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■ Thinking Physics 22.1
On a business trip to Australia, you take along your U.S.-made compass that you
used in your Boy Scout days. Does this compass work correctly in Australia?
Reasoning Using the compass in Australia presents no problem. The north pole of
the magnet in the compass will be attracted to the south magnetic pole near the
north geographic pole, just as it was in the United States. The only difference in the
magnetic field lines is that they have an upward component in Australia, whereas
they have a downward component in the United States. When you hold the compass in a horizontal plane, it cannot detect the vertical component of the field,
however; it only displays the direction of the horizontal component of the magnetic
field. ■
EXAMPLE 22.1
An Electron Moving in a Magnetic Field
An electron in a television picture tube moves toward
the front of the tube with a speed of 8.0 106 m/s
along the x axis (Fig. 22.6). The neck of the tube is surrounded by a coil of wire that creates a magnetic field
of magnitude 0.025 T, directed at an angle of 60 to the
x axis and lying in the xy plane. Calculate the magnetic
force on and acceleration of the electron.
z
–e
60°
y
B
Solution Using Equation 22.2, we find the magnitude
of the magnetic force:
FB q vB sin (1.60 1019 C)(8.0 106 m/s)(0.025 T)(sin 60)
2.8 1014 N
:
v B is in the positive z direction (from the
Because :
:
right-hand rule) and the charge is negative, FB is in the
negative z direction.
Once we have determined the magnetic force, we
have a Chapter 4 problem because the electron is a
particle under a net force and the acceleration is
determined from Newton’s second law. The mass of the
electron is me 9.1 1031 kg, and so its acceleration is
a
v
FB
2.8 1014 N
3.1 1016 m/s2
me
9.1 1031 kg
in the negative z direction.
x
FB
FIGURE 22.6
:
(Example 22.1) The magnetic force FB on the elec:
tron is in the negative z direction when :
v and B lie
in the xy plane.
22.3 MOTION OF A CHARGED PARTICLE
IN A UNIFORM MAGNETIC FIELD
In Section 22.2, we found that the magnetic force acting on a charged particle moving in a magnetic field is perpendicular to the velocity of the particle and that, consequently, the work done on the particle by the magnetic force is zero. Consider
now the special case of a positively charged particle moving in a uniform magnetic
field when the initial velocity vector of the particle is perpendicular to the field. Let
us assume that the direction of the magnetic field is into the page. Active Figure
22.7 shows that the particle moves in a circular path whose plane is perpendicular
to the magnetic field.
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■ Thinking Physics 22.2
Suppose a uniform magnetic field exists in a finite region of space as in Figure 22.9.
Can you inject a charged particle into this region and have it stay trapped in the
region by the magnetic force?
v
FB
Particle
motion
FIGURE 22.9 (Thinking Physics
22.2) A positively charged particle
enters a region of magnetic field
directed out of the page.
Reasoning Consider separately the components of the particle velocity parallel and
perpendicular to the field lines in the region. For the component parallel to the
field lines, no force is exerted on the particle and it continues to move with the parallel component until it leaves the region of the magnetic field. Now consider the
component perpendicular to the field lines. This component results in a magnetic
force that is perpendicular to both the field lines and the velocity component. As
discussed earlier, if the force acting on a charged particle is always perpendicular to
its velocity, the particle moves in a circular path. Thus, the particle follows half of a
circular arc and exits the field on the other side of the circle, as shown in Figure
22.9. Therefore, a particle injected into a uniform magnetic field cannot stay
trapped in the field region. ■
A Proton Moving Perpendicular to a Uniform Magnetic Field
A proton is moving in a circular orbit of radius 14.0 cm
in a uniform 0.350-T magnetic field directed perpendicular to the velocity of the proton.
A Find the translational speed of the proton.
B Find the period of the circular motion of the
proton.
Solution From Equation 22.5,
T
2mp
Solution From Equation 22.3, we find that
v
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MAGNETIC FORCES AND MAGNETIC FIELDS
Magnetic field
region (out of page)
EXAMPLE 22.2
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qBr
(1.60 1019 C)(0.350 T)(14.0 102 m)
mp
1.67 1027 kg
qB
2 (1.67 1027 kg)
(1.60 1019 C)(0.350 T)
1.87 107 s
4.69 106 m/s
INTERACTIVE
EXAMPLE 22.3
Bending an Electron Beam
A Assuming that the magnetic field is perpendicular
to the beam, what is the magnitude of the field?
Solution The drawing in Active Figure 22.7 and the
photograph in Figure 22.10 help us conceptualize the
circular motion of the electrons. We categorize this
(Courtesy of Henry Leap and Jim Lehman)
In an experiment designed to measure the strength of
a uniform magnetic field, electrons are accelerated
from rest (by means of an electric field) through a
potential difference of 350 V. After leaving the region
of the electric field, the electrons enter a magnetic field
and travel along a curved path because of the magnetic
force exerted on them. The radius of the path is
measured to be 7.50 cm. Figure 22.10 shows such a
curved beam of electrons.
FIGURE 22.10
(Interactive Example 22.3) The bending of an
electron beam in a magnetic field.
problem as one in which we will use our understanding
of uniform circular motion along with our knowledge
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APPLICATIONS INVOLVING CHARGED PARTICLES MOVING IN A MAGNETIC FIELD ❚
of the magnetic force. Looking at Equation 22.3, we see
that we need the speed v of the electron if we are to
find the magnetic field magnitude, and v is not given.
Consequently, we must find the speed of the electron
based on the potential difference through which it is
accelerated. We can therefore also categorize this problem as one in which we must apply conservation of
mechanical energy for an isolated system. We begin to
analyze the problem by finding the electron speed. For
the isolated electron – electric field system, the loss of
potential energy as the electron moves through the
350-V potential difference appears as an increase in the
kinetic energy of the electron. Because K i 0 and
Kf 12mev 2, we have
K U 0
:
√
1
2
2 m ev
(e) V 0
√
2e V
2(1.60 1019 C)(350 V)
me
9.11 1031 kg
7
1.11 10 m/s
v
Now, using Equation 22.3, we find that
B
m ev
(9.11 1031 kg)(1.11 107 m/s)
er
(1.60 1019 C)(0.075 m)
8.4 104 T
B What is the angular speed of the electrons?
Solution Using Equation 22.4, we find that
v
1.11 107 m/s
1.5 108 rad/s
r
0.075 m
To finalize this problem, note that the angular speed
can be written as (1.5 108 rad/s)(1 rev/2 rad)
2.4 107 rev/s. The electrons travel around the circle 24 million times per second! This very high speed is
consistent with what we found in part A.
By logging into PhysicsNow at
www.pop4e.com and going to Interactive Example 22.3, you
can investigate the relationship between the radius of the
circular path of the electrons and the magnetic field.
22.4 APPLICATIONS INVOLVING CHARGED PARTICLES
MOVING IN A MAGNETIC FIELD
:
v in the presence of an electric field E and a magnetic
A charge moving with velocity :
:
:
:
v B. The total
field B experiences both an electric force q E and a magnetic force q :
force, called the Lorentz force, acting on the charge is therefore the vector sum,
:
:
:
F q E q:
v B
[22.6]
In this section, we look at three applications involving particles experiencing the
Lorentz force.
Velocity Selector
In many experiments involving moving charged particles, it is important to have particles that all move with essentially the same velocity. That can be achieved by applying a combination of an electric field and a magnetic field oriented as shown in
Active Figure 22.11a. A uniform electric field is directed vertically downward (in the
plane of the page in Active Fig. 22.11a), and a uniform magnetic field is applied
perpendicular to the electric field (into the page in Active Fig. 22.11a). Particles moving through this region will experience the Lorentz force, given by Equation 22.6.
:
v B is upward and the
For a positively charged particle, the magnetic force q :
:
electric force q E is downward. When the magnitudes of the two fields are chosen
so that qE qvB, the particle is in equilibrium (Active Fig. 22.11b) and moves in
a straight horizontal line through the region of the fields. From qE qvB we
find that
v
E
B
735
[22.7]
Only those particles having this speed are undeflected as they move through the
perpendicular electric and magnetic fields and pass through a small opening at the
end of the device. The magnetic force exerted on particles moving faster than this
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MAGNETIC FORCES AND MAGNETIC FIELDS
these points:
b
:
FB I
a
:
d:
s B
[22.12]
When this integration is carried out, the magnitude of the magnetic field and the
direction of the field relative to the vector d :
s may vary from point to point.
QUICK QUIZ 22.3 A wire carries current in the plane of this paper toward the top of the
page. The wire experiences a magnetic force toward the right edge of the page. What is
the direction of the magnetic field causing this force? (a) in the plane of the page
and toward the left edge (b) in the plane of the page and toward the bottom edge
(c) upward out of the page (d) downward into the page
■ Thinking Physics 22.3
In a lightning stroke, negative charge rapidly moves from a cloud to the ground. In
what direction is a lightning stroke deflected by the Earth’s magnetic field?
Reasoning The downward flow of negative charge in a lightning stroke is equivalent to an upward-moving current. Thus, the vector d :
s is upward, and the magnetic field vector has a northward component. According to the cross product of
the length element and magnetic field vectors (Eq. 22.11), the lightning stroke
would be deflected to the west. ■
EXAMPLE 22.4
Force on a Semicircular Conductor
A wire bent into the shape of a semicircle of radius R
forms a closed circuit and carries a current I. The circuit lies in the xy plane, and a uniform magnetic field is
present along the positive y axis as in Figure 22.18. Find
the magnetic force on the straight portion of the wire
and on the curved portion.
Solution The force on the straight portion of the wire
has a magnitude F1 IB 2IRB because 2R and
:
:
the wire is perpendicular to B. The direction of F1 is out
:
:
:
of the paper because B is outward. (That is, is to
the right, in the direction of the current, and so by the
:
:
rule of cross products, B is outward.)
B
θ
ds
R
θ
dθθ
I
To find the magnetic force on the curved part, we
:
first write an expression for the magnetic force d F2 on
:
s . If is the angle between B and d :
s in
the element d :
:
Figure 22.18, the magnitude of d F2 is
:
dF 2 I d :
s B IB sin ds
To integrate this expression, we express ds in terms of .
Because s R , ds Rd , and the expression for dF 2
can be written as
dF 2 IRB sin d
To obtain the total magnetic force F 2 on the curved
portion, we integrate this expression to account for
contributions from all elements. Note that the direction
of the magnetic force on every element is the same: into
:
s B is inward). Therefore, the
the paper (because d :
:
resultant magnetic force F2 on the curved wire must
also be into the paper. Integrating dF 2 over the limits
0 to (i.e., the entire semicircle) gives
F 2 IRB
0
sin d IRB cos 0
IRB(cos cos 0) IRB(1 1) 2IRB
:
FIGURE 22.18
(Example 22.4) The net force on a closed current
loop in a uniform magnetic field is zero. For the
loop shown here, the force on the straight portion
is 2IRB and out of the page, whereas the force on
the curved portion is 2IRB and into the page.
Because F 2 2IRB and the vector F2 is directed
into the paper and because the force on the straight
wire has magnitude F 1 2IRB and is out of the paper,
we see that the net magnetic force on the closed loop is
zero.
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MAGNETIC FORCES AND MAGNETIC FIELDS
cancel each other and produce no torque because they have the same line of ac:
:
tion. The magnetic forces F2 and F4 acting on sides and , however, both
produce a torque about an axis through the center of the loop. Referring to Active
:
Figure 22.20, we note that the moment arm of F2 about this axis is (b/2) sin .
:
Likewise, the moment arm of F4 is also (b/2) sin . Because F 2 F4 IaB, the net
torque has the magnitude
µ
b
b
sin F4
sin 2
2
b
b
(IaB)
sin (IaB)
sin IabB sin 2
2
IAB sin F2
A
I
where A ab is the area of the loop. This result shows that the torque has its
maximum value IAB (Eq. 22.13) when the field is parallel to the plane of the
loop ( 90) and is zero when the field is perpendicular to the plane of the loop
( 0). As we see in Active Figure 22.20, the loop tends to rotate in the direction
of decreasing values of (i.e., so that the normal to the plane of the loop rotates
toward the direction of the magnetic field). A convenient vector expression for the
torque is
■ Magnetic moment of a current
loop
:
:
IA B
:
FIGURE 22.21
Right-hand rule for
determining the direction of the
:
vector A. The direction of the
:
magnetic moment is the same as
:
the direction of A.
[22.14]
:
where A, a vector perpendicular to the plane of the loop (Active Fig. 22.20), has a
:
magnitude equal to the area of the loop. The sense of A is determined by the righthand rule illustrated in Figure 22.21. When the four fingers of the right hand are
curled in the direction of the current in the loop, the thumb points in the direction
:
:
:
of A. The product I A is defined to be the magnetic dipole moment (often simply
called the “magnetic moment”) of the loop:
:
:
IA
[22.15]
The SI unit of magnetic dipole moment is the ampere-meter2 (A m2). Using this
definition, the torque can be expressed as
:
:
B
:
■ Torque on a current loop
[22.16]
:
Although the torque was obtained for a particular orientation of B with respect
to the loop, Equation 22.16 is valid for any orientation. Furthermore, although the
torque expression was derived for a rectangular loop, the result is valid for a loop of
any shape. Once the torque is determined, the motion of the coil can be modeled
as a rigid object under a net torque, which was studied in Chapter 10.
If a coil consists of N turns of wire, each carrying the same current and each
having the same area, the total magnetic moment of the coil is the product of the
:
:
number of turns and the magnetic moment for one turn, NI A. Thus, the
torque on an N-turn coil is N times greater than that on a one-turn coil.
A common electric motor consists of a coil of wire mounted so that it can rotate
in the field of a permanent magnet. The torque on the current-carrying coil is used
to rotate a shaft that drives a mechanical device such as the power windows in your
car, your household fan, or your electric hedge trimmer.
EXAMPLE 22.5
The Magnetic Moment and Torque on a Coil
A rectangular coil of dimensions 5.40 cm 8.50 cm
consists of 25 turns of wire. The coil carries a current of
15.0 mA.
A Calculate the magnitude of its magnetic moment.
Solution The magnitude of the magnetic moment of a
current loop is IA (Eq. 22.15). In this case,
A (0.054 0 m)(0.085 0 m) 4.59 103 m2.
Because the coil has 25 turns and assuming that each
turn has the same area A, we have
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THE BIOT – SAVART LAW ❚
coil NIA (25)(15.0 103 A)(4.59 103 m2)
1.72 103 Am2
743
Solution The torque is given by Equation 22.16,
:
:
:
:
:
B. In this case, B is perpendicular to coil , so
coil B (1.72 103 A m2)(0.350 T)
B Suppose a uniform magnetic field of magnitude
0.350 T is applied parallel to the plane of the loop.
What is the magnitude of the torque acting on the loop?
6.02 104 Nm
22.7 THE BIOT – SAVART LAW
In the previous sections, we investigated the result of placing an object in an existing magnetic field. When a moving charge is placed in the field, it experiences a
magnetic force. A current-carrying wire placed in the field also experiences a magnetic force; a current loop in the field experiences a torque.
Now we shift our thinking and investigate the source of the magnetic field.
Oersted’s 1819 discovery (Section 22.1) that an electric current in a wire deflects a
nearby compass needle indicates that a current acts as a source of a magnetic field.
From their investigations on the force between a current-carrying conductor and a
magnet in the early 19th century, Jean-Baptiste Biot and Félix Savart arrived at an
expression for the magnetic field at a point in space in terms of the current that
produces the field. No point currents exist comparable to point charges (because
we must have a complete circuit for a current to exist). Hence, we must investigate
the magnetic field due to an infinitesimally small element of current that is part of
a larger current distribution. Suppose the current distribution is a wire carrying a
steady current I as in Figure 22.22. The Biot – Savart law says that the magnetic field
:
d B at point P created by an element of infinitesimal length ds of the wire has the
following properties:
:
• The vector d B is perpendicular both to d :
s (which is in the direction of the current) and to the unit vector r̂ directed from the element toward P.
:
• The magnitude of d B is inversely proportional to r 2, where r is the distance
from the element to P.
:
• The magnitude of d B is proportional to the current I and to the length ds of the
element.
:
• The magnitude of d B is proportional to sin , where is the angle between d :
s
and r̂ .
The Biot – Savart law can be summarized in the following compact form:
:
d B km
I d:
s r̂
r2
[22.17]
where k m is a constant that in SI units is exactly 107 T m/A. The constant k m is
usually written 0/4, where 0 is another constant, called the permeability of free
space:
0
k m 107 Tm/A
4
0 4 k m 4 107 Tm/A
d Bout P
r
I
r̂
θ ds
r̂
FIGURE 22.22 The magnetic field
:
d B at a point P due to a current I
through a length element d :
s is given
by the Biot – Savart law. The field is
out of the page at P and into the page
at P . (Both P and P are in the plane
of the page.)
PITFALL PREVENTION 22.2
THE BIOT – SAVART LAW When you
are applying the Biot – Savart law, it
is important to recognize that the
magnetic field described in these
calculations is the field due to a
given current-carrying conductor.
This magnetic field is not to be
confused with any external field that
may be applied to the conductor
from some other source.
[22.18]
[22.19]
■ Permeability of free space
[22.20]
■ Biot – Savart law
Hence, the Biot – Savart law, Equation 22.17, can also be written
:
dB 0 I d :
s r̂
4
r2
×P′
dBin
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I
r
B
FIGURE 22.23 The right-hand
rule for determining the direction of
the magnetic field surrounding a
long, straight wire carrying a current.
Note that the magnetic field lines
form circles around the wire. The
magnitude of the magnetic field at a
distance r from the wire is given by
Equation 22.21.
It is important to note that the Biot – Savart law gives the magnetic field at a
point only for a small length element of the conductor. We identify the product
:
I d:
s as a current element. To find the total magnetic field B at some point due to a
conductor of finite size, we must sum contributions from all current elements mak:
ing up the conductor. That is, we evaluate B by integrating Equation 22.20 over the
entire conductor.
There are two similarities between the Biot – Savart law of magnetism and Equation 19.7 for the electric field of a charge distribution, and there are two important
differences. The current element I ds produces a magnetic field, and the charge element dq produces an electric field. Furthermore, the magnitude of the magnetic
field varies as the inverse square of the distance from the current element, as does
the electric field due to a charge element. The directions of the two fields are quite
different, however. The electric field due to a charge element is radial; in the
:
case of a positive point charge, E is directed away from the charge. The magnetic
field due to a current element is perpendicular to both the current element
and the radius vector. Hence, if the conductor lies in the plane of the page, as in
:
Figure 22.22, d B points out of the page at the point P and into the page at P . Another important difference is that an electric field can be a result either of a single
charge or a distribution of charges, but a magnetic field can only be a result of a
current distribution.
Figure 22.23 shows a convenient right-hand rule for determining the direction
of the magnetic field due to a current. Note that the field lines generally encircle
the current. In the case of current in a long, straight wire, the field lines form circles that are concentric with the wire and are in a plane perpendicular to the wire.
If the wire is grasped in the right hand with the thumb in the direction of the cur:
rent, the fingers will curl in the direction of B .
Although the magnetic field due to an infinitely long, current-carrying wire can
be calculated using the Biot – Savart law (Problem 22.52), in Section 22.9 we use a
different method to show that the magnitude of this field at a distance r from the
wire is
B
■ Magnetic field due to a long,
straight wire
0I
2 r
[22.21]
QUICK QUIZ 22.4 Consider the current in the length of wire shown in Figure 22.24.
Rank the points A, B, and C, in terms of magnitude of the magnetic field due to the
current in the length element d :
s shown, from greatest to least.
B
C
A
ds
I
FIGURE 22.24 (Quick Quiz 22.4) Where is the magnetic field the greatest?
INTERACTIVE
EXAMPLE 22.6
Magnetic Field on the Axis of a Circular Current Loop
Consider a circular loop of wire of radius R located in
the yz plane and carrying a steady current I as in Figure
22.25. Calculate the magnetic field at an axial point P a
distance x from the center of the loop.
Solution In this situation, note that any element d :
s is
perpendicular to r̂. Furthermore, all elements around
the loop are at the same distance r from P, where
:
r 2 x 2 R 2. Hence, the magnitude of d B due to the
element d :
s is
dB 0I d :
s r̂ I
ds
0
4
r2
4 (x 2 R 2)
:
[22.22]
The direction of the magnetic field d B due to the
element d :
s is perpendicular to the plane formed by r̂
:
and d :
s as in Figure 22.25. The vector d B can be resolved
into a component dBx , along the x axis, and a component
dBy, which is perpendicular to the x axis. When the components dBy are summed over the whole loop, the result
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THE BIOT – SAVART LAW ❚
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where we have used that ds 2R (the circumference of the loop).
To find the magnetic field at the center of the loop,
we set x 0 in Equation 22.23. At this special point, we
have
ds
θ r̂
O
dB
r
x
P
dBx
x
B
(Interactive Example 22.6) The geometry for calculating the magnetic field at a point P lying on
the axis of a current loop. By symmetry, the total
:
field B is along this axis.
dB cos 0I
4
0IR
4(x 2 R 2)3/2
ds 0IR 2
2x 3
(for x R)
B
[22.24]
[22.25]
0 2 x 3
[22.26]
This result is similar in form to the expression
for the electric field due to an electric dipole,
E ke(2qa)/y 3 ke p/y 3 (Example 19.3), where p is the
electric dipole moment. The pattern of the magnetic
field lines for a circular loop is shown in Figure 22.26a.
For clarity, the lines are drawn only for one plane that
contains the axis of the loop. Note that the field-line
pattern is axially symmetric and looks like the pattern
around a bar magnet, shown in Figure 22.26c.
ds cos x2 R2
and the integral must be taken over the entire loop.
Because , x, and R are constants for all elements of the
loop and because cos R/(x 2 R 2)1/2, we obtain
Bx (at x 0)
Because the magnitude of the magnetic dipole
moment of the loop is defined as the product of the
current and the area (Eq. 22.15), I(R 2), we can
express Equation 22.25 in the form
is zero. That is, by symmetry, any element on one side of
the loop sets up a component dBy that cancels the component set up by an element diametrically opposite it.
For these reasons, the resultant field at P must be
along the x axis and can be found by integrating the
components dBx dB cos , where this expression is
:
obtained from resolving the vector d B into its compo:
nents as shown in Figure 22.25. That is, B Bx î , where
Bx 0I
2R
It is also interesting to determine the behavior of the
magnetic field far from the loop, that is, when x is large
compared with R. In this case, we can ignore the term R 2
in the denominator of Equation 22.23 and we find that
θ
I
FIGURE 22.25
B
dBy
R
z
745
By logging into PhysicsNow at
0 IR 2
[22.23]
2
2(x R 2)3/2
www.pop4e.com and going to Interactive Example 22.6, you
can explore the field for different loop radii.
N
(© Richard Megna, Fundamental Photographs)
N
I
S
(a)
(b)
S
(c)
FIGURE 22.26 (Interactive Example 22.6) (a) Magnetic field lines surrounding a current loop. (b) Magnetic field lines surrounding a current
loop displayed with iron filings. (c) Magnetic field lines surrounding a bar magnet. Note the similarity between this line pattern
and that of a current loop.
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AMPÈRE’S LAW ❚
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The following examples illustrate some symmetric current configurations for which
Ampère’s law is useful.
EXAMPLE 22.7
The Magnetic Field Created by a Long Current-Carrying Wire
A long, straight wire of radius R carries a steady current
I that is uniformly distributed through the cross-section
of the wire (Fig. 22.31). Calculate the magnetic field a
distance r from the center of the wire in the regions
r R and r R.
Solution As mentioned in Section 22.7, we could use
the Biot – Savart law to solve this problem, but Ampère’s
law provides a much simpler solution. For r R, let us
choose path 1 in Figure 22.31, a circle of radius r cen:
tered on the wire. From symmetry, we see that B must
be constant in magnitude — condition 1 — and parallel
s — condition 2 — at every point on this circle. Beto d :
cause the total current passing through the plane of the
circle is I, Ampère’s law applied to the circular path
gives
:
Bd:
s B
ds B(2r) 0I
0I
B
2r
Here the current I passing through the plane of the
circle is less than the total current I. Because the current is uniform over the cross-section of the wire, the
fraction of the current enclosed by the circle of radius
r R must equal the ratio of the area r 2 enclosed by
circular path 2 and the cross-sectional area R 2 of the
wire:
r 2
I
I
R 2
I Following the same procedure as for circular path 1, we
apply Ampère’s law to circular path 2:
:
Bd:
s B(2r) 0I 0
B
(for r R)
which is the result (Eq. 22.21) referred to in
Section 22.7.
Now consider the interior of the wire, where r R.
We choose the circular path 2 shown in Figure 22.31.
r2
I
R2
2RI r
0
2
Rr I
2
2
(for r R) [22.30]
The magnitude of the magnetic field versus r for this
configuration is sketched in Figure 22.32. Note that
inside the wire B : 0 as r : 0. This result is similar in
form to that of the electric field inside a uniformly
charged rod.
B
I
1
B∝r
R
B ∝ 1/r
2
r
ds
R
FIGURE 22.31
EXAMPLE 22.8
(Example 22.7) A long, straight wire of radius R
carrying a steady current I uniformly distributed
across the wire. The magnetic field at any point
can be calculated from Ampère’s law using a circular path of radius r, concentric with the wire.
FIGURE 22.32
r
(Example 22.7) Magnitude of the magnetic field
versus r for the wire described in Figure 22.31. The
field is proportional to r inside the wire and varies
as 1/r outside the wire.
The Magnetic Field Created by a Toroid
A device called a toroid (Fig. 22.33) is often used to
create an almost uniform magnetic field in some
enclosed area. The device consists of a conducting wire
wrapped around a ring (a torus) made of a nonconduct-
ing material. For a toroid having N closely spaced turns
of wire and air in the torus, calculate the magnetic field
in the region occupied by the torus, a distance r from
the center.
44560_22_p727-764
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❚ CHAPTER 22
9:38 AM
Page 750
g
B
ds
a
r
b
c
I
FIGURE 22.33
pp
MAGNETIC FORCES AND MAGNETIC FIELDS
Solution To calculate the field inside the toroid, we
:
evaluate the line integral of B d :
s over the circular amperian loop of radius r in the plane of Figure 22.33. By
symmetry, we see that conditions 1 and 2 apply: the
magnetic field is constant in magnitude on this circle
:
and tangent to it, so B d :
s B ds .
Furthermore, note that the closed path surrounds a
circular area through which N loops of wire pass, each
of which carries a current I. The right side of Equation
22.29 is therefore 0NI in this case. Ampère’s law
I
p
(Example 22.8) A toroid consisting of many turns
of wire wrapped around a doughnut-shaped structure (called a torus). If the coils are closely spaced,
the field in the interior of the toroid is tangent to
the dashed circle and varies as 1/r. The dimension
a is the cross-sectional radius of the torus. The
field outside the toroid is very small and can be described by using the amperian loop at the right
side, perpendicular to the page.
applied to the circular path gives
:
Bd:
s B
B
ds B(2r) 0NI
0NI
2r
[22.31]
This result shows that B varies as 1/r and hence is
nonuniform within the coil. If r is large compared with
the cross-sectional radius a of the torus, however, the
field is approximately uniform inside the coil.
For an ideal toroid in which the turns are closely
spaced, the external magnetic field is close to zero. It is
not exactly zero, however. In Figure 22.33, imagine the
radius r of the amperian loop to be either smaller than
b or larger than c. In either case, the loop encloses zero
:
net current, so B d :
s 0. We might be tempted to
:
claim that this expression proves that B 0, but it does
not. Consider the amperian loop on the right side of
the toroid in Figure 22.33. The plane of this loop is perpendicular to the page and the toroid passes through
the loop. As charges enter the toroid as indicated by
the current directions in Figure 22.33, they work their
way counterclockwise around the toroid. Thus, a current passes through the perpendicular amperian loop!
This current is small, but it is not zero. As a result, the
toroid acts as a current loop and produces a weak external field of the form shown in Figure 22.26a. The rea:
son that B d :
s 0 for the amperian loops of radius
r b and r c in the plane of the page is that the
field lines are perpendicular to d :
s (condition 3), not
:
because B 0 (condition 4).
22.10 THE MAGNETIC FIELD OF A SOLENOID
A solenoid is a long wire wound in the form of a helix. If the turns are closely
spaced, this configuration can produce a reasonably uniform magnetic field
throughout the volume enclosed by the solenoid, except close to its ends. Each of
the turns can be modeled as a circular loop, and the net magnetic field is the vector
sum of the fields due to all the turns.
If the turns are closely spaced and the solenoid is of finite length, the field lines
are as shown in Figure 22.34a. In this case, the field lines diverge from one end and
converge at the opposite end. An inspection of this field distribution exterior to the
solenoid shows a similarity to the field of a bar magnet (Fig. 22.34b). Hence, one
end of the solenoid behaves like the north pole of a magnet and the opposite end
behaves like the south pole. As the length of the solenoid increases, the field within
it becomes more and more uniform. When the solenoid’s turns are closely spaced
and its length is large compared with its radius, it approaches the case of an ideal solenoid. For an ideal solenoid, the field outside the solenoid is negligible and the
field inside is uniform. We will use the ideal solenoid as a simplification model for a
real solenoid.
If we consider the amperian loop perpendicular to the page in Figure 22.35, surrounding the ideal solenoid, we see that it it encloses a small current as the charges
in the wire move coil by coil along the length of the solenoid. Thus, there is a