Sources of the Magnetic Field

30.1 The Biot–Savart Law

30.2 The Magnetic Force Between Two Parallel Conductors

30.3 Ampère’s Law

30.4 The Magnetic Field of a Solenoid

30.5 Magnetic Flux

30.6 Gauss’s Law in Magnetism

30.7 Displacement Current and the General Form of Ampère’s Law

30.8 Magnetism in Matter

30.9 The Magnetic Field of the Earth

A proposed method for launching future payloads into space is the use of rail guns, in which projectiles are accelerated by means of magnetic forces. This photo shows the firing of a projectile at a speed of over 3 km/s from an experimental rail gun at Sandia National Research Laboratories, Albuquerque, New Mexico. (Defense Threat Reduction Agency [DTRA])

I n the preceding chapter, we discussed the magnetic force exerted on a charged particle moving in a magnetic field. To complete the description of the magnetic interaction, this chapter explores the origin of the magnetic field—moving charges. We begin by showing how to use the law of Biot and Savart to calculate the magnetic field produced at some point in space by a small current element. Using this formalism and the principle of superposition, we then calculate the total magnetic field due to various current distributions. Next, we show how to determine the force between two current-carrying conductors, which leads to the definition of the ampere. We also introduce Ampère’s law, which is useful in calculating the magnetic field of a highly symmetric configuration carrying a steady current.

This chapter is also concerned with the complex processes that occur in magnetic materials. All magnetic effects in matter can be explained on the basis of atomic magnetic moments, which arise both from the orbital motion of electrons and from an intrinsic property of electrons known as spin.

dB out P

30.1 The Biot–Savart Law

I rˆ θ ds

Shortly after Oersted’s discovery in 1819 that a compass needle is deflected by a

rˆ

current-carrying conductor, Jean-Baptiste Biot (1774–1862) and Félix Savart

× P′

(1791–1841) performed quantitative experiments on the force exerted by an

dB in

electric current on a nearby magnet. From their experimental results, Biot and Figure 30.1 The magnetic field Savart arrived at a mathematical expression that gives the magnetic field at some

d B at a point due to the current I point in space in terms of the current that produces the field. That expression is through a length element ds is given by the Biot–Savart law. The

based on the following experimental observations for the magnetic field d B at a direction of the field is out of the point P associated with a length element d s of a wire carrying a steady current I

page at P and into the page at P%. (Fig. 30.1):

• The vector dB is perpendicular both to ds (which points in the direction of the current) and to the unit vector rˆ directed from ds toward P.

▲ PITFALL PREVENTION

• The magnitude of dB is inversely proportional to r 2 , where r is the distance from

d s to P.

30.1 The Biot–Savart Law

• The magnitude of dB is proportional to the current and to the magnitude ds of the The magnetic field described by length element ds. the Biot–Savart law is the field due to a given current-carrying • The magnitude of dB is proportional to sin !, where ! is the angle between the

conductor. Do not confuse this vectors ds and rˆ.

field with any external field that may be applied to the conductor

These observations are summarized in the mathematical expression known today from some other source. as the Biot–Savart law:

# 0 Id s ! ˆr

d B" 2 (30.1)

Biot–Savart law

CHAPTER 30• Sources of the Magnetic Field

where # 0 is a constant called the permeability of free space:

Permeability of free space

(30.2) Note that the field dB in Equation 30.1 is the field created by the current in only a

# 0 " 4$ & 10 ' 7 T(m/A

small length element ds of the conductor. To find the total magnetic field B created at some point by a current of finite size, we must sum up contributions from all current elements I ds that make up the current. That is, we must evaluate B by integrating Equation 30.1:

where the integral is taken over the entire current distribution. This expression must be handled with special care because the integrand is a cross product and therefore a vector quantity. We shall see one case of such an integration in Example 30.1.

Although we developed the Biot–Savart law for a current-carrying wire, it is also valid for a current consisting of charges flowing through space, such as the electron beam in a television set. In that case, ds represents the length of a small segment of space in which the charges flow.

Interesting similarities exist between Equation 30.1 for the magnetic field due to a current element and Equation 23.9 for the electric field due to a point charge. The magnitude of the magnetic field varies as the inverse square of the distance from the source, as does the electric field due to a point charge. However, the direc- tions of the two fields are quite different. The electric field created by a point charge is radial, but the magnetic field created by a current element is perpendicular to both the length element ds and the unit vector rˆ, as described by the cross product in Equation 30.1. Hence, if the conductor lies in the plane of the page, as shown in Figure 30.1, d B points out of the page at P and into the page at P %.

Another difference between electric and magnetic fields is related to the source of the field. An electric field is established by an isolated electric charge. The Biot–Savart law gives the magnetic field of an isolated current element at some point, but such an isolated current element cannot exist the way an isolated electric charge can. A current element must be part of an extended current distribution because we must have a complete circuit in order for charges to flow. Thus, the Biot–Savart law (Eq. 30.1) is only the first step in a calculation of a magnetic field; it must be followed by an integration over the current distribution, as in Equation 30.3.

Quick Quiz 30.1 Consider the current in the length of wire shown in Figure 30.2. Rank the points A, B, and C, in terms of magnitude of the magnetic field due to the current in the length element shown, from greatest to least.

ds I

Figure 30.2 (Quick Quiz 30.1) Where is the magnetic field the greatest?

SECTION 30.1• The Biot–Savart Law

Example 30.1 Magnetic Field Surrounding a Thin, Straight Conductor Interactive

Consider a thin, straight wire carrying a constant current product is simply the magnitude of ds, which is the length

I and placed along the x axis as shown in Figure 30.3. dx. Substitution into Equation 30.1 gives Determine the magnitude and direction of the magnetic # 0 I field at point P due to this current. dx sin ! d B " (dB)ˆk "

r 2 ˆk Solution From the Biot–Savart law, we expect that the

Because all current elements produce a magnetic field magnitude of the field is proportional to the current in

in the kˆ direction, let us restrict our attention to the the wire and decreases as the distance a from the wire to

magnitude of the field due to one current element, which is point P increases. We start by considering a length

element ds located a distance r from P. The direction of #

0 I dx sin ! dB "

the magnetic field at point P due to the current in this

element is out of the page because ds ! ˆr is out of the To integrate this expression, we must relate the variables !, page. In fact, because all of the current elements I ds lie

x, and r. One approach is to express x and r in terms of !. in the plane of the page, they all produce a magnetic field

From the geometry in Figure 30.3a, we have directed out of the page at point P. Thus, we have the

r" " a csc !

direction of the magnetic field at point P, and we need

only find the magnitude. Taking the origin at O and

sin !

letting point P be along the positive y axis, with k ˆ being a Because tan ! " a/(' x) from the right triangle in Figure unit vector pointing out of the page, we see that

30.3a (the negative sign is necessary because ds is located at

d s ! ˆr " " d s ! ˆr " k " (dx sin !) ˆ k ˆ

a negative value of x), we have

x " 'a cot !

where "ds ! ˆr" represents the magnitude of ds ! rˆ. Because rˆ is a unit vector, the magnitude of the cross

Taking the derivative of this expression gives

dx " a csc 2 ! d!

Substitution of Equations (2) and (3) into Equation (1) gives ds  = dx

# 0 I a csc 2 ! sin ! d! #

dB " "

a an expression in which the only variable is !. We now obtain the magnitude of the magnetic field at point P by integrat-

rˆ ing Equation (4) over all elements, where the subtending θ

2 x as defined in Figure 30.3b: 1 to ! ds

angles range from !

4$a !

I B"

sin ! d! "

(cos ! 1 ' cos ! 2 ) (30.4)

! 1 4$a

(a)

We can use this result to find the magnetic field of any straight current-carrying wire if we know the geometry and P

hence the angles ! 1 and ! 2 . Consider the special case of an infinitely long, straight wire. If we let the wire in Figure 30.3b become infinitely long, we see that ! 1 " 0 and ! 2 " $ for length elements ranging between positions x " ' ) and x " * ). Because (cos ! 1 ' cos ! 2 ) " (cos 0 ' cos $) " 2,

Equation 30.4 becomes

Equations 30.4 and 30.5 both show that the magnitude of Figure 30.3 (Example 30.1) (a) A thin, straight wire carrying a

current I. The magnetic field at point P due to the current in the magnetic field is proportional to the current and each element ds of the wire is out of the page, so the net field at

decreases with increasing distance from the wire, as we

expected. Notice that Equation 30.5 has the same mathe- for determining the net field. When the wire is infinitely long,

point P is also out of the page. (b) The angles ! 1 and ! 2 used

matical form as the expression for the magnitude of the ! 1 " 0 and ! 2 " 180°.

electric field due to a long charged wire (see Eq. 24.7).

At the Interactive Worked Example link at http://www.pse6.com, you can explore the field for different lengths of wire.

CHAPTER 30• Sources of the Magnetic Field

I The result of Example 30.1 is important because a current in the form of a long, straight wire occurs often. Figure 30.4 is a perspective view of the magnetic

field surrounding a long, straight current-carrying wire. Because of the symmetry of the wire, the magnetic field lines are circles concentric with the wire and lie in planes perpendicular to the wire. The magnitude of B is constant on any circle of radius a and is given by Equation 30.5. A convenient rule for determining the direction of B is to grasp the wire with the right hand, positioning the thumb along the direction of the current. The four fingers wrap in the direction of the magnetic

a field.

B Another observation we can make in Figure 30.4 is that the magnetic field line shown has no beginning and no end. It forms a closed loop. This is a major difference

Figure 30.4 The right-hand rule between magnetic field lines and electric field lines, which begin on positive charges for determining the direction of the

and end on negative charges. We will explore this feature of magnetic field lines magnetic field surrounding a long,

further in Section 30.6.

straight wire carrying a current. Note that the magnetic field lines form circles around the wire.

Example 30.2 Magnetic Field Due to a Curved Wire Segment

Calculate the magnetic field at point O for the current-

of length ds:

carrying wire segment shown in Figure 30.5. The wire consists of two straight portions and a circular arc of radius

# 0 I ds

R, which subtends an angle !. The arrowheads on the wire 2 R indicate the direction of the current.

dB "

Because I and R are constants in this situation, we can easily integrate this expression over the curved path AC:

Solution The magnetic field at O due to the current in the straight segments AA% and CC% is zero because ds is parallel

ds " # 0 4$R I 2 ! 4$R 2 s" ! (30.6)

# 0 I # to rˆ along these paths; this means that ds ! rˆ " 0. Each I

B"

length element ds along path AC is at the same distance R 4$R from O, and the current in each contributes a field element

d B directed into the page at O. Furthermore, at every where we have used the fact that s " R! with ! measured in point on AC, ds is perpendicular to rˆ; hence, "ds ! ˆr" " ds.

radians. The direction of B is into the page at O because Using this information and Equation 30.1, we can find the

d s ! rˆ is into the page for every length element. magnitude of the field at O due to the current in an element

What If? What if you were asked to find the magnetic field

A′

at the center of a circular wire loop of radius R that carries a current I? Can we answer this question at this point in our A understanding of the source of magnetic fields?

rˆ

ds

Answer Yes, we can. We argued that the straight wires in

Figure 30.5 do not contribute to the magnetic field. The

O θ only contribution is from the curved segment. If we imagine increasing the angle !, the curved segment will become a

full circle when ! " 2$. Thus, we can find the magnetic

C field at the center of a wire loop by letting ! " 2$ in

I Equation 30.6:

# 0 I # 0 I Figure 30.5 (Example 30.2) The magnetic field at O due to

2R the current in the curved segment AC is into the page. The

4$R

contribution to the field at O due to the current in the two We will confirm this result as a limiting case of a more straight segments is zero.

general result in Example 30.3.

Example 30.3 Magnetic Field on the Axis of a Circular Current Loop Interactive

Consider a circular wire loop of radius R located in the yz Solution In this situation, every length element ds is per- plane and carrying a steady current I, as in Figure 30.6.

pendicular to the vector rˆ at the location of the element. Calculate the magnetic field at an axial point P a distance x

Thus, for any element, "ds ! ˆr " " (ds) (1)sin 90+ " ds. Fur- from the center of the loop.

thermore, all length elements around the loop are at the

SECTION 31.1• The Biot–Savart Law

B"B x i ˆ where

% dB cos ! " % 4$ x 2 *R 2

0 I ds cos !

ds

and we must take the integral over the entire loop. Because θ ˆr

!, x, and R are constants for all elements of the loop and

2 dB 1/2

because cos ! " R/(x *R 2 ) , we obtain

dB

4$(x 2 *R 2 ) 3/2 % ds " 2(x 2 *R 2

where we have used the fact that # ds " 2$R (the circumfer-

dB x

ence of the loop).

Figure 30.6 (Example 30.3) Geometry for calculating the To find the magnetic field at the center of the loop, we set magnetic field at a point P lying on the axis of a current loop.

x " 0 in Equation 30.7. At this special point, therefore, By symmetry, the total field B is along this axis.

(at x " 0)

which is consistent with the result of the What If? feature magnitude of dB due to the current in any length element

same distance r from P, where r 2 "x 2 *R 2 . Hence, the

in Example 30.2.

d s is The pattern of magnetic field lines for a circular current loop is shown in Figure 30.7a. For clarity, the lines are drawn

# 0 I " d s ! ˆr " #

" 0 I ds

for only one plane—one 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 30.7c.

The direction of d B is perpendicular to the plane formed by rˆ and d s, as shown in Figure 30.6. We can resolve this

What If? What if we consider points on the x axis very far vector into a component dB x along the x axis and a

from the loop? How does the magnetic field behave at these component dB y perpendicular to the x axis. When the

distant points?

components dB y are summed over all elements around the loop, the resultant component is zero. That is, by sym-

Answer In this case, in which x ,, R, we can neglect the metry the current in any element on one side of the loop

term R 2 in the denominator of Equation 30.7 and obtain sets up a perpendicular component of d B that cancels the

# 0 IR 2 perpendicular component set up by the current through

3 ,, (for x R) (30.9)

B$

the element diametrically opposite it. Therefore, the resul-

2x

tant field at P must be along the x axis and we can find it Because the magnitude of the magnetic moment # of the by integrating the components dB x " dB cos ! . That is,

loop is defined as the product of current and loop area (see

© Richard Megna, Fundamental Photographs

(c) Figure 30.7 (Example 30.3) (a) Magnetic field lines surrounding a current loop.

(a)

(b)

(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.

CHAPTER 30• Sources of the Magnetic Field

This result is similar in form to the expression for the express Equation 30.9 as

Eq. 29.10), # " I($R 2 ) for our circular loop. We can

electric field due to an electric dipole, E " k e (2qa/y 3 ) (see

# 0 # B$ Example 23.6), where 2qa " p is the electric dipole moment

2$ x 3 (30.10)

as defined in Equation 26.16.

At the Interactive Worked Example link at http://www.pse6.com, you can explore the field for different loop radii.

30.2 The Magnetic Force Between Two Parallel Conductors

In Chapter 29 we described the magnetic force that acts on a current-carrying conduc- tor placed in an external magnetic field. Because a current in a conductor sets up its own magnetic field, it is easy to understand that two current-carrying conductors exert magnetic forces on each other. Such forces can be used as the basis for defining the ampere and the coulomb.

Consider two long, straight, parallel wires separated by a distance a and carrying !

1 currents I 1 and I 2 in the same direction, as in Figure 30.8. We can determine the force

I 1 exerted on one wire due to the magnetic field set up by the other wire. Wire 2, which

B 2 carries a current I 2 and is identified arbitrarily as the source wire, creates a magnetic field

F 1 a B 2 at the location of wire 1, the test wire. The direction of B 2 is perpendicular to wire 1, as 2

I 2 shown in Figure 30.8. According to Equation 29.3, the magnetic force on a length ! of wire 1 is F 1 "I 1 "! B 2 . Because " is perpendicular to B 2 in this situation, the magnitude

a of F

1 is F 1 "I 1 !B 2 . Because the magnitude of B 2 is given by Equation 30.5, we see that

2$a ' 2$a

Active Figure 30.8 Two parallel

F "I !B "I

wires that each carry a steady current exert a magnetic force on

The direction of F 1 is toward wire 2 because " ! B 2 is in that direction. If the field set each other. The field B 2 due to the

up at wire 2 by wire 1 is calculated, the force F 2 acting on wire 2 is found to be equal in current in wire 2 exerts a magnetic

magnitude and opposite in direction to F 1 . This is what we expect because Newton’s

third law must be obeyed. wire 1. The force is attractive if the 1 When the currents are in opposite directions (that is, when currents are parallel (as shown)

force of magnitude F 1 "I !B

1 2 on

one of the currents is reversed in Fig. 30.8), the forces are reversed and the wires repel and repulsive if the currents are

each other. Hence, parallel conductors carrying currents in the same direction antiparallel.

attract each other, and parallel conductors carrying currents in opposite direc-

At the Active Figures link

tions repel each other.

at http://www.pse6.com, you

Because the magnitudes of the forces are the same on both wires, we denote the

can adjust the currents in the

magnitude of the magnetic force between the wires as simply F B . We can rewrite this

wires and the distance between

magnitude in terms of the force per unit length:

them to see the effect on the force.

2$a

The force between two parallel wires is used to define the ampere as follows:

Definition of the ampere

When the magnitude of the force per unit length between two long parallel wires that carry identical currents and are separated by 1 m is 2 & 10 ' 7 N/m, the current

in each wire is defined to be 1 A.

1 Although the total force exerted on wire 1 is equal in magnitude and opposite in direction to the total force exerted on wire 2, Newton’s third law does not apply when one considers two

small elements of the wires that are not exactly opposite each other. This apparent violation of Newton’s third law and of the law of conservation of momentum is described in more advanced treatments on electricity and magnetism.

SECTION 30.3• Ampère’s Law

The value 2 & 10 ' 7 N/m is obtained from Equation 30.12 with I 1 "I 2 " 1 A and

a " 1 m. Because this definition is based on a force, a mechanical measurement can be used to standardize the ampere. For instance, the National Institute of Standards and Technology uses an instrument called a current balance for primary current measure- ments. The results are then used to standardize other, more conventional instruments, such as ammeters.

The SI unit of charge, the coulomb, is defined in terms of the ampere:

When a conductor carries a steady current of 1 A, the quantity of charge that flows through a cross section of the conductor in 1 s is 1 C.

In deriving Equations 30.11 and 30.12, we assumed that both wires are long compared with their separation distance. In fact, only one wire needs to be long. The equations accurately describe the forces exerted on each other by a long wire and a straight parallel wire of limited length !.

Andre-Marie Ampère

French Physicist (1775–1836)

Quick Quiz 30.2 Ampère is credited with the

For I 1 " 2 A and I 2 " 6 A in Figure 30.8, which is true:

discovery of electromagnetism—

(a) F 1 " 3F 2 , (b) F 1 "F 2 /3, (c) F 1 "F 2 ?

the relationship between electric currents and magnetic fields. Ampère’s genius, particularly in

Quick Quiz 30.3 A loose spiral spring carrying no current is hung from the

mathematics, became evident ceiling. When a switch is thrown so that a current exists in the spring, do the coils move

by the time he was 12 years old;

(a) closer together, (b) farther apart, or (c) do they not move at all? his personal life, however, was

filled with tragedy. His father, a wealthy city official, was guillotined during the French Revolution, and his wife died young, in 1803. Ampère died at

30.3 Ampère’s Law

the age of 61 of pneumonia. His judgment of his life is clear from the epitaph he chose for his

Oersted’s 1819 discovery about deflected compass needles demonstrates that a gravestone: Tandem Felix current-carrying conductor produces a magnetic field. Figure 30.9a shows how

(Happy at Last). (Leonard de this effect can be demonstrated in the classroom. Several compass needles

Selva/CORBIS) are placed in a horizontal plane near a long vertical wire. When no current is

ds

© Richard Megna, Fundamental Photographs

Active Figure 30.9 (a) When no current is present in the wire, all compass needles

At the Active Figures link

point in the same direction (toward the Earth’s north pole). (b) When the wire carries

at http://www.pse6.com, you

a strong current, the compass needles deflect in a direction tangent to the circle, which

can change the value of the

is the direction of the magnetic field created by the current. (c) Circular magnetic field

current to see the effect on the

lines surrounding a current-carrying conductor, displayed with iron filings.

compasses.

CHAPTER 30• Sources of the Magnetic Field

present in the wire, all the needles point in the same direction (that of the Earth’s magnetic field), as expected. When the wire carries a strong, steady current, the needles all deflect in a direction tangent to the circle, as in Figure 30.9b. These observations demonstrate that the direction of the magnetic field produced by the current in the wire is consistent with the right-hand rule described in Figure 30.4. When the current is reversed, the needles in Figure 30.9b also reverse.

Because the compass needles point in the direction of B, we conclude that the lines of B form circles around the wire, as discussed in the preceding section. By symmetry, the magnitude of B is the same everywhere on a circular path centered on the wire and lying in a plane perpendicular to the wire. By varying the current and distance a from the wire, we find that B is proportional to the current and inversely proportional to the distance from the wire, as Equation 30.5 describes.

Now let us evaluate the product B ( d s for a small length element d s on the circular path defined by the compass needles, and sum the products for all elements over the closed circular path. 2 Along this path, the vectors d s and B are parallel at each point (see Fig. 30.9b), so B ( d s " B ds. Furthermore, the magnitude of B is constant on this circle and is given by Equation 30.5. Therefore, the sum of the products B ds over the closed path, which is equivalent to the line integral

▲ PITFALL PREVENTION

of B ( d s, is

% B(ds " B % ds " (2$ r) " #

30.2 Avoiding Problems

with Signs

2$r

When using Ampère’s law, apply the following right-hand rule. Point your thumb in the direction

where #ds " 2$r is the circumference of the circular path. Although this result was of the current through the amper-

calculated for the special case of a circular path surrounding a wire, it holds ian loop. Your curled fingers then

for a closed path of any shape (an amperian loop) surrounding a current that exists point in the direction that you

in an unbroken circuit. The general case, known as Ampère’s law, can be stated should integrate around the loop

as follows:

in order to avoid having to define the current as negative.

The line integral of B ( ds around any closed path equals # 0 I, where I is the total steady current passing through any surface bounded by the closed path.

% B(d s"#

Ampère’s law

0 I (30.13)

Ampère’s law describes the creation of magnetic fields by all continuous current configurations, but at our mathematical level it is useful only for calculating the magnetic field of current configurations having a high degree of symmetry. Its use is similar to that of Gauss’s law in calculating electric fields for highly symmetric charge distributions.

d 5A

1A

c Quick Quiz 30.4 Rank the magnitudes of # B ( ds for the closed paths in Figure 30.10, from least to greatest.

b a × 2A

2 You may wonder why we would choose to do this. The origin of Ampère’s law is in nineteenth century science, in which a “magnetic charge” (the supposed analog to an isolated electric

charge) was imagined to be moved around a circular field line. The work done on the charge was Figure 30.10 (Quick Quiz 30.4)

related to B ( ds, just as the work done moving an electric charge in an electric field is related to Four closed paths around three

E ( d s. Thus, Ampère’s law, a valid and useful principle, arose from an erroneous and abandoned current-carrying wires.

work calculation!

SECTION 30.3• Ampère’s Law

Quick Quiz 30.5 Rank the magnitudes of # B ( ds for the closed paths in Figure 30.11, from least to greatest.

abcd

Figure 30.11 (Quick Quiz 30.5) Several closed paths near a single current-carrying wire.

Example 30.4 The Magnetic Field Created by a Long Current-Carrying Wire

A long, straight wire of radius R carries a steady current I

circle is I, Ampère’s law gives

that is uniformly distributed through the cross section of the wire (Fig. 30.12). Calculate the magnetic field a

% distance r from the center of the wire in the regions r - R B ( ds " B % ds " B(2$r) " # 0 I

and r . R.

Solution Figure 30.12 helps us to conceptualize the wire

B"

(for r - R) (30.14)

2$r

and the current. Because the wire has a high degree of symmetry, we categorize this as an Ampère’s law problem. For the r - R case, we should arrive at the same result

which is identical in form to Equation 30.5. Note how we obtained in Example 30.1, in which we applied

much easier it is to use Ampère’s law than to use the the Biot–Savart law to the same situation. To analyze the

Biot–Savart law. This is often the case in highly symmetric problem, let us choose for our path of integration circle 1

situations.

in Figure 30.12. From symmetry, B must be constant in Now consider the interior of the wire, where r . R. magnitude and parallel to ds at every point on this circle.

Here the current I % passing through the plane of circle 2 is Because the total current passing through the plane of the

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 circle 2 must equal the ratio of the area $r 2

enclosed by circle 2 to the cross-sectional area $R 2 of the

1 I wire: 3

Following the same procedure as for circle 1, we apply Figure 30.12 (Example 30.4) A long, straight wire of radius R

Ampère’s law to circle 2:

carrying a steady current I uniformly distributed across the cross section of the wire. The magnetic field at any point can be calculated from Ampère’s law using a circular path of radius r,

% B(d s " B(2$r) " # 0 I

& concentric with the wire. I R 2 '

3 Another way to look at this problem is to realize that the current enclosed by circle 2 must

equal the product of the current density J " I/$R 2 and the area $r 2 of this circle.

CHAPTER 30• Sources of the Magnetic Field

# I B"

(for r . & 2$R '

R)

To finalize this problem, note that this result is similar

B∝r

in form to the expression for the electric field inside a

B ∝ 1/r

uniformly charged sphere (see Example 24.5). The magni- tude of the magnetic field versus r for this configuration is plotted in Figure 30.13. Note that inside the wire, B : 0 as r : 0. Furthermore, we see that Equations 30.14 and

30.15 give the same value of the magnetic field at r " R, demonstrating that the magnetic field is continuous at the

Figure 30.13 (Example 30.4) Magnitude of the magnetic field surface of the wire.

versus r for the wire shown in Figure 30.12. The field is propor- tional to r inside the wire and varies as 1/r outside the wire.

Example 30.5 The Magnetic Field Created by a Toroid

A device called a toroid (Fig. 30.14) is often used to create an so that the total current through the loop is NI. Therefore, almost uniform magnetic field in some enclosed area. The

the right side of Equation 30.13 is # 0 NI in this case. device consists of a conducting wire wrapped around a ring

Ampère’s law applied to the circle gives (a torus) made of a nonconducting material. For a toroid having N closely spaced turns of wire, calculate the magnetic

B(d s"B % % ds " B(2$r) " # 0 NI

field in the region occupied by the torus, a distance r from the center.

# 0 NI

(30.16) Solution To calculate this field, we must evaluate #B ( ds

B"

2$r

over the circular amperian loop of radius r in the plane of Figure 30.14. By symmetry, we see that the magnitude of the

This result shows that B varies as 1/r and hence is nonuniform field is constant on this circle and tangent to it, so B ( ds "

in the region occupied by the torus. However, if r is very

B ds. Furthermore, the wire passes through the loop N times, large compared with the cross-sectional radius a of the torus, then the field is approximately uniform inside the torus.

For an ideal toroid, in which the turns are closely spaced,

B the external magnetic field is close to zero. It is not exactly zero, however. In Figure 30.14, imagine the radius r of the amperian loop to be either smaller than b or larger than c. In

ds either case, the loop encloses zero net current, so # B ( ds " 0.

We might be tempted to claim that this proves that B " 0, but

b c it does not. Consider the amperian loop on the right side

I a of the toroid in Figure 30.14. The plane of this loop is perpen- dicular to the page, and the toroid passes through the loop. As

charges enter the toroid as indicated by the current directions

I in Figure 30.14, they work their way counterclockwise around the toroid. Thus, a current passes through the perpendicular

Figure 30.14 (Example 30.5) A toroid consisting of many amperian loop! This current is small, but it is not zero. As a turns of wire. If the turns are closely spaced, the magnetic field

in the interior of the torus (the gold-shaded region) is tangent result, the toroid acts as a current loop and produces a weak to the dashed circle and varies as 1/r. The dimension a is the

external field of the form shown in Figure 30.7. The reason cross-sectional radius of the torus. The field outside the toroid

that # B ( d s " 0 for the amperian loops of radius r . b and is very small and can be described by using the amperian loop

r , c in the plane of the page is that the field lines are perpen- at the right side, perpendicular to the page.

dicular to ds, not because B " 0.

Example 30.6 Magnetic Field Created by an Infinite Current Sheet

So far we have imagined currents carried by wires of small due to an infinite sheet of charge does not depend on cross section. Let us now consider an example in which a

distance from the sheet. Thus, we might expect a similar current exists in an extended object. A thin, infinitely large

result here for the magnetic field.

sheet lying in the yz plane carries a current of linear current To evaluate the line integral in Ampère’s law, we density J s . The current is in the y direction, and J s represents

construct a rectangular path through the sheet, as in Figure the current per unit length measured along the z axis. Find

30.15. The rectangle has dimensions ! and w, with the sides the magnetic field near the sheet.

of length ! parallel to the sheet surface. The net current in the plane of the rectangle is J s ! . We apply Ampère’s law over

Solution This situation is similar to those involving Gauss’s the rectangle and note that the two sides of length w do not law (see Example 24.8). You may recall that the electric field

contribute to the line integral because the component of B

SECTION 30.3• Ampère’s Law

hence the field should not vary from point to point. The

J s (out of paper)

only choices of field direction that are reasonable in this situation are either perpendicular or parallel to the sheet. However, a perpendicular field would pass through the

B current, which is inconsistent with the Biot–Savart law. Assuming a field that is constant in magnitude and parallel

to the plane of the sheet, we obtain

Figure 30.15 (Example 30.6) End view of an infinite current This result shows that the magnetic field is independent of sheet lying in the yz plane, where the current is in the y direc-

distance from the current sheet, as we suspected. The tion (out of the page). This view shows the direction of B on

expression for the magnitude of the magnetic field is similar both sides of the sheet. in form to that for the magnitude of the electric field due to

an infinite sheet of charge (Example 24.8): along the direction of these paths is zero. By symmetry, the

magnetic field is constant over the sides of length ! because

E"

every point on the infinitely large sheet is equivalent, and

Example 30.7 The Magnetic Force on a Current Segment Wire 1 in Figure 30.16 is oriented along the y axis and

from wire 1 (see Eq. 30.14) is

carries a steady current I 1 . A rectangular loop located to the

B" 0 I 1 ('ˆ k) Find the magnetic force exerted by wire 1 on the top wire of

right of the wire and in the xy plane carries a current I 2 .

2$x

length b in the loop, labeled “Wire 2” in the figure. where the unit vector ' kˆ is used to indicate that the field

due to the current in wire 1 at the position of ds points into You may be tempted to use Equation 30.12 the page. Because wire 2 is along the x axis, ds " dx iˆ, and to obtain the force exerted on a small segment of length

Solution

we find that

dx of wire 2. However, this equation applies only to two parallel wires and cannot be used here. The correct

d F # 0 I 1 I 2 dx " # 0 [ˆi ! ('ˆk )] I 1 I 2 B dx " approach is to consider the force exerted by wire 1 on a

ˆj

2$ x small segment ds of wire 2 by using Equation 29.4. This force is given by d F B " I d s ! B, where I " I and B is

2$x

Integrating over the limits x " a to x " a * b gives

the magnetic field created by the current in wire 1 at the

F B " 0 I 1 I position of ds. From Ampère’s law, the field at a distance x 2 ln x ˆj

a*b

& a ' Wire 1 B Wire 2

The force on wire 2 points in the positive y direction, as

indicated by the unit vector jˆ and as shown in Figure 30.16.

× ds ×

What If? What if the wire loop is moved to the left in Figure 30.16 until a " 0? What happens to the magnitude of the

force on the wire?

Answer The force should become stronger because the

loop is moving into a region of stronger magnetic field.

a b Equation (1) shows that the force not only becomes

stronger but the magnitude of the force becomes infinite as

a : 0! Thus, as the loop is moved to the left in Figure 30.16,

the loop should be torn apart by the infinite upward force Figure 30.16 (Example 30.7) A wire on one side of a rectangu-

on the top side and the corresponding downward force on lar loop lying near a current-carrying wire experiences a force.

the bottom side! Furthermore, the force on the left side is

CHAPTER 30• Sources of the Magnetic Field

toward the left and should also become infinite. This is

A similar situation occurs when we re-examine the larger than the force toward the right on the right side

magnetic field due to a long straight wire, given by Equation because this side is still far from the wire, so the loop should

30.5. If we could move our observation point infinitesimally

be pulled into the wire with infinite force! close to the wire, the magnetic field would become infinite! Does this really happen? In reality, it is impossible for

But in reality, the wire has a radius, and as soon as we enter

a : 0 because both wire 1 and wire 2 have finite sizes, so the wire, the magnetic field starts to fall off as described by that the separation of the centers of the two wires is at least

Equation 30.15 — approaching zero as we approach the the sum of their radii.

center of the wire.

30.4 The Magnetic Field of a Solenoid

A solenoid is a long wire wound in the form of a helix. With this configuration, a reasonably uniform magnetic field can be produced in the space surrounded by the turns of wire—which we shall call the interior of the solenoid—when the solenoid carries a current. When the turns are closely spaced, each can be approximated as a circular loop, and the net magnetic field is the vector sum of the fields resulting from all the turns.

Figure 30.17 shows the magnetic field lines surrounding a loosely wound solenoid. Exterior

Note that the field lines in the interior are nearly parallel to one another, are uniformly distributed, and are close together, indicating that the field in this space is strong and almost uniform.

If the turns are closely spaced and the solenoid is of finite length, the magnetic field lines are as shown in Figure 30.18a. This field line distribution is similar to that surrounding a bar magnet (see Fig. 30.18b). 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 interior field becomes more uniform and the exterior field becomes weaker. An ideal solenoid is approached when the turns are closely spaced and the length is much greater than the radius of the turns. Figure

30.19 shows a longitudinal cross section of part of such a solenoid carrying a current I. In this case, the external field is close to zero, and the interior field is uniform over a great volume.

Interior

Figure 30.17 The magnetic field lines for a loosely wound solenoid.

Henry Leap and Jim Lehman

(a)

(b)

Figure 30.18 (a) Magnetic field lines for a tightly wound solenoid of finite length, carrying a steady current. The field in the interior space is strong and nearly uniform. Note that the field lines resemble those of a bar magnet, meaning that the solenoid effectively has north and south poles. (b) The magnetic field pattern of a bar magnet, displayed with small iron filings on a sheet of paper.

SECTION 30.4• The Magnetic Field of a Solenoid

Figure 30.19 Cross-sectional view of an ideal solenoid, × 4 where the interior magnetic field is uniform and the

exterior field is close to zero. Ampère’s law applied to the circular path near the bottom whose plane is

perpendicular to the page can be used to show that

there is a weak field outside the solenoid. Ampère’s law applied to the rectangular dashed path in the plane of the page can be used to calculate the magnitude of the interior field.

If we consider the amperian loop perpendicular to the page in Figure 30.19, surrounding the ideal solenoid, we see that 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 nonzero magnetic field outside the solenoid. It is a weak field, with circular field lines, like those due to a line of current as in Figure 30.4. For an ideal solenoid, this is the only field external to the solenoid. We can eliminate this field in Figure 30.19 by adding a second layer of turns of wire outside the first layer, with the current carried along the axis of the solenoid in the opposite direction compared to the first layer. Then the net current along the axis is zero.

We can use Ampère’s law to obtain a quantitative expression for the interior magnetic field in an ideal solenoid. Because the solenoid is ideal, B in the interior space is uniform and parallel to the axis, and the magnetic field lines in the exterior space form circles around the solenoid. The planes of these circles are perpendicular to the page. Consider the rectangular path of length ! and width w shown in Figure 30.19. We can apply Ampère’s law to this path by evaluating the integral of B ( d s over each side of the rectangle. The contribution along side 3 is zero because the magnetic field lines are perpendicular to the path in this region. The contributions from sides 2 and 4 are both zero, again because B is perpendicular to d s along these paths, both inside and outside the solenoid. Side 1 gives a contribution to the integral because along this path B is uniform and parallel to d s. The integral over the closed rectangular path is therefore

B(d s "

B(d s"B ds " B!

path 1 path 1

The right side of Ampère’s law involves the total current I through the area bounded by the path of integration. In this case, the total current through the rectan- gular path equals the current through each turn multiplied by the number of turns. If N is the number of turns in the length !, the total current through the rectangle is NI. Therefore, Ampère’s law applied to this path gives

% B(d s " B! " #

Magnetic field inside a solenoid

where n " N/! is the number of turns per unit length.

CHAPTER 30• Sources of the Magnetic Field

We also could obtain this result by reconsidering the magnetic field of a toroid (see Example 30.5). If the radius r of the torus in Figure 30.14 containing N turns is much greater than the toroid’s cross-sectional radius a, a short section of the toroid approximates a solenoid for which n " N/2$r. In this limit, Equation 30.16 agrees with Equation 30.17.

Equation 30.17 is valid only for points near the center (that is, far from the ends) of a very long solenoid. As you might expect, the field near each end is smaller than the value given by Equation 30.17. At the very end of a long solenoid, the magnitude of the field is half the magnitude at the center (see Problem 32).

Quick Quiz 30.6 Consider a solenoid that is very long compared to the radius. Of the following choices, the most effective way to increase the magnetic field in the interior of the solenoid is to (a) double its length, keeping the number of turns per unit length constant, (b) reduce its radius by half, keeping the number of turns per unit length constant, (c) overwrapping the entire solenoid with an additional layer of current-carrying wire.

dA θ

B 30.5 Magnetic Flux

The flux associated with a magnetic field is defined in a manner similar to that used to define electric flux (see Eq. 24.3). Consider an element of area dA on an arbitrarily

Figure 30.20 The magnetic flux through an area element dA is

shaped surface, as shown in Figure 30.20. If the magnetic field at this element is B, the B ( d A " B d A cos !, where d A is a

magnetic flux through the element is B ( dA, where dA is a vector that is perpendicular vector perpendicular to the

to the surface and has a magnitude equal to the area dA. Therefore, the total magnetic surface.

flux 1 B through the surface is

Definition of magnetic flux

B(d

Consider the special case of a plane of area A in a uniform field B that makes an angle ! with dA. The magnetic flux through the plane in this case is

(30.19) If the magnetic field is parallel to the plane, as in Figure 30.21a, then ! " 90° and the

1 B " BA cos !

flux through the plane is zero. If the field is perpendicular to the plane, as in Figure 30.21b, then ! " 0 and the flux through the plane is BA (the maximum value).

The unit of magnetic flux is T ( m 2 , which is defined as a weber (Wb); 1 Wb "

1T(m 2 .

dA

dA

At the Active Figures link

(a)

(b)

at http://www.pse6.com, you

Active Figure 30.21 Magnetic flux through a plane lying in a magnetic field.

can rotate the plane and

(a) The flux through the plane is zero when the magnetic field is parallel to the plane

change the value of the field to

surface. (b) The flux through the plane is a maximum when the magnetic field is

see the effect on the flux.

perpendicular to the plane.

SECTION 30.6• Gauss’s Law in Magnetism

Example 30.8 Magnetic Flux Through a Rectangular Loop Interactive

A rectangular loop of width a and length b is located near a The factor 1/r indicates that the field varies over the loop, long wire carrying a current I (Fig. 30.22). The distance

and Figure 30.22 shows that the field is directed into the between the wire and the closest side of the loop is c. The

page at the location of the loop. Because B is parallel to dA wire is parallel to the long side of the loop. Find the total

at any point within the loop, the magnetic flux through an magnetic flux through the loop due to the current in the

area element dA is

wire.

! From Equation 30.14, we know that the magni- ! 2$r

tude of the magnetic field created by the wire at a distance r To integrate, we first express the area element (the tan from the wire is

region in Fig. 30.22) as dA " b dr. Because r is now the only # I variable in the integral, we have

B" 0

2$ ! c r 2$ ( c

What If? Suppose we move the loop in Figure 30.22 very

I far away from the wire. What happens to the magnetic flux? r ×

Answer The flux should become smaller as the loop moves into weaker and weaker fields.

As the loop moves far away, the value of c is much larger

than that of a, so that a/c : 0. Thus, the natural logarithm in Equation (1) approaches the limit

ln & 1*

c ' 9: