# Problem 2-1 - 物理學系

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18 Οκτ 2013 (πριν από 4 χρόνια και 6 μήνες)

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Fundamentals of Physics

Ha
lliday & Resnic

98

2

1

12

0401

0409

.

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.

<

>

(a) Eq. 28
-

(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
.

<

>

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
.

<

>

We apply

to solve for
:

Fundamentals of Physics

Ha
lliday & Resnic

98

2

2

12

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.

<

>

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

<

>

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

98

2

3

12

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

<

>

(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.

<

>

(a) From

we get

(b) From

we get

(c) The “orbital” frequency is

Fundamentals of Physics

Ha
lliday & Resnic

98

2

4

12

(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.

<

>

(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?

<

>

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

98

2

5

12

（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.

<

>

(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

98

2

6

12

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.

<

>

(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.

<

>

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

Fundamentals of Physics

Ha
lliday & Resnic

98

2

7

12

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

<

>

(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?

<

>

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

98

2

8

12

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

<

>

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

symmetric, diverging magnetic field. The magnetic field at the

ring is everywhere of the
Fundamentals of Physics

Ha
lliday & Resnic

98

2

9

12

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

<

>

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
lliday & Resnic

98

2

10

12

<

>

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.

<

>

(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

98

2

11

12

（圖
28
-
5
1

<

>

(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?

<

>

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
lliday & Resnic

98

2

12

12

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

<

>

(a) Using Eq.
2
8
-
35 and Figure
28
-
23, we have

.

Then, Eq. 28
-
38 gives

.

(b) Using the fact that
, Eq. 28
-