Physics For Scientists And Engineers 6E - part 236

30.6 Gauss’s Law in Magnetism

In Chapter 24 we found that the electric flux through a closed surface surrounding a net
charge is proportional to that charge (Gauss’s law). In other words, the number of electric
field lines leaving the surface depends only on the net charge within it. This property is
based on the fact that electric field lines originate and terminate on electric charges.

The situation is quite different for magnetic fields, which are continuous and form

closed loops. In other words, magnetic field lines do not begin or end at any point—as
illustrated in Figures 30.4 and 30.23. Figure 30.23 shows the magnetic field lines of a bar
magnet. Note that for any closed surface, such as the one outlined by the dashed line in
Figure 30.23, the number of lines entering the surface equals the number leaving the
surface; thus, the net magnetic flux is zero. In contrast, for a closed surface surrounding
one charge of an electric dipole (Fig. 30.24), the net electric flux is not zero.

Gauss’s law in magnetism states that

SECTION 3 0.6 • Gauss’s Law in Magnetism

941

A rectangular loop of width a and length b is located near a
long  wire  carrying  a  current  I (Fig.  30.22).  The  distance
between  the  wire  and  the  closest  side  of  the  loop  is  c.  The
wire  is  parallel  to  the  long  side  of  the  loop.  Find  the  total
magnetic  flux  through  the  loop  due  to  the  current  in  the
wire.

Solution From Equation 30.14, we know that the magni-
tude of the magnetic field created by the wire at a distance r
from the wire is

B "

2$r

The factor 1/r indicates that the field varies over the loop,
and Figure 30.22 shows that the field is directed into the
page at the location of the loop. Because

B is parallel to dA

at any point within the loop, the magnetic flux through an
area element dA is

To integrate, we first express the area element (the tan

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

What If?

Suppose we move the loop in Figure 30.22 very

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

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

and we find that 1

0 as we expected.

1 *

ln(1 * 0) " ln(1) " 0

1 *

(1)

a * c

a*c

ln r

(

a*c

B dA "

2$r

Example 30.8 Magnetic Flux Through a Rectangular Loop

Interactive

Figure 30.22 (Example 30.8) The magnetic field due to the wire

carrying a current I is not uniform over the rectangular loop.

At the Interactive Worked Example link at http://www.pse6.com, you can investigate the flux as the loop parameters change.

the net magnetic flux through any closed surface is always zero:

(30.20)

B(d

A " 0

Gauss’s law in magnetism

This statement is based on the experimental fact, mentioned in the opening of
Chapter 29, that isolated magnetic poles (monopoles) have never been detected and
perhaps do not exist. Nonetheless, scientists continue the search because certain theo-
ries that are otherwise successful in explaining fundamental physical behavior suggest
the possible existence of monopoles.

30.7 Displacement Current and the General

Form of Ampère’s Law

We  have  seen  that  charges  in  motion  produce  magnetic  fields.  When  a  current-
carrying  conductor  has  high  symmetry,  we  can  use  Ampère’s  law  to  calculate  the
magnetic  field  it  creates.  In  Equation  30.13,

# B ( ds " #

I, the line integral is over

any closed path through which the conduction current passes, where the conduction
current is defined by the expression I " dq/dt. (In this section we use the term
conduction current to refer to the current carried by the wire, to distinguish it from a
new type of current that we shall introduce shortly.) We now show that

Ampère’s law

in this form is valid only if any electric fields present are constant in time.
Maxwell recognized this limitation and modified Ampère’s law to include time-varying
electric fields.

We can understand the problem by considering a capacitor that is being charged as

illustrated in Figure 30.25. When a conduction current is present, the charge on the
positive plate changes but no conduction current exists in the gap between the plates. Now
consider the two surfaces S

and S

in Figure 30.25, bounded by the same path P.

Ampère’s law states that

# B ( ds around this path must equal #

I, where I is the total

current through any surface bounded by the path P.

When the path P is considered as bounding S

# B ( ds " #

I because the

conduction current passes through S

. When the path is considered as bounding S

however,

# B ( ds " 0 because no conduction current passes through S

. Thus, we

have a contradictory situation that arises from the discontinuity of the current!
Maxwell solved this problem by postulating an additional term on the right side

942

CHAPTE R 3 0 • Sources of the Magnetic Field

Figure 30.23 The magnetic field lines of a bar magnet form

closed loops. Note that the net magnetic flux through a closed

surface surrounding one of the poles (or any other closed surface)

is zero. (The dashed line represents the intersection of the surface

with the page.)

–

Figure 30.24 The electric field lines surrounding an electric

dipole begin on the positive charge and terminate on the negative

charge. The electric flux through a closed surface surrounding one

of the charges is not zero.

Path P

–q

Figure 30.25 Two surfaces S

and

near the plate of a capacitor are

bounded by the same path P. The

conduction current in the wire

passes only through S

. This leads

to a contradiction in Ampère’s law

that is resolved only if one postu-

lates a displacement current

through S

of Equation 30.13, which includes a factor called the

displacement current I

defined as

(30.21)

where 0

is the permittivity of free space (see Section 23.3) and

is the

electric flux (see Eq. 24.3).

As the capacitor is being charged (or discharged), the changing electric field

between  the  plates  may  be  considered  equivalent  to  a  current  that  acts  as  a
continuation  of  the  conduction  current  in  the  wire.  When  the  expression  for  the
displacement  current  given  by  Equation  30.21  is  added  to  the  conduction  current
on  the  right  side  of  Ampère’s  law,  the  difficulty  represented  in  Figure  30.25  is
resolved.  No  matter  which  surface  bounded  by  the  path  P is  chosen,  either  a
conduction current or a displacement current passes through it. With this new term
I

, we can express the general form of Ampère’s law (sometimes called the

Ampère–Maxwell law) as

(30.22)

We can understand the meaning of this expression by referring to Figure 30.26. The
electric flux through surface S

, where A is the area of the

capacitor plates and E is the magnitude of the uniform electric field between
the plates. If q is the charge on the plates at any instant, then E " q/(0

A). (See

Section 26.2.) Therefore, the electric flux through S

is simply

Hence, the displacement current through S

(30.23)

That is, the displacement current I

through S

is precisely equal to the conduction

current I through S

EA "

!E(d

A " 23

B(d

s " #

(I * I

) " #

I * #

!E(d

) 0

SECTION 3 0.7 • Displacement Current and the General Form of Ampère’s Law

943

Ampère–Maxwell law

Displacement current

Displacement in this context does not have the meaning it does in Chapter 2. Despite the

inaccurate implications, the word is historically entrenched in the language of physics, so we

continue to use it.

Strictly speaking, this expression is valid only in a vacuum. If a magnetic material is present,

one must change

and

on the right-hand side of Equation 30.22 to the permeability

(see

Section 30.8) and permittivity

characteristic of the material. Alternatively, one may include a

magnetizing current I

on the right hand side of Equation 30.22 to make Ampère’s law fully

general. On a microscopic scale, I

is as real as I.

–q

Figure 30.26 Because it exists only in the wires attached to the capacitor plates, the

conduction current I " dq/dt passes through S

but not through S

. Only the

displacement current I

/dt passes through S

. The two currents must be equal

for continuity.

944

CHAPTE R 3 0 • Sources of the Magnetic Field

A sinusoidally varying voltage is applied across an 8.00-#F
capacitor. The frequency of the voltage is 3.00 kHz, and the
voltage amplitude is 30.0 V. Find the displacement current
in the capacitor.

Solution The angular frequency of the source, from
Equation 15.12, is given by 4 " 2$f " 2$(3.00 & 10

Hz) "

1.88 & 10

. Hence, the voltage across the capacitor in

terms of t is

V " 5V

max

sin 4t " (30.0 V) sin(1.88 & 10

We can use Equation 30.23 and the fact that the charge
on the capacitor is q " C 5V to find the displacement

By considering surface S

, we can identify the displacement current as the source of

the magnetic field on the surface boundary. The displacement current has its physical
origin in the time-varying electric field. The central point of this formalism is that

Quick Quiz 30.7

In an RC circuit, the capacitor begins to discharge. During

the discharge, in the region of space between the plates of the capacitor, there is
(a) conduction current but no displacement current, (b) displacement current but no
conduction current, (c) both conduction and displacement current, (d) no current of
any type.

Quick Quiz 30.8

The capacitor in an RC circuit begins to discharge. During

the discharge, in the region of space between the plates of the capacitor, there is (a) an
electric field but no magnetic field, (b) a magnetic field but no electric field, (c) both
electric and magnetic fields, (d) no fields of any type.

magnetic fields are produced both by conduction currents and by time-varying
electric fields.

current:

The displacement current varies sinusoidally with time and
has a maximum value of 4.52 A.

(4.52 A) cos(1.88 & 10

(8.00 & 10

[(30.0 V) sin(1.88 & 10

t)]

∆V

) " C

(

∆V

)

Example 30.9 Displacement Current in a Capacitor

This result was a remarkable example of theoretical work by Maxwell, and it
contributed to major advances in the understanding of electromagnetism.

30.8 Magnetism in Matter

The magnetic field produced by a current in a coil of wire gives us a hint as to what
causes certain materials to exhibit strong magnetic properties. Earlier we found that a
coil like the one shown in Figure 30.18 has a north pole and a south pole. In general,
any current loop has a magnetic field and thus has a magnetic dipole moment, includ-
ing the atomic-level current loops described in some models of the atom.

The Magnetic Moments of Atoms

We begin our discussion with a classical model of the atom in which electrons move
in circular orbits around the much more massive nucleus. In this model, an
orbiting electron constitutes a tiny current loop (because it is a moving charge),

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