
z
component), but these forces are (by
the right

hand rule) in the ±
y
directions and are thus unable to produce a torque about the
y
axis. Consequently, the t
orque derives completely from the force exerted on the
straight

segment located at
x
= 0.050 m, which has length
L
= 0.10 m and is shown in
Figure 28

47 carrying current in the
–
y
direction. Now, the
B
z
component will produce a
force on this straight

segme
nt which points in the
–
x
direction (back towards the hinge)
and thus will exert no torque about the hinge. However, the
B
x
component (which is equal
to
B
cos
where
B
= 0.50 T and
= 30°) produces a force equal to
NiLB
x
which points (by
the right

hand ru
le) in the +
z
direction. Since the action of this force is perpendicular to
the plane of the coil, and is located a distance
x
away from the hinge, then the torque has
magnitude
Since
the direction of the torque
is
–
y
.
In unit

vector notation, the torque is
An alternative way to do this problem is through the use of Eq. 28

37. We do not show
those details here, but note that the magnetic moment vector (a necessary part of Eq.
28

37) ha
s magnitude
and points in the
–
z
direction. At this point, Eq. 3

30 may be used to obtain the result for
the torque vector.
【
Problem
28

49
】
Figure 28

48 shows a wire ring of radius a _ 1.8 cm
that is perpendicular to the general dir
ection of
a radially
symmetric, diverging magnetic field. The magnetic field at the
ring is everywhere of the
Fundamentals of Physics
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學年度第
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same magnitude
, and its
direction at the ring everywhere
makes an angle
with
a
normal to t
he plane of the ring.
The twisted lead wires have no
effect on the problem. Find the
magnitude of the force the field
exerts on the ring if the ring carries
a current
.
（圖
28

48
）
<
解
>
：
Consider an infinitesimal segment of the loop
, of length
ds
. The magnetic field is
perpendicular to the segment, so the magnetic force on it has magnitude
dF = iB ds
. The
horizontal component of the force has magnitude
and points inward toward the center of the loop. The
vertical component has magnitude
and points upward. Now, we sum the forces on all the segments of the loop. The horizontal
component of the total force vanishes, since each segment of wire can be paired with
another, diametrical
ly opposite, segment. The horizontal components of these forces are
both toward the center of the loop and thus in opposite directions. The vertical component
of the total force is
We note that
i, B
, and
have the same value for ever
y segment and so can be factored from
the integral.
【
Problem
28

53
】
Figure 28

50 shows a
wood cylinder of mass
and length
,
with
of wire
wrapped around it longitudinally,
so that the plane of the
wire coil contains
the long central
axis of the cylinder. The
cylinder is released on a plane
inclined at an angle
to
the
horizontal, with the plane of
the coil parallel to the incline
plane. If there i
s a vertical uniform
magnetic field of magnitude
0.500 T, what is the least current
through the coil that
keeps the
cylinder from rolling down the plane?
（圖
28

50
）
Fundamentals of Physics
Ha
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東海大學物理系
98
學年度第
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學期
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頁，共
12
頁
<
解
>
：
We use Eq. 28

37 where
is the magneti
c dipole moment of the wire loop and
is the
magnetic field, as well as Newton’s second law. Since the plane of the loop is parallel to the
incline the dipole moment is normal to the incline. The forces acting on the cylinder are the
f
orce of gravity
mg
, acting downward from the center of mass, the normal force of the
incline
F
N
, acting perpendicularly to the incline through the center of mass, and the force of
friction
f
, acting up the incline at the point of contact. We take the
x
axi
s to be positive down
the incline. Then the
x
component of Newton’s second law for the center of mass yields
For purposes of calculating the torque, we take the axis of the cylinder to be the axis of
rotation. The magnetic field produ
ces a torque with magnitude
B
sin
, and the force of
friction produces a torque with magnitude
fr
, where
r
is the radius of the cylinder. The first
tends to produce an angular acceleration in the counterclockwise direction, and the second
tends to produce
an angular acceleration in the clockwise direction. Newton’s second law for
rotation about the center of the cylinder,
=
I
, gives
Since we want the current that holds the cylinder in place, we set
a
= 0 and
= 0, and use
one equa
tion to eliminate
f
from the other. The result is
The loop is rectangular
with two sides of length
L
and two of length 2
r
, so its area is
A
= 2
rL
and the dipole moment
is
Thus,
and
【
Problem
28

55
】
A circular coil of 160 turns has a radius of 1.90 cm.
(a) Calculate the current that results in a
magnetic dipole
moment of magnitude
. (b) Find the maximum
magnitude of the torque
that the coil, carrying this current,
can experience in a uniform 35.0 mT magnetic field.
<
解
>
：
(a) The magnitude of the magnetic dipole moment is given by
, where
N
is the
number of turns,
i
is the current in each turn, and
A
is
the area of a loop. In this case the
loops are circular, so
A =
r
2
, where
r
is the radius of a turn. Thus
(b) The maximum torque occurs when the dipole moment is perpendicular to the field (or
the plane of the loop is parallel to the
field). It is given by
【
Problem
28

59
】
Two concentric, circular
wire loops, of radii
and
, are located in
an xy plane;
each carries a
clockwise current of 7.00 A
(Fig. 28

51).
(a) Find the magnitude
of the net magnetic
dipole
moment of the system.
(b) Repeat for reversed current
in the inner loop.
Fundamentals of Physics
Ha
lliday & Resnic
東海大學物理系
98
學年度第
2
學期
第
11
頁，共
12
頁
（圖
28

5
1
）
<
解
>
：
(a) The magnitude of the magnetic moment vector is
(b) Now,
【
Problem
28

63
】
A wire of length 25.0 cm carrying a current of 4.51 mA
is to be formed into a circular coil and
placed in a uniform
magnetic field
of magnitude 5.71 mT. If the torque on the
coil from the
field is maximized, what are (a) the
angle
between
and the coil’s magnetic dipole moment and
(b) the
number of turns in the coil? (c) What is the magnitude of that
maximum torque?
<
解
>
：
If
N
closed loops are formed from the wire of length
L
, the circumference of eac
h loop is
L
/
N
,
the radius of each loop is
, and the area of each loop is
(a)
For maximum torque, we orient the plane of the loops parallel to the magnetic field, so
the dipole moment is perpendi
cular
(i.e., at a
angle)
to the field.
(b)
The magnitude of the torque is then
To maximize the torque, we take
the number of turns
N
to have the smallest possible
value, 1. Then
.
(c) The magnitude of the maximum torque is
【
Problem
28

65
】
The coil in Fig. 28

55 carries
current
in the
direction indicated, is parallel to
an xz plane,
has 3.00 turns and
an area of
, and
lies in a uniform magnetic field
.
Fundamentals of Physics
Ha
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東海大學物理系
98
學年度第
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學期
第
12
頁，共
12
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What are (a) the magnetic
potential energy of the coil
–
magnetic field system and (b) the
magnetic
torque (in unit

vector notation) on the coil?
（圖
28

55
）
<
解
>
：
(a) Using Eq.
2
8

35 and Figure
28

23, we have
.
Then, Eq. 28

38 gives
.
(b) Using the fact that
, Eq. 28

37 leads to
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