02 Rao Elements 2004 ch2

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C H A P T E R 2
Maxwell’s Equations
in Integral Form
In Chapter 1, we learned the simple rules of vector algebra and familiarized

ourselves with the basic concepts of fields in general, and then introduced elec-
tric and magnetic fields in terms of forces on charged particles. We now have the
necessary background to introduce the additional tools required for the under-
standing of the various quantities associated with Maxwell’s equations and then
discuss Maxwell’s equations. In particular, our goal in this chapter is to learn
Maxwell’s equations in integral form as a prerequisite to the derivation of their
differential forms in the next chapter. Maxwell’s equations in integral form gov-
ern the interdependence of certain field and source quantities associated with
regions in space, that is, contours, surfaces, and volumes. The differential forms
of Maxwell’s equations, however, relate the characteristics of the field vectors at
a given point to one another and to the source densities at that point.
Maxwell’s equations in integral form are a set of four laws resulting from
several experimental findings and a purely mathematical contribution. We shall,
however, consider them as postulates and learn to understand their physical sig-
nificance as well as their mathematical formulation. The source quantities in-
volved in their formulation are charges and currents. The field quantities have
to do with the line and surface integrals of the electric and magnetic field vec-
tors. We shall therefore first introduce line and surface integrals and then con-
sider successively the four Maxwell’s equations in integral form.
2.1 THE LINE INTEGRAL

To introduce the line integral, let us consider in a region of electric field E the Line integral
movement of a test charge q from the point A to the point B along the path C as
shown in Fig. 2.1(a). At each and every point along the path the electric field
exerts a force on the test charge and, hence, does a certain amount of work in
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78 Chapter 2 Maxwell’s Equations in Integral Form
E
En an
B B
Ej a ln
j
C E3 lj
E1 E2 a3
a
a1 2 l
A A l2 3
l1
(b)
(a)
FIGURE 2.1
For evaluating the total amount of work done in moving a test charge along a path C from
point A to point B in a region of electric field.
moving the charge to another point an infinitesimal distance away.To find the total
amount of work done from A to B, we divide the path into a number of infinitesi-
mal segments ¢l 1, ¢l 2, ¢l 3, Á ¢l n, as shown in Fig. 2.1(b), find the infinitesimal
amount of work done for each segment, and then add up the contributions from all
the segments. Since the segments are infinitesimal in length, we can consider each
of them to be straight and the electric field at all points within a segment to be the
same and equal to its value at the start of the segment.
If we now consider one segment, say, the jth segment, and take the compo-
nent of the electric field for that segment along the length of that segment, we ob-
tain the result Ej cos aj, where aj is the angle between the direction of the electric
field vector Ej at the start of that segment and the direction of that segment. Since
the electric field intensity has the meaning of force per unit charge, the electric
force along the direction of the jth segment is then equal to qEj cos aj. To obtain
the work done in carrying the test charge along the length of the jth segment, we
then multiply this electric force component by the length ¢lj of that segment.
Thus, for the jth segment, we obtain the result for the work done by the electric
field as
¢Wj = qEj cos aj ¢lj (2.1)
If we do this for all the infinitesimal segments and add up all the contributions,
we get the total work done by the electric field in moving the test charge from A
to B along the path C to be
WAB = ¢W1 + ¢W2 + ¢W3 + Á + ¢Wn

= qE1 cos a1 ¢l1 + qE2 cos a2 ¢l2 + qE3 cos a3 ¢l3 + Á
+ qEn cos an ¢ln (2.2)
n
= q a Ej cos aj ¢lj
j=1
n
= q a 1Ej21¢lj2 cos aj
j=1
2.1 The Line Integral 79
y
B
(1, 1, 0)
10
y
y x2 j 2 0.01
lj
j
y x2
3 (j 1)20.01
1 2
A x x
0 1 (j 1)0.1 j0.1
(a) (b)
FIGURE 2.2
(a) Division of the path y = x2 from A(0, 0, 0) to B(1, 1, 0) into 10 segments. (b) Length
vector corresponding to the jth segment of part (a) approximated as a straight line.
Using the dot product operation between two vectors, we obtain

n
WAB = q a Ej # ¢l j (2.3)
j=1
For a numerical example, let us consider the electric field given by
E = yay
and determine the work done by the field in the movement of 3 mC of charge
from the point A(0, 0, 0) to the point B(1, 1, 0) along the parabolic path
y = x2, z = 0 shown in Fig. 2.2(a).
For convenience, we shall divide the path into 10 segments having equal
projections along the x-axis, as shown in Fig. 2.2(a). We shall number the seg-
ments 1, 2, 3, Á , 10. The coordinates of the starting and ending points of the jth
segment are as shown in Fig. 2.2(b). The electric field at the start of the jth seg-
ment is given by
Ej = 1j - 1220.01a y
The length vector corresponding to the jth segment, approximated as a straight
line connecting its starting and ending points, is
¢l j = 0.1a x + [j2 - 1j - 122]0.01a y = 0.1a x + 12j - 120.01a y
The required work is then given by

10
WAB = 3 * 10-6 a Ej # ¢l j
j=1
10
= 3 * 10-6 a [1j - 1220.01a y] # [0.1a x + 12j - 120.01ay]
j=1
10
= 3 * 10 -10 a 1j - 12212j - 12
j=1
= 3 * 10 [0 + 3 + 20 + 63 + 144 + 275 + 468 + 735

-10
+ 1088 + 1539]
= 3 * 10 -10 * 4335 J = 1.3005 mJ
The result that we have just obtained for WAB is approximate, since we di-
vided the path from A to B into a finite number of segments. By dividing it into
larger and larger numbers of segments, we can obtain more and more accurate
results. In the limit that n : q , the result converges to the exact value. The
summation in (2.3) then becomes an integral, which represents exactly the work
done by the field and is given by
B
WAB = q E # dl (2.4)
LA
The integral on the right side of (2.4) is known as the line integral of E from A to
B, along the specified path.
Evaluation of We shall illustrate the evaluation of the line integral by computing the
line integral exact value of the work done by the electric field in the movement of the 3 mC
charge for the path in Fig. 2.2(a). To do this, we note that at any arbitrary point
on the curve y = x2, z = 0,
dy = 2x dx dz = 0
so that the differential length vector tangential to the curve is given by
dl = dx ax + dy a y + dz a z
= dx ax + 2x dx ay
The value of E # dl at the point is
E # dl = yay # 1dx ax + 2x dx a y2
= x2a y # 1dx a x + 2x dx ay2
= 2x3 dx
Thus, the required work is given by

11,1,02 1
WAB = q E # dl = 3 * 10 -6 2x3 dx
L10,0,02 L0
2x4 1
= 3 * 10-6 c d = 1.5 mJ
4 0
Note that we have evaluated the line integral by using x as the variable of inte-
gration. Alternatively, using y as the variable of integration, we obtain
E # dl = yay # 1dx a x + dy ay2

= y dy
11,1,02 1
WAB = q E # dl = 3 * 10 -6 y dy
L10,0,02 L0
y2 1
= 3 * 10 -6 c d = 1.5 mJ
2 0
Thus, the integration can be performed with respect to x or y (or z in the three-
dimensional case). What is important, however, is that the integrand must be ex-
pressed as a function of the variable of integration and the limits appropriate to
that variable must be employed.
Returning now to (2.4) and dividing both sides by q, we note that the line Voltage
integral of E from A to B has the physical meaning of work per unit charge defined
done by the field in moving the test charge from A to B. This quantity is known
as the voltage between A and B along the specified path and is denoted by the
symbol VAB, having the units of volts. Thus,
B
VAB = E # dl (2.5)
LA
When the path under consideration is a closed path, that is, one that has no
beginning or ending, such as a rubber band, as shown in Fig. 2.3, the line integral
is written with a circle associated with the integral sign in the manner AC E # dl.
The line integral of a vector around a closed path is known as the circulation of
that vector. In particular, the line integral of E around a closed path is the work
per unit charge done by the field in moving a test charge around the closed path.
It is the voltage around the closed path and is also known as the electromotive
force. We shall now consider an example of evaluating the line integral of a vector
around a closed path.
FIGURE 2.3
Closed path C in a region of electric field.
Example 2.1 Evaluation of line integral around a closed path

Let us consider the force field
F = xay
and evaluate A C F # dl, where C is the closed path ABCDA shown in Fig. 2.4.
Noting that
B C D A
F # dl = F # dl + F # dl + F # dl + F # dl
CABCDA LA LB LC LD
we simply evaluate each of the line integrals on the right side and add them up to obtain
the required quantity.
First, we observe that since the entire closed path lies in the z = 0 plane, dz = 0
and dl = dx ax + dy ay for all four straight lines. Then for the side AB,
y = 1, dy = 0, dl = dx ax + 102ay = dx ax
F # dl = 1xay2 # 1dx ax2 = 0
B
F # dl = 0
LA
For the side BC,
x = 3, dx = 0, dl = 102ax + dy ay = dy ay
F # dl = 13ay2 # 1dy ay2 = 3 dy
C 5
F # dl = 3 dy = 12
LB L1
For the side CD,
y = 2 + x, dy = dx, dl = dx ax + dx ay
F # dl = 1xay2 # 1dx ax + dx ay2 = x dx
D 1
F # dl = x dx = -4
LC L3
y
(3, 5)
C
(1, 3)
D
FIGURE 2.4
A B
For evaluating the line integral of a vector (1, 1) (3, 1)
field around a closed path. x
For the side DA,
x = 1, dx = 0, dl = 102ax + dy ay
F # dl = 1ay2 # 1dy ay2 = dy
A 1
F # dl = dy = -2
LD L3
Finally,
F # dl = 0 + 12 - 4 - 2 = 6
CABCDA
In this example, we found that the line integral of F around the closed Conservative
path C is nonzero. The field is then said to be a nonconservative field. For a non- vs. noncon-
conservative field, the line integral between two points, say, A and B, is depen- servative
dent on the path followed from A to B. To show this, let us consider the two fields
paths ACB and ADB, as shown in Fig. 2.5. Then we can write
F # dl = F # dl + F # dl
CACBDA LACB LBDA
(2.6)
= F # dl - F # dl
LACB LADB
It can be easily seen that if AACBDAF # dl is not equal to zero, then 1ACB F # dl is
not equal to 1ADB F # dl. The two integrals are equal only if AACBDAF # dl is equal
to zero, which is the case for conservative fields. Examples of conservative fields
are Earth’s gravitational field and the static electric field. An example of non-
conservative fields is the time-varying electric field. Thus, in a time-varying elec-
tric field, the voltage between two points A and B is dependent on the path
followed to evaluate the line integral of E from A to B, whereas in a static elec-
tric field, the voltage, more commonly known as the potential difference, be-
tween two points A and B is uniquely defined because the line integral of E
from A to B is independent of the path followed from A to B.
D C
FIGURE 2.5
B Two different paths from point A to point B.
K2.1. Line integral; Line integral of E; Voltage; Line integral around a closed path;
Circulation; Line integral of E around a closed path; Electromotive force; Con-
servative vs. nonconservative fields.
D2.1. For each of the curves (a) y = x2, z = 0, (b) x2 + y2 = 2, z = 0, and (c) y =
sin 0.5px, z = 0 in a region of electric field E = yax + xay, find the approxi-
mate value of the work done by the field in carrying a charge of 1 mC from the
point (1, 1, 0) to the neighboring point on the curve, whose x coordinate is 1.1,
by evaluating E # ¢l along a straight line path.
Ans. (a) 0.31 mJ; (b) -0.0112 mJ; (c) 0.0877 mJ.
D2.2. For F = y1ax + ay2, find 1 F # dl for the straight-line paths between the follow-
ing pairs of points from the first point to the second point: (a) (0, 0, 0) to (2, 0, 0);
(b) (0, 2, 0) to (2, 2, 0); and (c) (2, 0, 0) to (2, 2, 0).
Ans. (a) 0; (b) 4; (c) 2.
2.2 THE SURFACE INTEGRAL
Surface To introduce the surface integral, let us consider a region of magnetic field and
integral an infinitesimal surface at a point in that region. Since the surface is infinitesi-
mal, we can assume the magnetic flux density to be uniform on the surface, al-
though it may be nonuniform over a wider region. If the surface is oriented
normal to the magnetic field lines, as shown in Fig. 2.6(a), then the magnetic
flux (webers) crossing the surface is simply given by the product of the surface
area (meters squared) and the magnetic flux density 1Wb/m22 on the surface,
that is, B ¢S. If, however, the surface is oriented parallel to the magnetic field
lines, as shown in Fig. 2.6(b), there is no magnetic flux crossing the surface. If
the surface is oriented in such a manner that the normal to the surface makes
an angle a with the magnetic field lines as shown in Fig. 2.6(c), then the amount
of magnetic flux crossing the surface can be determined by considering that the
Normal
Normal
B B B
S
a
S S
Normal
(a) (b) (c)
FIGURE 2.6
Infinitesimal surface ¢S in a magnetic field B oriented (a) normal to the field. (b) parallel to
the field, and (c) with its normal making an angle a to the field.
2.2 The Surface Integral 85
component of B normal to the surface is B cos a and the component tangential

to the surface is B sin a. The component of B normal to the surface results in a
flux of 1B cos a2 ¢S crossing the surface, whereas the component tangential to
the surface does not contribute at all to the flux crossing the surface. Thus, the
magnetic flux crossing the surface in this case is 1B cos a2 ¢S. We can obtain
this result alternatively by noting that the projection of the surface onto the
plane normal to the magnetic field lines is ¢S cos a. Hence, the magnetic flux
crossing the surface ¢S is the same as that crossing normal to the area ¢S cos a,
that is, B1¢S cos a2 or 1B cos a2 ¢S.
To aid further in the understanding of this concept, imagine raindrops
falling vertically downward uniformly. If you hold a rectangular loop horizon-
tally, the number of drops falling through the loop is simply equal to the area of
the loop multipled by the density (number of drops per unit area) of the drops.
If the loop is held vertically, no rain falls through the loop. If the loop is held at
some angle to the horizontal, the number of drops falling through the loop is the
same as that which would fall through another (smaller) loop, which is the pro-
jection of the slanted loop on to the horizontal plane.
Let us now consider a large surface S in the magnetic field region, as
shown in Fig. 2.7. The magnetic flux crossing this surface can be found by divid-
ing the surface into a number of infinitesimal surfaces ¢S1, ¢S2, ¢S3, Á , ¢Sn,
applying the result just obtained for each infinitesimal surface, and adding up
the contributions from all the surfaces. To obtain the contribution from the jth
surface, we draw the normal vector to that surface and find the angle aj between
the normal vector and the magnetic flux density vector Bj associated with that
surface. Since the surface is infinitesimal, we can assume Bj to be the value of B
at the centroid of the surface, and we can also erect the normal vector at that
point. The contribution to the total magnetic flux from the jth infinitesimal sur-
face is then given by
¢cj = Bj cos aj ¢Sj (2.7)
Normal
Bj
a nj
aj
Sj
S FIGURE 2.7
Division of a large surface S in a
magnetic field region into a number of
infinitesimal surfaces.
where the symbol c represents magnetic flux. The total magnetic flux crossing
the surface S is then given by
[c]S = ¢c1 + ¢c2 + ¢c3 + Á + ¢cn

= B1 cos a1 ¢S1 + B2 cos a2 ¢S2 + B3 cos a3 ¢S3 + Á
+ Bn cos an ¢Sn (2.8)
n
= a Bj cos aj ¢Sj
j=1
n
= a Bj1¢Sj2 cos aj
j=1
Using the dot product operation between two vectors, we obtain

n
[c]S = a Bj # ¢Sj anj (2.9)
j=1
where a nj is the unit vector normal to the surface ¢Sj. Furthermore, by using the
concept of an infinitesimal surface vector as one having magnitude equal to the
area of the surface and direction normal to the surface, that is,
¢Sj = ¢Sj a nj (2.10)
we can write (2.9) as

n
[c]S = a Bj # ¢Sj (2.11)
j=1
For a numerical example, let us consider the magnetic field given by
B = 3xy2a z Wb/m2
and determine the magnetic flux crossing the portion of the xy-plane lying be-
tween x = 0, x = 1, y = 0, and y = 1. For convenience, we shall divide the sur-
face into 25 equal areas, as shown in Fig. 2.8 (a). We shall designate the squares
as 11, 12, Á , 15, 21, 22, Á , 55, where the first digit represents the number of
the square in the x-direction and the second digit represents the number of the
square in the y-direction. The x- and y-coordinates of the midpoint of the ijth
square are 12i - 120.1 and 12j - 120.1, respectively, as shown in Fig. 2.8(b). The
magnetic field at the center of the ijth square is then given by
Bij = 312i - 1212j - 1220.001a z
Since we have divided the surface into equal areas and since all areas are in the
xy-plane,
¢Sij = 0.04 a z for all i and j

(2j 1)0.1
0 1 j
y
11 12
i
21 (2i 1)0.1
ij
1 55
(1, 1, 0)
x
(a) (b)
FIGURE 2.8
(a) Division of the portion of the xy-plane lying between x = 0, x = 1, y = 0,
and y = 1 into 25 squares. (b) Area corresponding to the ijth square.
The required magnetic flux is then given by

5 5
[c]S = a a Bij # ¢Sij
i=1 j=1
5 5
= a a 312i - 1212j - 1220.001a z # 0.04a z
i=1 j=1
5 5
= 0.00012 a a 12i - 1212j - 122
i=1 j=1
= 0.0001211 + 3 + 5 + 7 + 9211 + 9 + 25 + 49 + 812
= 0.495 Wb
The result that we have just obtained for [c]S is approximate since we
have divided the surface S into a finite number of areas. By dividing it into larg-
er and larger numbers of squares, we can obtain more and more accurate re-
sults. In the limit that n : q , the result converges to the exact value. The
summation in (2.11) then becomes an integral that represents exactly the mag-
netic flux crossing the surface and is given by
[c]S = B # dS (2.12)
LS
where the symbol S associated with the integral sign denotes that the integration
is performed over the surface S. The integral on the right side of (2.12) is known
as the surface integral of B over S. The surface integral is a double integral since
dS is equal to the product of two differential lengths.
We shall illustrate the evaluation of the surface integral by computing Evaluation of
the exact value of the magnetic flux crossing the surface in Fig. 2.8(a). To do surface
this, we note that at any arbitrary point on the surface, the differential surface integral
vector is given by
dS = dx dy az
The value of B # dS at the point is
B # dS = 3xy2 a z # dx dy a z
= 3xy2 dx dy
Thus, the required magnetic flux is given by
[c]S = B # dS
LS
1 1
= 3xy2 dx dy = 0.5 Wb
Lx = 0 Ly = 0
When the surface under consideration is a closed surface, the surface integral
is written with a circle associated with the integral sign in the manner AS B # dS. A
closed surface is one that encloses a volume. Hence, if you are anywhere in that
volume, you can get out of it only by making a hole in the surface, and vice versa.A
simple example is the surface of a balloon inflated and tied up at the mouth. The
surface integral of B over the closed surface S is simply the magnetic flux
emanating from the volume bounded by the surface. Thus, whenever a closed sur-
face integral is evaluated, the unit vectors normal to the differential surfaces are
chosen to be pointing out of the volume, so as to give the outward flux of the
vector field, unless specified otherwise. We shall now consider an example of eval-
uating AS B # dS.
Example 2.2 Evaluation of a closed surface integral

Let us consider the magnetic field
B = 1x + 22ax + 11 - 3y2ay + 2zaz
and evaluate AS B # dS , where S is the surface of the cubical box bounded by the planes
x = 0 x = 1
y = 0 y = 1
z = 0 z = 1
as shown in Fig. 2.9.

Noting that
B # dS = B # dS + B # dS + B # dS + B # dS
CS Labcd Lefgh Ladhe Lbcgf
+ B # dS + B # dS
Laefb Ldhgc
d c
1
h g
1
y
a b
FIGURE 2.9
1
e f
For evaluating the surface integral of a vector field
x over a closed surface.
we simply evaluate each of the surface integrals on the right side and add them up to ob-
tain the required quantity. In doing so, we recognize that since the quantity we want is
the magnetic flux out of the box, we should direct the unit normal vectors toward the
outside of the box. Thus, for the surface abcd,
x = 0, B = 2ax + 11 - 3y2ay + 2zaz, dS = -dy dz ax

B # dS = -2 dy dz
1 1
B # dS = 1-22 dy dz = -2
Labcd Lz = 0 Ly = 0
For the surface efgh,
x = 1, B = 3ax + 11 - 3y2ay + 2zaz, dS = dy dz ax

B # dS = 3 dy dz
1 1
B # dS = 3 dy dz = 3
Lefgh Lz = 0 Ly = 0
For the surface adhe,
y = 0, B = 1x + 22ax + 1ay + 2zaz, dS = -dz dx ay

B # dS = -dz dx
1 1
B # dS = 1-12 dz dx = -1
Laehd Lx = 0 Lz = 0
For the surface bcgf,
y = 1, B = 1x + 22ax - 2ay + 2zaz, dS = dz dx ay

B # dS = -2 dz dx
1 1
B # dS = 1-22 dz dx = -2
Lbfgc Lx = 0 Lz = 0
For the surface aefb,
z = 0, B = 1x + 22ax + 11 - 3y2ay + 0az, dS = -dx dy az

B # dS = 0
B # dS = 0
Laefb
For the surface dhgc,
z = 1, B = 1x + 22ax + 11 - 3y2ay + 2az, dS = dx dy az

B # dS = 2 dx dy
1 1
B # dS = 2 dx dy = 2
Ldhgc Ly = 0 Lx = 0
Finally,
B # dS = -2 + 3 - 1 - 2 + 0 + 2 = 0
CS
K2.2. Surface integral; Surface integral of B; Magnetic flux; Surface integral over a
closed surface.
D2.3. Given B = 1yax - xay2 Wb/m2, find by evaluating B # ¢S the approximate
absolute value of the magnetic flux crossing from one side to the other side of an in-
finitesimal surface of area 0.001 m2 at the point (1, 2, 1) for each of the following
orientations of the surface: (a) in the x = 1 plane; (b) on the surface 2x2 + y2 = 6;
and (c) normal to the unit vector 1312ax + ay + 2az2.
Ans. (a) 2 * 10 -3 Wb; (b) 11/122 * 10-3 Wb; (c) 10-3 Wb.
D2.4. For the vector field A = x1ax + ay2, find the absolute value of 1 A # dS over the
following plane surfaces: (a) square having the vertices at (0, 0, 0), (0, 2, 0), (0, 2,
2), and (0, 0, 2); (b) square having the vertices at (2, 0, 0), (2, 2, 0), (2, 2, 2), and (2,
0, 2); (c) square having the vertices at (0, 0, 0), (2, 0, 0), (2, 0, 2), and (0, 0, 2); and
(d) triangle having the vertices at (0, 0, 0), (2, 0, 0), and (0, 0, 2).
Ans. (a) 0; (b) 8; (c) 4; (d) 43.
2.3 FARADAY’S LAW

In the preceding two sections, we introduced the line and surface integrals. We
are now ready to consider Maxwell’s equations in integral form. The first equa-
tion, which we shall discuss in this section, is a consequence of an experimental
finding by Michael Faraday in 1831 that time-varying magnetic fields give rise to
electric fields and, hence, it is known as Faraday’s law. Faraday discovered that
when the magnetic flux enclosed by a loop of wire changes with time, a current
is produced in the loop, indicating that a voltage or an electromotive force, ab-
breviated as emf, is induced around the loop. The variation of the magnetic flux
can result from the time variation of the magnetic flux enclosed by a fixed loop
2.3 Faraday’s Law 91
dS
FIGURE 2.10
For illustrating Faraday’s law.
or from a moving loop in a static magnetic field or from a combination of the

two, that is, a moving loop in a time-varying magnetic field.
In mathematical form, Faraday’s law is given by Statement of
Faraday’s law
d
E # dl = - B # dS (2.13)
CC dt LS
where S is a surface bounded by the closed path C, as shown in Fig. 2.10. In

words, Faraday’s law states that the electromotive force around a closed path is
equal to the negative of the time rate of change of the magnetic flux enclosed by
that path. There are certain procedures and observations of interest pertinent to
the application of (2.13). We shall discuss these next.
1. The magnetic flux on the right side is to be evaluated in accordance with Right-hand
the right-hand screw rule (R.H.S. rule), a convention that is applied consistently screw rule
for all electromagnetic field laws involving integration over surfaces bounded by
closed paths. The right-hand screw rule consists of imagining a right-hand screw
being turned around the closed path, as illustrated in Fig. 2.11 for two opposing
senses of paths, and using the resulting direction of advance of the screw to
evaluate the surface integral. The application of this rule to the geometry of
(a) (b)
FIGURE 2.11
Right-hand screw rule convention employed in the formulation of
electomagnetic field laws.
z z
R R
C C
O y O y
Q Q
P P
x x
(a) (b)
FIGURE 2.12
(a) A plane surface and (b) a combination of three plane surfaces, bounded by the closed
path C.
Fig. 2.10 means that in evaluating the surface integral of B over S, the normal
vector to the differential surface dS should be directed as shown in that figure.
2. In evaluating the surface integral of B, any surface S bounded by C
can be employed. For example, if the loop C is a planar loop, it is not necessary
to consider the plane surface having the loop as its perimeter. One can consid-
er a curved surface bounded by C or any combination of plane (or plane and
curved) surfaces which together are bounded by C, and which is sometimes a
more desirable choice. To illustrate this point, consider the planar loop PQRP
in Fig. 2.12 (a). The most obvious surface bounded by this loop is the plane
surface PQR inclined to the coordinate planes. Now imagine this plane sur-
face to be an elastic sheet glued to the perimeter and pushed in toward the
origin so as to conform to the coordinate planes. Then we obtain the combina-
tion of the plane surfaces OPQ, OQR, and ORP, as shown in Fig. 2.12(b),
which together constitute a surface also bounded by the loop. To evaluate the
surface integral of B for the surface in Fig. 2.12(a), we need to make use of the
dS vector on that slant surface. On the other hand, for the geometry in Fig.
2.12(b), we can use the (simpler) dS vectors associated with the coordinate
planes. The fact that any surface S bounded by a closed path C can be em-
ployed to evaluate the magnetic flux enclosed by C implies that the magnetic
flux through all such surfaces is the same in order for the emf around C to be
unique. As we shall learn in Section 2.4, it is a fundamental property of the
magnetic field that the magnetic flux is the same through all surfaces bounded
by a given closed path.
3. The closed path C on the left side need not represent a loop of wire,
but can be an imaginary contour. It means that the time-varying magnetic flux
induces an electric field in the region and this results in an emf around the
closed path. If a wire is placed in the position occupied by the closed path, the
emf will produce a current in the loop simply because the charges in the wire
are constrained to move along the wire.
4. The minus sign on the right side together with the right-hand screw Lenz’s law
rule ensures that Lenz’s law is always satisfied. Lenz’s law states that the sense
of the induced emf is such that any current it produces tends to oppose the
change in the magnetic flux producing it. It is important to note that the in-
duced emf acts to oppose the change in the flux and not the flux itself. To clari-
fy this, let us consider that the flux is into the paper and increasing with time.
Then the induced emf acts to produce flux out of the paper. On the other hand,
if the same flux is decreasing with time, then the induced emf acts to produce
flux into the paper.
5. If the loop C contains more than one turn, such as in an N-turn coil, Faraday’s law
then the surface S bounded by the periphery of the loop takes the shape of a for N-turn
spiral ramp, as shown in Fig. 2.13 (a) for N equal to 2. This surface can be visual- coil
ized by taking two paper plates, cutting each of them along a radius, as shown in
Figs. 2.13(b) and (c), and joining the edge BO of the plate in (c) to the edge A¿O
of the plate in (b). For a tightly wound coil, this is equivalent to the situation in
which N separate, identical, single-turn loops are stacked so that the emf in-
duced in the N-turn coil is N times that induced in one turn. Thus, for an N-turn
coil,
dc
emf = -N (2.14)
dt
where c is the magnetic flux computed as though the coil is a one-turn coil.
We shall now consider two examples to illustrate the determination of in-

duced emf using Faraday’s law, the first involving a stationary loop in a time-
varying magnetic field and the second involving a moving conductor in a static
magnetic field.
O
O O
B B
A A B
A A B
(a) (b) (c)
FIGURE 2.13
For illustrating the surface bounded by a loop containing two turns.
Example 2.3 Induced emf around a rectangular loop in a time-varying

magnetic field
Stationary A time-varying magnetic field is given by
loop in a
time-varying B = B0 cos vt ay
magnetic field
where B0 is a constant. It is desired to find the induced emf around the rectangular loop C
in the xz-plane bounded by the lines x = 0, x = a, z = 0, and z = b, as shown in Fig. 2.14.
Choosing dS = dx dz ay in accordance with the right-hand screw rule and using
the plane surface S bounded by the loop, we obtain the magnetic flux enclosed by the
loop to be
b a
c = B # dS = B0 cos vt ay # dx dz ay
LS Lz = 0 Lx = 0
b a
= B0 cos vt dx dz = abB0 cos vt
Lz = 0 Lx = 0
Note that since the magnetic flux density is uniform and normal to the plane of the loop, this
result could have been obtained by simply multiplying the area ab of the loop by the compo-
nent B0 cos vt of the flux density vector. The induced emf around the loop is then given by
d
E # dl = - B # dS
CC dt LS
d
= - [abB0 cos vt] = abB0v sin vt
dt
The time variations of the magnetic flux enclosed by the loop and the induced
emf around the loop are shown in Fig. 2.15. It can be seen that when the magnetic flux
enclosed by the loop into the paper is decreasing with time, the induced emf is positive,
thereby producing a clockwise current if the loop were a wire. This polarity of the cur-
rent gives rise to a magnetic field directed into the paper inside the loop and, hence, acts
to oppose the decrease of the magnetic flux enclosed by the loop. When the magnetic
flux enclosed by the loop into the paper is increasing with time, the induced emf is neg-
ative, thereby producing a counterclockwise current around the loop. This polarity of
the current gives rise to a magnetic field directed out of the paper inside the loop and
hence acts to oppose the increase of the magnetic flux enclosed by the loop. These ob-
servations are consistent with Lenz’s law.
y x 0
z
C
z 0 z=b
S dS
FIGURE 2.14 xa

B0 cos vt ay
Rectangular loop in the xz-plane situated in a
time-varying magnetic field. x
abB0
vt
0 p 2p 3p
emf
abB0 v
vt
0 p 2p 3p FIGURE 2.15
Time variations of magnetic flux c
enclosed by the loop of Fig. 2.14 and the
resulting induced emf around the loop.
Example 2.4 Induced emf around an expanding loop in a uniform static

magnetic field
A rectangular loop of wire with three sides fixed and the fourth side movable is situated Moving
in a plane perpendicular to a uniform magnetic field B = B0az, as illustrated in Fig. 2.16. conductor in
The movable side consists of a conducting bar moving with a velocity v0 in the y-direction. a static
It is desired to find the emf induced around the closed path C of the loop. magnetic field
Letting the position of the movable side at any time t be y0 + v0 t, considering
dS = dx dy az in accordance with the right-hand screw rule, and using the plane surface
S bounded by the loop, we obtain the magnetic flux enclosed by the loop to be
B # dS = B0az # dx dy az
LS LS
l y0 + v0 t
= B0 dx dy
Lx = 0 Ly = 0
= B0l1y0 + v0 t2
Note that this result could also have been obtained as the product of the area of the loop
l1y0 + v0t2 and the flux density B0, because of the uniformity of the flux density within
B
C
l x S dS v0ay
z FIGURE 2.16
y
Rectangular loop of wire with a movable side
situated in a uniform magnetic field.
the area of the loop and its perpendicularity to the plane of the loop. The emf induced
around C is given by
d
E # dl = - B # dS
CC dt LS
d
= - [B0 l1y0 + v0 t2]
dt
= -B0 lv0
Note that if the bar is moving to the right, the induced emf is negative and produces a
current in the sense opposite to that of C. This polarity of the current is such that it gives
rise to a magnetic field directed out of the paper inside the loop. The flux of this magnet-
ic field is in opposition to the flux of the original magnetic field and hence tends to
oppose the increase in the magnetic flux enclosed by the loop. On the other hand, if the
bar is moving to the left, v0 is negative, the induced emf is positive, and produces current
in the same sense as that of C. This polarity of current is such that it gives rise to a mag-
netic field directed into the paper inside the loop. The flux of this magnetic field is in aug-
mentation to the flux of the original magnetic field and hence tends to oppose the
decrease in the magnetic flux enclosed by the loop. These observations are once again
consistent with Lenz’s law.
It is also of interest to note that the induced emf can also be interpreted as being
due to the electric field induced in the moving bar by virtue of its motion perpendicular
to the magnetic field. Thus, a charge Q in the bar experiences a force F = Qv B or
Qv0 ay B0a z = Qv0 B0ax. To an observer moving with the bar, this force appears as an
electric force due to an electric field F>Q = v0B0a x. Viewed from inside the loop, this
electric field is in the counterclockwise sense. Hence, the induced emf, which is the line
integral of E along the bar, is given by
l l
v0B0 ax # dx ax = v0B0 dx = v0 B0 l
Lx = 0 Lx = 0
in the counterclockwise sense (i.e., opposite to C), consistent with the result deduced
from Faraday’s law. This concept of induced emf is known as the motional emf concept,
which is employed widely in the study of electromechanics.
In the two examples we just discussed, we have implicitly illustrated the

principles behind two of the practical applications of Faraday’s law. These are
pertinent to the reception of radio and TV signals using a loop antenna and
electromechanical energy conversion.
Principle of That the arrangement considered in Example 2.3 illustrates the principle
loop antenna of a loop antenna can be seen by noting that if the loop C were in the xy-plane
or in the yz-plane, no emf would be induced in it since the magnetic flux densi-
ty is then parallel to the plane of the loop and no flux is enclosed by the loop.
In fact, for any arbitrary orientation of the loop, only that component of B nor-
mal to the plane of the loop contributes to the magnetic flux enclosed by the
loop and, hence, to the emf induced in the loop. Thus, for a given magnetic field,
the voltage induced in the loop varies as the orientation of the loop is changed,
with the maximum occurring when the loop is in the plane perpendicular to the
magnetic field. Pocket AM radios generally contain a type of loop antenna con-
sisting of many turns of wire wound around a bar of magnetic material, and TV
receivers generally employ a single-turn circular loop for UHF channels.
Thus, for maximum signals to be received, the AM radios and the TV loop
antennas need to be oriented appropriately. Another point of interest evi-
dent from Example 2.3 is that the induced emf is proportional to v, the radian
frequency of the source of the magnetic field. Hence, for the same voltage to
be induced for a given amplitude B0 of the magnetic flux density, the area of
the loop times the number of turns is inversely proportional to the frequency.
What is undesirable for one purpose can sometimes be used to advantage Locating a
for another purpose. The fact that no voltage is induced in the loop antenna radio
when the magnetic field is parallel to the plane of the loop is useful for locating transmitter
the transmitter of a radio wave. Since the magnetic field of an incoming radio
wave is perpendicular to its direction of propagation, no voltage is induced in
the loop when its axis is along the direction of the transmitter. For a transmitter
on Earth’s surface, it is then sufficient to use two spaced vertical loop antennas
and find their orientations for which no signals are received. By then producing
backward along the axes of the two loop antennas, as shown by the top view in
Fig. 2.17, the location of the transmitter can be determined.
That the arrangement considered in Example 2.4 is a simple example of Electro-
an electromechanical energy converter can be seen by recognizing that in view mechanical
of the current flow in the moving bar, the bar is acted on by a magnetic force. energy
Since for positive v0, the current flows in the loop in the sense opposite to that conversion
of C and hence in the positive x-direction in the moving bar, and since the mag-
netic field is in the z-direction, the magnetic force is exerted in the ax a z or
-ay-direction. Thus, to keep the bar moving, an external force must be exerted
in the +a y-direction, thereby requiring mechanical work to be done by an ex-
ternal agent. It is this mechanical work that is converted into electrical energy in
the loop.
What we have just discussed is the principle of generation of electric power Principle of
by linear motion of a conductor in a magnetic field. Practical electric generators rotating
are of the rotating type. The principle of a rotating generator can be illustrated generator
by considering a rectangular loop of wire situated symmetrically about the z-axis
and rotating with angular velocity v around the z-axis in a constant magnetic
field B = B0a x, as shown in Fig. 2.18(a). Then noting from the view in Fig.
2.18(b) that the magnetic flux c enclosed by the loop at any arbitrary value of
time is the same as that enclosed by its projection onto the yz-plane at that time,
we obtain c = B0A cos vt, where A is the area of the loop and the situation
Loop 1
Transmitter
FIGURE 2.17
Top view of an arrangement consisting
of two loop antennas for locating a
Loop 2 transmitter of radio waves.
z
v
B B0ax
y
z
y B
x
(a) (b)
FIGURE 2.18
For illustrating the principle of a rotating generator.
shown in Fig. 2.18(a) is assumed for t = 0. The emf induced in the loop is
-dc/dt, or vB0 A sin vt. Thus, the rotating loop in the constant magnetic field
produces an alternating voltage. The same result can be achieved by a stationary
loop in a rotating magnetic field. In fact, in most generators, a stationary mem-
ber, or stator, carries the coils in which the voltage is induced, and a rotating
member, or rotor, provides the magnetic field. As in the case of the arrangement
of Example 2.4, a certain amount of mechanical work must be done to keep the
loop rotating. It is this mechanical work, which is supplied by the prime mover
(such as a turbine in the case of a hydroelectric generator or the engine of an au-
tomobile in the case of its alternator) turning the rotor, that is converted into
electrical energy.
Magnetic There are numerous other applications of Faraday’s law, but we shall dis-
levitation cuss only one more before we conclude this section. This is the phenomenon of
magnetic levitation, a basis for rapid transit systems employing trains that hover
over their guideways and do not touch the rail, among other applications. Mag-
netic levitation arises from a combination of Faraday’s law and Ampère’s force
law. It can be explained and demonstrated through a series of simple experi-
ments, culminating in a current-carrying coil lifting up above a metallic plate, as
described in the following:
1. Consider a pair of coils (30 to 50 turns of No. 26 wire of about 4-in. di-
ameter) attached to nails on a piece of wood, as shown in Fig. 2.19. Set to zero
the output of a variable power supply obtained by connecting a variac to the
110-V ac mains. Connect one output terminal (A) of the variac to the begin-
nings (C1 and C2) of both coils and the second output terminal (B) to the ends
(D1 and D2) of both coils so that currents flow in the two coils in the same sense.
Apply some voltage to the coils by turning up the variac and note the attraction
between the coils. Repeat the experiment by connecting A to C1 and D2 and B
to C2 and D1, so that currents in the two coils flow in opposite senses, and note
repulsion this time. What we have just described is Ampère’s force law at work.
If the currents flow in the same sense, the magnetic force is one of attraction,
C2 D2
C1 D1
Coil No. 2
Variac
Fuse Coil No. 1

A
B
110 V AC
FIGURE 2.19
Experimental setup for demonstration of Ampère’s force law, Faraday’s law, and the principle
of magnetic levitation.
and if the currents flow in opposite senses, it is one of repulsion, as shown in

Figs. 2.20(a) and (b), respectively, for straight wires, for the sake of simplicity.1
2. Connect coil No. 2 to the variac and coil No. 1 to an oscilloscope to ob-
serve the induced voltage in coil No. 1, thereby demonstrating Faraday’s law.
Note the change in the induced voltage as the variac voltage is changed. Note
also the change in the induced voltage by keeping the variac voltage constant
and moving coil No. 1 away from coil No. 2 and/or turning it about the vertical.
3. Connect coil No. 2 to the variac and leave coil No. 1 open-circuited. Ob-
serve that no action takes place as the variac voltage is applied to coil No. 2. This
is because although a voltage is induced in coil No. 1, no current flows in it.
I I I
FIGURE 2.20
B F F B F B B F
For explaining (a) force of attraction for
I currents flowing in the same sense and
(b) force of repulsion for currents flowing
(a) (b) in opposite senses.
1
See L. Pearce Williams, “André-Marie Ampère,” Scientific American, January 1989, pp. 90–97, for an
interesting account of Ampère’s experiments involving helical and spiral coils.
Aluminum
Plate
To 110 V AC
Through Coil
Variac
FIGURE 2.21
Setup for demonstrating magnetic
levitation.
Now short circuit coil No. 1 and repeat the experiment to note repulsion. This
is due to the induced voltage in coil No. 1 causing a current flow in it in the
sense opposite to that in coil No. 2, and, hence, is a result of the combination of
Faraday’s law and Ampère’s force law. That the force is one of repulsion can be
deduced by writing circuit equations and showing that the current in the short-
circuited coil does indeed flow in the sense opposite to that in the excited coil.
However, it can be explained with the aid of physical reasoning as follows.
When both coils are excited in the same sense in part (1) of the demonstration,
the magnetic flux linking each coil is the sum of two fluxes in the same sense,
due to the two currents. When the two coils are excited in opposite senses, the
magnetic flux linking each coil is the algebraic sum of two fluxes in opposing
senses, due to the two currents. Therefore, for the same source voltage and for
the same pair of coils, the currents that flow in the coils in the second case have
to be greater than those in the first case, for the induced voltage in each coil to
equal the applied voltage. Thus, the force of repulsion in the second case is
greater than the force of attraction in the first case. Consider now the case of
one of the coils excited by source voltage, say, Vg, and the other short-circuited.
Then the situation can be thought of as the first coil excited by Vg>2 and Vg>2 in
series, and the second coil excited by Vg>2 and -Vg>2 in series, thereby result-
ing in a force of attraction and a force of repulsion. Since the force of repulsion
is greater than the force of attraction, the net force, according to superposition,
is one of repulsion.
4. Now to demonstrate actual levitation, place a smaller coil (about 30
turns of No. 28 wire of about 2-in. diameter) on a heavy aluminum plate
15 in. * 5 in. * 12 in.2, as shown in Fig. 2.21. Applying only the minimum neces-
sary voltage and turning the variac only momentarily to avoid overheating, pass
current through the coil from the variac to see the coil levitate. This levitation is
due to the repulsive action between the current in the coil and the induced cur-
rents in the metallic plate. Since the plate is heavy and cannot move, the alter-
native is for the coil to lift up.
K2.3. Faraday’s law; Right-hand screw rule; Lenz’s law; Faraday’s law for N-turn coil;
Motional emf concept; Principle of loop antenna; Electromechanical energy
conversion; Rotating generator; Magnetic levitation.
D2.5. Given B = B01sin vt ax - cos vt ay2 Wb/m2, find the induced emf around each
of the following closed paths: (a) the rectangular path from (0, 0, 0) to (0, 1, 0) to
(0, 1, 1) to (0, 0, 1) to (0, 0, 0); (b) the triangular path from (1, 0, 0) to (0, 1, 0) to
2.4 Ampère’s Circuital Law 101
(0, 0, 1) to (1, 0, 0); and (c) the rectangular path from (0, 0, 0) to (1, 1, 0) to (1, 1, 1)
to (0, 0, 1) to (0, 0, 0).
vB0
Ans. (a) -vB0 cos vt V; (b) - cos1vt - p>42 V;
12
(c) - 12 vB0 cos1vt + p>42 V.
D2.6. A square loop lies in the xy-plane forming the closed path C connecting the
points (0, 0, 0), (1, 0, 0), (1, 1, 0), (0, 1, 0), and (0, 0, 0), in that order. A magnetic
field B exists in the region. From considerations of Lenz’s law, determine
whether the induced emf around the closed path C at t = 0 is positive, negative,
or zero for each of the following magnetic fields, where B0 is a positive constant:
(a) B = B0 taz; (b) B = B0 cos 12pt + 60°2 az; and (c) B = B0e -t a z.
2
Ans. (a) negative; (b) positive; (c) zero.

D2.7. For B = B0 cos vt az Wb/m2, find the induced emf around the following closed
paths: (a) the closed path comprising the straight lines successively connecting
the points (0, 0, 0), (1, 0, 0), (1, 1, 0), (0, 1, 0), (0, 0, 0.001), and (0, 0, 0); (b) the
closed path comprising the straight lines successively connecting the points
(0, 0, 0), (1, 0, 0), (1, 1, 0), (0, 1, 0), (0, 0, 0.001), (1, 0, 0.001), (1, 1, 0.001), (0, 1,
0.001), (0, 0, 0.002), and (0, 0, 0) with a slight kink in the straight line at the point
(0, 0, 0.001) to avoid touching the point; and (c) the closed path comprising the
helical path r = 1> 1p, f = 1000pz from 11> 1p, 0, 02 to 11> 1p, 0, 0.012 and
the straight-line path from 11> 1p, 0, 0.012 to 11> 1p, 0, 02 with slight kinks to
avoid touching the helical path.
Ans. (a) vB0 sin vt V; (b) 2vB0 sin vt V; (c) 5vB0 sin vt V.
2.4 AMPÈRE’S CIRCUITAL LAW

In the preceding section, we introduced Faraday’s law, one of Maxwell’s equa-
tions, in integral form. In this section, we introduce another of Maxwell’s
equations in integral form. This equation, known as Ampère’s circuital law, is a
combination of an experimental finding of Oersted that electric currents gen-
erate magnetic fields and a mathematical contribution of Maxwell that time-
varying electric fields give rise to magnetic fields. It is this contribution of
Maxwell that led to the prediction of electromagnetic wave propagation even
before the phenomenon was discovered experimentally. In mathematical
form, Ampère’s circuital law is analogous to Faraday’s law and is given by
B # d
dl = [Ic]S + e E # dS (2.15)
CC m0 dt LS 0
where S is a surface bounded by C.

B #
The quantity dl on the left side of (2.15) is the line integral of the
C 0
C m
vector field B/m0 around the closed path C. We learned in Section 2.1 that the
quantity AC E # dl has the physical meaning of work per unit charge associated
with the movement of a test charge around the closed path C. The quantity
B #
dl does not have a similar physical meaning. This is because magnetic
CC 0m
force on a moving charge is directed perpendicular to the direction of motion of
the charge, as well as to the direction of the magnetic field, and hence does not
do work in the movement of the charge. The vector B>m0 is known as the “mag-
netic field intensity vector” and is denoted by the symbol H. By recalling from
(1.78) that B has the units of [(permeability)(current)(length)] per [1distance22],
we note that the quantity H has the units of current per unit distance or amp/m.
This gives the units of current or amp to AC H # dl. In analogy with the name
“electromotive force” for AC E # dl, the quantity AC H # dl is known as the “mag-
netomotive force,” abbreviated as mmf.
The quantity [Ic]S on the right side of (2.15) is the current due to flow of
free charges crossing the surface S. It can be a convection current such as due to
motion of a charged cloud in space, or a conduction current due to motion of
charges in a conductor. Although [Ic]S can be filamentary current, surface cur-
rent, or volume current, or a combination of these, it is formulated in terms of
the volume current density vector, J, in the manner
[Ic]S = J # dS (2.16)
LS
Just as the surface integral of the magnetic flux density vector B 1Wb/m22 over
a surface S gives the magnetic flux (Wb) crossing that surface, the surface inte-
gral of J 1A/m22 over a surface S gives the current (A) crossing that surface.
The quantity 1S e0E # dS on the right side of (2.15) is the flux of the vector
field e0E crossing the surface S. The vector e0E is known as the “displacement
vector” or the “displacement flux density vector” and is denoted by the sym-
bol D. By recalling from (1.62) that E has the units of (charge) per [(permit-
tivity)1distance22], we note that the quantity D has the units of charge per unit
area or C>m2. Hence the quantity 1S e0E # dS, that is, the displacement flux,
d d
has the units of charge, and the quantity e E # dS has the units of (charge)
dt LS 0 dt
or current and is known as the “displacement current.” Physically, it is not a cur-
rent in the sense that it does not represent the flow of charges, but mathemati-
cally it is equivalent to a current crossing the surface S.
Statement of Replacing B>m0 and e0E in (2.15) by H and D, respectively, and using
Ampère’s (2.16), we rewrite Ampère’s circuital law as
circuital law
d
H # dl = J # dS + D # dS (2.17)
CC LS dt LS
In words, (2.17) states that the magnetomotive force around a closed path C is
equal to the algebraic sum of the current due to flow of charges and the displace-
ment current bounded by C. The situation is illustrated in Fig. 2.22.
J, D
C
dS
FIGURE 2.22
For illustrating Ampère’s
circuital law.
As in the case of Faraday’s law, there are certain procedures and observa-
tions pertinent to the application of (2.17). These are as follows.
1. The surface integrals on the right side of (2.17) are to be evaluated in accor-
dance with the R.H.S. rule, which means that for the geometry of Fig. 2.22,
the normal vector to the differential surface dS should be directed as shown
in the figure.
2. In evaluating the surface integrals, any surface S bounded by C can be em-
ployed. However, the same surface must be employed for the two surface
integrals. It is not correct to consider two different surfaces to evaluate the
two surface integrals, although both surfaces may be bounded by C.
Observation 2 implies that for the mmf around C to be unique, the sum of
the two currents (current due to flow of charges and displacement current)
through all possible surfaces bounded by C is the same. Let us now consider two
surfaces S1 and S2 bounded by the closed paths C1 and C2, respectively, as shown
in Fig. 2.23, where C1 and C2 are traversed in opposite senses and touch each
other so that S1 and S2 together form a closed surface. The situation may be
imagined by considering the closed surface to be that of a potato and C1 and C2
to be two rubber bands around the potato.
Applying Ampère’s circuital law to C1 and S1 and noting that dS1 is chosen
in accordance with the R.H.S. rule, we have
d
H # dl = J # dS1 + D # dS1 (2.18a)
CC1 LS1 dt LS1
dS1 dS2
FIGURE 2.23
C1 C2
Two closed paths C1 and C2 touching each
S1 S2
other and bounding the surfaces S1 and S2,
respectively, which together form a closed
surface.
Similarly, applying Ampère’s circuital law to C2 and S2 and noting again that dS2
is chosen in accordance with the R.H.S. rule, we have
d
H # dl = J # dS2 + D # dS2 (2.18b)
CC2 LS1 dt LS2
Now adding (2.18a) and (2.18b), we obtain
d
0 = J # dS + D # dS (2.19)
CS1 + S2 dt CS1 + S2
where the left side results from the fact that C1 and C2 are actually the same
path but traversed in opposite senses, so that the two line integrals are the neg-
atives of each other.
Since the closed surface S1 + S2 can be of any size and shape, we can gen-
eralize (2.19) to write
d
J # dS + D # dS = 0
CS dt CS
or
d
D # dS = - J # dS (2.20)
dt CS CS
Thus, the displacement current emanating from a closed surface is equal to the
current due to charges flowing into the volume bounded by that closed surface.
Capacitor An important example of the property given by (2.20) at work is in a ca-
circuit pacitor circuit, as shown in Fig. 2.24. In this circuit, the time-varying voltage
source sets up a time-varying electric field between the plates of the capacitor
Capacitor
Plates
S
I(t) I(t)
FIGURE 2.24
Capacitor circuit for illustrating that the
displacement current from one plate to the other is
equal to the wire current. V(t)
and directed from one plate to the other. Therefore, one can talk about dis-
placement current crossing a surface between the plates. According to (2.20) ap-
plied to a closed surface S enclosing one of the plates, as shown in the figure,
d
D # dS = I1t2 (2.21)
dt CS
where I(t) is the current (due to flow of charges in the wire) drawn from the
voltage source. Neglecting fringing effects and assuming that the electric field is
normal to the plates and uniform, we have, from (2.21),
1DA2 = I1t2
d d
D # dS = (2.22)
dt CS dt
where A is the area of each plate. Thus, where the wire current ends on one of
the plates, the displacement current takes over and completes the circuit to the
second plate.
Let us now return to Ampère’s circuital law (2.17) and examine it together Radiation
with Faraday’s law (2.13). To do this, we repeat the two laws from antenna
d
E # dl = - B # dS (2.23)
CC dt LS
d
H # dl = J # dS + D # dS (2.24)
CC LS dt LS
and observe that time-varying electric and magnetic fields are interdependent,
since according to Faraday’s law (2.23), a time-varying magnetic field produces
an electric field, whereas according to Ampère’s circuital law (2.24), a time-
varying electric field gives rise to a magnetic field. In addition, Ampère’s cir-
cuital law tells us that an electric current generates a magnetic field. These
properties from the basis for the phenomena of radiation and propagation of
electromagnetic waves. To provide a simplified, qualitative explanation of radi-
ation from an antenna, we begin with a piece of wire carrying a time-varying
current, I(t), as shown in Fig. 2.25. Then, the time-varying current generates a
time-varying magnetic field H(t), which surrounds the wire. Time-varying elec-
tric and magnetic fields, E(t) and H(t), are then produced in succession, as
shown by two views in Fig. 2.25, thereby giving rise to electromagnetic waves.
Thus, just as water waves are produced when a rock is thrown in a pool of water,
electromagnetic waves are radiated when a piece of wire in space is excited by a
time-varying current.
K2.4. Ampère’s circuital law; Magnetic field intensity; Magnetomotive force; Dis-
placement flux density; Displacement current; Capacitor circuit; Radiation from
an antenna.
I(t)
E
H
FIGURE 2.25
E
Two views of a simplified depiction of
electromagnetic wave radiation from a
piece of wire carrying a time-varying
current.
2
D2.8. For E = E0te -t az in free space, find the displacement current crossing an area
of 0.1 m2 in the xy-plane from the -z-side to the +z-side for each of the follow-
ing values of t: (a) t = 0; (b) t = 1> 12 s; and (c) t = 1 s.
Ans. (a) 0.1 e0 E0 A; (b) 0; (c) -0.1e -1e0E0 A.
D2.9. Three point charges Q11t2, Q21t2, and Q31t2 situated at the corners of an equi-
lateral triangle of sides 1 m are connected to each other by wires along the
sides of the triangle. Currents of I A and 3I A flow from Q1 to Q2 and Q1 to Q3,
respectively. The displacement current emanating from a spherical surface of ra-
dius 0.1 m and centered at Q2 is -2I A. Find the following: (a) the current flow-
ing from Q2 to Q3; (b) the displacement current emanating from the spherical
surface of radius 0.1 m and centered at Q1; and (c) the displacement current em-
anating from the spherical surface of radius 0.1 m and centered at Q3.
Ans. (a) 3I A; (b) -4I A; (c) 6I A.
2.5 GAUSS’ LAWS

In the previous two sections, we learned two of the four Maxwell’s equations.
These two equations have to do with the line integrals of the electric and mag-
netic fields around closed paths. The remaining two Maxwell’s equations are
pertinent to the surface integrals of the electric and magnetic fields over closed
surfaces. These are known as Gauss’ laws.
2.5 Gauss’ Laws 107
Gauss’ law for the electric field states that the displacement flux emanating Gauss’ law
from a closed surface S is equal to the charge contained within the volume V for the
bounded by that surface. This statement, although familiarly known as Gauss’ electric field
law, has its origin in experiments conducted by Faraday. In mathematical form,
it is given by
D # dS = [Q]V (2.25)
CS
The quantity [Q]V is the charge contained within the volume V bounded by S.
Although [Q]V can be a point charge, surface charge, or volume charge, or a
combination of these, it is formulated as the volume integral of the volume
charge density r, that is, in the manner
[Q]V = r dv (2.26)
LV
The volume integral is a triple integral since dv is the product of three dif- Evaluation of
ferential lengths. For an illustration of the evaluation of a volume integral, let us volume
consider integral
r = 1x + y + z2 C/m3
and the cubical volume V bounded by the planes x = 0, x = 1, y = 0, y = 1,

z = 0, and z = 1. Then the charge Q contained within the cubical volume is
given by
1 1 1
Q = r dv = 1x + y + z2 dx dy dz
LV Lx = 0 Ly = 0 Lz = 0
1 1
z2 1
= cxz + yz + d dx dy
Lx = 0 Ly = 0 2 z=0
1 1
ax + y + b dx dy
1
=
Lx = 0 Ly = 0 2
1
y2 y 1
= cxy + + d dx
Lx = 0 2 2 y=0
1
= 1x + 12 dx
Lx = 0
x2 1
= c + xd
2 x=0
3
= 2 C
We may now write Gauss’ law for the electric field (2.25) in the manner
D # dS = r dv (2.27)
CS LV
where we recall that

D = e0E
and it is understood that 1V r dv, although formulated in terms of the volume

charge density r, represents the algebraic sum of all free charges contained
within V. The situation is illustrated in Fig. 2.26.
Gauss’ law Gauss’ law for the magnetic field is analogous to Gauss’ law for the elec-
for the tric field and is given by
magnetic field
B # dS = 0 (2.28)
CS
In words, (2.28) states that the magnetic flux emanating from a closed surface is
equal to zero. In physical terms, (2.28) signifies that magnetic charges do not exist
and magnetic flux lines are closed. Whatever magnetic flux enters (or leaves) a
certain part of a closed surface must leave (or enter) through the remainder of
the closed surface, as illustrated in Fig. 2.27.
This property of the magnetic field is sometimes useful in the computation
of magnetic flux crossing a given surface (which is not closed). For example, to
find the magnetic flux crossing the slanted plane surface S1 in Fig. 2.28, it is not
necessary to evaluate formally the surface integral of B over that surface. Since
the slant surface S1 and the three surfaces S2, S3, and S4 in the coordinate planes
together form a closed surface, the required flux is the same as the net flux
crossing the surfaces S2, S3, and S4. In fact, the net flux crossing the surfaces
S2, S3, and S4 is the same as that crossing any nonplanar surface having the same
periphery as that of S1. Thus, as already pointed out in Section 2.3, it is a funda-
mental property of the magnetic field that the magnetic flux is the same through
dS
V S
FIGURE 2.26
For illustrating Gauss’ law
for the electric field.
2.5 Gauss’ Laws 109
dS
S
FIGURE 2.27
For illustrating Gauss’ law for the
magnetic field.
S1
S2
S3
S4 FIGURE 2.28
Slanted plane surface S1 and surface S2, S3,
x and S4 in the coordinate planes.
all surfaces bounded by a closed path, and hence any surface S bounded by
closed path C can be used in Faraday’s law.
In view of the foregoing discussion, it can be seen that Gauss’ law for the
magnetic field is not independent of Faraday’s law. To show this mathematical-
ly, we consider the geometry shown in Fig. 2.23 and apply Faraday’s law to the
two closed paths to write
d
E # dl = - B # dS1
CC1 dt LS1
d
E # dl = - B # dS2
CC2 dt LS2
Adding the two equations, we obtain
d
0 = - B # dS
dt CS1 + S2
or
B # dS = constant with time (2.29)

CS1 + S2
Since there is no experimental evidence that the right side of (2.29) is nonzero,
it follows that
B # dS = 0
CS
where we have replaced S1 + S2 by S.
K2.5. Volume integral; Gauss’ law for the electric field; Gauss’ law for the magnetic
field.
D2.10. Several types of charge are located, in Cartesian coordinates, as follows: a point
charge of 1 mC at 11, 1, -1.52, a line charge of uniform density 2 mC/m along the
straight line from 1-1, -1, -12 to (3, 3, 3), and a surface charge of uniform den-
sity -1 mC/m2 on that part of the plane x = 0 between z = -1 and z = 1. Find
the displacement flux emanating from each of the following closed surfaces:
(a) surface of the cubical box bounded by the planes x = ;2, y = ;2, and
z = ;2; (b) surface of the cylindrical box of radius 2 m, having the z-axis as its
axis and lying between z = -2 and z = 2; and (c) surface of the octahedron hav-
ing its vertices at (3, 0, 0), 1-3, 0, 02, 10, 3, 02, 10, -3, 02, (0, 0, 3), and 10, 0, -32.
Ans. (a) 3.3923 mC; (b) 1.3631 mC; (c) -3.0718 mC.
D2.11. Magnetic fluxes of absolute values c1, c2, and c3 cross three surfaces S1, S2, and
S3, respectively, constituting a closed surface S. If c1 + c2 + c3 = c0, find the
smallest of c1, c2, and c3 for each of the following cases: (a) c1, c2, and c3 are in
arithmetic progression; (b) 1>c1, 1>c2, and 1>c3 are in arithmetic progression;
and (c) ln c1, ln c2, and ln c3 are in arithmetic progression.
1 1 1
Ans. (a) c0; (b) c0; (c) c0.
6 2 + 212 3 + 15
2.6 THE LAW OF CONSERVATION OF CHARGE

Law of Just as Gauss’s law for magnetic field is not independent of Faraday’s law,
Conservation Gauss’ law for the electric field is not independent of Ampère’s circuital law in
of Charge view of the law of conservation of charge. The law of conservation of charge
states that the net current due to flow of charges emanating from a closed surface
S is equal to the time rate of decrease of the charge within the volume V bounded
by S. It is given in mathematical form by
d
J # dS = - r dv (2.30)
CS dt LV
As illustrated in Fig. 2.29, this law follows from the property that electric charge
is conserved. If the charge in a given volume is decreasing with time at a certain
rate, there must be a net outflow of the charge at the same rate. Since current is
defined to be the rate of flow of charge, (2.30) then follows. As in the case of
(2.17), it is understood that AS J # dS in (2.30), although formulated in terms of J,
represents the algebraic sum of all currents due to flow of charges crossing S.
2.6 The Law of Conservation of Charge 111
r(t)
dS
V S
FIGURE 2.29
For illustrating the law of
conservation of charge.
Comparing (2.20) and (2.30), we obtain
d d
D # dS = r dv
dt CS dt LV
a D # dS -
d
r dvb = 0 (2.31)
dt CS LV
D # dS - r dv = constant with time
CS LV
Since there is no experimental evidence that the right side of (2.31) is nonzero,
it follows that
D # dS = r dv
CS LV
Thus, since (2.20) follows from Ampère’s circuital law, Gauss’ law for the elec-
tric field follows from Ampère’s circuital law with the aid of the law of conser-
vation of charge.
We shall now illustrate the combined application of Gauss’ law for the
electric field, the law of conservation of charge, and Ampère’s circuital law by
means of an example.
Example 2.5 Combined application of several of Maxwell’s equations

in integral form
Let us consider current I A flowing from a point charge Q(t) at the origin to infinity
along a semi-infinitely long straight wire occupying the positive z-axis, and find AC H # dl,
where C is a circular path of radius a lying in the xy-plane and centered at the point
charge, as shown in Fig. 2.30.
Considering the hemispherical surface S bounded by C, and above the xy-plane, as
shown in Fig. 2.30, and applying Ampère’s circuital law, we obtain
d
H # dl = I + D # dS (2.32)
CC dt LS
a
y
Q(t)
FIGURE 2.30 C
Semi-infinitely long wire of current I, with a point
charge Q(t) at the origin. x
From Gauss’ law for the electric field, the displacement flux emanating from a spherical
surface centered at the point charge is equal to Q. In view of the spherical symmetry of
the electric field about the point charge, half of the flux goes through the hemispherical
surface. Thus,
Q
D # dS = (2.33)
LS 2
From the law of conservation of charge applied to a spherical surface centered at the
point charge,
dQ
I = - (2.34)
dt
Substituting (2.33) into (2.32) and then using (2.34), we obtain
a b
d Q
H # dl = I +
CC dt 2
1 dQ
= I +
2 dt
= I + 1-I2
1
2
I
=
2
It should be noted that the same result holds for any contour C lying in any plane pass-
ing through the origin and surrounding the point charge Q(t) and the wire in the right-
hand sense as seen looking along the positive z-axis.
K2.6. Law of conservation of charge.

D2.12. Three point charges Q11t2, Q21t2, and Q31t2 are situated at the vertices of a tri-
angle and are connected by means of wires carrying currents. A current I A
2.7 Application to Static Fields 113
flows from Q1 to Q2 and 3I A flows from Q2 to Q3. The charge Q3 is increasing

with time at the rate of 5I C/s. Find the following: (a) dQ1>dt; (b) dQ2>dt; and (c)
the current flowing from Q1 to Q3.
Ans. (a) -3I C>s; (b) -2I C>s; (c) 2I A.
2.7 APPLICATION TO STATIC FIELDS

Collecting together Faraday’s law (2.13), Ampere’s circuital law (2.17), Gauss’
law for the electric field (2.27), and Gauss’ law for the magnetic field (2.28), we
have the four Maxwell’s equations in integral form given by
d
E # dl = - B # dS (2.35a)
CC dt LS
d
H # dl = J # dS + D # dS (2.35b)
CC LS dt LS
D # dS = r dv (2.35c)
CS LV
B # dS = 0 (2.35d)
CS
whereas the law of conservation of charge is given by
d
J # dS = - r dv (2.36)
CS dt LV
For static fields, that is, for d>dt = 0, Maxwell’s equations in integral form Maxwell’s
become equations in
integral form
for static
E # dl = 0 (2.37a) fields
CC
H # dl = J # dS (2.37b)
CC LS
D # dS = r dv (2.37c)
CS LV
B # dS = 0 (2.37d)
CS
whereas the law of conservation of charge becomes
J # dS = 0 (2.38)
CS
It can be immediately seen from (2.37a)–(2.37d) that the interdependence

between the electric and magnetic fields no longer exists. Equation (2.37a) tells
us simply that the static electric field is a conservative field. Similarly, (2.37d)
tells us that the magnetic flux is the same through all surfaces bounded by a
closed path. On the other hand, (2.37c) and (2.37b) enable us to find the static
electric and magnetic fields for certain time-invariant charge and current distri-
butions, respectively. These distributions must be such that the resulting electric
and magnetic fields possess symmetry to be able to replace the integrals on the
left sides of (2.37c) and (2.37b) by algebraic expressions involving the compo-
nents of electric and magnetic fields, respectively.
In addition, in the case of (2.37b), the current on the right side must be
uniquely given for a given closed path C, which property is ensured by (2.38).
An example in which this current is uniquely given is that of the infinitely long
wire in Fig. 2.31(a). This is because the current crossing all possible surfaces
bounded by the closed path C is equal to I since the wire, being infinitely long,
pierces through all such surfaces. This can also be seen in a different manner by
imagining the closed path to be a rigid loop and visualizing that the loop can-
not be moved to one side of the wire without cutting the wire. On the other
hand, if the wire is finitely long, as shown in Fig. 2.31(b), it can be seen that for
some surfaces bounded by C, the wire pierces through the surface, whereas for
some other surfaces, it does not. Alternatively, a rigid loop occupying the
closed path can be moved to one side of the wire without cutting the wire. Thus,
for this case, there is no unique value of the wire current enclosed by C and
hence (2.37b) cannot be used to determine H. The problem here is that (2.38)
is not satisfied, since for current to flow in the finitely long wire, there must be
time-varying charges at the two ends, thereby giving rise to time-varying elec-
tric field. Hence, a displacement current exists in addition to the wire current
From I To
(a)
C
FIGURE 2.31
For illustrating that the current enclosed by a closed
path C is uniquely given in (a) but not in (b). (b)
such that the algebraic sum of the two currents crossing all surfaces bounded
by C is the same and requires the use of (2.17).
We shall now illustrate the application of (2.37c) and (2.37b) by means of
some examples.
Example 2.6 Electric field due to an infinitely long line charge using
Gauss’ law
Let us consider charge distributed uniformly with density rL0 C/m along the z-axis and D due to a
find the electric field due to the infinitely long line charge using (2.37c). line charge
Let us consider the closed surface S of a cylinder of radius r, with the line charge as
its axis and extending from z = 0 to z = l, as shown in Fig. 2.32. Then according to (2.37c),
D # dS = rL0 l (2.39)
CS
Although this result is valid for any closed surface enclosing the portion of the line
charge from z = 0 to z = l, we have chosen the particular surface in Fig. 2.32 to be able
to reduce the surface integral of D in (2.37c), and hence in (2.39), to an algebraic quanti-
ty. To do this, we note the following:
(a) In view of the uniform charge density, the entire line charge can be thought of as
the superposition of pairs of equal point charges located at equal distances above
and below any given point on the z-axis. Hence the field due to the entire line
charge has only a radial component independent of f and z.
(b) In view of (a), the contribution to the closed surface integral from the top and bot-
tom surfaces of the cylindrical box is zero.
Thus, we have
D = Dr1r2ar
Line
Charge
r df dz ar
y
r
f FIGURE 2.32
For the determination of electric field due to an
x infinitely long line charge of uniform density
rL0 C/m.
and
2p l
D # dS = Dr1r2ar # r df dz ar
CS Lf = 0 Lz = 0 (2.40)
= 2prlDr1r2
Comparing (2.39) and (2.40), we obtain
2prlDr1r2 = rL0 l
Dr1r2 =
rL0
2pr
rL0
D = a (2.41)
2pr r
The field varies inversely with the radial distance away from the line charge.
Example 2.7 Electric field due to a spherical volume charge using

Gauss’ law
D due to a Let us consider charge distributed uniformly with density r0 C/m3 in the spherical region
spherical r … a, as shown by the cross-sectional view in Fig. 2.33, and find the electric field due to
volume the spherical charge by using (2.37c).
charge As in Example 2.6, we shall once again choose a surface S that enables the re-
placement of the surface integral in (2.37c) by an algebraic quantity. To do this, we note
from considerations of symmetry, and of the spherical charge as a superposition of point
charges, that D possesses only an r-component dependent on r only. Thus,
D = Dr1r2ar
ra
x
a
FIGURE 2.33
For the determination of electric field due r a
to a spherical charge of uniform density
r0 C/m3.
Dr
1
3
r 0a
1
r 0a FIGURE 2.34
12
r Variation of Dr with r for the
0 a 2a 3a spherical charge of Fig. 2.33.
Choosing, then, a spherical surface of radius r centered at the origin, we obtain
p 2p
D # dS = Dr1r2ar # r2 sin u du df ar
CS Lu = 0 Lf = 0 (2.42)
= 4pr2Dr1r2
Noting that the charge exists only for r 6 a, and with uniform density, we obtain the
charge enclosed by the spherical surface to be
4
pr3r
r dv = e 43 3 0
for r … a
(2.43)
LV 3 pa r0 for r Ú a
Substituting (2.42) and (2.43) into (2.37c), we get
4
pr3r0
4pr2Dr1r2 = e 43 3
for r … a
3 pa r0 for r Ú a
r0 r
for r … a
Dr1r2 = d
3
r0a3
for r Ú a
3r2
r0 r
a for r … a
3 r
D = d (2.44)
r0a3
ar for r Ú a
3r2
The variation of Dr with r is shown plotted in Fig. 2.34.
Example 2.8 Magnetic field due to cylindrical wire of current using

Ampere’s circuital law
Let us consider current flowing with uniform density J = J0 a z A/m2 in an infinitely long H due to a
solid cylindrical wire of radius a with its axis along the z-axis, as shown by the cross- cylindrical
sectional view in Fig. 2.35. We wish to find the magnetic field everywhere using (2.37b). wire of current
f C
r a
x
a
FIGURE 2.35
For the determination of magnetic field due r a
to an infinitely long solid cylindrical wire of
uniform current density J0 a z A/m2.
The current distribution can be thought of as the superposition of infinitely long

filamentary wires parallel to the z-axis. Then in view of the symmetry about the z-axis
and from the nature of the magnetic field due to an infinitely long wire given by (1.79),
we can say that the required H has only a f component dependent on r only. Thus,
H = Hf1r2af
Choosing, then, a circular closed path C of radius r lying in the xy-plane and centered at
the origin, we obtain
2p
H # dl = Hf1r2af # r df af
CC Lf = 0 (2.45)
= 2prHf1r2
Considering the plane surface bounded by C, and noting that the current exists only for
r 6 a, we obtain the current enclosed by the closed path to be
r 2p
J0 az # r dr df az for r … a
Lr = 0 Lf = 0
J # dS = d a 2p
LS
J0 az # r dr df az for r Ú a
Lr = 0 Lf = 0 (2.46)
J0pr2
= e
for r … a
J0pa2 for r Ú a
Substituting (2.45) and (2.46) into (2.37b), we get
J0pr2
2prHf = e
for r … a
J0pa2 for r Ú a
J0r
for r … a
2
Hf = d
J0 a2
for r Ú a
2r
Summary 119
Hf
1 J0a
2
1 J0a FIGURE 2.36

4
Variation of Hf with r for the
r cylindrical wire of current of
0 a 2a 3a Fig. 2.35.
J0 r
a for r … a
2 f
H = d 2 (2.47)
J0a
a for r Ú a
2r f
The variation of Hf with r is shown plotted in Fig. 2.36.
K2.7. Maxwell’s equations in integral form for static fields; Uniqueness of current en-
closed by a closed path; D due to symmetrical charge distributions; H due to
symmetrical current distributions.
D2.13. Charge is distributed with uniform density r0 C>m3 inside a regular solid of edges
a. Find the displacement flux emanating from one side of the solid for each of the
following shapes of the solid: (a) tetrahedron; (b) cube; and (c) octahedron.
Ans. (a) 0.0295r0 a 3 C; (b) 0.1667r0a3 C; (c) 0.0589r0 a3 C.
D2.14. The cross section of an infinitely long solid wire having the z-axis as its axis is a
regular polygon of sides a. Current flows in the wire with uniform density
J0a z A/m2. Find the line integral of H along one side of the polygon and tra-
versed in the sense of increasing f for each of the following shapes of the poly-
gon: (a) equilateral triangle; (b) square; and (c) octagon.
Ans. (a) 0.1443a2J0 A; (b) 0.25a2J0 A; (c) 0.6036a2J0 A.
SUMMARY
We first learned in this chapter how to evaluate line and surface integrals of vec-
tor quantities, and then we introduced Maxwell’s equations in integral form.
These equations, which form the basis of electromagnetic field theory, are given
as follows in words and in mathematical form:
Faraday’s law. The electromotive force around a closed path C is equal to the
negative of the time rate of change of the magnetic flux enclosed by that path;
that is,
d
E # dl = - B # dS (2.48)
CC dt LS
Ampère’s circuital law. The magnetomotive force around a closed path C is

equal to the sum of the current enclosed by that path due to the actual flow of
charges and the displacement current due to the time rate of change of the dis-
placement flux enclosed by that path; that is,
d
H # dl = J # dS + D # dS (2.49)
CC LS dt LS
Gauss’ law for the electric field. The displacement flux emanating from a closed
surface S is equal to the charge enclosed by that surface; that is,
D # dS = r dv (2.50)
CS LV
Gauss’ law for the magnetic field. The magnetic flux emanating from a closed
surface S is equal to zero; that is,
B # dS = 0 (2.51)
CS
An auxiliary equation, the law of conservation of charge, is given by
d
J # dS = - r dv (2.52)
CS dt LV
In words, (2.52) states that the current due to flow of charges emanating from a
closed surface is equal to the time rate of decrease of the charge enclosed by
that surface.
In using (2.48)–(2.52), we recall that
D = e 0E (2.53)
B
H = (2.54)
m0
In evaluating the right sides of (2.48) and (2.49), the normal vectors to the sur-
faces must be chosen such that they are directed in the right-hand sense, that is,
toward the side of advance of a right-hand screw as it is turned around C. In
(2.50), (2.51), and (2.52), it is understood that the surface integrals are evaluated
so as to find the flux outward from the volume bounded by the surface. We also
learned that (2.51) is not independent of (2.48) and that (2.50) follows from
(2.49) with the aid of (2.52).
Finally, we discussed several applications of Maxwell’s equations, includ-
ing the computation of static electric and magnetic fields due to symmetrical
charge and current distributions, respectively.
Review Questions 121
REVIEW QUESTIONS
Q2.1. How do you find the work done in moving a test charge by an infinitesimal dis-
tance in an electric field? What is the amount of work involved in moving the
test charge normal to the electric field?
Q2.2. What is the physical interpretation of the line integral of E between two points
A and B?
Q2.3. How do you find the approximate value of the line integral of a vector field
along a given path? How do you find the exact value of the line integral?
Q2.4. Discuss conservative versus nonconservative fields, giving examples.
Q2.5. How do you find the magnetic flux crossing an infinitesimal surface?
Q2.6. What is the magnetic flux crossing an infinitesimal surface oriented parallel to
the magnetic flux density vector? For what orientation of the infinitesimal sur-
face relative to the magnetic flux density vector is the magnetic flux crossing the
surface a maximum?
Q2.7. How do you find the approximate value of the surface integral of a vector field
over a given surface? How do you find the exact value of the surface integral?
Q2.8. Provide physical interpretations for the closed surface integrals of any two vec-
tors of your choice.
Q2.9. State Faraday’s law.
Q2.10. What are the different ways in which an emf is induced around a loop?
Q2.11. Discuss the right-hand screw rule convention associated with the application of
Faraday’s law.
Q2.12. To find the induced emf around a planar loop, is it necessary to consider the
magnetic flux crossing the plane surface bounded by the loop? Explain.
Q2.13. What is Lenz’s law?
Q2.14. Discuss briefly the motional emf concept.
Q2.15. How would you orient a loop antenna to obtain maximum signal from an inci-
dent electromagnetic wave that has its magnetic field directed along the
north–south line?
Q2.16. State three applications of Faraday’s law.
Q2.17. State Ampère’s circuital law.
Q2.18. What is displacement current? Compare and contrast displacement current
with current due to flow of charges.
Q2.19. Is it meaningful to consider two different surfaces bounded by a closed path to
compute the two different currents on the right side of Ampère’s circuital law to
find A H # dl around the closed path?
Q2.20. Discuss the relationship between the displacement current emanating from a
closed surface and the current due to flow of charges emanating from the same
closed surface.
Q2.21. Give an example involving displacement current.
Q2.22. Discuss briefly the principle of radiation from a wire carrying a time-varying
current.
Q2.23. State Gauss’ law for the electric field.
Q2.24. How do you evaluate a volume integral?
Q2.25. State Gauss’ law for the magnetic field.
Q2.26. What is the physical interpretation of Gauss’ law for the magnetic field?
Q2.27. Discuss the dependence of Gauss’ law for the magnetic field on Faraday’s law.
Q2.28. State the law of conservation of charge.
Q2.29. How is Gauss’ law for the electric field dependent on Ampère’s circuital law?
Q2.30. Summarize Maxwell’s equations in integral form for time-varying fields.
Q2.31. Summarize Maxwell’s equations in integral form for static fields.
Q2.32. Are static electric and magnetic fields interdependent? Explain.
Q2.33. Discuss briefly the application of Gauss’ law for the electric field to determine
the electric field due to charge distributions.
Q2.34. When can you say that the current in a wire enclosed by a closed path is unique-
ly defined? Give two examples.
Q2.35. Give an example in which the current in a wire enclosed by a closed path is not
uniquely defined. Is it correct to apply Ampère’s circuital law for the static case
in such a situation? Explain.
Q2.36. Discuss briefly the application of Ampère’s circuital law to determine the mag-
netic field due to current distributions.
PROBLEMS
Section 2.1
P2.1. Evaluation of line integral in Cartesian coordinates. For the vector field F =
11, 1, 12
yax - zay + xaz, find 110, 0, 02 F # dl for each of the following paths from (0, 0, 0)
to (1, 1, 1): (a) x = y = z and (b) x = y = z3.
P2.2. Evaluation of line integral around a closed path in Cartesian coordinates. Given
F = xyax + yzay + zxaz, find AC F # dl, where C is the closed path compris-
ing the straight lines from (0, 0, 0) to (1, 1, 1), from (1, 1, 1) to (1, 1, 0), and
from (1, 1, 0) to (0, 0, 0).
P2.3. Evaluation of line integral in Cartesian coordinates. For the vector field F =
11, 2p, 12
cos y a x - x sin y ay, find 110, 0, 02 F # dl in each of the following ways: (a) along
the straight-line path between the two points; (b) along the curved path
x = z = sin 1y>42 between the two points; and (c) without choosing any partic-
ular path. Is the vector field conservative or nonconservative? Explain.
P2.4. Evaluation of line integral around closed path in cylindrical coordinates. Given
A = 2r sin f ar + r2af + zaz in cylindrical coordinates, find AC A # dl, where C
is the closed path comprising the straight line from (0, 0, 0) to (1, 0, 0), the circu-
lar arc from (1, 0, 0) to 11, p>2, 02 through 11, p>4, 02, the straight line from
11, p>2, 02 to 11, p>2, 12, and the straight line from 11, p>2, 12 to (0, 0, 0).
P2.5. Evaluation of line integral in spherical coordinates. Given A = e -r1cos u ar +
sin u au2 + r sin u af in spherical coordinates, find 1 A # dl for each of the fol-
lowing paths: (a) straight-line path from (0, 0, 0) to (2, 0, 0); (b) circular arc from
12, 0, p>42 to 12, p>2, p>42 through 12, p>4, p>42; and (c) circular arc from
12, p>6, 02 to 12, p>6, p>22 through 12, p>6, p>42.
Problems 123
Section 2.2
P2.6. Evaluation of a closed surface integral in Cartesian coordinates. Given A =
x2yzax + y2zxay + z2xyaz, evaluate AS A # dS, where S is the surface of the cubi-
cal box bounded by the planes x = 0, x = 1, y = 0, y = 1, z = 0, and z = 1.
P2.7. Evaluation of a closed surface integral in Cartesian coordinates. Given A =
1x2y + 22ax + 3ay - 2xyzaz, evaluate AS A # dS, where S is the surface of the
rectangular box bounded by the planes x = 0, x = 1, y = 0, y = 2, z = 0, and
z = 3.
P2.8. Evaluation of a closed surface integral in cylindrical coordinates. Given A =
r cos f ar - r sin f af in cylindrical coordinates, evaluate AS A # dS, where S is
the surface of the box bounded by the plane surfaces f = 0, f = p>2,
z = 0, z = 1, and the cylindrical surface r = 2, 0 6 f 6 p>2.
P2.9. Evaluation of a closed surface integral in spherical coordinates. Given A =
r2ar + r sin u au in spherical coordinates, find AS A # dS, where S is the surface
of that part of the spherical volume of radius unity and lying in the first octant.
Section 2.3
P2.10. Induced emf around a closed path in a time-varying magnetic field. Find the in-
duced emf around the rectangular closed path C connecting the points (0, 0, 0),
(a, 0, 0), (a, b, 0), (0, b, 0), and (0, 0, 0), in that order, for each of the following
magnetic fields:
B0a2
1x + a22
(a) B = e -taz
px
(b) B = B0 sin cos vt az
a
P2.11. Induced emf around a moving loop in a static magnetic field. A magnetic field is
given in the xz-plane by B = 1B0>x2ay Wb/m2, where B0 is a constant. A rigid
rectangular loop is situated in the xz-plane and with its corners at the points
1x0, z02, 1x0, z0 + b2, 1x0 + a, z0 + b2, and 1x0 + a, z02. If the loop is moving in
that plane with a velocity v = v0ax m/s, where v0 is a constant, find by using Fara-
day’s law the induced emf around the loop in the sense defined by connecting the
above points in succession. Discuss your result by using the motional emf concept.
P2.12. Induced emf around a closed path in a time-varying magnetic field. A magnet-
ic field is given in the xz-plane by B = B0 cos p1x - v0t2 ay Wb/m2. Consider a
rigid square loop situated in the xz-plane with its vertices at (x, 0, 1), (x, 0, 2),
1x + 1, 0, 22, and 1x + 1, 0, 12. (a) Find the expression for the emf induced
around the loop in the sense defined by connecting the above points in succes-
sion. (b) What would be the induced emf if the loop is moving with the velocity
v = v0a x m/s instead of being stationary?
P2.13. Induced emf around a swinging loop in a static magnetic field. A rigid rectan-
gular loop of metallic wire is hung by pivoting one side along the x-axis, as
shown in Fig. 2.37. The loop is free to swing about the pivoted side without fric-
tion under the influence of gravity and in the presence of a uniform magnetic
field B = B0a z Wb/m2. If the loop is given a slight angular displacement and re-
leased, show that the emf induced around the closed path C of the loop is ap-
proximately equal to B0 abv, where v is the angular velocity of swing of the loop
y
a
x a g
b
C
FIGURE 2.37
For Problem P2.13. v
toward the vertical. Does the loop swing faster or slower than in the absence of
the magnetic field? Explain.
P2.14. A conducting bar rolling down inclined rails in a uniform static magnetic field.
A rigid conducting bar of length L, mass M, and electrical resistance R rolls
without friction down two parallel conducting rails that are inclined at an angle
a with the horizontal, as shown in Fig. 2.38. The rails are of negligible resistance
and are joined at the bottom by another conductor, also of negligible resistance,
so that the total resistance of the loop formed by the rolling bar and the three
other sides is R. The entire arrangement is situated in a region of uniform static
magnetic field B = B0az Wb/m2, directed vertically downward. Assume the bar
to be rolling down with uniform velocity v parallel to the rails under the influ-
ence of Earth’s gravity (acting in the positive z-direction) and the magnetic
force due to the current in the loop produced by the induced emf. Show that v is
equal to 1MgR>B20L22 tan a sec a.
g L
v
g
a y
FIGURE 2.38 x
For Problem P2.14. z
P2.15. Induced emf around a revolving loop in a static magnetic field. A rigid rectan-
gular loop of base b and height h situated normal to the xy-plane and with its
sides pivoted to the z-axis revolves about the z-axis with angular velocity
v rad/s in the sense of increasing f, as shown in Fig. 2.39. Find the induced emf
around the closed path C of the loop for each of the following magnetic fields:
(a) B = B0ay Wb/m2 and (b) B = B01yax - xay2 Wb/m2. Assume the loop to
be in the xz-plane at t = 0.
Problems 125
z
v
h
C
y
f
b FIGURE 2.39
x For Problem P2.15.
P2.16. Induced emf around a loop in a time-varying magnetic field for several cases.
A rigid rectangular loop of area A is situated in the xz-plane and symmetri-
cally about the z-axis, as shown in Fig. 2.40, in a region of magnetic field
B = B01sin vt ax + cos vt ay2 Wb/m2. Find the induced emf around the closed
path C of the loop for each of the following cases: (a) the loop is stationary;
(b) the loop revolves around the z-axis in the sense of increasing f with uni-
form angular velocity of v rad/s; and (c) the loop revolves around the z-axis in
the sense of decreasing f with uniform angular velocity of v rad/s. For parts
(b) and (c), assume that the loop is in the xz-plane at t = 0.
A
y
f
FIGURE 2.40
x For Problem P2.16.
Section 2.4
P2.17. Application of Ampere’s circuital law in integral form. Given that H =
;H01t < 2m0e0 z22a y and D = 2m0e0 H01t < 2m0e0 z22a x for z
0, find
the current due to flow of charges enclosed by the rectangular closed path from
(0, 0, 1) to (0, 1, 1) to 10, 1, -12 to 10, 0, -12 to (0, 0, 1).
P2.18. Application of Ampere’s circuital law in integral form. A current density due to
flow of charges is given by J = -1xax + yay + z2a z2 A/m2. Find the displace-
ment current emanating from each of the following closed surfaces: (a) the sur-
face of the cubical box bounded by the planes x = ;2, y = ;2, and z = ;2,
and (b) the surface of the cylindrical box bounded by the surfaces r = 1, z = 0,
and z = 2.
P2.19. Finding rms value of current drawn from voltage source connected to a capaci-
tor. A voltage source connected to a parallel-plate capacitor by means of wires
sets up a uniform electric field of E = 180 sin 2p * 106t sin 4p * 106t V/m be-
tween the plates of the capacitor and normal to the plates. Assume that no field
exists outside the region between the plates. If the area of each plate is 0.1 m2
and the medium between the plates is free space, find the root-mean-square
value of the current drawn from the voltage source.
P2.20. Finding rms value of current drawn from voltage source connected to a capaci-
tor. Assume that the time variation of the electric field in Problem P3.19 is as
shown in Fig. 2.41. Find and plot versus time the current drawn from the voltage
source. What is the root-mean-square value of the current?
E, V/m
180
FIGURE 2.41
–3 –2 –1 0 1 2 3 4 5 6 t, s For Problem P2.20.
Section 2.5
P2.21. Finding displacement flux emanating from a surface enclosing charge. For each
of the following charge distributions, find the displacement flux emanating from the
surface enclosing the charge: (a) r1x, y, z2 = r013 - x2 - y2 - z22 for the cubi-
cal box bounded by x = ;1, y = ;1, and z = ;1; and (b) r1x, y, z2 = r01xyz2
for x 7 0, y 7 0, z 7 0, and x2 + y2 + z2 6 1.
P2.22. Finding displacement flux emanating from a surface enclosing charge. For each
of the following charge distributions, find the displacement flux emanating from
2
the surface enclosing the charge: (a) r1r, f, z2 = r0 e -r for r 6 1, 0 6 z 6 1 in
cylindrical coordinates; and (b) r1r, u, f2 = 1r0>r2 sin u for r 6 1, 0 6 u 6 p>2
2
in spherical coordinates.
P2.23. Application of Gauss’ law for the magnetic field in integral form. Using the
property that AS B # dS = 0, find the absolute value of the magnetic flux crossing
that portion of the surface y = sin x bounded by x = 0, x = p, z = 0, and z = 1
for B = B01yax - xay2 Wb/m2.
Section 2.6
P2.24. Application of the law of conservation of charge. Given J = 1xax + yay +
zaz2 A/m2, find the time rate of decrease of the charge contained within each of
the following volumes: (a) volume bounded by the planes x = 0, x = 1, y = 0,
y = 1, z = 0, and z = 1; (b) volume bounded by the cylinders r = 1 and r = 2
and the planes z = 0 and z = 1; and (c) volume bounded by the spherical sur-
faces r = 1 and r = 2 and the conical surface u = p>3.
P2.25. Combined application of several of Maxwell’s equations in integral form. Cur-
rent I flows along a straight wire from a point charge Q11t2 located at the origin
to a point charge Q21t2 located at (0, 0, 1). Find the line integral of H along the
square closed path having the vertices at (1, 1, 0), 1-1, 1, 02, 1-1, -1, 02, and
11, -1, 02 and traversed in that order.
P2.26. Combined application of several of Maxwell’s equations in integral form. Cur-
rent I flows along a straight wire from a point charge Q11t2 at the origin to a
point charge Q21t2 at the point (2, 2, 2). Find the line integral of H around the
Review Problems 127
triangular closed path having the vertices at (3, 0, 0), (0, 3, 0), and (0, 0, 3) and
traversed in that order.
Section 2.7
P2.27. Application of Gauss’ law for the electric field in integral form and symmetry.
Charge is distributed with density r1x, y, z2 in a cubical box bounded by the
planes x = ;1 m, y = ;1 m, and z = ;1 m. Find the displacement flux ema-
nating from one side of the box for each of the following cases: (a) r1x, y, z2 =
13 - x2 - y2 - z22 C/m3 and (b) r1x, y, z2 = 2 ƒ xyz ƒ C/m3.
P2.28. Electric field due to a cylindrical charge distribution using Gauss’ law. Charge
2
is distributed with density r0e -r C/m3 in the cylindrical region r 6 1. Find D
everywhere.
P2.29. Electric field due to a spherical charge distribution using Gauss’ law. Charge is
distributed with uniform density r0C/m3 in the region a 6 r 6 2a in spherical
coordinates. Find D everywhere and plot Dr versus r.
P2.30. Application of Ampere’s circuital law in integral form and symmetry. Current
flows with density J(x, y) in an infinitely long thick wire having the z-axis as its
axis. The cross section of the wire in the xy-plane is the square bounded by
x = ;1 m and y = ;1 m. Find the line integral of H along one side of the
square and traversed in the sense of increasing f for each of the following cases:
(a) J1x, y2 = 1 ƒ x ƒ + ƒ y ƒ 2az A/m2 and (b) J1x, y2 = x2y 2az A/m2.
P2.31. Magnetic field due to a solid wire of current using Ampere’s circuital law. Cur-
rent flows with density J = J01r>a2az A/m2 along an infinitely long solid cylin-
drical wire of radius a having the z-axis as its axis. Find H everywhere and plot
Hf versus r.
P2.32. Magnetic field for a coaxial cable using Ampere’s circuital law. A coaxial cable
consists of an inner conductor of radius 3a and an outer conductor of inner ra-
dius 4a and outer radius 5a. Assume the cable to be infinitely long and its axis to
be along the z-axis. Current I flows with uniform density in the +z-direction in
the inner conductor and returns with uniform density in the -z-direction in the
outer conductor. Find H everywhere and plot Hf versus r.
REVIEW PROBLEMS
R2.1. Determination of a specified static vector field to be a conservative field. Show
that the vector field given by
F = cos u sin f ar - sin u sin f au + cot u cos u af
is a conservative field. Then find the value of 1 F # dI from the point 11, p>6, p>32
to the point 14, p>3, p>62.
R2.2. Induced emf around an expanding loop in a nonuniform static magnetic field. In
Fig. 2.42, a rectangular loop of wire with three sides fixed and the fourth side
movable is situated in a plane perpendicular to a nonuniform magnetic field
B = B0yaz Wb/m2, where B0 is a constant. The position of the movable side is
varied with time in the manner y = y0 + a cos vt, where a 6 y0. Find the in-
duced emf around the closed path C of the loop.Verify that Lenz’s law is satisfied.
Show also that the induced emf consists of two frequency components, v and 2v.
B
l
x
z
FIGURE 2.42 y
For Problem R2.2.
R2.3. Finding amplitude of current from sinusoidal voltage source connected to a ca-
pacitor. A voltage source is connected by means of wires to a parallel-plate ca-
pacitor made up of circular plates of radii a in the z = 0 and z = d planes and
having their centers on the z-axis.The electric field between the plates is given by
pr
E = E0 sin cos vt az for r 6 a
2a
Find the amplitude of the current drawn from the voltage source, assuming the
region between the plates to be free space and that no field exists outside this
region.
R2.4. Combined application of several of Maxwell’s equations in integral form. Cur-
rent I flows along a straight wire from a point charge Q11t2 located at one of the
vertices of a cube to a point charge Q21t2 at the center of the cube. Find the ab-
solute value of the line integral of H around the periphery of one of the three
sides of the cube not containing the vertex at which Q1 is located.
R2.5. Electric field due to a spherical charge distribution using Gauss’ law. Charge is
distributed with density r = r01r>a22, where r0 is a constant, in the spherical re-
gion r 6 a. Find D everywhere and plot Dr versus r.
R2.6. Magnetic field in the hollow region of wire bounded by two parallel cylindrical
surfaces. Current flows axially with uniform density J0 A/m2 in the region be-
tween two infinitely long parallel, cylindrical surfaces of radii a and b 16a2,
and with their axes separated by the vector distance c, where ƒ c ƒ 6 1a - b2.
Find the magnetic field intensity in the current-free region inside the cylindri-
cal surface of radius b.

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qxd 12/18/03 3:08 PM Page 77

In Chapter 1, we learned the simple rules of vector algebra and familiarized

2.1 THE LINE INTEGRAL

78 Chapter 2 Maxwell’s Equations in Integral Form

¢Wj = qEj cos aj ¢lj (2.1)

WAB = ¢W1 + ¢W2 + ¢W3 + Á + ¢Wn

2.1 The Line Integral 79

Using the dot product operation between two vectors, we obtain

For a numerical example, let us consider the electric field given by

¢l j = 0.1a x + [j2 - 1j - 122]0.01a y = 0.1a x + 12j - 120.01a y

The required work is then given by

80 Chapter 2 Maxwell’s Equations in Integral Form

= 3 * 10 [0 + 3 + 20 + 63 + 144 + 275 + 468 + 735

so that the differential length vector tangential to the curve is given by

The value of E # dl at the point is

Thus, the required work is given by

2.1 The Line Integral 81

E # dl = yay # 1dx a x + dy ay2

82 Chapter 2 Maxwell’s Equations in Integral Form

Example 2.1 Evaluation of line integral around a closed path

For the side BC,

For the side CD,

2.1 The Line Integral 83

For the side DA,

84 Chapter 2 Maxwell’s Equations in Integral Form

2.2 THE SURFACE INTEGRAL

(a) (b) (c)

2.2 The Surface Integral 85

component of B normal to the surface is B cos a and the component tangential

¢cj = Bj cos aj ¢Sj (2.7)

86 Chapter 2 Maxwell’s Equations in Integral Form

[c]S = ¢c1 + ¢c2 + ¢c3 + Á + ¢cn

Using the dot product operation between two vectors, we obtain

¢Sj = ¢Sj a nj (2.10)

we can write (2.9) as

For a numerical example, let us consider the magnetic field given by

Bij = 312i - 1212j - 1220.001a z

¢Sij = 0.04 a z for all i and j

2.2 The Surface Integral 87

The required magnetic flux is then given by

88 Chapter 2 Maxwell’s Equations in Integral Form

The value of B # dS at the point is

Thus, the required magnetic flux is given by

Example 2.2 Evaluation of a closed surface integral

B = 1x + 22ax + 11 - 3y2ay + 2zaz

as shown in Fig. 2.9.

2.2 The Surface Integral 89

x = 0, B = 2ax + 11 - 3y2ay + 2zaz, dS = -dy dz ax

For the surface efgh,

x = 1, B = 3ax + 11 - 3y2ay + 2zaz, dS = dy dz ax

For the surface adhe,

y = 0, B = 1x + 22ax + 1ay + 2zaz, dS = -dz dx ay

For the surface bcgf,

y = 1, B = 1x + 22ax - 2ay + 2zaz, dS = dz dx ay

90 Chapter 2 Maxwell’s Equations in Integral Form

For the surface aefb,

z = 0, B = 1x + 22ax + 11 - 3y2ay + 0az, dS = -dx dy az

For the surface dhgc,