sin Q, = AB X
X
+ AB. y
y
(112)
(113)
6 1.2.1
Vectors EXAMPLE:
WORK
DONE BY A
FORCE
A simple physical example of the scalar product is the work clone by a force F acting through a displacement s: W = F • s, as in Fig. 14.
Figure 14 The work done by a force F whose point of application moves by a distance s is (F cos
1.3
VECTOR
PRODUCT
The vector, or cross product, of two vectors is a vector whose direction is perpendicular to the plane containing the two initial vectors and whose magnitude is the product of the magnitudes of those vectors and the sine of the angle between them. We indicate the vector product thus: A x B = C.
(114)
C = \AB sin (cp  0),
(115)
The magnitude of C is
with
1.3 Vector Product
7
rotates the firstnamed vector (A) into the secondnamed (B) through the smaller angle. The commutative rule is not followed for the vector product, since inverting the order of A and B inverts the direction of C: (116)
A x B =  ( B x A). The distributive rule, however, is followed for any three vectors: A x (B + C) = (A x B) + (A x C).
(117)
This will be shown in Prob. 17. From the definition of the vector product it follows that i x i = 0,
j x j = 0,
k x k
(118)
= 0,
and, for the usual righthanded coordinate systems, such as that of Fig. 12, i x j = k,
j x k = i,
k x i = j ,
j x i = — k , and so on.
(119)
Writing out the vector product of A and B in terms of the components, A x B = (A i + A ] + A±) x
y
= (A B y
i A
x
B
x
z
 A B )\ z
y
j
k
A
y
A
B
B
y
x (B \ + B \ x
y
+ (A._B  A B )\ X
X
Z
+ B k),
(120)
z
+ (A B x
y
 A B )k, y
x
(121)
(122)
:
:
We can check this result for the two vectors of Fig. 13 by expanding sin ((p — 9) and noting that the vector product is in the positive z direction.
1.3.1
EXAMPLES:
TORQUE,
AREA
OF A
PARALLELOGRAM
A good physical example of the vector product is the torque T produced by a force F acting with a moment arm r about a point O, as in Fig. 15, where T = r x F. A second example is the area of a parallelogram, as in Fig. 16, where the area S = A x B. The area is thus represented by a vector perpendicular to the surface.
Figure 16 The area of the parallelogram is A x B = S. The vector S is normal to the parallelogram.
THE TIME
DERIVATIVE
We shall often be concerned with the rates of change of scalar and vector quantities with both time and space coordinates, and thus with the time and space derivatives.
9
Figure 17 A vector A, its increment AA, and their components.
The time derivative of a vector quantity is straightforward. In a time Af, a vector A, as in Fig. 17, may change by AA, which in general represents a change both in magnitude and in direction. O n dividing AA by At and taking the limit in the usual way, we arrive at the definition of the
time derivative dA/dt: dA dt
=
H
m
A,^O
Ajt + AO  A(r) At
= Hm ^ i
+ A^j + A ^ Af
Af^O
dt
dt
(
1
2
4
)
tit
The time derivative of a vector is thus equal to the vector sum of the time derivatives of its components.
1.4.1
EXAMPLES:
POSITION.
VELOCITY.
AND
ACCELERATION
The time derivative of the position r of a point is its velocity v, and the time derivative of v is the acceleration a. See Prob. 110.
10
THE
Vectors
GRADIENT
Let us consider a scalar quantity that is a continuous and differentiable function of the coordinates and has the value /' at a certain point in space, as in Fig. 18. For example, / could be the electrostatic potential V. We wish to know how / changes over the distance dl measured from that point. Now
df = % dx + 4 dy = A • dl. cx oy
(126)
where A = % i + % j , < .\cy
and dl = dx i + dy j .
(127)
The vector A, whose components are the rates of change of / with distance along the coordinate axes, is called the gradient of the scalar quantity / . The operation on the scalar / defined by the term gradient is indicated by the symbol V, called "del". Thus A = \f.
Figure 18 The quantity /, a function of position, changes from / to /' + df over the distance dl.
(128)
1.5 The Gradient
11
In three dimensions, . .
V =
.
.3 , C + j — + k—. dy dz
d
C.Y
d29)
The partial differentiations indicated are carried out on whatever scalar quantity stands to the right of the V symbol. Thus
ox
dy
(130)
dz
and
IVI
+
+
df
1/2
d31)
Thus, df = V/ • dl = V/' dl cos 9,
(132)
where 9 is the angle between the vectors V / and dl. We now ask what direction one should choose for dl in order that df be maximum. The answer is: the direction in which 9 = 0, that is, the direction of Vf. The gradient of / is thus a vector whose magnitude and direction are those of the maximum space rate of change of / . The gradient of a scalar function at a given point is a vector having the following properties: 1) Its components are the rates of change of the function along the directions of the coordinate axes. 2) Its magnitude is the maximum rate of change of the function with distance. 3) Its direction is that of the maximum rate of change of the function. 4) It points toward larger values of the function. The gradient is a vector pointfunction derived from a scalar pointfunction.
12
1.5.1
Vectors
EXAMPLES:
TOPOGRAPHIC
MAP. ELECTRIC
FIELD
INTENSITY
Fig. 19 shows the gradient of the elevation on a topographic map. In the next chapter we shall see that, in an electrostatic field, the electric field intensity E is equal to minus the gradient of the electric potential V: E = — W.
Figure 19 Topographic m a p of a hill. The numbers shown give the elevation E in meters. The gradient of E is the slope of the hill at the point considered, and it points toward an mcrease in elevation: V£ = (dE/dx)i + (dE/dy)\. The arrow shows \E at one point where the elevation is 400 meters.
1.6
THE SURFACE
INTEGRAL
Consider the xy plane. T o locate a point in this plane, one requires two coordinates, x and y. N o w consider a given spherical surface, as in Fig. 110.
13 z
y x
Figure 110 Spherical surface of radius r. The two coordinates 8 and q> suffice to determine the position of a point P on the surface. How do these coordinates compare with the latitude and the longitude at the surface of the earth?
T o fix a point we again need two coordinates, 0 and q>. Of course we could use the three coordinates x, y, z. But those three coordinates are redundant, on the given sphere, since x + y + z = r . So we only need two coordinates, 0 and (p. 2
2
2
2
This is a general rule: one always needs to specify two coordinates to identify a point on a given surface. If we require the area of the given surface, we must integrate the element of area, say dx dy, over both variables. We then have a surface integral. More generally, an integral evaluated over two variables (that are not necessarily space coordinates) is called a double integral. Suppose the surface carries a charge. Then, to find the total charge, we must integrate the surface charge density, multiplied by the element of area, over both variables. So, a surface integral is a double integral evaluated over the two coordinates that are required to fix a point on a given surface. If the surface is situated in the xy plane, the surface integral is of the form
14
Vectors
where f{x,y) is a function of x and y, and where the limits a and b on x may be functions of y. The limits c and d are constants. We first evaluate the integral between the braces. This gives either a constant or a function of y. Then this constant, or function, is integrated over y. The double integral is therefore an integral within an integral. As we shall see in the next example, one arrives at the same result by inverting the order of integration and writing the double integral in the form
where the limits of integration are different: the limits c' and a" can now be functions of x, while a' and b' are constants. We have shown braces in the above expressions so as to indicate the manner in which a double integral is evaluated, but these braces are never used in practice.
EXAMPLE:
AREA
OF A
CIRCLE
It is always advisable to test one's skill at understanding and using new concepts by applying them to extremely simple cases. So, let us calculate the area of a circle by means of a surface integral. So as to simplify the calculation, we shall find the area of the top righthand sector of the circle of Fig. 111, and then multiply by 4. The element of area dA is dx dy. We proceed as follows. We first slide dA from left to right, to generate the shaded strip in the figure. Then we slide the strip from y = 0 to y = R, while adjusting its length correctly. This generates the 90degree sector. Let us first find an integral for the strip. The strip has a height dy. It starts at x = 0 and ends at the edge of the circle, where x
2
+ v =
R.
2
1
(133)
Thus, at the righthand end of the strip. X
and the strip has an area
= (R* 
v) 2
(134)
15
Figure 111 (a) The infinitesimal element of area tlx dy sweeps from the vaxis to the periphery of the circle to generate the shaded strip, (b) The strip now sweeps from the .vaxis to the top of the circle to generate the topright quadrant.
The total area A of the circle is given by A 4
crfr""*}*
(135)
(136) Setting y/R = sin 8, then cos 0.
dy = R cos 0 dO.
(137)
Also, when y = 0, 0 = 0, and when y = R, 8 = n/2. Then — = R 4 J"
R c o s 8d8 =  R , 4 2
2
/I = TtR ,
(138) (139)
2
as expected. If we used a vertical strip, it would extend from y = 0
to
y = (R
2

x) . 2 112
(140)
16
Vectors
Then we would have that A
=
4
,,
.
P
( / V ( / A
Jv = o J, = o
,
(141)
'
= \ l (R  x ) ' dx. 2
2
1
(142)
2
This integral is similar to that of Eq. 136 and the area of the circle is again
EXAMPLE:
ELECTRICALLY
CHARGED
nR . 2
DISK
If we have an electrically charged disk with a surface charge density a given by, say, a = Ky .
(143)
2
then, using a vertical strip and integrating over the complete circle, the total charge is (144)
J
+
R i " v ^ + (R * )" ") K — 1 _ 3_ 3
2
2
2
(145)
(146) Using a table of integrals/ we find that Q = 4
v
1.7
THE VOLUME
KR . 4
(147)
INTEGRAL
In a volume integral, there are three variables, say x, y, z in Cartesian coordinates, or r, 0, (p in polar coordinates (Fig. 110). In the m o r e general case where the three variables can be of any nature, say velocities, or voltages, etc., one has a triple
integral.
Here again, one integrates with respect t o each variable in succession, starting with the innermost integral. * See, for example, Dwight, Table of Integrals and Other Mathematical
Data, No. 350.03.
1.7 The Volume Integral
1.7.1
EXAMPLE:
VOLUME
OF A
17
SPHERE
T o calculate the volume of a sphere, we first find the element of volume. Referring to Fig. 112, this is (rdO)(r sin 9 dq>){dr). Then
Let us integrate, first with respect to cp, then with respect to 0, and finally with respect to r. This means that we first rotate the element of volume about the vertical axis to generate a ring. Then we slide the ring from 9 = 0 to 8 = n, while keeping the radius r and dr constant. This generates a spherical shell of radius r and thickness dr. Then we vary r from 0 to R to generate the sphere. So V = r " r p=
Jr=0
=
"2nr
2
sin 0 dO dr,
Je = 0
(149) '
'
(150)
(151)
Figure 112 Element of volume in spherical coordinates.
18
Vectors
This is a particularly simple case because the limits of integration are all constants. Indeed, the triple integral of Eq. 148 is really the product of three integrals. In Prob. 119 we shall repeat the calculation in Cartesian coordinates. The limits of integration will then be functions, instead of being constants.
1.8
FLUX It is often necessary to calculate the flux of a vector through a surface. For example, one might wish to calculate the magnetic flux through the iron core of an electromagnet. The flux of a vector A through an infinitesimal surface da is (152)
c/O = A • da.
where the vector da is normal to the surface. The flux d
J>
da.
F o r a closed surface the vector da is taken to point
1.8.1
EXAMPLE:
FLUID
(153) outward.
FLOW
Let us consider fluid flow. We define a vector pv, p being the fluid density and v the fluid velocity at a point. The flux of pv through any closed surface is the net rate at which mass leaves the volume bounded by the surface. In an incompressible fluid this flux is zero.
1.9
DIVERGENCE The outward flux O of a vector A through a closed surface can be calculated either from the above equation or as follows. Let us consider an infinitesimal volume dx, dy, dz and the vector A, as in Fig. 113, whose
19
Figure 113 Element of volume dx, dy, dz around a point P, where the vector A has the value illustrated by the arrow.
components A , A A, are functions of the coordinates x, y, z. We consider an infinitesimal volume and firstorder variations of A. The value of A at the center of the righthand face can be taken to be the average value over that face. Through the righthand face of the volume element, the outgoing flux is x
x
dHf = U x
c
+ ^ ^ \ d y d z ,
(154)
since the normal component of A at the righthand face is the xcomponent of A at that face. At the lefthand face,
c^
L
=  ^  ^ ^ j d y d z .
(155)
The minus sign before the parenthesis is necessary here because, A i being inward at this face and da being outward, the cosine of the angle between x
20
Vectors
the two vectors is — 1. The net outward flux through the two faces is then PA PA = — dx dy dz = —^ dx,
d$> + m, R
OX
ox
(156)
where dx is the volume of the infinitesimal element. If we calculate the net flux through the other pairs of faces in the same manner, we find the total outward flux for the element of volume dx to be
\ ox
Py
dz J
Suppose now that we have two adjoining infinitesimal volume elements and that we add the flux through the bounding surface of the first volume to the flux through the bounding surface of the second. At the c o m m o n face the fluxes are equal in magnitude but opposite in sign, and they cancel as in Fig. 114. The flux from the first volume plus that from the second is the flux through the bounding surface of the combined volumes. To extend this calculation to a finite volume, we sum the individual fluxes for each of the infinitesimal volume elements in the finite volume, and the total outward flux is
At any given point in the volume, the quantity PA
X
— 
ox
H
PA
V
dy
PA,
I
Pz
is thus the outgoing flux per unit volume. We call this the divergence of the vector A at the point. The divergence of a vector pointfunction is a scalar pointfunction. According to the rules for the scalar product,
v
.
A
=
8A, ox
+
8A dy l
+
dA dz lt
( 1
.
5 9 )
where the operator V is defined as in Eq. 129. The operator V is not a vector, of course, but it is convenient to use the notation of the scalar product to indicate the operation that is carried out.
21
Figure 114 At the common face, the flux of A out oidz opposite in sign, to the flux of A out of dz •
l
is equal in magnitude, but
2
1.10
THE DIVERGENCE
THEOREM
The total outward flux of Eq. 158 is also equal to the surface integral of the normal outward component of A. Thus
This is the divergence theorem. N o t e that the integral on the left involves only the values of A on the surface 5, whereas the two integrals on the right involve the values of A throughout the volume T enclosed by S. The divergence theorem is a generalization of the fundamental theorem of the calculus (see Prob. 120). We can now redefine the divergence of the vector A as follows. If the volume T is allowed to shrink sufficiently, so that V • A does not vary appreciably over it, then J A • da = (V • A),
(t>0),
s
(161)
and V • A = lim  f A • da. r^O
T
(162)
J
S
The divergence is thus the outward flux per unit volume, as the volume t approaches zero.
22
1.10.1
Vectors
EXAMPLES:
INCOMPRESSIBLE
FLUID,
EXPLOSION
In an incompressible fluid, V • (pv) is everywhere equal to zero, since the outward mass flux per unit volume is zero. Within an explosion, V • (pv) is positive.
1.11
THE LINE
INTEGRAL
The integral
/>•„,. evaluated from the point a to the point b on some specified curve, is a line integral. Each element of length dl on the curve is multiplied by the local value of A according to the rule for the scalar product. These products are then summed to obtain the value of the integral. A vector field A is said to be conservative if the line integral of A • dl around any closed curve is zero: A • dl = 0.
(163)
The circle on the integral sign indicates that the path of integration is closed. We shall find in Sec. 2.5 that an electrostatic field is conservative.
/.//./
EXAMPLE:
WORK
DONE BY A
FORCE
The work done by a force F acting from a to b along some specified path is W  £
F • dl,
(164)
where both F and dl must be known functions of the coordinates if the integral is to be evaluated analytically. Let us calculate the work done by a force F that is in the y direction and has a magnitude proportional to y, as it moves around the circular path from a to b in Fig. 115. Since F = kyj W =
S!
and F > d l
dl = dx i + dy j,
= 5?ydy
=
kr /2. 2
(165) (166)
23
1.12
THE
CURL
Let us calculate the value of the line integral of A • dl in the more general case where it is not zero. F o r an infinitesimal element of path dl in the xyplane, and from the definition of the scalar product, A • dl = A dx + A dy. x
(167)
y
Thus, for any closed path in the xyplane and for any A, A • dl = j>A dx + &A x
y
dy.
(168)
N o w consider the infinitesimal path in Fig. 116. There are two contributions to the first integral on the righthand side of the above equation, one at y — {dy/2) and one at y + (dy/2):
24
y
Figure 116 Closed, rectangular path in the xrplane. centered on the point P(x, y, 0), where the vector A has the components A , A . The integration around the path is performed in the direction of the arrows. x
y
There is a minus sign before the second term because the path element at y + (cly/2) is in the negative x direction. Therefore
A dx
= ~dydx. dy
(170)
j>A dy = ^dxdy,
(171)
x
Similarly,
y
and
Adl =
ox
for the infinitesimal path of Fig. 116.
[^?£*)dxdy dy
1.12 The Curl
25
If we set dA g = ^ ox
dA
y
x
d73)
3
dy
then
A'A\
= g da,
(174)
3
where da = dx dy is the area enclosed on the xyplane by the infinitesimal path. N o t e that the above equation is correct only if the line integral is evaluated in the positive direction in the xyplane, that is, in the direction in which one would have to turn a righthand screw to make it advance in the positive direction of the zaxis. This is known as the righthand screw rule. Let us now consider g and the other two symmetrical quantities as the components of a vector 3
/ dA,
dA.
\dy
dz
dA I
+
x
cA\
dz
dx
. ^ /dA
y
I
dx
dA ^ x
dv
which may be written as
V x A
(175)
We shall call this vector the curl of A. The quantity g is then its z component. If we consider the element of area as a vector da pointing in the direction of advance of a righthand screw turned in the direction chosen for the line integral, then da = da k and 3
A d l = (V x A )  d a .
(176)
This means that the line integral of A • dl around the edge of an element of area da is equal to the scalar product of the curl of A by this
26
Vectors
element of area, as long as we observe the above sign convention. This is a general result that applies to any element of area da, whatever its orientation. Thus
(V x A)„ = l i m ^ c f A  d l : so S J
(177)
the component of the curl of a vector normal to a surface S at a given point is equal to the line integral of the vector a r o u n d the b o u n d a r y of the surface, divided by the area of the surface when this area approaches zero a r o u n d the point.
EXAMPLE:
WATER
VELOCITY
IN A
STREAM
Let us consider a stream in which the velocity v is proportional to the distance from the bottom. We set the zaxis parallel to the direction of flow, and the xaxis perpendicular to the stream bottom, as in Fig. 117. Then i\ = 0,
r = 0,
(178)
v
We shall calculate the curl from Eq. 177. F o r (V x v) we choose a path parallel to the yzplane. In evaluating v
vdl around such a path we note that the contributions are equal and opposite on the parts parallel to the zaxis, hence (V x v) = 0. Likewise, (V x v) = 0. For the j'component we choose a path parallel to the xzplane and evaluate the integral around it in the sense that would advance a righthand screw in the positive y direction. O n the parts of the path parallel to the xaxis, v • dl = 0 since v and dl are perpendicular. O n the bottom part of the path, at a distance x from the yzplane, v
2
v • dl = cx Az,
(179)
v • dl = — c(x + Ax) Az.
(180)
whereas at (x + Ax)
£ For the whole path,
dl
c Ax Az
(181)
2~
Figure 117 The velocity v in a viscous fluid is assumed to be in the direction of the raxis and proportional to the distance .x from the bottom. Then V x v = — c'y
and the ycomponent of the curl is
(V x v)j, = lim s~o
§ v • dl
: Ax A ;
S
Ax Az
(182)
Calculating V x v directly from Eq. 175,
i V x v
j
k
d
f
0
dx
r.v
fz
0
0
cx
(183)
which is the same result as above.
1.13
STOKES'S
THEOREM
Equation 176 is true only for a path so small that V x A can be considered constant over the surface da bounded by the path. What if the path C is so large that this condition is not met? The equation can be extended readily to arbitrary paths. We divide the surface—any surface bounded by the
28
Figure 118 An arbitrary surface bounded by the curve C. The sum of the line integrals around the curvilinear squares shown is equal to the line integral around C.
path of integration in question—into elements of area da^ d a , and so forth, as in Fig. 118. F o r any one of these small areas, 2
A d I , = (V x A )  d a , ,
(184)
We add the lefthand sides of these equations for all the da's, and then we add all the righthand sides. The sum of the lefthand sides is the line integral around the external boundary, since there are always two equal and opposite contributions to the sum along every common side between adjacent da's. The sum of the righthand side is merely the integral of (V x A) • da over the finite surface. Thus, for any surface S bounded by a closed curve C, A • dl = J (V x A) • da. s
This is Stokes's theorem. Figure 118 illustrates the sign convention.
(185)
1.14 The Laplacian
EXAMPLE:
CONSERVATIVE
VECTOR
29
FIELDS
Stokes's theorem can help us to understand conservative vector fields. Under what condition is a vector field conservative? In other words, under what condition is the line integral of A • dl around an arbitrary path equal to zero? From Stokes's theorem, the line integral of A • dl around an arbitrary closed path is zero if and only if V x A = 0 everywhere. This condition is met if A = V/. Then £
1
1
df
8
dx'
(186)
dz
and 8A (V x A). = — i dy Z
dA, —1 dz
df
df
dy cz
cz dy
2
2
= 0,
(187)
and so on for the other components of the curl. The field of A is therefore conservative, if A can be expressed as the gradient of some scalar function / .
1.14
THE
LAPLACIAN
The divergence of the gradient is of great importance in electromagnetism. Since
\f
= f i dx
+ f i dy
+ fk, dz
dx
cy
(188)
then
dz*
(189)
The divergence of the gradient is the sum of the second derivatives with respect to the rectangular coordinates. The quantity V • V / is abbreviated to \ f, 2
and is called the Laplacian
operator.
o f / . The operator V is called the 2
Laplace
30
1.14.1
Vectors
EXAMPLE:
THE LAPLACIAN
OF THE ELECTRIC
POTENTIAL
We shall see in Sec. 3.4 that the Luplacian of the electric potential is proportional to the space charge density.
1.15
SUMMARY In Cartesian coordinates a vector quantity is written in the form A = A i + A, j + A k. x
(//)
z
where i, j . k are unit vectors directed along the x, y, z axes respectively. The magnitude of the vector A is the scalar A = (A
+ A]. + A ) .
2
(12)
2 112
Vectors can be added and subtracted: A + B = (A
x
A  B = (A
x
+ B )i + (A x
+ B )j + (A
y
z
+ B )k. :
(13)
z
 B._)k.
(14)
y
 B )\ + (A,  B )\ + (A x
y
The scalar product (Fig. 13) is A • B = AB cos (
X
+ A B, y
(15)
+ A B._. Z
(111)
It is commutative, A • B = B • A,
(16)
A • (B + C) = A • B + A • C.
(17)
and distributive,
The vector product (Fig. 13) is A x B = C
(114)
1.15 Summary
31
with C = /4Bsin (
0)\.
(115)
Another equivalent definition of the vector product is i
j
k
A x B
(/22) B
B.
v
t
B.
The vector product follows the distributive rule A x (B + C) = (A x B) + (A x C),
(//7)
but not the commutative rule: A x B =  ( B x A). The time derivative of A is dA
dA
dA, i + — j dt dt r
dt
dA. , + — k. dt
(125)
The del operator is defined as follows:
V
0
. 0
. C
ox
dy
cz
(129)
and the gradient of a scalar function / is
cx
oz
(130)
The gradient gives the maximum rate of change of / with distance at the point considered, and it points toward larger values off. The surface integral is a c o m m o n type of double integral. It is used for integrating over two coordinates, say x and y. It is composed of an integral within an integral. The inner integral is evaluated first.
32
Vectors
The volume integral is one form of triple integral. It is used for integrating functions of three space coordinates, say x, y, z. In this case, we have an integral that is within an integral that is within an integral. The innermost integral is evaluated first, leaving a double integral, and so on. The flux O of a vector A through a surface S is
j :
A • da.
(153)
F o r a closed surface the vector da points outward. The divergence of A, V •A
dA,
dA,.
dx
+
dy
dA, +
dz
(159)
is the outward flux of A per unit volume at the point considered. The divergence theorem states that
j * V • A dx = J* A da, s
(160)
where x is the volume bounded by the surface S. The line integral
J>over a specified curve is the sum of the terms A • dl for each element dl of the curve between the points a and b. For a closed curve C that bounds a surface S, we have Stokes's theorem: A • dl =
J
s
(V x A) • da.
(185)
where
V x A
is the curl of the vector function A.
•
J
k
d
d
d
dx
dy
cz
/L
A
A,
r
(175)
Problems
33
The Laplacian is the divergence of the gradient: df =  L ox 2
V • V/' = \ f 2
' df + ^4 oy 2
cf + U. dz 2
(759)
PROBLEMS 11
Show that the two vectors A = 9i + j — 6k and B = 4i — 6j + 5k are perpendicular to each other.
12
Show that the angle between the two vectors A = 2i + 3j + k and B = i — 6j + k is 130.5°.
13
The vectors A, B, C are coplanar. Show graphically that A • (B + C) = A • B + A • C.
14
If A and B are adjacent sides of a parallelogram, C = A + B and D = A — B are the diagonals, and 0 is the angle between A and B, show that (C + D ) = 2(A + B ) and that ( C  D ) = 4AB cos 9. 2
2
15
2
2
2
2
Consider two unit vectors a and b, as shown in Fig. 119. Find the trigonometric relations for the sine and cosine of the sum and difference of two angles from the values of a • b and a x b. Solution:
First, a = cos a i + sin a j ,
b = cos fix + sin [S j .
(1) (2)
N o w we can write the product a • b in two different ways. From Eq. 111. a • b = cos a cos [1 + sin a sin /?,
(3)
and, from Eq. 15, a • b = cos (a  /?).
(4)
cos (a — ji) = cos a cos j) + sin a sin /?.
(5)
Thus
If we set \V =
/?,
cos (a + /?') = cos a cos ft' — sin a sin fi'.
(6)
34
Figure 119 Two unit vectors, a and b, situated in the xvplane. See Prob. 15.
Similarly, from Eq. 122,
a x b
cos a
sin a
0
cos p
sin p
0
= (cos a sin p — cos p sin a)k
(7)
(S)
and, from Eqs. 114 and 115, a x b = —sin (a — p)k.
(9)
Thus sin (ot — P) = sin a cos p — cos a sin p.
(10)
Setting again p' = — p, sin (x + P') = sin a cos p' + cos a sin P'.
(ID
16
Show that the magnitude of (A x B) • C is the volume of parallelepiped whose edges are A, B, C, and show that (A x B) • C = A • (B x C).
17
Show that A x ( B + C) = A x B +
18
Show that a x (b x c) = b(a • c)  c(a • b).
19
If r • (dr/</>) = 0, show that r = constant.
110
AxC.
A gun fires a bullet at a velocity of 500 meters per second and at an angle of 30" with the horizontal. Find the position vector r, the velocity vector v, and the acceleration vector a of the bullet, t seconds after the gun is fired. Sketch the trajectory and show the three vectors for some time t.
35 V
Figure 120 Two points P' and P in space, with the vector r and the unit vector r . We shall often use such pairs of points, with P' at the source, and P at the point where the field is calculated. For example, one might calculate the electric field intensity at P with the element of charge at P'. See Prob. 113. t
///
Let r be the radius vector from the origin of coordinates to any point, and let A be a constant vector. Show that V(A • r) = A.
112
Show that (A • V)r = A.
113
The vector r is directed from P'(x',y',z') to P(.v.y.r), as in Fig. 120. a) If the point P is fixed and the point P' is allowed to move, show that the gradient of (\/r) under these conditions is
where rj is the unit vector along r. Show that this is the maximum rate of change of b) Show similarly that, if P ' is fixed and P is allowed to move,
114
The components of a vector A are
I.  V
ry
—y
where /'is a function of x, y, :. Show that A = r x Y/;
115
A • r = 0.
and
A • Y/' = O.
a) Show that V • r = 3. b) What is the flux of r through a spherical surface of radius al
36
II
Figure 121 Truncated cylinder. See Prob. 118.
116
Show that V • (/A) = / V A + A • v./; where / is a scalar function and A is a vector function.
117
The vector A = 3.vi + yj + 2zk, a n d / = x + y + : . a) Show that V • (/A) at the point (2,2,2) is 120, by first calculating / A and then calculating its divergence. b) Calculate V • (/A) by first finding V/'and V • A. and then using the identity of Prob. 116 above. c) If x, y, : are measured in meters, what are the units of V • (/A)?
118
Calculate the volume of the portion of cylinder shown in Fig. 121. Set the xyplane on the base of the cylinder and the origin on the axis. Write the element of volume of height /; as h dx dy, with
119
Calculate the volume of a sphere of radius R in Cartesian coordinates.
120
Consider a function f(x).
2
2
2
The fundamental theorem of the calculus states that
Let A = /(x)i, and consider a cylindrical volume i parallel to the xaxis, extending from x = a to x = b, as in Fig. 1 22. Show that the divergence theorem, applied to the field A and the volume T, yields the fundamental theorem of the calculus in the form stated above.
37
Figure 122 Cylindrical volume parallel to the xaxis, extending from x = a to x = b. See Prob. 120.
121
The gravitational forces exerted by the sun on the planets are always directed toward the sun and depend only on the distance r. This type of field is called a central force field. Find the potential energy at a distance r from a center of attraction when the force varies as 1/r . Set the potential energy equal to zero at infinity. 2
122
Show that the gravitational field is conservative. Disregard the curvature of the earth and assume that g is a constant. What if one takes into account the curvature of the earth and the fact that g decreases with altitude? Is the gravitational field still conservative?
123
Let P and Q be any two paths in space that have the same end points a and b. Show that, if A is a conservative field, J>dl = J>dl. over P
124
over Q
Let A be a conservative field and P a fixed point in space. Let 0
f(x,y,z) = ^
A • dl,
where the line integral is calculated along any path joining P Show that A = V/. 125
0
to (x,y,z).
The azimuthal force exerted on an electron in a certain betatron (Prob. 119) is proportional to J ' . Show that the force is nonconservative. 0
4
38
Vectors
126
A vector field is defined by A = f(r)r. Show that V x A is equal to zero.
127
Show that V x (/A) = (Yf) x A + / ( V x A), where / is scalar and A is a vector.
128
Show that V • (A x D; = D • (V x A) — A • (V x D), where A and D are any two vectors.
129
Show that V • V x A = 0.
130
One of the four Maxwell equations states that V x E = —dB/dt, where E and B are respectively the electric field intensity in volts per meter and the magnetic induction in teslas at a point. Show that the line integral of E • dl is 20.0 microvolts over a square 100 millimeters on the side when B is 2.00 x 10" f tesla in the direction normal to the square, where t is the time in seconds. 3
131
In Sec. 114 we defined the Laplacian of a scalar pointfunction / . It is also useful to define the Laplacian of a vector pointfunction A: V A = VAi 2
+ \A)
2
+
2
x
y
\ A k. 2
z
Show that V (V/) = V(V /). 2
132
2
Show that
(1)
V x V x A = V(V • A) — V A. 2
We shall need this result in Chapter 20. Solution: Let us find the .vcomponents of these three quantities. Then we can find the y and zcomponents by symmetry. Since
V x V x A 8A
K
i
J
8
8
c
dx
dy
dz
Z
dA
dA
dy
dz
dz
y
x
12]
dA
8A,
dA,
dx
dx
8y
z
then its .vcomponent is dA
d_ fdA,
8A
dy \ dx
~8y
:
x
The .vcomponent of V(V • A) is 8
(cA
dx V dx
x
dA
y
dA
z
~dz~
(3)
Problems
39
while that of V A is 2
dA
8A
8A
tx
cy
cz*
2
2
x
2
X
X
Then we should have that f /l
dA
3*4,
2
2
t
5y
cy fx
d A, 2
2
—
fz
v
I 2
f /L 2
fx dy
~
fz fx 8 A, fx fz
8A
8A
dy
fz
2
2
X
fx
2
2
2
2
X
(5)
which is correct. The ycomponent of the given equation can be found from Eq. 5 by replacing x by y, y by z, and z by x. This gives another identity, and similarly for the zcomponent.
CHAPTER
2
FIELDS OF ST A TIONARY ELECTRIC CHARGES: I Coulomb''s Law, Electric Field Intensity Electric Potential V
2.1
COULOMB'S
2.2
THE ELECTRIC
FIELD
2.3
THE PRINCIPLE
OF
2.4
THE FIELD CHARGE
2.5 2.5.1
THE ELECTRIC POTENTIAL Example: The Electric Dipole
2.6
SUMMARY PROBLEMS
LAW INTENSITY
E
SUPERPOSITION
OF AN EXTENDED DISTRIBUTION V
E,
We begin our study of electric and magnetic phenomena by investigating the fields of stationary electric charges. We shall start with Coulomb's law and deduce from it, in this chapter, the electric field intensity and the electric potential for any distribution of stationary electric charges.
COULOMB'S
LAW
It is found experimentally that the force between two stationary electric point charges Q and Q (a) acts along the line joining the two charges, (b) is proportional to the product Q Q , and (c) is inversely proportional to the square of the distance r separating the charges. If the charges are extended, the situation is more complicated in that the "distance between the charges" has no definite meaning. Moreover, the presence of Q can modify the charge distribution within Q , and vice versa, leading to a complicated variation of force with distance. We thus have Coulomb's law for stationary point charges: a
b
a
b
b
a
F
"
b
" 4 ^ ^ 
r
i
'
( 2 _ 1 )
where F is the force exerted by Q on Q , and r is a unit vector pointing in the direction from Q to Q , as in Fig. 21. The force is repulsive if Q and Q are of the same sign; it is attractive if they are of different signs. As usual, F is measured in newtons, Q in coulombs, and r in meters. The constant e ah
a
a
h
b
t
a
b
0
42
fa*
Figure 21 Charges Q and Q separated by a distance r. The force exerted on Q by Q is F and is in the direction of ^ along the line joining the two charges. a
a
b
h
ah
is the permittivity
of free space: e
0
= 8.854 187 82 x 1 < T
12
farad/meter.
+
(22)
Coulomb's law applies to a pair of point charges in a vacuum. It also applies in dielectrics and conductors if F is the direct force between Q and Q , irrespective of the forces arising from other charges within the medium. Substituting the value of e into Coulomb's law, a b
a
b
0
F ^ 9 x l 0 #
r
r
i
,
(23)
where the factor of 9 is accurate to about one part in 1000; it should really be 8.987 551 79. The C o u l o m b forces in nature are enormous when compared with the gravitational forces. The gravitational force between two masses m and m separated by a distance r is a
h
F = 6.672 0 x \Q m m lr . ll
2
a
b
(24)
* From Eq. 21, e is expressed in coulombs squared per newtonsquare meter. However, a coulomb squared per newtonmeter, or a coulomb squared per joule, is a farad, as we shall see in Sec. 4.3. 0
2.3 The Principle of Superposition
43
The gravitational force on a proton at the surface of the sun (mass = 2.0 x 1 0 kilograms, radius = 7.0 x 10 meters) is equal to the electric force between a proton and one microgram of electrons, separated by a distance equal to the sun's radius. It is remarkable indeed that we should not be conscious of these enormous forces in everyday life. The positive and negative charges carried respectively by the proton and the electron are very nearly, if not exactly, the same. Experiments have shown that they differ, if at all, by at most one part in 1 0 . Ordinary matter is thus neutral, and the enormous Coulomb forces prevent the accumulation of any appreciable quantity of charge of either sign. 3 0
8
22
2.2
THE ELECTRIC
FIELD INTENSITY
E
We think of the force between the point charges Q and Q in Coulomb's law as the product of Q and the field of Q , or vice versa. We define the electric field intensity E to be the force per unit charge exerted on a test charge in the field. Thus the electric field intensity due to the point charge Q is a
a
b
b
a
E
"
=
7 r = 7% >
<">
r
Qb
2
5
47re r 0
The electric field intensity is measured in volts per meter. The electric field intensity due to the point charge Q is the same, whether the test charge Q is in the field or not, even if Q is large compared to Q„. a
h
2.3
THE PRINCIPLE
OF
b
SUPERPOSITION
If the electric field is produced by more than one charge, each one produces its own field, and the resultant E is simply the vector sum of the individual E's. This is the principle of superposition.
44
Fields of Stationary
Electric Charges: I
THE FIELD OF AN EXTENDED CHARGE DISTRIBUTION F o r an extended charge distribution,
EJLf./**'.
(26)
The meaning of the various terms under the integral sign is illustrated in Fig. 22: p is the electric charge density at the source point (x',y',z')\ rj is a unit vector pointing from the source point to the field point (x,y,z) where E is calculated; r is the distance between these two points; dr' is the element of volume dx' dy' dz'. If there exist surface distributions of charge, then we must add a surface integral, with the volume charge density p replaced by the surface charge density a and the volume x' replaced by the surface S'. The electric charge density inside macroscopic bodies is of course not a smooth function of x', y', z'. However, nuclei and electrons are so small and so closely packed, that one can safely use an average electric charge density p, as above, to calculate macroscopic fields. When the electric field is produced by a charge distribution that is disturbed by the introduction of a finite test charge Q', E is the force per unit charge as the magnitude of the test charge Q' tends to zero:
E = lim
(27)
2.5 The Electric Potential V
THE ELECTRIC
POTENTIAL
45
V
Consider a test point charge Q' that can be moved about in an electric field. The work W required to move it at a constant speed from a point P to a point P along a given path is t
2
(28) The negative sign is required to obtain the work done against the field. Here again, we assume that Q' is so small that the charge distributions are not appreciably disturbed by its presence. If the path is closed, the total work done is (29) Let us evaluate this integral. To simplify matters, we first consider the electric field produced by a single point charge Q. Then
(210)
Figure 23 shows that the term under the integral on the right is simply dr/r or — d(l/r). The sum of the increments of (1/r) over a closed path is zero, since r has the same value at the beginning and at the end of the path. Then the line integral is zero, and the net work done in moving a point charge Q' around any closed path in the field of a point charge Q, which is fixed, is zero. If the electric field is produced, not by a single point charge Q, but by some fixed charge distribution, then the line integrals corresponding to each individual charge of the distribution are all zero. Thus, for any distribution of fixed charges, 2
(211) An electrostatic field is therefore conservative (Sec. 1.11). It can be shown that this important property follows from the sole fact that the C o u l o m b force is a central force (see Prob. 121).
46
Then, from Stokes's theorem (Sec. 1.13), at all points in space, V x E = 0.
(212)
E = W,
(213)
and we can write that
where V is a scalar point function, since V x \V = 0. We can thus describe an electrostatic field completely by means of the function V(x,y,z), which is called the electric potential. The negative sign is required in order that the electric field intensity E point toward a decrease in potential, according to the usual convention. It is important to note that V is not uniquely defined; we can add to it any quantity that is independent of the coordinates without affecting E in any way. As shown in Prob. 123, the work done in moving a test charge at a constant speed from a point P to a point P is independent of the path. We must remember that we are dealing here with electrosfaf/cs. If there were moving charges present, V x E would not necessarily be zero, and W would then describe only part of the electric field intensity E. We shall investigate these more complicated phenomena later on, in Chapter 11. {
2
47
According to Eq. 213, E d l =  V K  d l = —cIV. Then, for any two points P and P x
V 2
2
(214)
as in Fig. 2  4 ,
V, =  J ^ E d l = £
Edl.
(215)
Note that the electric field intensity E(x,y,z) determines only the difference between the potentials at two different points. When we wish to speak of the electric potential at a given point, we must therefore arbitrarily define the potential in a given region of space to be zero. It is usually convenient to choose the potential at infinity to be zero. Then the potential V at the point 2 is V = f" E d l .
(216)
The work W required to bring a charge Q' from a point at which the potential is defined to be zero to the point considered is VQ'. Thus V is WjQ' and can be defined to be the work per unit charge. The potential V is expressed in joules per coulomb, or in volts.
48
Figure 25 Equipotential surface and lines of force near a point charge.
When the field is produced by a single point charge Q, the potential at a distance r is (217)
It will be observed that the sign of the potential V is the same as that of Q. The principle of superposition applies to the electric potential V as well as to the electric field intensity E. The potential V at a point P due to a charge distribution of density p is therefore
(218)
where r is the distance between the point P and the element of charge p dx'. The points in space that are at a given potential define an equipotential surface. F o r example, an equipotential surface about a point charge is a concentric sphere as in Fig. 25. We can see from Eq. 213 that E is everywhere normal to the equipotential surfaces (Sec. 1.5).
2.5 The Electric Potential V
49
If we join endtoend infinitesimal vectors representing E, we get a curve in space—called a line of force—that
is everywhere normal to the
equipotential surfaces, as in Fig. 25. The vector E is everywhere tangent to a line of force.
2.5.1
EXAMPLE:
THE ELECTRIC
DIPOLE
The electric ciipole shown in Fig. 26 is one type of charge distribution that is encountered frequently. We shall return to it in Chapter 6. The electric dipole consists of two charges, one positive and the other negative, of the same magnitude, separated by a distance s. We shall calculate V and E at a distance r that is large compared to s. At P, (219)
+ Q<
s
Q
Figure 26 The two charges + Q and — Q form a dipole. The electric potential at P is the sum of the potentials due to the individual charges. The vector s points from — Q to +Q, and r, is a unit vector pointing from the origin to the point of observation P.
50
Fields of Stationary
Electric Charges: I
where rz = r +

(220)
+ rs cos 0
, s y 1 + (— ~lr
s +  cos 0 r
(221)
(222)
r I 1 +  cos 0 r
(223) 1 +  cos 0 r 1  — cos 0.* 2r
(224)
1 + — cos 9, 2r
(225)
Similarly, r r
b
and V =
4ne r
2
cos 0.
(226)
0
This expression is valid for r » s . It is interesting to note that the potential due to a dipole falls off as 1/r , whereas the potential from a single point charge varies only as 1/r. This comes from the 3
3
2
* Remember that n(n  1) , n(n  l)(n  2) , a + (1 + df = 1 + na + —r.—a +  £ 2' 3! 1
3
If n is a positive integer, the series stops when the coefficient is zero. Try, for example, n = 1 and n = 2. If n is not a positive integer, the series converges for a < 1. This series is very often used when a « 1. Then 2
(1 + a)" * 1 + na. For example. (1 + a) * 1 + 2a. 2
(1 + aC"
2
« 1   , etc.
51
Figure 27 Lines of force (arrows) and equipotential lines for the dipole of Fig. 26. The dipole is vertical, at the center of the figure, with the positive charge close to and above the negative charge. In the central region the lines come too close together to be shown.
fact that the charges of a dipole appear close together for an observer some distance away, and that their fields cancel more and more as the distance r increases. We define the dipole moment p = Qs as a vector whose magnitude is Qs and that is directed from the negative to the positive charge. Then V = 
. 2
(227)
Figure 27 shows equipotential lines for an electric dipole. Equipotential surfaces are generated by rotating equipotential lines around the vertical axis.
52
Fields of Stationary
Electric Charges: I
Let us now calculate E at the point P(x.O.r) in the plane y = 0 as in Fig. 26. We have that p
cos 0
p
=
K
4
: ?
,
2
"
2
8
)
and. at P(.\,0,r), cV £
3p xz 3r> sin 0 cos 0 = — ?  — = —! , cx 4ne r 4ne r
=
5

9
V
3 P
dy
4ne
(229)
3
0
0
y
= 0.
:
r
0
,230)
since y = 0, and 8V _ p E =  — = y— <3z 47ie dz 4ne
(1 I 3 \r\
r
0
0
p 4ne
3j  , r 2
(231)
5
3 c o s f) — 1 2
(232)
r
3
0
Figure 27 shows lines offeree for the electric dipole.
SUMMARY It is found empirically that the force exerted by a point charge Q on a point charge Q is a
b
4ne
0
r
2
where /• is the distance between the charges and r, is a unit vector pointing from Q to Q . This is Coulomb's law. We consider the force ¥ as being the product of Q by the electric field intensity due to Q , a
b
a b
b
a
E^T^Sr,, or vice versa.
(25)
Problems
According to the principle
of superposition,
53
two or more electric
field intensities acting at a given point add vectorially. For an extended charge distribution,
E =
1 4ne
cm)
J*' r
2
0
The electrostatic field is conservative, V x E = 0,
(272)
E = \V,
(213)
hence
where 1 47T€A 4ne 0
is the electric potential;
cp J
z
dx' r
p dx' is the element of charge contained within the
element of volume dx', and r is the distance between this element and the point where V is calculated.
PROBLEMS 2JE
1
COULOMB'S LAW a) Calculate the electric field intensity that would be just sufficient to balance the gravitational force of the earth on an electron. b) If this electric field were produced by a second electron located below the first one, what would be the distance between the two electrons? The charge on an electron is —1.6 x 10" coulomb and its mass is 9.1 x 1 0 " kilogram. 1 9
3 1
22E
SEPARATION OF PHOSPHATE FROM QUARTZ Crushed Florida phosphate ore consists of particles of quartz mixed with particles of phosphate rock. If the mixture is vibrated, the quartz becomes charged +
The letter E indicates that the problem is relatively easy.
54
Figure 28 Electrostatic separation of phosphate from quartz in crushed phosphate ore. The vibrating feeder VF charges the phosphate particles positively, and the quartz particles negatively. See Prob. 22.
negatively and the phosphate positively. The phosphate can then be separated out as in Fig. 28. Over what minimum distance must the particles fall if they must be separated by at least 100 millimeters? Set £ = 5 x 10 volts per meter and the specific charge equal to 1 0 " coulomb per kilogram. 5
23E
5
ELECTRIC FIELD INTENSITY A charge + Q is situated at x = a, y = 0, and a charge — Q at x = — a, y = 0. Calculate the electric field intensity at the point x = 0, y = a.
Problems
24
55
ELECTRIC FIELD INTENSITY A circular disk of charge has a radius R and carries a surface charge density a. Find the electric field intensity £ at a point P on the axis at a distance a. What is the value of E when a « Rl You can solve this problem by calculating the field at P due to a ring of charge of radius /• and width dr, and then integrating from r = 0 to r = R. CATHODERAY TUBE The electron beam in the cathoderay tube of an oscilloscope is deflected vertically and horizontally by two pairs of parallel deflecting plates that are maintained at appropriate voltages. As the electrons pass between one pair, they are accelerated by the electric field, and their kinetic energy increases. Thus, under steadystate conditions, we achieve an increase in the kinetic energy of the electrons without any expenditure of power in the deflecting plates, as long as the beam does not touch the plates. Could this phenomenon be used as the basis of a perpetual motion machine?
26E
CATHODERAY TUBE In a certain cathoderay tube, the electrons are accelerated under a difference of potential of 5 kilovolts. After being accelerated, they travel horizontally over a distance of 200 millimeters. Calculate the downward deflection over this distance caused by the gravitational force. An electron carries a charge of —1.6 x 1 0 " coulomb and has a mass of 9.1 x 1 0 " kilogram. 1 9
3 1
27E
MACROSCOPIC PARTICLE GUN It is possible to obtain a beam of fine particles in the following manner. Figure 29 shows a parallelplate capacitor with a hole in the upper plate. If a particle of dust, say, is introduced in the space between the plates, it sooner or later comes into contact with one of the plates and acquires a charge of the same polarity. It then flies over to the opposite plate, and the process repeats itself. The particle oscillates back and forth, until it is lost either at the edges or through the central hole. O n e can thus obtain a beam emerging from the hole by admitting particles steadily into the capacitor. In order to achieve high velocities, the gun must, of course, operate in a vacuum. Beams of macroscopic particles are used for studying the impact of micrometeorites. Now it has been found that a spherical particle of radius R lying on a charged plate acquires a charge Q = 1.65 x 4ne e R E 2
r
0
0
coulomb,
where £ is the electric field intensity in the absence of the particle, and e is the relative permittivity of the particle (Sec. 6.7). We assume that the particle is at least slightly conducting. Assuming that the plates are 10 millimeters apart and that a voltage difference of 15 kilovolts is applied between them, find the velocity of a spherical particle 1 0
r
56
Figure 29 Device for producing a beam of fine particles. See Prob. 27 micrometer in diameter as it emerges from the hole in the upper plate. Assume that the particle has the density of water and that e = 2. r
28E
ELECTROSTATIC SPRAYING When painting is done with an ordinary spray gun, part of the paint escapes deposition. The fraction lost depends on the shape of the surface, on drafts, etc., and can be as high as 80%. The use of the ordinary spray gun in largescale industrial processes would therefore result in intolerable waste and pollution. The efficiency of spray painting can be increased to nearly 100%, and the pollution reduced by a large factor, by charging the droplets of paint, electrically, and applying a voltage difference between the gun and the object to be coated. It is found that, in such devices, the droplets carry a specific charge of roughly one coulomb per kilogram. Assuming that the electric field intensity in the region between the gun and the part is at least 10 kilovolts per meter, what is the minimum ratio of the electric force to the gravitational force?
29
THE RUTHERFORD EXPERIMENT In 1906, in the course of a historic experiment that demonstrated the small size of the atomic nucleus, Rutherford observed that an alpha particle (Q = 2 x 1.6 x 1 0 " coulomb) with a kinetic energy of 7.68 x 10 electronvolts making a headon collision with a gold nucleus (Q = 79 x 1.6 x 1 0 " coulomb) is repelled backwards. The electronvolt is the kinetic energy acquired by a particle carrying one electronic charge when it is accelerated through a difference of potential of one volt: 1 9
6
1 9
1 electronvolt = 1.6 x 1 0 "
1 9
joule.
a) What is the distance of closest approach at which the electrostatic potential energy is equal to the initial kinetic energy? Express your result in femtometers ( 1 0 " meter). 1 5
57
Figure 210 Cylindrical electrostatic analyser. A particle P can reach the collector C only if it is positive and if it satisfies the equation given in Prob. 211. Otherwise, it stops on one of the cylindrical electrodes. For a given chargetomass ratio Q/m, one can select different velocities by changing V. A circle with the letter I represents an ammeter, while a circle with a letter V represents a voltmeter.
10"
2 7
b) What is the maximum force of repulsion? c) What is the maximum acceleration in
ELECTROSTATIC SEED SORTING Normal seeds can be separated from discolored ones and from foreign objects by means of an electrostatic seedsorting machine that operates as follows. The seeds are observed by a pair of photocells as they fall one by one inside a tube. If the color is not right, voltage is applied to a needle that deposits a charge on the seed. The seeds then fall between a pair of electrically charged plates that deflect the undesired ones into a separate bin. One such machine can sort dry peas at the rate of 100 per second, or about 2 tons per 24hour day. a) If the seeds are dropped at the rate of 100 per second, over what distance must they fall if they must be separated by 20 millimeters when they pass between the photocells? Neglect air resistance. b) Assuming that the seeds acquire a charge of 1.5 x 1 0 " coulomb, that the deflecting plates are parallel and 50 millimeters apart, and that the potential difference 9
58
Fields of Stationary
Electric Charges: I
between them is 25 kilovolts, how far should the plates extend below the charging needle if the charged seeds must be deflected by 40 millimeters on leaving the plates? Assume that the charging needle and the t o p of the deflecting plates are close to the photocell. 21 IE
CYLINDRICAL ELECTROSTATIC ANALYSER Figure 210 shows a cylindrical electrostatic analyser or velocity selector. It consists of a pair of cylindrical conductors separated by a radial distance of a few millimeters. Show that, if the radial distance between the cylindrical surfaces of average radius R is a, and if the voltage between them is V, then the particles collected at / have a velocity (Q \m
VRV a
12
J
PARALLELPLATE ANALYSER Figure 211 shows a parallelplate analyser. The instrument serves to measure the energy distribution of the charged particles emitted by a source. It is often used for sources of electrons. It has been used, for example, to investigate the energy distribution of electrons in the plasma formed at the focus of a lens illuminated by a powerful laser. If the source emits particles of various energies, then, for a given value of V and for a given geometry, the collector receives only those particles that have the
Figure 211 Parallelplate analyser for measuring the energy distribution of particles emerging from a source S. The particles follow a ballistic trajectory in the uniform electric field between the two plates. Particles of the correct energy and, of course, of the correct polarity are collected at C. The other particles hit either one of the two parallel plates. The polarities shown apply to electrons or to negative ions. See Prob. 212.
Problems
59
corresponding energy. O n e therefore observes the energy spectrum of the particles by measuring the collector current as a function of V. As we shall see below, we assume that the particles all carry the same charge. In the figure, particles of charge Q have a kinetic energy QV . This type of analyser is relatively simple to build. It can also be, at the same time, compact and accurate. For example, in one case, the electrodes measured 40 m m x 115 mm, with b = 15 mm a n d a = 50 mm. With slit widths of 0.25 mm, the detector current as a function of V for monoenergetic electrons gave a sharp peak as in Fig. 212 having a width at half maximum of only 1%. Q
a) Find a for a particle of mass m and charge Q. Solution: We can treat the particle as a projectile of mass m, charge Q, and initial velocity v . with Q
1\
,
(1)
V , = 2
QV , 0
r J ^)
.
2
0
in
(2,
The particle has a vertical acceleration, downwards, equal to QE/m, or Q(V/b)
Then the time of flight, from slit to slit, is Do sin 8 2vr.mb 7 = 2 — = ——sinfl, Q(V/b)/m QV
(3)
and the distance a is given by a = ( r cos 8)T = n
0
=
AbV
0
V
2vhnb QV
sin 6 cos 8 =
sin 8 cos 0,
2bV V
0
sin 28.
(4)
(5)
Note that, for a given value of V , a is independent of both m and Q. It is proportional to V . If all the particles carry the same charge Q, a is proportional to their energy QV . 0
0
0
b) The optimum value of 0 may be defined as that for which the horizontal distance a traveled by the particle is maximum. One can then tolerate a slightly divergent beam at the entrance slit, with an average 8 of 0 . Find ()„„,. opI
60
AV Figure 212 The current / at the collector C of Fig. 211, as a function of V.
2±
Solution: Since the distance a must be maximum, da/dO must be zero. N o w sin 20 has a maximum at 0 = 45 , so 6 , = 45°. Note that, since a is maximum at 0 = 45 this is also the condition for maximum resolution. op
c) Now find V as a function of V, for given values of a and b, at 0 . 0
Solution:
opl
From Eq. 5, with 0 = 45°,
V d) Find da/dV for 0 = 0
Solution:
0
(6)
V.
2b
0 . opl
At 0 = fl„ ,. n
da
2b
'IK,
~V'
(7)
e) Find the minimum value of V as a function of V at fl . 0
opl
Solution: In all the above equations. V is always divided by b. T o find V we use the fact that V must be sufficient to reduce the vertical velocity to zero:
(8)
QV >  m(v sin 4 5 ) , c
2
0
V >
m 2QV 1 0
2Q
m
2
V
0
(9)
Comparing now with Eq. 6, we see that alb must be smaller than 4. This analyser is practical for energies Q K less than, say, 10 kiloelectronvolts. It would be inconvenient to use voltages V larger than a few kilovolts. For higher energies one would use the analyser of Prob. 211. 0
Problems
213
61
CYLINDRICAL AND PARALLELPLATE ANALYSERS COMPARED Show that both the cylindrical analyser of Prob. 211 and the parallelplate analyser of Prob. 212 measure the ratio (1 2)mv
2
Q where m is the mass of the particle, v its velocity, and Q its charge. 214
ION
THRUSTER Synchronous satellites describe circular orbits above the equator at the altitude where their angular velocity is equal to that of the earth. Since these satellites are extremely costly to build and to launch, they must remain operational for several years. However, it is impossible to adjust their initial position and velocity with sufficient accuracy to maintain them fixed with respect to the earth for long periods of time. Moreover, they are perturbed by the moon, and their antennas must remain constantly directed towards the earth. Synchronous satellites are therefore equipped with thrusters whose function is to exert the small forces necessary to keep them properly oriented and on their prescribed orbits. Even then, they keep wandering about their reference position. For example, their altitude can vary by as much as 15 kilometers. The thrust is m'v, where m is the mass of propellant ejected per unit time and v is the exhaust velocity with respect to the satellite. Clearly, m should be small. Then v should be large. However, as we shall see, too large a r could lead to an excessive power consumption. So the choice of velocity depends on the maximum allowable value for m' and on the available power. One way of achieving large values of v is to eject a beam of charged particles, as in Fig. 213. The propellant is ionized in an ion source and ejected as a positive ion beam. By itself, the beam cannot exert a thrust, for the following reason. Initially, the net charge on the satellite is zero. When it ejects positive particles, it acquires a negative charge. The particles are thus held back by the electrostatic force, and, once the satellite is sufficiently charged, they fall back on the satellite. To achieve a thrust, the ions must be neutralized on leaving the satellite. This is achieved by means of a heated filament, as in Fig. 213. The filament emits electrons that are attracted by the positively charged beam. a) The current / of positive ions in the beam is carried by particles of mass m and charge ne, where e is the electronic charge. Show that the thrust is
b) What is the value o f f for a 0.1 ampere beam of protons with V = 50kilovolts?
62
S
Figure 213 Schematic diagram of an ion thruster. The propellant is admitted at P and ionized in S: A is a beamshaping electrode and B is the accelerating electrode, maintained at a voltage V with respect to S; the heated filament F injects electrons into the ion beam to make it neutral. Electrode B is part of the outer skin of the satellite. See Prob. 214.
c) If we call P the power IV spent in accelerating the beam, show that
Note that, since F = (2m'P) , for given values of m and P, the thrust is independent of the chargetomass ratio of the ions. Also, since F is 2P/v, the thrust is inversely proportional to v, for a given power expenditure P. Finally, the last expression shows that, again for a given P, it is preferable to use heavy ions carrying a single charge (» = 1) and to use a low accelerating voltage V. d) If the filament is left unheated and if the beam current is 0.1 ampere, how long does it take the body of the satellite to attain a voltage of 50 kilovolts? Assume that the satellite is spherical and that it has a radius of 1 meter. 1
COLLOID
2
THRUSTER
Before reading this solved problem you should first read Prob. 214. Figure 214 shows a socalled colloid thruster. A conducting fluid is pumped into a hollow needle maintained at a high positive voltage with respect to the satellite's outer skin. Microscopic droplets form around the edge of the opening, where the electric field intensity and the surface charge density are extremely high.
63
Figure 214 Colloid thruster. The hollow needle TV ejects a beam of positively charged, highvelocity droplets. Electrode B and filament F are as in Fig. 213. See Prob. 215.
These droplets, carrying specific charges Q/m as large as 4 x 10 coulombs per kilogram, are accelerated in the electric field and form a highenergy jet. As a rule, several needles are grouped together so as to obtain a larger thrust. 4
a) Calculate the thrust exerted by a single needle for the specific charge given above, when V = 10 kilovolts and / = 5 microamperes. Use the results of Prob. 214. Solution: '2m
QV.
1/2
P.
ID 1/2
10 x 5 x 1 0 " , 4
V4 x 10* x 1 0
4
6
= 3.5 micronewtons.
(2) (3)
b) Calculate the mass of fluid ejected per second. Solution: The mass in ejected per second is m/Q times the charge ejected per second, which is 5 x 1 0 " coulomb: 6
tri = 5 x 10 ' 4 x 10 = 1.3 x 1 0 " 6
4
1 0
kilogram/second.
(4)
c) The thrust is so small that it is not practical to measure it directly. The method used is illustrated in Fig. 215. The thruster is first put into operation. Then switch S is opened and the voltage IR across R is observed on an oscilloscope, as a function of the time. The curve has the shape shown in Fig. 216a. Calculate the thrust as a function of V, /, D, T. Solution: The thrust is ) = IV, v 2
= 21V
T
V = 2 — IT. D D
(5)
(6)
Figure 215 Method used for measuring the thrust of the colloid thruster of Fig. 214. The current / produced by the charged droplets collected on the hemispherical electrode gives a voltage IR that can be observed as a function of the time on an oscilloscope.
d) If there are several types of droplets, the curve of J as a function of t has several plateaus as in Fig. 216b. How can one calculate the thrust in this case? Solution: In the last expression for F in Eq. 6, we see that F is 2V/D times the area under the curve of Fig. 216a. Thus, in the case of Fig. 216b,
F = 2[ Idt. D J°
(7)
m
T
t (a)
t (b)
Figure 216 (a) Current / as a function of the time in the setup of Fig. 215, starting at the instant when switch S is opened, when all the droplets are of the same size and carry equal charges, (b) If there are several types of droplets of different sizes or charges, the curve of / as a function of t has one plateau for each type.
CHAPTER
3
FIELDS OF STATIONARY ELECTRIC CHARGES: II Gauss's Law, Poissons Uniqueness Theorem
and Laplace's
Equations,
3.1 3.1.1 3.1.2
SOLID ANGLES The Angle Subtended by a Curve at a Point The Solid Angle Subtended by a Closed Surface at a Point
3.2 3.2.1 3.2.2
GAUSS'S Example: Example:
3.3 3.3.1 3.3.2 3.3.3
CONDUCTORS Example: Hollow Conductor Example: Isolated Charged Conducting Plate Example: Pair of Parallel Conducting Plates Carrying Charges of Equal Magnitudes and Opposite Signs
3.4 3.4.1
POISSON'S EQUATION Example: Flat Ion Beam
3.5
LAPLACE'S
3.6
THE
3.7 3.7.1
IMAGES Example: Point Charge Near an Infinite Conducting Plate
3.8
LAW Infinite Sheet of Charge Spherical Charge
EQUATION
UNIQUENESS
SUMMARY PROBLEMS
THEOREM Grounded
Now that we have discussed the two fundamental concepts of electric field intensity E and electric potential V, we shall study Gauss's law, Poisson's equation, and the uniqueness theorem. This will give us three powerful methods for calculating electrostatic fields. But first we must study solid angles.
3.1 3.1.1
SOLID THE
ANGLES
ANGLE
SUBTENDED
BY A CURVE
AT A
POINT
Consider the curve C and the point P of Fig. 31. They are situated in a plane. We wish to find the angle oc subtended by C at P. This is
(31) The integrals are evaluated over the curve C: for the second one, we multiply each element of length dl by sinW and divide by r; then we sum the results. The arc c subtends the same angle a at P. If now the curve is closed, as in Fig. 32a, with P inside C, the angle subtended at P by the small circle of radius b is 27rb/b, or 2n radians. The angle subtended by C at any inside point P is therefore also 2n radians. If P is outside C, as in Fig. 32b, then the segments C and C subtend angles of the same magnitude but of opposite signs, since sin 0 is positive on C and negative on C . Thus the angle subtended by C at any outside point is zero. x
t
2
2
67
Figure 31 A curve C and a point P situated in a plane. The angle subtended by C at P is a. The element dl' is the component of dl in the direction perpendicular to the radius vector.
Figure 32 [a) Closed curve C and a point P situated inside. The angle subtended by C at P is the same as that subtended by the small circle, namely 2n radians. (/)) When the point P is outside C. the angle subtended by C at P is zero because segments Cj on the right and C on the left of the points where the tangents to the curve go through P subtend equal and opposite angles. 2
3.1.2
THE
SOLID
SURFACE
ANGLE AT
A
SUBTENDED
BY
A
CLOSED
POINT
Figure 33 shows a closed surface S and a point P situated inside. The small cone intersects an element of area da on S. This small cone defines the solid angle subtended by da at P.
68
Figure 33 A closed surface S and a point P situated inside. The cone defines the solid angle subtended by the element of area da at P. The solid angle subtended at P by the complete surface S is the same as that subtended by the small sphere of radius b, namely 4JI steradians.
By definition, this solid angle is the area da, projected on a plane perpendicular to the radius vector, and divided by r : 2
da cos 9
dQ =
da
(32)
where 0 is the angle between the radius vector r and the vector da, normal to the surface and pointing outward. Solid angles are expressed in steradians. The total solid angle subtended by S at P is da cos 0
Q
(33)
This solid angle is also the solid angle subtended at P by the small sphere s of radius a. Thus the solid angle subtended by S at any inside point is Q = Ana /a 2
2
= An steradians.
Note that a solid angle is dimensionless, like an ordinary angle.
(34)
If the point P is situated outside the surface as in Fig. 34, then rj • da is positive over S and negative over S , and the total solid angle subtended at any outside point is zero. The situation remains unchanged if the surface is convoluted in such a way that a line drawn from P cuts the surface at more than one point. The total solid angle subtended by a closed surface is still 4n at an inside point, and zero at an outside point. Y
GAUSS'S
2
LAW
Gauss's law relates the flux of E through a closed surface to the total charge enclosed within the surface. By using this law one can find the electric field of simple charge distributions with a minimum of effort. Let us suppose that a point charge Q is located at P in Fig. 33. We can calculate the flux of E through the closed surface as follows. By definition, the flux of E through the element of area da is (35)
70
Fields of Stationary Electric Charges: II
where T • da is the projection of da on a plane normal to r . Then 1
t
E • da =
Ane
dO,
(36)
0
where da is the element of solid angle subtended by da at the point P. To find the total flux of E, we integrate over the whole surface S. Since the point P is inside the surface, by hypothesis, the integral of da is An and J* E • da = Q/e . s
(37)
0
If more than one point charge resides within S, the fluxes add algebraically, and the total flux of E leaving the volume is equal to the total enclosed charge divided by e . If the charge enclosed by the surface S' is distributed over a finite volume, the total enclosed charge is 0
Q = J*, P dz\
(38)
where p is the charge density and x is the volume enclosed by the surface S'. Then
f, E • da' = 
f, p dr'.
(39)
This is Gauss's law stated in integral form. Applying the divergence theorem (Sec. 1.10) to the lefthand side,
J \ V • E dr' = — £ p dr'.
(310)
Since this equation is valid for any closed surface, the integrands must be equal and, at any point in space, V  E = p/e . 0
(311)
This is Gauss's law stated in differential form. It concerns the derivatives of E with respect to the coordinates, and not E itself.
3.2 Gauss's Law
3.2.1
EXAMPLE:
INFINITE
SHEET
OF
71
CHARGE
Figure 35 shows an infinite sheet of charge, of surface charge density a. On each side, E = a/2e . 0
Figure 35 Portion of an infinite sheet of charge of density a coulombs per square meter. The imaginary box of crosssection A encloses a charge a A. The flux of E leaving the box is 2EA. Therefore, E = c / 2 e . 0
EXAMPLE:
SPHERICAL
CHARGE
Let us calculate E, both inside and outside a spherical charge of radius R and uniform density p, as in Fig. 36. The electric field intensity E is a function of the distance r from the center 0 of the sphere to the point considered. We shall use the subscript o to indicate that we are dealing with the field outside the charge distribution, and the subscript / inside. The total charge Q is (4/3)7iK p. Consider an imaginary sphere of radius r > R, concentric with the charged sphere. We know that E must be radial. Then, according to the integral form of Gauss's law. 3
0
4nr E 2
0
=
Q/e , 0
(312)
Figure 36 Uniform spherical charge. We can calculate the electric field at a point P outside the sphere, as well as at a point P' inside the sphere, by using Gauss's law.
0
R
Figure 37 In the case of a uniform spherical charge distribution, the electric field intensity E rises linearly from the center to the surface of the sphere and falls off as the inverse square of the distance outside the sphere. On the other hand, the electric potential falls from a maximum at the center in parabolic fashion inside the sphere, and as the inverse first power of the distance outside.
and
E „ = — %
(313)
as if the total charge Q were situated at the center of the sphere. If the charge were not distributed with spherical symmetry, E would not be uniform over the imaginary sphere. Gauss's law would then only give the average value of the normal component of E„ over the sphere. To calculate E, at an internal point, we draw an imaginary sphere of radius r through the point P. Symmetry requires again that E, be radial: thus 0
47rc £,. = (4/3)jtr p/e ,
(314)
E, = p r / 3 e .
(315)
2
3
0
and
0
Figure 37 shows E and V as functions of the radial distance r. Note that E = E„ at ;• = R. This is required by Gauss's law. since the charge contained within a t
3.3 Conductors
73
spherical shell of zero thickness at r = R is zero. Note also that V = V at r = R. A discontinuity in V would require an infinite electric field intensity, since E is dV/dr. t
0
CONDUCTORS A conductor can be denned as a material inside which charges can flow freely. Since we are dealing with electrostatics, we assume that the charges have reached their equilibrium positions and are fixed in space. Then, inside a conductor, there is zero electric field, and all points are at the same potential. If a conductor is placed in an electric field, charges flow temporarily within it so as to produce a second field that cancels the first one at every point inside the conductor. Coulomb's law applies within conductors, even though the net field is zero. Gauss's law, V • E = p/e , is also valid. Then, since E = 0, the charge density p must be zero. As a corollary, any net static charge on a conductor must reside on its surface. At the surface of a conductor, E must be normal, for if there were a tangential component of E, charges would flow along the surface, which would be contrary to our hypothesis. Then, according to Gauss's law, just outside the surface, E = c / e , as in Fig. 38, a being the surface charge density. It is paradoxical that one should be able to express the electric field intensity at the surface of a conductor in terms of the local surface charge density a alone, despite the fact that the field is of course due to all the charges, whether they are on the conductor or elsewhere. 0
0
EA
Figure 38 Portion of a charged conductor carrying a surface charge density a. The charge enclosed by the imaginary box is a da. There is zero field inside the conductor. Then, from Gauss's law, E = ff/e . 0
74
Fields of Stationary
EXAMPLE:
Electric Charges: II
HOLLOW
CONDUCTOR
A hollow conductor, as in Fig. 39, has a charge on its inner surface that is equal in magnitude and opposite in sign to any charge that may be enclosed within the hollow. This is readily demonstrated by considering a Gaussian surface that lies within the conductor and that encloses the hollow. Since E is everywhere zero on this surface, the total enclosed charge must be zero.
Figure 39 Crosssection of a hollow conductor. A charge Q in the hollow induces a total charge  Q on the inner surface. If the conductor has a zero net charge, a charge + Q is induced on the outer surface.
3.3.2
EXAMPLE:
ISOLATED
CHARGED
CONDUCTING
PLATE
An isolated charged conducting plate of infinite extent must carry equal charge densities a on its two faces, as in Fig. 310a. Outside, £ = 2(a/2e ) = a/e , while, inside the plate, the fields of the two surface distributions cancel and £ = 0. 0
EXAMPLE: CHARGES
0
PAIR OF PARALLEL CONDUCTING PLATES CARRYING OF EQUAL MAGNITUDES AND OPPOSITE SIGNS
If one has a pair of parallel conducting plates, as in Fig. 310b, carrying charges of egual magnitude and opposite signs, the charges position themselves as in the figure and £ = 0 everywhere except in the region between the plates.
75
:
a •
+
* "T"
• •
I
• a £ =0 E=0 + £ = —
TO
"
/•: = ()
£ =0
i
(a)
l
l
©
(b)
Figure 310 (a) Charged conducting plate carrying equal surface charge densities a on each side. T h e field inside the plate is zero, {b) Pair of parallel plates carrying surface charge densities +a and —
3A
POISSON'S
EQUATION
According to Gauss's law, Eq. 311, V • E = p/e . 0
Now, since E =
W,
from Eq. 213, „,„ SV 2
This is Poisson's
8V = — dx 2
dV dV +  3 + — dy dz 2
2
y
p =  A e
(316)
0
equation.
Poisson's equation often serves as the starting point for calculating electrostatic fields. It states that, within a constant factor, the charge density p is equal to the sum of the second derivatives of V with respect to x, y, z.
3.4.1
EXAMPLE:
FLAT
ION
BEAM
Figure 311 shows a flat ion beam situated between two grounded parallel conducting plates. A complete study of the electric and magnetic fields of the beam would be quite elaborate, but we shall disregard the motion of the ions and limit ourselves to the electric field, in the simple case where p is uniform inside the beam
76
p=0P
p=0
Figure 311 Flat ion beam of uniform space charge density p, between two grounded conducting plates.
and where edge effects are negligible. We shall use the subscript i on E and V inside the beam, and the subscript o outside. Inside the beam, d V,
tf
~dx
e'
2
T
Et =

(317)
0
dV,
px
=
dx
.4.
6A
(318)
where A is a constant of integration. By symmetry, E, = 0 at x = 0. Therefore A = 0; inside the beam, E, =
(319)
px/e . 0
+ B,
F = —~x
2
(320)
where B is another constant of integration. Now consider a cylindrical volume whose upper and lower faces are outside the beam, at equal distances from the plane x = 0, as in Fig. 311. The magnitude of E„ is the same at both faces. Let S be the area of one face. According to Gauss's law. 2£„S = 2aSp/e ,
(321)
0
£„ = ap/e
0
£
„ = ap/e
0
x > a,
(322)
x < a.
(323)
3.4 Poisson's Equation
77
Therefore, outside the beam. apx
—— + C
V„ =
x > a.
apx V. = — + D
x <
a,
(324)
(325)
where C and D are again constants of integration. Since V = 0 both at x = b and at x = — b, C = D =
apb/e
(326)
0
and ap V„ = —(b
x)
x > a.
(327)
x < —a.
(328)
N o w there can be no discontinuity of V at the surface of the beam, for otherwise E = — \V would be infinite. Then Eqs. 320 and 327 must agree at x = a:
^a> 2e
ap
B
+
(b  a).
(329)
0
B =
ap I,
a
—\b
2
(330)
Equating the V's from Eqs. 320 and 328 at x = a gives the same result. Finally, inside the beam,
r
p V: = — e
x + a 2 2
0
£, =
h
^
(331) (332)
px/e . 0
Above, for x > a.
K =
°Abx),
E = ap/e , 0
0
(333) (334)
78
Figure 312 The electric field intensity E and potential F a s functions of .v, between the two plates in Fig. 311. and below, for x < — a, V = ^(b
+ x),
0
(335)
E = ap/e . 0
(336)
0
The curves for V and E as functions of x are shown in Fig. 312.
3.5
LA PL A CE 'S EQ UA TION When p = 0, Poisson's equation becomes \V 2
= 0.
(337)
This is Laplace s equation. This equation applies not only to electrostatics but also to heat conduction, hydro and aerodynamics, elasticity, and so on.
3.6
THE UNIQUENESS
THEOREM
According to the uniqueness theorem, Poisson's equation has only one solution V(x,y,z), for a given charge density p(x,y,z) and for given boundary conditions. 1
f
For a proof of this theorem, see our Electromagnetic
Fields and Waves, p. 142.
3.7 Images
79
This is an important theorem. If, somehow, by intuition or by analogy, one finds a function V that satisfies both Poisson's equation and the given boundary conditions, then it is the correct V.
3.7
IMAGES The method of images involves the conversion of an electric field into another equivalent electric field that is simpler to calculate. With this method, one modifies the field in such a way that, for the region of interest, both the charge distribution and the boundary conditions are conserved. The method is a remarkable application of the uniqueness theorem. It is best explained by means of an example.
EXAMPLE: POINT CHARGE CONDUCTING PLATE
NEAR
AN INFINITE
GROUNDED
Consider a point charge Q at a distance D from an infinite conducting plate connected to ground, as in Fig. 313a. This plate may be taken to be at zero potential. It is clear that if we remove the grounded conductor and replace it by a charge — Q at a distance D behind the plane, then every point of the plane will be equidistant from Q and from — Q, and will thus be at zero potential. The field to the left of the conducting sheet is therefore unaffected. The charge — Q is said to be the image of the charge Q in the plane. The potential V at a point P(x,y) as in Fig. 313b, is given by Q
(338)
where + y)
r = (x
2
2 112
and
f = [(2D  x)
2
+ y ] ' . 2
1
The components of the electric field intensity at P are those of dV Qx Q(2D 47ie A,. = — 47re —— = —w H , n
n
BV Qy *te — =  3 dv r 0
x)
2
(339)
W:
,
Qy 73r
The lines of force and the equipotentials are shown in Fig. 314.
(340)
(341)
80
(a)
(b)
Figure 313 (a) Point charge Q near a grounded conducting plate, (b) The conducting plate has been replaced by the image charge — Q to calculate the field at P. At the surface of the conducting plate, x = D, r = /•', E = 2QD/4ne r\ x
0
E, = 0,
(3421
and the induced charge density is  e E . In this particular case, the surface charge density is negative, and the electric field intensity points to the right at the surface of the conducting plate. 0
x
Figure 314 Lines of force (identified by arrows) and equipotentials for a point charge near a grounded conducting plate. Equipotentials and lines of force near the charge cannot be shown because they get too close together. Equipotential surfaces are generated by rotating the figure about the axis designated by the curved arrow. The image field to the right of the conducting plate is indicated by broken lines.
3.8 Summary
81
It will be observed that image charges are located outside the region where we calculate the field. In this case we require the field in the region to the left of the plate, whereas the image is to the right. N o w what is the electrostatic force of attraction between the charge Q and the grounded plate? It is obviously the same as between two charges Q and —Q separated by a distance 2D, since Q cannot tell whether it is in the presence of a point charge —Q or of a grounded plate. The force on Q is thus Q /4ne {2D) . This is always true. The force between a charge and a conductor is always given correctly by the Coulomb force between the charge and its image charges. 2
2
0
SUMMARY The solid angle subtended by a closed surface S at a point P is 4n steradians if P is situated inside S, and zero if P is situated outside. Gauss's law follows from Coulomb's law. It can be stated either in integral or in differential form: (39)
where S' is the surface that encloses r', or V •E = Poisson's
/e .
fi
(311)
0
equation, \V
= p/co,
2
then follows from Eq. 311 and from E = — W. When the charge density is zero we have Laplace's \V 2
= 0.
(316)
equation, (337)
According to the uniqueness theorem, for a given p and for a given set of boundary conditions, there is only one possible potential satisfying Poisson's equation. The method of images is one remarkable application of the uniqueness theorem. For example, the field of a point charge Q in front of an infinite
82
Fields of Stationary
Electric Charges: II
grounded conducting plate is the same as if the plate were replaced by a charge — Q at the position of the image of Q. The m e t h o d of images gives the correct field only outside the region where the images are situated.
PROBLEMS 31E
ANGLE SUBTENDED BY A LINE AT A POINT What is the angle 0 subtended at a point P by a line of length a situated at a distance b as in Fig. 315?
Figure 315 Line of length a at a distance b from a point P. See Prob. 31. 32
SOLID
ANGLE SUBTENDED BY A DISK AT A POINT What is the solid angle subtended at a point P by a circular area of radius a at a distance b as in Fig. 316? Hint: Consider first the solid angle subtended at P by a ring of radius /• and width dr. and then integrate.
Figure 316 Circular area of radius a situated at a distance b from the point P. See Prob. 32.
33E
GAUSS'S LAW Could one use Gauss's law to calculate the electric field intensity near a dipole?
34E
SURFACE DENSITY OF ELECTRONS ON A CHARGED BODY The maximum electric field intensity that can be maintained in air is 3 x 10 volts per meter. a) Calculate the surface charge density that will produce this field. 6
Problems
83
b) Consider a metal with an interatomic spacing of 0.3 nanometer. What is the approximate number of atoms per square meter? c) How many extra electrons are required, per atom, to give a surface charge density of the above magnitude? 35
THE ELECTRIC FIELD IN A NUCLEUS a) Calculate the electric field intensity in volts per meter at the surface of an iodine nucleus (53 protons and 74 neutrons). b) Find the electric potential at the center of the nucleus. Assume that the charge density is uniform and that the radius of the nucleus is 1.25 x 1 0 j 4 meter, where A is the total number of particles in the nucleus.  1 5
1 / 3
36E
THE SPACE DERIVATIVES OF E , £ , E List all the relations between the various space derivatives of the three components of E, using Gauss's law and the fact that an electrostatic field is conservative.
37
PHYSICALLY IMPOSSIBLE FIELDS An electric field points everywhere in the z direction. a) What can you conclude about the values of the partial derivatives of E with respect to x, to y. and to z, (i) if the volume charge density p is zero, (ii) if the volume charge density is not zero? b) Sketch lines of force for some fields that are possible and for some that are not.
x
v
z
z
PROTON
BEAM
A 1.00microampere beam of protons is accelerated through a difference of potential of 10,000 volts. a) Calculate the volume charge density in the beam, once the protons have been accelerated, assuming that the current density is uniform within a diameter of 2.00 millimeters, and zero outside this diameter. (The current density in an ion beam is in fact largest in the center and falls off gradually with increasing radius.) Solution: The charge per unit length k is equal to the current I, divided by the velocity v of the protons: ;. =
i/ . v
in
Also, (l/2)mi' = eV, 2
(2)
where m is the proton mass, e is the proton charge, and V is the accelerating voltage. Thus,
(2eVjm)
m
\2eVj
84
Figure 317 Cylindrical surface surrounding the ion beam of Prob. 38 for calculating the magnitude of E using Gauss's law.
and = 2.300 x 1 0 " coulomb/meter ,
p = )JnR
2
7
3
(4)
where R is the radius of the beam. b) Calculate the radial electric field intensity, both inside and outside the beam. Solution: Outside the beam, E is radial. Applying Gauss's law to a cylindrical surface as in Fig. 317, 2nr(e E ) 0
t
= /.,
(5)
E„ = X/2ne r,
(6)
0
/ 2ne
/ m Y' 1 2
\2eVj H
0
r'
= 1.299 x 10" /r volts/meter,
(8)
2
= 12.99 volts/meter at r = 1 0 " meter. 3
(9)
Inside the beam, we find E by applying Gauss's law to a cylindrical surface of radius r < R. Then E = nr p/2tie r 2
i
0
= pr/2e ,
(10)
0
= 1.299 x 10 r volts/meter,
(11)
4
= 12.99 volts/meter at r = 1 0 " meter. 3
(12)
c) Draw a graph of the radial electric field intensity for values of r ranging from zero to 10 millimeters. Solution:
See Fig. 318.
85
Figure 318 Potential V and electric field intensity E as functions of the distance r from the axis for the proton beam of Prob. 38. The beam has a radius of 1 millimeter.
d) Now let the beam be situated on the axis of a grounded cylindrical conducting tube with an inside radius of 10.0 millimeters. Are the above values of E still valid? Solution: They are still valid: £„ depends solely on "/. and r, and £, on p and r. It is the value of V that is affected by the presence of the tube. e) Draw a graph of V inside the tube. Solution: Outside the beam, K = 0 a t r = 1 0 "
2
meter and beyond. Then
V„ = J ' ° " E„ dr = j ' " " 1.299 x 1 0 " dr/r, 2
r
(13)
= 1.299 x 1 0 " [ l n ;  ] " ,
(14)
= 1.299 x 1 0 " [ l n 1 0 "  In r ] ,
(15)
= (5.982 + 1.299 In r ) 1 0 " volt,
(16)
= 2.991 x 1 0 " volt at r = 1 0 " meter.
(17)
1 0
2
2
r
2
2
2
2
3
86
Fields of Stationary Electric Charges: II
Inside the beam, \ = 2.991 x 1CT +
(18)
2
11
(19)
2.991 x 1 ( T + 6.495 x 1 0 ( 1 ( T  r ) volt, 2
3
6
2
(20)
= 36.41 x 1 0 " volt at r = 0. 3
Figure 318 shows V as a function of r.
39
ION
BEAM An ion beam of uniform density p passes between a pair of parallel plates, one at x = 0, at zero potential, and one at x = a, at the potential V . Neglect the motion of the ions, as well as edge effects. The beam completely fills the space between the plates. Find V and the transverse £ as functions of x. 0
310
A UNIFORM AND A NONUNIFORM FIELD Consider a pair of parallel conducting plates, one at x = 0 that is grounded, and one at x = 0.1 meter that is maintained at a potential of 100 volts. a) Express V as a function of x when p = 0. Show the equipotentials at V = 0, 10, 2 0 , . . . 100 volts. Draw a curve of £ as a function of x. b) D o the same for the case where p/e is uniform and equal to 10 . 4
0
311
VACUUM DIODE In a vacuum diode, electrons are emitted by a heated cathode and collected by an anode. In the simplest case, the electrodes are plane, parallel, and close together, so edge effects can be neglected. Since the electrons move at a finite velocity, the space charge density is not zero. It is given by
P =
~
where V is the voltage at the anode (the voltage at the cathode is zero); s is the distance between the electrodes; and x is the distance of a point to the cathode. So, at x = 0, V = 0, while, at x = s, V = V . This peculiar charge distribution comes from the fact that the electrons are accelerated in the electric field, which itself depends on the charge density, and hence on the electron velocity. We assume that the current is limited by the space charge, and not by electron emission at the heated cathode. a) Use Poisson's equation to find V as a function of x. b) At any point, the current density J is pv, where v is the velocity of the electrons at that point. Use the value of pv at x = s to find J. 0
0
Problems
312
87
IMAGES Two charges + Q and — Q are separated by a distance a and are at a distance b from a large conducting sheet as in Fig. 319. Find the x and y components of the force exerted on the righthand charge.
+Q
Figure 319 Pair of charges near a conducting surface. See Prob. 312. 313
IMAGES A charge Q is situated near a conducting plate forming a right angle as in Fig. 320. Find the electric field intensity at P. There are three images of equal magnitudes.
Figure 320 Point charge near a conductor forming a right angle. See Prob. 313. 314
315
IMAGES A point charge Q is situated at a distance D from an infinite conducting plate connected to ground, as in Fig. 313. a) Find the surface charge density on the conductor as a function of the radius. b) Show that the total charge on the conductor is —Q.
ANTENNA
IMAGES
Antennas are often mounted near conducting surfaces. One example is the whip antenna shown in Fig. 184. Figure 321 shows schematically three antennas, or portions of antennas, near conducting surfaces. The arrows show the direction of current flow at a given instant.
Figure 321 Three antennas situated above conducting planes, and their images. The arrows show the direction of the currents in the antennas, at a given instant. See Prob. 315.
Draw figures showing the direction of current flow in the images. Solution: (a) Consider a positive charge moving downward in the vertical antenna. Its image is a negative charge that moves upward. Now a negative charge moving upward gives a downward current. So the current in the image is as shown in Fig. 322a. b) Similarly, a positive charge moving to the right has a negative image also moving to the right. The current in the image is in the opposite direction, as shown in Fig. 322b. c) A positive charge moving to the right and downward has a negative image moving to the right and upward. The image current is in the direction shown in Fig. 322c.
(a)
(b)
:
:
^
_
(<)
"
Figure 322 Relation between the current directions in the antennas and in their images.
CHAPTER
4
FIELDS OF STATIONARY ELECTRIC CHARGES: III Capacitance,
Energy, and Forces
4.1 4.1.1
CAPACITANCE OF AN ISOLATED Example: Isolated Spherical Conductor
4.2 4.2.1 4.2.2 4.2.3
CAPACITANCE BETWEEN TWO Example: ParallelPlate Capacitor Capacitors Connected in Parallel Capacitors Connected in Series
4.3
POTENTIAL
4.4 4.4.1
ENERGY Example:
4.5 4.5.1 4.5.2
FORCES ON CONDUCTORS Example: Electric Force for E = 3 x 10 volts per meter ParallelPlate Capacitor
4.6
ENERGY
CONDUCTOR CONDUCTORS
OF A CHARGE
DENSITY IN AN ELECTRIC Isolated Spherical Conductor
FIELD
6
SUMMARY PROBLEMS
DISTRIBUTION
In this chapter we shall complete our study of the electric fields of stationary electric charges in a vacuum. We shall start with the concept of capacitance. This will lead us to the energy stored in an electric field, and this energy, in turn, will give us the force exerted on a conductor situated in an electric field.
CAPACITANCE
OF AN ISOLATED
CONDUCTOR
Imagine a conductor situated a long distance away from any other body. As charge is added to it, its potential rises, the magnitude of the change in potential being proportional to the amount of charge added and depending on the geometrical configuration of the conductor. The ratio C = Q/V is called the capacitance by definition.
(41)
of the isolated conductor. A capacitance is positive,
The unit of capacitance is the coulomb per volt or
EXAMPLE:
ISOLATED
SPHERICAL
farad.
CONDUCTOR
An isolated spherical conductor of radius R carries a total surface charge Q. Then its capacitance is
91
Figure 41 Two conductors carrying equal and opposite charges. The integral of E • dl from one conductor to the other, along any path, gives the potential difference V. The capacitance between the conductors is Q/V.
4.2
CAPACITANCE
BETWEEN
TWO
CONDUCTORS
We have just defined the capacitance of an isolated conductor. We can proceed in a similar way to define the capacitance between two conductors, as in Fig. 41. Initially, both conductors are uncharged. Let us assume that one of the conductors is grounded, so that its potential is kept equal to zero. We then gradually add small charges to the other until it carries a charge Q. In the process, charges of opposite sign are attracted to the grounded conductor until it carries a charge —Q. At that point the grounded conductor is still at zero potential, while the other conductor is at a potential V. The capacitance between the two conductors is again defined as in Eq. 4  1 : C = Q/V.
(43)
The capacitance between two conductors is also positive, by definition. In other words, if we establish a potential difference V between two conductors by giving them charges +Q and — Q, then the capacitance between the conductors is Q/V.
4.2.1
EXAMPLE:
PARALLELPLATE
CAPACITOR
Figure 42 shows a parallelplate capacitor with the spacing s grossly exaggerated. We neglect edge effects. We have a uniform positive charge density er on the top
92
Figure 42 Parallelplate capacitor. The lower end of the small cylindrical figure is situated inside the lower plate where E = 0.
surface of the lower plate and a uniform negative charge density — a on the bottom surface of the upper plate. The electric field intensity E points upward. Applying Gauss's law to a cylindrical volume as in the figure, the outward flux of E is Eu. where a is the crosssection of the cylinder. Since the enclosed charge is (jo. then Ea = aa/e , 0
E = a'e ,
(44)
0
so that V = Es = as/e .
(45)
0
Now Q = a A, where A is the area of one plate, and C = Q/V = e A/s.
(46)
0
This is the capacitance between two parallel conducting plates of area A separated by a distance s.
4.2.2
CAPACITORS
CONNECTED
IN
PARALLEL
Figure 43a shows two capacitors C, and C
2
connected in parallel and
carrying charges Q a n d Q . We wish to find the value of C in Fig. 43b that l
2
will give the same V for the same charge Q + Q .This x
2
Qi + Q = c,v + c v = cv. 2
2
is simple: (47)
93 VQ
(a)
(b)
Figure 43 The single capacitor C has the same capacitance as the two capacitors C, and C connected in parallel. 2
Then C = C, + C .
(48)
2
So, the capacitance of two capacitors connected in parallel is the sum of the two capacitances. If we have several capacitors in parallel, the total capacitance is the sum of the individual capacitances.
4.2.3
CAPACITORS
CONNECTED
IN
SERIES
Figure 44a shows two capacitors C, and C connected in series. We first connect points A and B to ground, temporarily. This makes the voltages V and V zero. The capacitor plates are uncharged. We now disconnect A and B from ground. 2
l
2
We then apply a charge Q to A. We assume that the voltmeters draw zero current. N o charge can flow into or out of B. A charge — Q leaves the upper plate of C and goes to the lower plate of Cy. This leaves a charge + Q on the top plate of C . Then the lower plate of C, takes on a charge —Q, as in the figure. 2
2
Then Vi =
QIC,,
V = 2
Q/C . 2
(49)
94
VQ
VQ
(b)
(a)
Figure 44 The capacitor C has the same capacitance as the two capacitors C, and C connected in series. 2
N o w what should be the value of C in Fig. 44b to give a voltage V equal to K, + V with the same charge Q'l We must have 2
Q
Q
Q
c
c
c
c
c ,
V =  = — + —, x
(410)
2
or +
(4lD c
2
C =
(412)
If we have three capacitors in series, I
* . j . J _ + _L
(
4,3)
C^Cj
=
C C 2
3
+ C C, + 3
c,c
2
N o w the inverse of the capacitance, 1/C, is called the elastance. So, if capacitors are connected in series, the total elastance is the sum of the individual elastances.
4.3 Potential Energy of a Charge Distribution
95
Note that, when capacitors are connected in parallel, the voltages are the same. When capacitors are connected in series, the charges are the same.
POTENTIAL CHARGE
ENERGY OF A DISTRIBUTION
We can find a general expression for the potential energy of a charge distribution by considering the energy expended in charging a pair of conductors, as in Sec. 4.2. When we add a small charge dQ to the lefthand conductor, we must expend an energy dW = V dQ =  dQ,
(415)
where Q is the charge already on the conductor. Then, to charge the conductor to a potential V = QIC, we must supply an energy
W = F %dQ = % = QV =  CV . Jo c • 2C 2 2 C
1
2
(416)
The reason for the factor of onehalf should be clear from the reasoning that we have used to arrive at it. It is that, on the average, the potential V at the time of arrival of a charge dQ is just onehalf the final potential. More generally, if we have a number of charged conducting bodies, the stored energy is the sum of similar terms for each one:
W = WQ V, i
^
(417)
i
F o r a continuous charge distribution of density p(x',y',z') occupying a volume t ' , we can replace Q by p clr' and the summation by an integration: :
W
=  { v p d f ,
(418)
where V and p are both functions of the coordinates in the general case.
96
4.4
Fields of Stationary Electric Charges: III
ENERGY DENSITY ELECTRIC FIELD
IN AN
The energy stored in a parallelplate capacitor is therefore
\
c r
''HO'
l 4
•
l 9 ,
= Q e £ ) As,
(420)
2
0
where As is the volume occupied by the field. The stored energy may therefore be calculated by associating with each point an energy density [e /2)E joules per cubic meter. This is a general result. The energy associated with a charge distribution, that is, the energy required to assemble it, starting with the charge spread over an infinite volume, is 2
0
W = ^ [ E dr, 2 Jt
(421)
2
where the volume t extends to all the region where the field under consideration differs from zero. The energy stored in an electric field can therefore be expressed, either as in Eq. 418, or as in Eq. 421. N o t e that the terms under the integral signs in these two equations are unrelated. For example, at a point where p = 0, Vp is zero but E is usually not zero. 2
4.4.1
EXAMPLE:
ISOLATED
SPHERICAL
CONDUCTOR
For the isolated spherical conductor of radius R and carrying a charge Q, W = f
£ dr.
(422)
2
Using a spherical shell of radius r and thickness dr as element of volume dr. W = f2 r 9 JR 2 J* V/W
47tr dr. 2
2
(423)
Q
2
(424) This energy is jQV, orjCV , 2
where Cis the capacitance of the sphere, as in Sec. 4.1.1.
97
a
Figure 45 The local electric charge density at the surface of a conductor gives rise to oppositely directed electric field intensities
0
0
FORCES
ON
CONDUCTORS
An element of charge a da on the surface of a conductor experiences the electric field of all the other charges in the system and is therefore subject to an electric force. In a static field, this force must be perpendicular to the surface of the conductor, for otherwise the charges would move along the surface. Since the charge a da is b o u n d to the conductor by internal forces, the force acting on a da is transmitted to the conductor itself. T o calculate the magnitude of this force, we consider a conductor with a surface charge density a and a field E at the surface. F r o m Gauss's law, the electric field intensity just outside the conductor is
0
0
0
98
Fields of Stationary Electric Charges: III
The element of charge a da itself therefore produces exactly half the total field at a point outside, arbitrarily close to the surface. This is reasonable, for the nearby charge is more effective than the rest. N o w if a da produces half the field, then all the other charges must produce the other half, and the electric field intensity acting on a da must be ff/2e . The force dF on the element of area da of the conductor is therefore given by the product of its charge a da multiplied by the field of all the other charges: 0
dF = —o 2e
da,
(425)
0
and the force per unit area is 1 dF a — = — =  eE. da 2e 2 2
2
0
(426)
0
N o t e that this force is just equal to the energy density of Sec. 4.4 at the surface of the conductor. We can show that this is correct by using the method of virtual work. In this method we assume a small change in the system and apply the principle of conservation of energy. Let us imagine that the conducting bodies in the field are disconnected from their power supplies. Then, if the electric forces are allowed to perform mechanical work, they must do so at the expense of the electric energy stored in the field. Imagine that a small area a of a conductor is allowed to be pulled into the field by a small distance x. The mechanical work performed is ax times the the force per unit area. It is also equal to the energy lost by the field, which is ax times the energy density. Thus the force per unit area is equal to the energy density. If we use the same type of argument for a gas, we find that its energy density is equal to its pressure. N o t e that the electric force on a conductor always tends to pull the conductor into the field. In other words, an electric field exerts a negative pressure on a conductor. In order to visualize electric forces, it is useful to imagine that electric that they arrange themselves lines of force are real and under tension—also, in space as if they repelled each other.
99
F o r e x a m p l e , t h e lines of force b e t w e e n t h e p l a t e s of a c a p a c i t o r t e n d t o pull t h e p l a t e s t o g e t h e r a n d b u l g e o u t a t t h e e d g e s of t h e p l a t e s , as in F i g . 46.
4.5.1
EXAMPLE:
ELECTRIC
FORCE
FOR E = 3 x 10
6
VOLTS
PER
METER
The maximum electric field intensity that can be sustained in air at normal temperature and pressure is about 3 x 10 volts per meter. The force is then 40 newtons per square meter. 6
EXAMPLE:
PARALLELPLATE
CAPACITOR
Let us calculate the forces on the plates of a parallelplate capacitor of area S carrying a charge density + n on one plate and —
2
0
0
2
0
100
Figure 47 A charged parallelplate capacitor with its plates insulated. The plates are held in equilibrium by mechanical forces F that act on each plate in a direction tending to increase the separation s, and by electric forces F tending to decrease the separation.
s
m
e
SUMMARY If Q is the net charge carried by an isolated conductor, and if V is its potential, the ratio Q/V is called its capacitance. The capacitance between two isolated conductors is Q/V, where Q is the charge transferred from one to the other, and V is the resulting potential difference. The capacitance of two or more capacitors connected in parallel is the sum of the capacitances. The inverse of a capacitance is an elastance. The elastance of two or more capacitors connected in series is the sum of the elastances. The potential energy associated with a charge distribution can be written either as
(418) or as (421)
In the first integral, T' must be chosen to include all the charge distribution, and in the second, must include all regions of space where E is nonvanishing. The assignment of an energy density e E /2 to every point in space leads to the correct potential energy for the whole charge distribution. z
0
Finally, the force per unit area on a charged conductor is equal to cr /2e , or to the energy density in the electric field, e £ / 2 . 2
2
0
0
101
Figure 48 Parallelplate capacitor with seven plates. See Prob. 43.
PROBLEMS 41E
Show that e is expressed in farads per meter.
42E
THE EARTH'S ELECTRIC FIELD a) The earth has a radius of 6.4 x 10 meters. What is its capacitance? b) The earth carries a negative charge that gives a field of about 100 volts per meter at the surface. Calculate the total charge. c) Calculate the potential at the surface of the earth.
0
6
43E
PARALLELPLATE CAPACITOR Show that the capacitance of a parallelplate capacitor having TV plates, as in Fig. 48, is C = 8.85(N 
l)A/t picofarads,
where N is the number of plates, A is the area of one side of one plate, and t is the distance between the plates. One picofarad is 10" farad. 1 2
44E
PARALLELPLATE CAPACITOR Can you suggest reasonable dimensions for a 1 picofarad ( 1 0 " farad) airinsulated parallelplate capacitor that is to operate at a potential difference of 500 volts? 1 2
45
46
PARALLELPLATE CAPACITOR A parallelplate capacitor has plates of area S separated by a distance s. How is the capacitance affected by the introduction of an insulated sheet of metal of thickness s', parallel to the plates?
CYLINDRICAL
CAPACITOR
Calculate the capacitance per unit length C between two coaxial cylinders of radii R\ and R as in Fig. 49. Calculate C when R = 2R . 2
2
l
102
Figure 49 Cylindrical capacitor. See Prob. 46.
Solution: Let us assume that the inner conductor is positive and that it carries a charge of k coulombs per meter of length. Then the capacitance per unit length is k divided by the voltage V of the inner conductor with respect to the outer one, where
r ^ d r .
V = [ Ech= Rl
k
R
2ne
(2)
P,
0
(1)
Thus _
_2ne^ ln(P /R ) 2
For R
2
=
2R
lt
C
DROPLET
1
= 27T€ /ln 2 = 8.026 x 1 0 " 0
1 1
farad/meter.
(4)
GENERATOR
There are a number of devices that utilize charged droplets of liquid. Electrostatic spray guns (Prob. 28), colloid thrusters for spacecraft (Prob. 215). and inkjet printers (Prob. 417) are examples. As we shall see, drop size is controlled by the specific charge (coulombs per kilogram) carried by the fluid. a) When a charged droplet splits into two parts, electrostatic repulsion drives the droplets apart and the electrostatic energy decreases. Calculate the loss of electrostatic energy when a droplet of radius R carrying a charge Q splits into two equalsized droplets of charge Q/2 and radius R'. Assume that the droplets are repelled to a distance that is large compared to R'. Neglect evaporation.
Problems
103
Solution: Since 4 2 x nR'
4 = ~TIR\
3
(1)
then R' = R/2 .
(2)
113
Initially, the charge Q is at the potential of the surface of the drop of radius R and the electrostatic energy is QV
=
2
1
Q
2
(3)
2 4ne
0
After splitting, the electrostatic energy is (Q/2) 2" 2
2
3
(4)
8rce « 0
and the electrostatic energy gained is Q 1 AW = — ^ — KTT 
\ 1 = 0.3700
Q
2
(5)
b) The energy associated with surface tension is equal to a constant that is a characteristic of the liquid, multiplied by the surface area. The constant is called the surface tension of the liquid. The surface tension of water is 7.275 x 1 0 ~ joule per square meter. When a droplet splits, the surface area increases, and the energy associated with the surface tension increases proportionately. Calculate the gain in this energy when a droplet of water of radius R splits into two equalsized droplets. 2
Solution:
The increase in surface energy is
AW = (2 x 4nR'
2
 4nR )T 2
=
2 ' 2
V
= 0.2599 x 4 ; i R r = 0.2376K , 2
2
3

\ ]4nR T, 2
(6)
(7)
where T is the surface tension of water. c) A droplet of water has a radius of one micrometer and carries a specific charge of one coulomb per kilogram. Will the droplet split?
104
Fields of Stationary
Electric Charges:
III
Solution: The droplet will split if the total AM is negative. The charge Q is numerically equal to the mass, in this case: 7
4 , Q = 1000 x  nR coulombs.
(8)
Thus
[(4000rc/3)J? ]
3 2
AW =  0 . 3 7 0 0 
— — — + 0.2376K , Sne R
(9)
2
0
=  2 . 9 1 9 x 10 R 16
+ 0.2376K .
S
(10)
2
For R = 1 0 " , 6
AW =  2 . 9 1 9 x 10"
+ 0.2376 x 10"
1 4
1 2
joule.
(11)
The droplet will not split. d) What is the radius of the largest droplet of water that is stable with this specific charge? Solution: For a specific charge of one coulomb per kilogram, a droplet of water will not split spontaneously if its radius is such that 2.919 x 10 R 1 6
5
+ 0.2376K > 0. 2
or for radii smaller than 1/3
0.2376 2.919 x 10
= 2.012 x 1 0 " meter.
(12)
6
16
48E
ELECTROSTATIC ENERGY a) Two capacitors are connected in series as in Fig. 410a. Calculate the ratio of the stored energies. b) Calculate this ratio for capacitors connected in parallel as in Fig. 410b.
49E
ELECTROSTATIC ENERGY Two capacitors of capacitances C, and C have charges Q and Q , respectively. a) Calculate the amount of energy dissipated when they are connected in parallel. b) How is this energy dissipated? 2
410
x
2
PROTON BOMB a) Calculate the electric potential energy of a sphere of radius R carrying a total charge Q uniformly distributed throughout its volume.
105
+

O V o 
c,
Figure 410 Two capacitors: (a) connected in series, and (b) connected in parallel. See Prob. 48.
b) Now calculate the gravitational potential energy of a sphere of radius R' of total mass M. c) The moon has a mass of 7.33 x 1 0 kilograms and a radius of 1.74 x 10° meters. Calculate its gravitational potential energy. d) Imagine that you can assemble a sphere of protons with a density equal to that of water. What would be the radius of this sphere if its electric potential energy were equal to the gravitational potential energy of the moon? 22
411
ELECTROSTATIC MOTOR Can you suggest a rough design for an electrostatic motor? Draw a sketch, explain its operation, specify voltages and currents, and make a rough estimate of what its power would be.
412
ELECTROSTATIC PRESSURE A light spherical balloon is made of conducting material. It is suggested that it could be kept spherical simply by connecting it to a highvoltage source. The balloon has a diameter of 100 millimeters, and the maximum breakdown voltage in air is 3 x 10 volts per meter. a) What must be the voltage of the source if the electric force is to be as large as possible? b) What gas pressure inside the balloon would produce the same effect? 6
413
PARALLELPLATE CAPACITOR Calculate the force of attraction F between the plates of a parallelplate capacitor. Assume that the capacitor is connected to a battery supplying a constant voltage V. Use the method of virtual work, assuming a small increase ds in the spacing s between the plates, and set dW
B
+ dW
m
=
dW , e
106
Fields of Stationary
Electric Charges:
III
where dW is the work done by the battery, dW = Fds is the mechanical work done on the system, and dW is the increase in the energy stored in the electric field. You should find that B
m
e
I K 1 F =  e„ — S =2 .r 2
, e E S.
2
2
0
Note that one half of the energy supplied by the battery appears as mechanical work, and one half as an increase in the energy stored in the electric field. This is a general rule. 414
PARALLELPLATE CAPACITOR Calculate the force of attraction between the plates of a parallelplate capacitor, as in the preceding problem, assuming now that the capacitor is charged and disconnected from the battery. In this case, =
dW
m
dW . e
You should find the same result. 415
OSCILLATING PARALLELPLATE CAPACITOR Figure 411 shows a parallelplate capacitor whose upper plate of mass m is supported by a spring insulated from ground. A voltage V is applied to the upper plate. There is zero tension in the spring when the length of the air gap is x . Set m = 0.1 kilogram, x = 0.01 meter, k = 150 newtons per meter, V = 100 volts, and A = 0.01 square meter. Assume that the mass of the spring is negligible, and neglect edge effects. a) Set the various potential energies of the complete system equal to zero when the tension in the spring is zero, at x = x . Show that the potential energy is then 0
0
0
W = mg(x  x ) 0
+
I k(x
2
 x )
2
0
 \ e A V (2 \x

2
0
—
x
0
Draw a curve of W as a function of x between x = 1 0 ~ and x = 1 0 ~ meter. Use a logarithmic scale for the xaxis. b) Show that, at equilibrium, 6
mg + k(x  x ) + 0
i
~e AV /x 2
1
0
2
= 0.
You can arrive at this result in two different ways. There is stable equilibrium at x = 3.46 x 10" meter and unstable equilibrium at x = 2.93 x 1 0 " meter. c) If the system is brought to the point of stable equilibrium, and then disturbed slightly, it oscillates. For small oscillations, the restoring force F is proportional to the 3
eq
5
107
Figure 411 Parallelplate capacitor, with the lower plate fixed in position and the upper plate suspended by a spring insulated from ground. See Prob. 415.
displacement and we have a simple harmonic motion: (d x\
(dW\
2
where the derivatives must be evaluated in the neighborhood of .x = The circular frequency is o j = (K/m) , where
x. eq
111
Show that
OJ =
'k 
(e AV /x ) 2
0
m
3
eq
—
This gives a frequency of 6.16 hertz. 416
HIGHVOLTAGE GENERATOR Imagine the following simpleminded highvoltage generator. A parallelplate capacitor has one fixed plate that is permanently connected to ground, and one plate that is movable. When the plates are close together at the distance s, the capacitor is charged by a battery to a voltage V. Then the movable plate is disconnected from the battery and moved out to a distance ns. The voltage on this plate then increases to nV, if we neglect edge effects. Once the voltage has been raised to nV, the plate is discharged through a load resistance. a) Verify that there is conservation of energy. b) Can you suggest a rough design for such a highvoltage generator with a more convenient geometry?
108
QK
mm
N
Figure 412 Inkjet printer. A nozzle N forms a fine jet that breaks up into droplets inside electrode £*. A variable voltage V applied to the electrode induces a known charge on each droplet in succession. The droplets are deflected in the constant electric field between the deflecting plates and impinge on the paper form P. Uncharged droplets are collected at C. See Prob. 417.
417
INKJET PRINTER The inkjet printer is one solution to the problem of printing data at high speed. It operates more or less like an oscilloscope, except that it utilizes microscopic droplets of ink, instead of electrons. See Prob. 47. Its principle of operation is shown in Fig. 412. A nozzle produces a fine jet of conducting ink that separates into droplets inside a cylindrical electrode. The potential on the electrode is either positive or zero. At a given instant, the charge carried by a droplet that breaks off from the jet depends on the charge induced on the jet and hence on V. The droplets are then deflected as in an oscilloscope, except that here the deflecting field is constant, each droplet being deflected according to its charge. Uncharged droplets are collected in an ink sump. The jet operates continuously, and only a small fraction of the droplets is actually used. The deflection is vertical and the jet moves horizontally. The pressure in the nozzle is modulated at frequencies up to 0.5 megahertz, producing up to 5 x 10' droplets per second, with radii of the order of 10 micrometers. a) Let the radius of the jet be R, and the inside radius of electrode E be R What is the charge per unit length on the jet, neglecting end effects? b) The jet breaks up into larger droplets of radius 2R . What length of jet is required to form one droplet? c) Calculate the charge Q per droplet. 2
X
Problems
109
d) Calculate the specific charge in coulombs per kilogram for = 20 micrometers, R = 5 millimeters, V = 100 volts. Set the ink density equal to that of water (1000 kilograms per cubic meter). e) Calculate the velocity of the droplets if they are produced at the rate of 10' per second and if they are separated by a distance of 100 micrometers. f) Calculate the vertical deflection and the vertical velocity of a droplet upon leaving the deflecting plates, assuming a uniform transverse field of 10 volts per meter extending over a horizontal distance of 40 millimeters. 2
5
CHAPTER DIRECT
5 CURRENTS
IN ELECTRIC
CIRCUITS
5.1 5.1.1 5.1.2
CONDUCTION OF ELECTRICITY Conservation of Charge Charge Density in a Homogeneous Conductor
5.2 5.2.1
OHM'S LAW The Joule Effect
5.3 5.3.1 5.3.2
NONLINEAR RESISTORS Example: Incandescent Lamp Example: VoltageDependent Resistors
5.4
RESISTORS
CONNECTED
IN
5.5
RESISTORS
CONNECTED
IN
5.6 5.6.1
THE PRINCIPLE OF SUPERPOSITION Example: Simple Circuit with Two Sources
5.7 5.7.1
KIRCHOFF'S LAWS Example: The Branch Currents in a TwoMesh
5.8 5.8.1 5.8.2
SUBSTITUTION THEOREM Example: Resistor Replaced by a Battery Example: Battery Replaced by a Resistor
5.9 5.9.1
MESH Example:
5.10 5.10.1
DELTASTAR TRANSFORMATIONS Example: Transformation of a Symmetrical Symmetrical Star
5.11
VOLTAGE
5.12 5.12.1 5.12.2
THEVENIN'S THEOREM Example: Thevenin's Theorem Applied to a Simple Example and Review: The Wheatstone Bridge
5.13
SOLVING
5.14 5.14.1
TRANSIENTS IN RC CIRCUITS Example: Simple RC Circuit
5.15
PARALLEL
CURRENTS The Mesh Currents in a TwoMesh
AND CURRENT
A SIMPLE
SUMMARY PROBLEMS
SERIES
Circuit
Circuit Delta and of a
SOURCES
DIFFERENTIAL
Circuit
EQUATION
Until now we have limited ourselves to the study of the fields of stationary charges. We shall now consider the flow of electric current, and we shall develop methods for calculating the currents in complex arrangements of conductors. We shall not be concerned as yet with magnetic fields.
CONDUCTION
OF
ELECTRICITY
We have seen in Sec. 3.3 that, under static conditions, E is zero inside a conductor. However, if an electric field is maintained by an external source— for example, when one connects a length of copper wire between the terminals of a battery—then charge carriers drift in the field and there is an electric current. Within the conductor, the current density vector J points in the direction of flow of positive charge carriers; for negative carriers, J points in the direction opposite to the flow, as in Fig. 5  1 . The magnitude of J is the quantity of charge flowing through an infinitesimal surface perpendicular to the direction of flow, per unit area and per unit time. Current density is expressed in coulombs per square meter per second, or in amperes per square meter. In good conductors, such as copper, the charge carriers are the conduction electrons, and we have the situation shown in Fig. 5lb. Semiconductors contain either or both of two types of mobile charges, namely conduction electrons and holes. A hole is a vacancy left by the release of an electron by an atom. A hole can migrate in the material as if it were a positive particle. F o r ordinary conductors, the current density J is proportional to the electric field intensity E: J = crE,
(51)
112
^ 0 0 —
^  0
^ 0
^ 0
da
da
^ 0 0
^ 0

Figure 51 Current density vector J for (a) positive and (b) negative charge carriers.
Table 51
Conductor
Conductivity a (siemens per meter)
Aluminum
3.54 x 10
7
Brass (65.8 Cu, 34.2 Zn)
1.59 x 10
7
Chromium
3.8 x 10
Copper
5.80 x 10
7
Gold
4.50 x 10
7
Graphite
1.0 x 10
5
Magnetic iron
1.0 x 10
7
Mumetal (75 Ni, 2 Cr, 5 Cu, 18 Fe)
0.16 x 10
Nickel
1.3 x 10
Sea water
7
7
7
ss 5
Silver
6.15 x 10
Tin
0.870 x 10
Zinc
1.86 x 10
7
7
7
5.1 Conduction of Electricity
113
where a is the conductivity of the material. Conductivity is expressed in amperes per volt per meter, or in Siemens per meter. Table 51 gives the conductivities of a number of c o m m o n materials. The drift velocity v of conduction electrons is easily calculated. It is surprisingly low. If n is the number of conduction electrons per cubic meter and if e is the magnitude of the electron charge, then 11
J = nev.
(52)
In copper, there is one conduction electron per atom, and 8.5 x 1 0 atoms per cubic meter, so n = 8.5 x 1 0 electrons per cubic meter. F o r a current of one ampere in a wire having a crosssection of one square millimeter, J = 10 amperes per square meter and 2 8
2 8
6
10 8.5 x 1 0
2 8
6
x 1.6 x 10
x 7.4 x 10
5
meter/second,
(53)
or about 0.26 meter per hour}
5.1.1
CONSERVATION
OF
CHARGE
N o w consider a volume T bounded by a surface S inside conducting material. The integral of J • da over the surface is the charge flowing out of S in one second, or the charge lost by the volume % in one second. Then
l  Jtl 3
Aa=
pch
(5
"
4)
This is the law of conservation of charge. Using the divergence theorem (Sec. 1.10) on the lefthand side, J V j *
(55)
After Ernst Werner von Siemens (18161892). The word therefore takes a terminal s in the singular: one Siemens. The Siemens was formerly called a "mho." f
Then how is it that one can, say, turn on a light kilometers away, instantaneously? The moment one closes the switch, a guided electromagnetic wave is launched along the wire. This wave travels at the velocity of light and establishes the electric field in and around the wire. {
114
Direct Currents in Electric
Circuits
Since this equation is valid for any volume,
V•J =
—2.
(56)
dt This is again the law of conservation of charge, stated in differential form.
5.1.2
CHARGE
DENSITY
IN A HOMOGENEOUS
CONDUCTOR
Under steadystate conditions, the time derivative in Eq. 56 is zero. Then V • J = V •
(57)
If now we have a homogeneous conductor with a independent of the coordinates, V • E is zero. Thus, from Eq. 311, the net charge density inside a homogeneous conductor carrying a current under steadystate conditions is zero.
5.2
OHM'S
LA W
Figure 52 shows an element of conductor of length / and crosssection A. If a difference of potential V is maintained between its two ends, then J = I/A = oE = aV/l,
(58)
and aVA
—r~m
l
V
V
= R>
,5
"
9)
where R = 1/oA
(510)
is the resistance of the element. Resistances are expressed in volts per ampere or in ohms.
115
Figure 52 Element of conductor of crosssection A and length / carrying a current /.
V
Figure 53 The current / through a linear resistor is proportional to the voltage V across it.
Equation 59 shows that the current through a resistor is proportional to the voltage across it, as in Fig. 53. This is Ohms law. A resistor satisfying this law is said to be ohmic, or linear. Equation 51 is another, more general form of O h m ' s law. The current and the voltage are measured as in Fig. 54. It is assumed that the current flowing through the voltmeter V is negligible compared to that through R. That is usually the case.
VWV\A R Figure 54 Measurement of the current / through a resistor R. and of the voltage V across it. A circle marked 1 represents an ammeter, and a circle marked V, a voltmeter.
/16
5.2.1
Direct Currents in Electric
THE JOULE
Circuits
EFFECT
When a current flows through a conductor, the charge carriers gain kinetic energy by moving in the electric field. Their velocity never becomes large, however, because they can travel only a short distance before colliding with an atom or a molecule. The kinetic energy gained by the carriers thus serves to raise the temperature of the conducting medium. This phenomenon is called the Joule
effect.
The energy that is dissipated as heat in this way is easily calculated. Consider a cubic meter of conductor with E parallel to one edge. The current density J is the charge flowing through the cube in one second. The voltage difference across the cube is E. Then the kinetic energy gained by the carriers and lost to the conductor as heat in one second is JE, or
2
2
2
If one has a resistance R carrying a current L then the voltage across it is IR and the kinetic energy gained by the charge carriers in one second is I(IR). Then the power dissipated as heat is I R. 2
5.3
NONLINEAR
RESISTORS
Although O h m ' s law applies to ordinary resistors, it is by no means general.
EXAMPLE:
INCANDESCENT
LAMPS
Figure 55 shows the current as a function of voltage for a 60watt incandescent lamp. The resistance V/I is about 20 ohms at a few volts, and about 250 ohms at 100 volts. The reason is that, in the interval, the temperature of the tungsten filament has changed from 300 to 2400 kelvins. If the temperature of the filament were maintained constant by external means, say by immersing it in a constanttemperature oil bath, then it would follow Ohm's law.
EXAMPLE:
VOLTAGEDEPENDENT
RESISTORS
Voltagedependent resistors utilize ceramic semiconductors for which V = CI , where C and B are constants, and B is less than unity. This equation applies at room temperature. Figure 56 shows the currentvoltage curve for one particular type with C = 100 and B = 0.2. B
117
0
I
I
I
I
I
I
I
I
i
I
50 V (volts)
I
100
Figure 55 Currentvoltage relationship for a 60watt incandescent lamp.
0
25
50 75 V (volts)
100
Figure 56 Currentvoltage relationship for a certain type of voltagedependent resistor.
Voltagedependent resistors are used on telephone lines for shortcircuiting lightning to ground, for suppressing sparks, and so on, since their resistance decreases at large voltages. We shall be concerned henceforth only with linear resistors.
118
0 Figure 57 Three resistors connected in series.
5.4
RESISTORS
CONNECTED
IN
SERIES
Figure 57 shows three resistors connected in series. Since V = IRi
+ IR
+ IRi = HR, + R
2
2
+ R ) = IR . 3
S
(511)
the resistors have a total resistance R = R s
+ R
l
+ R.
2
(512)
3
The resistance of a set of resistors connected in series is the sum of the individual resistances.
5.5
RESISTORS
CONNECTED
IN
PARALLEL
In Fig. 58 the resistors are connected in parrallel. Then _ 7
=
V R
V +
x
R~
V _ +
2
~R ~
( 1 V
1 R~
+
3
2
+
1 \
R~J
 r;

(5 131
and the resistance R is given by p
1 R
R
1 Ri
p
_
1
1 R
R'
2
l23
R R
(514)
3
R
(515)
119
0V\AAA
O
WvAA R
3
Figure 58 Three resistors connected in parallel.
It is useful to remember that, for two resistances in parallel, R,R "
R
{
It is often convenient to use
2
+
(516)
R
2
conductance
G =
\/R,
(517)
instead of resistance. Then
Gi + G + G . 2
3
(518)
The conductance of a set of resistors connected in parallel is the sum of the individual conductances. Conductances are expressed in amperes per volt or in Siemens. We have already used the Siemens in Sec. 5.1.
THE PRINCIPLE
OF
SUPERPOSITION
The principle of superposition which we studied in Sec. 2.3 applies to electric circuits comprising sources and linear resistors: if a circuit comprises two or more sources, each source acts independently of the others and, at any point in the circuit, the current is the algebraic sum of the currents produced by the individual sources.
120
2C1
NAAA/^
ion Figure 59 The current flowing through the circuit is the algebraic sum of the current due to the 6volt battery, plus that due to the 12volt battery.
EXAMPLE:
SIMPLE
CIRCUIT
WITH
12V
TWO
SOURCES
Figure 59 shows a simple circuit comprising three resistors and two sources, all connected in series. The total resistance is 15 ohms. The 12volt source produces a counterclockwise current of 12/15 = 0.8 ampere, while the 6volt source produces a clockwise current of 0.4 ampere. The net current is 0.4 ampere, counterclockwise. The power dissipated as heat in the 2ohm resistor is (0.4) x 2 = 0.32 watt, and the total power dissipated in the circuit is (0.4) x 15 = 2.4 watts. 2
2
5.7
KIRCHOFF'S
LAWS
Let us consider a general circuit composed of resistors and voltage sources, for example, that of Fig. 510. Junctions such as A, B, C, D, E are called nodes; connections between nodes, such as AB, BD, DC .. ., are called branches; closed circuits, such as ABDE, or ABCDE, or BCD, are called meshes. Kirchoff's laws provide a method for calculating the branch currents when the values of the resistances and of the voltages supplied by the sources are known. Kirchoff's current law states that, at any given node, the sum of the incoming currents is equal to the sum of the outgoing currents. This is simply because charge does not accumulate at the nodes. For example, at A,
h = h+ h
(519)
121
Figure 510 Part of a complex circuit.
According to Kirchoff's voltage law, the voltage rise, or the voltage drop, around any given mesh is zero. For example, the voltage rise from A to B, to D, to £, and back to A, is clearly zero, so V,
+ /,/?,  IR 2
+ IR
2
5
5
+ IR 6
= 0.
b
(520)
T o find the branch currents, one writes down as many equations of the above two types as there are branches, and one solves the resulting system of simultaneous equations. The directions chosen for the currents are arbitrary. If the calculations give a positive value for, say, / , , then that current flows in the direction of the arrow. If the calculations give a negative value, then the current flows in the opposite direction.
EXAMPLE:
THE BRANCH
CURRENTS
IN A TWOMESH
CIRCUIT
For the circuit of Fig. 511, we have three unknowns, / , , I . 13. Hence we need three equations. At node A, 2
h = h + h
(521)
For the lefthand mesh, starting at B, clockwise. V — lR x
— IR 3
= 0.
(522)
122
i—AA/V
f
MA/—i t /
K
3
T
Figure 511 One can calculate I
lt
I , / , given .R and K 2
3
Similarly, for the righthand mesh, starting at B, clockwise, Z i?  7 P  I R 3
2
2
(523)
= 0.
Solving, we find that I, = 3F/5R,
SUBSTITUTION
/ , = 1 5K.
/
(524)
= 2F/5R.
3
THEOREM
Let us return to Fig. 510 and to Eq. 520. Consider the resistance R . The voltage drop across it is I R . Let us assume that I is really in the direction of the arrow. Then the upper end of R is positive. N o w let us substitute for R a voltage source as in Fig. 512. Equation 520 is unaffected. Similarly, the corresponding equation for the mesh BCD is unaffected. Then the currents 11, I , ^ 3 , and so forth, are unaffected. Thus, if a current I flows through a resistance R, one can replace the resistance by a voltage source IR of the same polarity without affecting any of the currents in the circuit. Inversely, if we have a voltage source V passing a current /, with the current entering the source at the positive terminal, one can replace the source by a resistance V/I without affecting any of the currents in the circuit. This is the substitution theorem. 2
2
2
2
2
2
2
5.8.1
EXAMPLE:
RESISTOR
REPLACED
BY A
BATTERY
In Fig. 513a, h = h + fR I R 2
+ IR
(525)
h,
(526)
= V,
2
(527)
+ 1,R = 0,
R
1
/,
/,
AA/V^—*—<
• —
+
+
+
j
1
R *
V

K/3
<> (6) Figure 513 The currents / , , I ,1 for the resistor R. 2
3
are not affected by substituting the battery F/3
124
Direct Currents in Electric
Circuits
giving 7j = 2V/3R,
I
2
= V/3R,
l
3
= VJ3R.
(528)
Using now the substitution theorem, we can replace the resistor on the right by a source I R = V/3, giving the circuit of Fig. 513b. Then 3
h = h+
h,
1,R + 1 R = V,
(530)
2
I R 2
(529)
+ (V/3) = 0.
(531)
Solving, we find the same values for J , 1 , / as before. Inversely, one can substitute the circuit of Fig. 513a for that of Fig. 513b without affecting J j , I , 7 . 1
2
EXAMPLE:
2
3
3
BATTERY
REPLACED
BY A
RESISTOR
In Sec. 5.6.1 we found that the current in Fig. 59 is counterclockwise. Since a current of 0.4 ampere flows into the positive terminal of the 6volt source, we can replace this source by a resistance of 6/0.4 = 1 5 ohms without affecting the current flowing in the circuit. Let us check. We now have a 12volt source and a total resistance of 15 + 10 + 3 + 2 = 30 ohms. Then the current is 12/30 = 0.4 ampere, as previously.
5.9
MESH
CURRENTS
If we use Kirchoff's laws as such, the number of unknown currents is equal to the number of branches. It is possible to reduce the number of unknowns in the following way. Instead of using branch currents as in Fig. 510, one uses mesh currents as in Fig. 514, since there are fewer mesh currents than branch currents. Kirchoff's current law is then automatically satisfied, and we need to apply only the voltage law. The advantage of using mesh currents comes from the fact that the time required to solve a set of simultaneous equations decreases rapidly as the number of equations is reduced. In the end, one must find the branch currents, but that is a simple matter, once the mesh currents are known.
125
Figure 514 Mesh currents in the circuit of Fig. 510.
5.9.1
EXAMPLE:
THE MESH
CURRENTS
IN A TWOMESH
CIRCUIT
In the circuit of Fig. 515 we have two meshes. Starting at B in both cases and proceeding in the clockwise direction, V — I R — (/„  I„)R = o,
(532)
= 0.
(533)
a
(/„  I )R  I 2R b
b
and 7=3175/?,
h=V/5R.
(534)
Note that we have only two unknowns, /„ and I . instead of the three we had previously. b
R
A
R
WW—f—vAAA/^—i
Figure 515 Mesh currents in the circuit of Fig. 511.
126
5.10
Direct Currents in Electric
DELTASTAR
Circuits
TRANSFORMATIONS
Figure 156 shows three nodes A, B, C, deltaconnected in part a and starconnected in part b. The nodes A, B, C form part of a larger circuit, and can be connected together either as in Fig. 516a or as in Fig. 516b. We shall see that, if certain relations are satisfied between the three resistances on the left and the three on the right, one circuit may be substituted for the other without disturbing the mesh currents. In fact, if one had two boxes, one containing a delta circuit, and the other containing a star circuit satisfying Eqs. 538 to 540, with only the terminals A, B, C showing on each box, there would be no way of telling which box contained the delta and which contained the star. The two circuits are thus said to be equivalent. This transformation is often used for simplifying the calculation of mesh currents. We could find R , R , R in terms of R , R , R , and inversely, by assuming the same mesh currents I , I , I in the two circuits, as in the figure, and then making the voltages V — V ,V  V , V  V in Fig. 516a equal to the corresponding voltages in Fig. 516b. This is the object of Prob. A
B
c
a
A
B
A
A
b
c
c
B
B
c
c
A
A
Figure 516 (a) Threenode circuit having the shape of a capital delta, (b) Fournode circuit having the shape of a star. Nodes A, B, C are part of a more extensive circuit. It is shown that, if either Eqs. 538 to 540 or Eqs. 544 to 546 are satisfied, the above two circuits are completely equivalent.
5.10 DeltaStar
Transformations
127
519. We shall use here another derivation that is somewhat less convincing, but shorter. We disconnect the delta and the star from the rest of the circuit. If the circuits are equivalent, then the resistance between A and B must be the same in b o t h :
RAB = „*<[V
l
R b
R +R + R a
h
= R
+ R
A
(535)
B
c
Similarly,
Note that, in the above equations, is associated with R R . R with i? /?„. and R with R R,,. Thus, simply by inspection, we have that h
c
c
B
c
0
R,. + R + R h
R
R
B
=
c
=
r
f°
,
R
D
(539)

<  °) 5
4
N o w we also need the inverse relationship. We could deduce it from the above three equations, but that would be rather long, and not very instructive. Instead, let us use the conductance between node A and the nodes B and C shortcircuited together. This gives
G
G = f> G
b +
e
+
r
VIA
+
°i
Oil +
• (/(•
(541.
128
Direct Currents in Electric
Circuits
Similarly,
G
c
+
G
"  G ,
R
+
G
G (G, +
G + G= a
G
+
c
b
G.4
r
\
B
, 5
GJ
A
+ G
'
r
+
O
.
"
4 2 )
,543)
c
Again, by inspection. GG B
G
A
C
(544)
+ G + G B
c
GG C
G
A
+ G + G B
GG A
G =
c
B
.
c
G
A
(545)
A
(546)
+ G + G B
c
EXAMPLE: TRANSFORMATION OF A SYMMETRICAL AND OF A SYMMETRICAL STAR
DELTA
In Fig. 516a, suppose R , R , R are all 100ohm resistances. Then, from Eqs. 538 to 540, the equivalent starconnected circuit is made up of three resistances of (100 x 100)/300 = 33 ohms each. If, instead, we had a starconnected circuit with resistances R , R , R of 100 ohms each, then the equivalent deltaconnected circuit would have three identical resistances with conductances given by Eqs. 544 to 546: 10" /(3/100) = 10" / 3 Siemens. The resistances in the delta would thus each have a resistance of 300 ohms. a
b
c
A
4
5.11
VOL TA GE A ND CURRENT
B
c
2
SO URCES
The ideal voltage source supplies a constant voltage that is independent of the current drawn, and hence of the load resistance. A good electronically voltagestabilized power supply is close to ideal, up to a specified current, beyond which the voltage falls as in Fig. 517. Similarly, the ideal current source supplies a constant current that is independent of the voltage across the load. Again, currentstabilized
129
O
I
Figure 517 Output voltage as a function of output current, for a voltage source. /
0
V
Figure 518 Output current as a function of output voltage, for a current source.
supplies are nearly ideal, but only up to a specified voltage beyond which the current decreases. This is shown in Fig. 518.
5.12
THEVENIN'S
THEOREM
It is found experimentally that the voltage supplied by a source decreases when the load current increases. This is true, even for voltagestabilized sources: the plateau in Fig. 517 is not really flat but slopes down slightly.
130
V
Q
0 •
I
(b)
(a)
Figure 519 (a) Any real source is equivalent to an ideal source V in series with a resistance R . The source is shown here connected to a load resistance R drawing a current /. The voltage V between the terminals is V — IR or IR. (b) The terminal voltage V as a function of the load current I. The slope is — Rj. 0
t
0
h
Thevenin's theorem states that any source—a flashlight battery, an oscillator, a voltage or currentstabilized power supply, etc.—is equivalent to an ideal voltage source V in series with a resistance R,, as in Fig. 519, where R, is called the output resistance, or the source resistance. An ideal voltage source, by definition, has a zero R,. In Fig. 519, the current I is F /(R, + R), while the voltage V is IR. The voltage V is smaller than V by the factor R/(R; + R), because of the voltage drop on R,. With R disconnected (R > o c ) , V is equal to F . The internal resistance R, is given by the slope of the curve of V as a function of I. If one increases / by decreasing R, then V decreases and R, is equal to lAK/A/l. For one particular electronically ro/fagestabilized power supply, the output voltage drops by one millivolt when the output current increases from zero to 100 milliamperes. Thus R, is 0.01 ohm. A small battery charger has an internal resistance that is of the order of one o h m : if the load current increases by one ampere, the output voltage drops by about one volt. A currairstabilized power supply has a large V and a large R . Its output current is V /R as long as R « R,, Thus its output voltage (F /R,)R is proportional to the load resistance R. If the load resistor is disconnected, then R is infinite and the output voltage rises to some limiting value that can be much smaller than V and that is fixed internally. In one particular currentstabilized supply, V is 50 kilovolts, R is 100 kilohms, and the limiting voltage is 50 volts. Thus, if V and R are known, one can predict the values of V and I for a given load resistance R. Thevenin's theorem therefore provides us with a simple model for describing the operation of a real source. 0
0
0
0
0
0
h
0
0
0
;
0
t
;
5.12 Thevenin's Theorem
Thevenin's theorem is also useful for calculating currents through circuits, as we shall see in the following examples.
EXAMPLE:
THEVENIN'S
THEOREM
APPLIED
TO A SIMPLE
131
flowing
CIRCUIT
Let us see how Thevenin's theorem applies to the circuit of Fig. 520a, where the part of the circuit to the left of the terminals is considered as a source feeding the load resistance R. Here V is a voltagestabilized source with an internal resistance that is negligible compared to Rj. The Thevenin equivalent of this source is shown in Fig. 520b. The voltage V is the voltage between the terminals of the circuit of Fig. 520a when the resistor R is removed. Also, the resistance R is the resistance measured at the terminals when R is removed, and when V is reduced to zero. This is simply R, and R in parallel. The current F /(Ri + R) in Fig. 520b is the same current one would find directly from Fig. 520a using mesh currents as in Sec. 5.9. x
0
t
x
2
0
figure 520 The circuit to the left of the terminals in (a) gives the same current I in the load resistance R as the circuit to the left of the terminals in (h).
EXAMPLE
AND REVIEW:
THE
WHEATSTONE
BRIDGE
Figure 521a shows a Wheatstone bridge, similar to that of Prob. 58, except that we have shown the resistances R and R of the source and of the detector, respectively, and that R is not infinite here. We wish to find the current in the detector, when the resistance of the upper lefthand branch is R(l + x), as a function of the current I in that branch. When x = 0, the bridge is balanced and the current through the detector is zero (Prob. 58). We could solve this problem by using either branch or mesh currents. Instead, we shall solve it in a more roundabout way that will use most of the tricks that we have learned in this chapter. s
d
d
132
Figure 521 (a) Wheatstone bridge circuit, showing the internal resistance of the source R and the resistance of the detector R . When x = 0, the bridge is balanced and no current flows through R . (b, c, d) Successive transformations of the circuit that are used in calculating the current through the detector as a function of x. s
d
d
a) Let the current through the resistance R(\ + x) be /. Then, using the substitution theorem, we can replace the resistance Rx by a source IRx and we have the circuit of Fig. 521b. b) We now have two sources, V and IRx. According to the principle of superposition, each one acts independently of the other. The currents due to V are now those for the balanced bridge. Thus the current through the detector is due only to the source IRx, and we may disregard V . We now have circuit c. s
s
5
c) Resistances R , bR, abR form a delta. If we use the deltastar we have circuit d. We now have a fifth node, £. s
transformation,
5.12 Thevenins
Theorem
133
Using the transformation rules, (bR)R
s
(ahR)R
n
+ abR'
bR +.R
S
(547)
s
= BR, where B is
s
abR + R + bR' = aBR,
(548)
R /R. 2
d) We can now use Thevenins theorem to find the current through the dectector We consider R as the load resistance, and all the rest of the circuit as a source. The opencircuit voltage K is that between D and B when the branch DB is cut. The current flowing around the outer branches is d
0
IRx R + R
+ Ri + aR'
2
and IRx R + R
+ K
2
+ aR (aR + R ).
IRx R + BR + aBR + aR = IRx
(549)
2
3
(aR + aBR),
(550)
ail + B) —, (1 + a)(\ + B)'
(551)
IRx.
1 + a
(552)
Note that this voltage is independent of R , of b, and of the quantity B of Eq. 548. We now require the resistance of R + R of Sec. 5.12, which is the resistance measured at a cut in the branch DB, when the source IRx is replaced by a shortcircuit. We can find this resistance from Fig. 52Id, but it is simpler to return to Fig. 521b. We imagine a source inserted in the branch DB, with both V and IRx replaced by shortcircuits. The bridge is balanced. Then a source in the branch DB gives zero current in the branch AC, so we may disregard the resistance R . So s
t
s
s
(R + bR)(aR + abR) R
'
+
R
> = < R
R
+
+
bR + aR
+
abR'
^
(1 + b)(\
R
d
+
o(l i(l + bb) ,1 + a R
(555)
134
Direct Currents in Electric
Circuits
and, finally, V
a IRx
0
h = R
d
/(l
+ a)
+ R, = R„ +, 0 („1 +[\„ fe)«/(l +, a ) ] ' r
d
u
/.V
(556) (557)
( 1 + ) a
(R /R) d
If we wish to maximize the ratio I jl, then R /R should be small, b should be small, and a large. These last two conditions are not at all critical and one can say, as a rule of thumb, that both a and b should be approximately equal to unity. d
5.13
SOLVING
A SIMPLE
d
DIFFERENTIAL
EQUATION
In the next section we shall have to solve the following equation: a  + bx = c, at
(558)
d X
where a, b, c are known constants, and x is an unknown function of f. This is a differential equation. Its general solution is composed of two parts: (a) the particular solution, obtained by conserving only the last term on the left, x i = c/b, plus (b) the complementary
function, x
where A is a constant of Thus
(559)
which is the solution for c = 0, or
= Ae ,
(560)
{h,a)t
2
integration.
x = x, + x
2
= + Ae . C
Wa)t
(561)
The numerical value of the constant of integration A is obtained, as a rule, from the known value of x at t = 0. It is easy to check this solution by substituting it in the original equation (Eq. 558). A uniqueness theorem states that this solution is the only possible one.
5.14 Transients in RC Circuits
5.14
TRANSIENTS
IN RC
135
CIRCUITS
W h e n a source is connected to a circuit comprising resistors and capacitors, transient
currents flow until the capacitors are charged. Such currents are
large, at first, and then decay exponentially with time.
5.14.1
EXAMPLE:
SIMPLE
RC
CIRCUIT
Figure 522 shows a capacitor C, initially uncharged, that is connected at time t = 0 to a voltage source V through a resistor R. We can find the voltage across the capacitor by using Kirchoff's voltage law. Adding up the voltage rises around the loop, starting at the lower lefthand corner, clockwise, Q
V — IR
c
0.
(562)
V
(563)
Since / is dQ/dt, R
dQ
1
dt
and, from the previous section, Q = CV +
Ae' . IRC
(564)
Since, by hypothesis, Q = 0 at t = 0, A = — C F a n d (565)
Q = CF(1 After a time r » RC, Q % CV. The voltage across the capacitor is V = F(l c
I.
(566)
Figure 523 shows V /V as a function of time for R = 1 megohm, C = 1 microfarad, RC = 1 second. c
Figure 522 An RC circuit.
136
1.0
2 3 t (seconds)
0
4
5
Figure 523 The voltage V across the capacitor C in Fig. 522 as a function of the time. It is assumed that V = 0 at t — 0 and that RC = 1. c
c
The product RC is called the time constant of the circuit. For t = RC, V /V = 1  1/e as 2/3. It will be shown in Prob. 523 that, if C is discharged through R, V = Ve~" . c
RC
c
5.75
SUMMARY The current density vector J points in the direction of flow of positive charge and is expressed in amperes per square meter. For any surface S bounding a volume t . (54) which is the law of conservation of charge. The net charge density inside a homogeneous conductor carrying a current under steadystate conditions is zero. F o r ordinary conductors, Ohm's law applies:
J =
(51)
where J is the current density, a is the conductivity, and E is the electric field intensity. The electric charge density inside a uniform currentcarrying conductor is zero under steadystate conditions.
5.15 Summary
137
O h m ' s law can also be stated as / = V/R,
(59)
where I is the current flowing through a resistance R, across which a voltage V is maintained. The power dissipated as heat by the Joule effect per cubic meter is J /a, and the power dissipated in a resistor R carrying a current I is I R. Resistors for which / is proportional to V are said to be linear, or ohmic. Other resistors are said to be nonlinear. The resistance of a set of resistors connected in series is equal to the sum of their individual resistances. The conductance of a set of resistors connected in parallel is equal to the sum of their individual conductances. The principle of superposition states that, as long as all the resistors in a circuit are linear, each source acts independently of the others and produces its own set of currents. Kirchoff's current law states that the algebraic sum of the currents entering a node, in an electric circuit, is equal to zero. His voltage law states that the sum of the voltage drops a r o u n d a mesh is equal to zero. Currents in the various branches of a circuit are usually calculated by using mesh currents and the voltage law. According to the substitution theorem, one can replace a resistance R in a circuit by a voltage source IR of the correct polarity without altering the currents flowing through the circuit or, inversely, if a current enters a voltage source V at its positive terminal, one can replace the source by a resistance V/I without changing the currents. O n e can transform the deltaconnected circuit of Fig. 516a into the starconnected circuit of Fig. 516b by means of the following equations: 2
2
RA
RhRr =
R
RB
R,
(538)
+
R
R„ + R„ +
R,
a
+ R
b
c
R,R„
R„R„ R„ + Rh +
R
r
(539)
(540)
138
Direct Currents in Electric
Circuits
Also, GG B
G
A
G
b
=
B
(544)
+
G
+
G
c
GG C
G
A
G, =
C
+ G
A
+ G
B
(545) c
GG A
B
G, + G + B
Gr
(546)
A voltage source supplies a voltage that is nearly independent of the current through the load; a current source supplies a current that is nearly independent of the voltage across the load. Thevenins
theorem states that any source may be represented by an
ideal voltage source, in series with a resistance called its output
resistance.
When a source is connected to a circuit comprising capacitors, transient currents flow temporarily in the circuit until all the capacitors are charged.
PROBLEMS 51E
CONDUCTION IN A NONUNIFORM MEDIUM Two plane parallel copper electrodes are separated by a plate of thickness s whose conductivity a varies linearly from a . near the positive electrode, to u + a near the negative electrode. Neglect edge effects. The current density is J. Find the electric field intensity in the conducting plate, as a function of the distance x from the positive electrode. 0
0
52E
RESISTIVE FILM A square film of Nichrome (an alloy of nickel and chromium) is deposited on a sheet of glass, and copper electrodes are then deposited on two opposite edges. Show that the resistance between the electrodes depends only on the thickness of the film and on its conductivity, as long as the film is square. This resistance is given in ohms per square. Nichrome films range approximately from 40 to 400 ohms per square.
53E
RESISTOJET Figure 524 shows the principle of operation of the resistojet, which is used as a thruster for correcting either the trajectory or the attitude of a satellite. See also Probs. 214, 215, and 1011.
139
Figure 524 Schematic diagram of a resistojet. The propellant gas P is admitted on the left, heated to a high temperature by the resistor, and exhausted through the nozzle. See Prob. 53.
Assuming complete conversion of the electric energy into kinetic energy, calculate the thrust for an input power of 3 kilowatts and a flow of 0.6 gram of hydrogen per second. 54E
JOULE
LOSSES A resistor has a resistance of 100 kilohms and a power rating of onequarter watt. What is the maximum voltage that can be applied across it?
55E
VOLTAGE DIVIDER Figure 525 shows a voltage divider or attenuator. A source supplies a voltage V, to the input, and a highresistance load (not shown) R » R is connected across P . The voltage on R and R is V . Show that 2
2
2
0
K
F
=
R
2
Ri + R
2
R
Figure 525 Voltage divider: V is R /(R + R ) times V . See Prob. 55. 0
2
1
2
t
140
Figure 526 Potentiometer circuit. The voltage V is known, and V is unknown. When the sliding contact is adjusted for zero current, V /V is equal to R /(R! + R ). See Prob. 56. a
b
b
2
56E
a
2
POTENTIOMETER Figure 526 shows a potentiometer circuit. Show that, when 7 = 0,
This circuit is extensively used for measuring the voltage supplied by a source V , without drawing any current from it. The circuit is used in stripchart recorders. In that case the current 7 is amplified to actuate the motor that displaces the pen, and simultaneously moves the tap on the resistance in a direction to decrease 7. This is one type of servomechanism. b
57E
SIMPLE CIRCUIT Show that, in Fig. 527, R  R V = — R + R 2
V
2
if the voltmeter draws a negligible amount of current. We shall use this result in Prob. 155.
R,
I — A W
AAA/—
+
+ Figure 527 See Prob. 57.
Ri
f
r
i
141
WHEATSTONE
BRIDGE
Figure 528 shows a circuit known as a Wheatstone bridge. As we shall see, the voltage V is zero when R1 Rl — = — • R$ R
(1)
4
Fhe bridge is then said to be balanced. As a rule, the bridge is used unbalanced, the voltage V being a measure of the unknown resistance, say R . The circuit is so widely used today that it is almost impossible to list all its applications. Some of the better known are the following. {
1. If one of the resistors is a temperaturesensitive resistor, or thermistor, the bridge serves as a thermometer, V being a measure of the temperature. 2. If one of the resistors is a temperaturesensitive wire, heated by the bridge current and immersed in a gas, V is a measure of the gas velocity, or of its turbulence. One then has a hotwire anemometer. 3. In the Pirani vacuum gauge the heated wire is in a partial vacuum and is cooled more or less by the residual gas, according to the pressure. 4. In certain gas analyzers, the heated wire is exposed to the unknown gas. Then V is a measure of the thermal conductivity of the gas. Such analyzers are used, in particular, for carbon dioxide. The thermal conductivity of carbon dioxide is roughly one half that of air (16.6, against 26.1 milliwatts per meterkelvin).
142
Direct Currents in Electric
Circuits
5. In flammablegas detectors the wire is heated sufficiently to ignite a sample of the gas contained in a small cell. The heat generated by the combusion increases the temperature and changes the resistance of the wire. The deflection of the voltmeter pointer is a measure of the flammability. 6. If one of the resistors is a fine wire whose resistance changes when elongated, the bridge serves as a strain gauge for measuring microscopic displacements or deformations. 7. If the strain gauge is attached to a diaphragm, one has a pressure
gauge.
a) Show that, when V = 0, Eq. 1 is satisfied. Solution:
Let us suppose that node C is grounded.
The voltage V being zero, there is zero current in the branch DB. The current in the lefthand side is V /(R + R ) and the voltage at D is s
t
3
V.
(2)
R, + R
3
R
4
Similarly, at B,
+ R
2
(3)
Since V = V , D
B
Ri + R R
3
R2 + R4
R
3
(4)
4
and, subtracting 1 on both sides, Ri
R
(5)
2
b) Suppose now that R, increases by the factor 1 + x as in Sec. 5.12.2. We require V/V as a function of x, where x can vary from —0.2 to +0.2. The voltmeter V draws a negligible current. Set s
R
3
= aR
lt
R
2
= bR
u
R
4
= abR
again as in Sec. 5.12.2. With x = 0, the bridge is balanced.
u
(6)
143
x
0.2
a 0 Figure 529 The ratio V/V for the Wheatstone bridge of Fig. 528, as a function of x and a. s
Find V/V as a function of x, a, b. Note that F/F turns out to be independent of b. s
Solution:
We now have that
F
I
R + K
R £ , ( 1 + x) + R
4
K RJR
4
3
4
R /i?, 3
2
1+(R /R )
1 + a
2
2
(1 + x) + (R /Ri)'
(8)
3
1 + x + a' ax
(1 + a)(l + a + x)
(9)
(10)
Figure 529 shows F/F, as a function of both x and a. c) For what value of a is F/F maximum, for a given value of x? This is the condition for maximum sensitivity. Figure 529 shows that this condition is =t 1.
144
Direct Currents in Electric
Solution:
Circuits
VjV is maximum when d(V/V )/da s
s
(1 + a)(l + a + x)x  axl(l
= 0. Thus we set
+ a + x) + (1 + a)]
(1 + a) (l + a + x ) 2
"
2
(1 + a)(l + a + x) = a(2 + 2a + x), 1 + a + x + a + a + ax = 2a + 2a 2
+ ax,
2
a = (1 + x ) ' . 1
1 1
'
(12) (13) (14)
2
Since x is small, the condition for maximum sensitivity is a * 1. as we had guessed. d) Draw curves of V/V as a function of x, for x = —0.2 to x = +0.2, and for a = 0.3, 1, and 3. s
0.05"
i /
/
11
1 1
1
1
0.2
V/
^
i
i i
i i
i i
0.2
/
/o.3
/
/1 0.05.
Figure 530 Ratio V/V for the Wheatstone bridge of Fig. 528 as a function of x, for a equal to 0.3, 1, 3. s
Problems
145
Note the slight nonlinearity of the curves. The nonlinearity decreases with increasing a. Solution:
See Fig. 530.
AMPLIFIER Figure 53la shows one element of an analog computer. amplify the input voltage by the factor — R /Ri '•
Its function is to
2
The triangular figure is an operational amplifier whose gain is — A. Such amplifiers have very high gains, of the order of 10 to 10 , and draw a negligible amount of current at their input terminals. The accuracy of the gain is limited only by the stability of the ratio R /Ri The drift in R /Ri due to aging, temperature changes, and so forth, can be smaller than the drift in A by orders of magnitude. a) You can show in the following way that the above equation is correct. First, you need an expression for the voltage V at the input to the amplifier. Since the amplifier draws essentially zero current at its input terminals, you can use the circuit of Fig. 531b. Then, setting V = — AV , you can show that 4
9
2
2
iA
0
K
2
l
i;
VNAA
R K,
_«2 R
2
if A » 1 and if A »
(b)
r I
„
f [ ( « , + RjFVTJ *
=
V,
The gain is — R /R
iA
T
ViA
I
Rz
t
R /R . 2
1 v
1
1
Figure 531 (a) Multiplying circuit. The triangle represents an amplifier whose gain is — A. The circuit has a gain of R /R!, as long as A » 1 and A » R /R . (b) Equivalent circuit for calculating the gain. See Prob. 59. 2
2
1
146
Direct Currents in Electric
Circuits
b) What is the minimum value of A if R, = 1000 ohms, R if V must be equal to — (R /R )V within one part in 1000?
2
0
2
l
= 2000 ohms, and
i
510E
G = el You are given a large number of 6ohm resistors. H o w can they be connected to give a circuit whose conductance is equal to e Siemens, where e is the base of the natural logarithms? Hint: Write out the series for e.
51 IE
TETRAHEDRON Six equal resistors R form the edges of a tetrahedron, as in Fig. 532. a) A battery is connected between A and B. Show that, by symmetry, points C and D will be at the same potential. That being so, there is zero current in branch CD, and we can remove it. b) What is the resistance between A and B?
C
B
512E
Figure 532 See Prob. 511.
CUBE Twelve equal resistances R form the edges of a cube, as in Fig. 533. a) A battery is connected between A and G. Can you find one set of three points that are at the same potential? Can you find another set? b) Now imagine that you have a sheet of copper connecting the first set, and another sheet of copper connecting the second set. This does not affect the currents in the resistors. You can now show that R
AG
= (5/6)R.
147
F
Figure 533 See Prob. 512.
F
G
H
5BE
Figure 534 See Prob. 513.
CUBE Twelve equal resistances R form the edges of a cube, as in Fig. 533. To find the resistance between A and C, we first flatten the cube as in Fig. 534. a) A battery is connected between A and C. Which four points are at the same potential? Which two branches can be removed without affecting the currents? b) Now show that R
AC
514
= (3/4)/?.
CUBE First, read Probs. 511 to 513. N o w show that, in Fig. 533, R
AD
= (1.4/2.4)R.
148
r^AAAfAAAf
550 V
+
R
~±r
S a aX a  ^ a a a > 20 km Figure 535 A 550volt source feeds a load resistance R through a 20kilometer line. The position of a fault R can be found from measurements of the current /, with and without a shortcircuit across R,. See Prob. 515. T
S
5/50*
LINE
FAULT
LOCATION
In Fig. 535. a 550volt source feeds a fixed load resistance R through a pair of copper (a = 5.8 x 10 Siemens per meter) wires 20 kilometers long. The wires are 3 millimeters in diameter. Suddenly, during a storm, the current I supplied by the source increases and the voltage V across R, falls. It is concluded that a fault has developed somewhere along the line, and that there is a shunt resistance R at some distance x from the source. If the load resistance R is disconnected, / is found to be 3.78 amperes. When the line terminals at the load are shortcircuited, / is 7.20 amperes. What is the distance x? X
7
S
T
516E
UNIFORM
RESISTIVE
NET
Figure 536 shows a twodimensional network composed of equal resistances. The outside nodes are either connected to batteries in some arbitrary way, or left unconnected. a) Use Kirchoff's current law to show that the potential at any inside node is equal to the average potential at the four neighboring nodes. F o r example. VA+V +V + M
V
C
L
4 b) What is the rule for a similar threedimensional network? 517
POTENTIAL
DIVIDER
Find the ratio VJV
T
in Fig. 537.
Problems marked D are relatively difficult.
149
Figure 536 Twodimensional network composed of equal resistances R. Arbitrary voltages are applied at the periphery. It is shown in Prob. 516 that the voltage on any inside node is equal to the average value of the voltages on the four neighboring nodes.
150
Figure 538 See Prob. 518.
518
SIMPLE CIRCUIT WITH TWO Show that, in Fig. 538,
SOURCES
R,
/ = 519
r(R, +
R) 2
DELTASTAR TRANSFORMATION Find the equations for either the deltastar or the stardelta transformation by assuming mesh currents, as in Fig. 516, and making the voltages V  V , V  F , V  V in part a the same as those in part b. Hint: Find one equation of the form A
c
B
B
c
A
(
•
)I
A
+ (
•
)IB + (
)Ic = 0.
Then, since it must be valid, whatever the values of the mesh currents, the parentheses must all be identically equal to zero. This should give you one of the equations of one set; the other two equations can be found by symmetry.
4 KQ
i Kn
3 Kn
KQ
o 5 Kn
Figure 539 See Prob. 520. The symbol Q stands for ohm.
Problems
520
52IE
151
DELTASTAR TRANSFORMATION Find the resistance of the circuit shown in Fig. 539. OUTPUT RESISTANCE OF A BRIDGE CIRCUIT Show that the output resistance of the bridge circuit shown in Fig. 540, as seen at the voltmeter V, is R. Assume that the source V has a zero output resistance. s
Figure 540 See Prob. 521.
522E
OUTPUT RESISTANCE OF AN AUTOMOBILE BATTERY It is found that the voltage at the terminals of a defective automobile battery drops from 12.5 to 11.5 volts when the headlights are turned on. What is the approximate value of the output resistance?
523E
DISCHARGING A CAPACITOR THROUGH A RESISTOR A resistance of 1 megohm is connected across a 1 microfarad capacitor initially charged to 100 volts. Calculate the voltage across the capacitor as a function of the time.
524E
RAMP
GENERATOR A constantcurrent source feeds a current / to a capacitor C. At t = 0 the capacitor voltage is zero. Find the voltage across C as a function of the time.
525
CHARGING A CAPACITOR THROUGH A RESISTOR A source supplying a voltage V charges a capacitor C through a resistance R. Calculate the energy supplied by the source, that dissipated by the resistor, and that stored in the capacitor, after an infinite time. You should find that one half of the energy is dissipated in R, and that the other half is stored in C.
526
RC
TRANSIENT In the circuit of Fig. 541, the capacitor is initially uncharged. a) At 1 = 0 the switch is closed. Find the voltage V as a function of t. b) After a long time the switch is opened. Find again V as a function of t, setting t = 0 at the moment the switch is opened.
152
Figure 541 See Prob. 526.
Figure 542 RC differentiating circuit. See Prob. 527.
527E
RC DIFFERENTIATING CIRCUIT Figure 542 shows an RC differentiating circuit. The load resistance connected at F> is large compared to R. Show that, if the voltage drop across R is negligible compared to that across C,
dV
t
K.RC^. Note that V„ « V,. See Prob. 529. 528E
DIFFERENTIATING A SQUARE WAVE A square wave as in Fig. 543 is applied to the RC differentiating circuit of Fig. 542. Sketch a curve of the output voltage as a function of the time.
Figure 543 Ideal square wave. In practice, the straight lines are portions of exponential curves. See Prob. 528.
529
DIFFERENTIATING
L
CIRCUIT
The circuit of Fig. 542 is simple and inexpensive, but V « V . Figure 544a shows a much superior, but more complex, differentiating circuit. Fhe triangle represents an operational amplifier as in Prob. 59. 0
t
153
(b) O
i
»
AAAo Figure 544 (a) Differentiating circuit using an amplifier, (b) Equivalent circuit used to calculate V . See Prob. 529.
t I;
I
I
1
0
Show that dV RC— , dt t
K =
1
as long as A » 1 and K /RC » \dVJdt\jA. Note that RC can be much larger than unity, so that V need not be much smaller than V . 0
0
Solution:
t
From Fig. 544b. V =®+V , t
dV,
V„=AV ,
iA
1 dQ
=
dt
(1)
iA
dV
_
u
C dt
dt
1
1
C
A
dV
a
dt'
Now V
i4
/ = —
 V
0
R
\
=
1 ,
R \
1 + — K,
(3)
1 dV .
(4)
A
1
so that dV — = t
dt
1 /
1\ 1+ 
RC \ 1 RC " *
VJRC,
0
IK,
I
A 1 / V A \RC
A dt dV" dt J' (6)
154
Direct Currents in Electric
Circuits
if A » 1 and if F/i?C » \dVJdt\/A. Thus, with this approximation, V„=RC^.
530
dV,
(7)
dt
RC INTEGRATING CIRCUIT Figure 545 shows an RC integrating circuit. The current through the load connected at F„ is negligible compared to dQ/dt at C. Show that, as long as the voltage across C is small compared to that across R,
As in Prob. 527, V « F It is assumed that V = 0 at t = 0. 0
531 E
0
INTEGRATING CIRCUIT A square wave as in Fig. 543 is applied to the RC integrating circuit of Fig. 545. Sketch a curve of the output voltage as a function of the time.
R
WV\A
Figure 545 RC integrating circuit. See Prob. 530.
532
°"
INTEGRATING CIRCUIT The integrating circuit shown in Fig. 546a performs integrations without the limitation F « V that applies to the circuit of Fig. 545. The triangle represents an amplifier as in Probs. 59 and 529. Analog computers make extensive use of the circuits of Figs. 531, 544, and 546. Show that 0
t
if A » 1 and if \V„\/RC « A\dVJdt\. Use the circuit of Fig. 546b and set
155
(a)
r
9
i;
I
Figure 546 (a) Integrating circuit using an amplifier, (b) Equivalent circuit for calculating V„. See Prob. 532. 533
T
V W
)—
c
1 y.
I
i
(b)
PULSECOUNTING CIRCUIT The circuit shown in Fig. 547 can be used either for counting pulses or for measuring a capacitance, either C\ or C , the other one being known. The pulse generator G produces / positive square pulses of amplitude V per second and 2
p
Cj « c , 2
/c, « c . 2
Diodes D and D pass currents only in the direction of the arrow. During a pulse, no current flows through diode D and capacitors C, and C charge in series. Between pulses, the terminals of G are effectively shorted and capacitor C discharges through diode D . Diode D is then nonconducting. a) Sketch a curve of V as a function of the time, for V « V , with V = 0 at f = 0. b) What is the value of F after a time f ? x
2
u
2
t
l
2
p
4^
D,
Jl , v
T
Figure 547 This circuit can be used either for measuring the repetition frequency of the current pulses originating in G or for measuring either C, or C . See Prob. 533. 2
CHAPTER
6
DIELECTRICS:
I
Electric Polarization P, Bound Charges, Gauss's Law, Electric Displacement D
6.1
THE ELECTRIC
6.2 6.2.1
THE BOUND CHARGE DENSITIES Example: Sheet of Dielectric
6.3
THE ELECTRIC FIELD OUTSIDE A DIELECTRIC Example: Sheet of Dielectric
6.3.1
POLA RIZA TION
p AND b
AND
a
h
INSIDE
6.4
THE DIVERGENCE GAUSS'S LAW
6.5
THE ELECTRIC
DISPLACEMENT
D
6.6
THE ELECTRIC
SUSCEPTIBILITY
Xe
6.7
THE RELATIVE
PERMITTIVITY
6.8 6.8.1 6.8.2 6.8.3
RELATION BETWEEN THE FREE CHARGE DENSITIES AND THE BOUND CHARGE DENSITIES The Volume Charge Densities p and p The Surface Charge Densities a and a Example: DielectricInsulated ParallelPlate Capacitor
6.9
POISSON'S
6.10
ELECT
6.11
OF E IN
P
f
(
AND LAPLACE'S
RETS
SUMMARY PROBLEMS
DIELECTRICS:
e
r
h
b
EQUATIONS
Dielectrics differ from conductors in that they have no free charges that can move through the material under the influence of an electric field. In dielectrics, all the electrons are b o u n d ; the only possible motion in an electric field is a minute displacement of positive and negative charges in opposite directions. The displacement is usually small compared to atomic dimensions. A dielectric in which this charge displacement has taken place is said to be polarized, and its molecules are said to possess induced dipole moments. These dipoles produce their own field, which adds to that of the external charges. The dipole field and the externally applied electric field can be comparable in magnitude. In addition to displacing the positive and negative charges, an applied electric field can also orient molecules that possess permanent dipole moments. Such molecules experience a torque that tends to align them with the field, but collisions arising from thermal agitation of the molecules tend to destroy the alignment. An equilibrium polarization is thus established in which there is, on the average, a net alignment. This is, in fact, a simplified view of dielectric behavior, because many solids, such as sodium chloride, for example, are not made up of molecules but of individual ions.
THE ELECTRIC
POLARIZATION
P
The electric polarization P is the dipole moment per unit volume at a given point. If p is the average electric dipole moment per molecule, and if N is the number of molecules per unit volume, then
p = yvp.
(61)
Kigure 61 Under the action of an electric field E, which is the resultant of an external field and of the field of the dipoles within the dielectric, positive and negative charges in the molecules are separated by an average distance s. In the process a net charge dQ = NQs • da' crosses the surface da', N being the number of molecules per unit volume and Q the positive charge in a molecule. The vector da' is perpendicular to the shaded surface. The circles indicate the centers of , charge for the positive and for the negative charges in one molecule.
THE BOUND
CHARGE
DENSITIES
p„ AND
a
b
We shall now show that the displacement of charge within the dielectric gives rise to net volume and surface charge densities p„=VP
and
a = P •n b
1 (
(62)
where rij is the normal to the surface, pointing outward. Let us first consider the surface density a . We imagine a small element of surface da' inside the dielectric, as in Fig. 61. Under the action of the field, positive and negative charges within the molecules separate by an average distance s. Positive charges cross the surface by moving in the direction of the field; negative charges cross it by moving in the opposite direction. For the purpose of our calculation, we consider the positive charges to be in the form of point charges Q and the negative charges in the form of point charges — Q. Furthermore, we consider the negative charges to be fixed and the positive charges to move a distance s. The amount of charge dQ that crosses da' is then just the total amount of positive charge within the imaginary parallelepiped shown in Fig. 61. The volume of this b
6.2 The Bound Charge Densities p and c b
b
159
parallelepiped is dx' = s • da'.
(63)
dQ = NQs • da',
(64)
and
where N is the number of molecules per unit volume and Qs is the dipole moment p of a molecule. Then dQ = P • da'.
(65)
If da' is on the surface of the dielectric material, dQ accumulates there as a surface distribution of density
(
7 = ^ 6
= Pn , 1
da
(66)
where n, is a unit vector, normal to the surface and pointing outward. The bound surface charge density a is thus equal to the normal component of P at the surface. We can show similarly that — V • P represents a volume density of charge. The net charge that flows out of a volume x across an element da' of its surface is P • da', as we found above. The net charge that flows out of the surface S' bounding x is thus h
Q = J* , P • da', g
(67)
and the net charge that remains within the volume x must be —Q. If p is the volume density of the charge remaining within this volume, then h
£, Pt dx =  Q =  J „ P • da' =  £ , (V • P) dx'. s
(68)
Since this equation must be true for all x, the integrands must be equal at every point, and the bound volume charge density is p = b
V P
(69)
160
Dielectrics:
I
We refer to p and a as bound charge densities, as distinguished from the free charge densities p and a . Bound charges are those that accumulate through the displacements that occur on a molecular scale in the polarization process. The other charges are called free charges. The conduction electrons in a conductor and the electrons injected into a dielectric with a highenergy electron beam are examples of free charges. b
b
f
EXAMPLE:
SHEET
OF
f
DIELECTRIC
If a sheet of dielectric is polarized uniformly in the direction normal to its surface, then the surface density ofbound charge cr is P. as in Fig. 62. Also. p = — V • P = 0. since P is independent of the coordinates. fc
6.3
THE ELECTRIC FIELD OUTSIDE INSIDE A DIELECTRIC
h
AND
Figure 63 shows a block of polarized dielectric material in which P is a function of position. We must find the electric field intensity that the dipoles in the dielectric produce at some point A, either inside or outside. Once obtained, this E can then be added to that produced by other charges to give the total E.
161
Figure 63 Block of dielectric with a dipole moment P per unit volume. The dipoles within the element of volume shown inside the block give rise to an electric potential dV at the point A.
Coulomb's law applies to any net accumulation of charge, regardless of other matter that may be present. Thus the field produced by the dielectric, both inside and outside, is the same as if the charge densities p and a were situated in a vacuum. b
6.3.1
EXAMPLE:
SHEET
OF
b
DIELECTRIC
In the polarized sheet of dielectric of Fig. 62. E = 0 outside and E = o /e pointing to the left inside, exactly as if the sheet were replaced by its surface charges. b
6.4
THE DIVERGENCE GA USS'S LA W
OF E IN
0
DIELECTRICS:
We have just seen that the electric field produced by the dipoles of a polarized dielectric can be calculated from the bound charge densities as if they were situated in a vacuum. Let us investigate the implications of this fact on Gauss's law (Sec. 3.2).
162
Dielectrics:
I
Gauss's law relates the flux of E through a closed surface to the charge Q enclosed within that surface: J E • da = £ V • E dx = (2/e . s
(610)
0
For dielectrics, Q includes bound as well as free charges:
Q = j(p
+ p )dx.
f
(611)
b
If we substitute this value of Q into Eq. 610 and equate the integrands of the volume integrals, then
e
0
This is Gauss's law in its more general form. It is one of Maxwell's four fundamental equations of electromagnetism. We shall find the other three in Sees. 8.2. 11.4, and 19.2. This one follows from (a) Coulomb's law. (b) the concept of electric field intensity, (c) the principle of superposition, and (d) the fact that the field of the dipoles situated in a polarized medium can be calculated from p and a . b
b
Note that we have implicitly assumed the existence of the space derivatives of E. These derivatives do not exist at the interface between two media: in such cases one must revert to the integral of E • da over a closed surface, which is equal to the total enclosed charge over e , according to Eq. 610. 0
THE ELECTRIC
DISPLACEMENT
D
Since p =  V • P. from Eq. 69, then b
V E = (p
f
 VP),
(613)
or V • (e E + P) = . 0
Pf
(614)
6.7 The Relative Permittivity
e
163
r
The vector e E + P is therefore such that its divergence depends only on the free charge density p . This vector is called the electric displacement and is designated by D : 0
f
D = e E + P.
(615)
0
Thus (616) In integral form, Gauss's law for D becomes
pjdc,
f D • da = f
(617)
and the flux of the electric displacement D through a closed surface is equal to the free charge enclosed by the surface. Note that the divergence of D, as well as the surface integral of D, are both unaffected by the bound charges.
6.6
THE ELECTRIC
SUSCEPTIBILITY
/
e
In most dielectrics the molecular charge separation is directly proportional to, and in the same direction as, E. Dielectrics that show this simple dependence of polarization on field are said to be linear and isotropic In practice, most dielectrics are homogenous, as well as linear and isotropic. The dipole moment per unit volume is then P = Np = x e E, e
(618)
0
where N is again the number of molecules per unit volume, and y is a dimensionless constant known as the electric susceptibility of the dielectric. e
6.7
THE RELA TIVE PERMITTIVITY
e
r
F o r a linear and isotropic dielectric we have, from Eqs. 615 and 618 that D = € (1 + yjE 0
= € e E = eE, 0
r
(619)
164
Table 61
Relative Permittivities of Dielectrics Near 25 C.
Frequency (hertz) 100
10
10"'
6
Bakelite
5.50
4.45
3.55
Butyl rubber
2.43
2.40
2.38
Fused silica
3.78
3.78
3.78
Lucite
3.20
2.63
2.57
Neoprene
6.70
6.26
4.0
Polystyrene
2.56
2.56
2.54
Steatite
6.55
6.53
6.51
Styrofoam
1.03
1.03
1.03
Teflon Water
2.1
2.1
2.08 34.
78.2
81.
where e, = 1 + l e = e / e
(620)
0
is a dimensionless constant larger than unity and is known as the relative permittivity of the material. In a vacuum, y = 0, e = 1. The relative permittivity was formerly called the dielectric constant. The quantity e is called the permittivity. e
r
The relative permittivity of dielectrics lies typically between 2 and 7. However, some nonlinear dielectrics have relative permittivities as high as 10 . Table 61 gives the relative permittivities of some c o m m o n dielectrics. It will be observed from the table that e is frequencydependent. We shall return to this phenomenon briefly in Sec. 7.7. 5
r
6.8 Relation Between the Free and Bound Charge Densities
6.8
6.8.1
165
RELATION BETWEEN THE FREE CHARGE DENSITIES AND THE BOUND CHARGE DENSITIES
THE
VOLUME
CHARGE
DENSIHES
p
AND p„
f
In a linear and isotropic dielectric, from Eqs. 615 and 619, D = e E + P = e e E. 0
r
(621)
0
and thus P = (e 
1 )e E = ^ —  D.
r
(622)
()
Taking now the divergence of both sides and using Eqs. 69 and 616,
p Note that p and p density is smaller than p •: b
f
b
=  ^  P f
(623)
have opposite signs, so that the total charge
s
P/ + Pt = Pf/e 
(624)
r
Also, if p if zero in a linear and isotropic dielectric, which is nearly always the case, then p is also zero. f
h
6.8.2
THE SURFACE
CHARGE
DENSITIES
AND o„
a
f
At the interface between a dielectric and a conductor, there is a bound surface charge density a on the dielectric, and a free surface charge density o on the conductor, as in Fig. 64. For steadystate conditions, there is zero electric field inside the conductor and. inside the dielectric, from Gauss's h
f
law.
€ E = Of + o , 0
h
D = eeE r
0
=
a. r
(625)
166
E,D
Figure 64 Interface between a dielectric and a conductor, with a positive charge density on the surface of the conductor.
Thus (T + a = F
(626)
a /e .
h
f
r
This equation is similar to Eq. 624.
6.8.3
EXAMPLE:
DIELECTRICINSULATED
PARALLELPLATE
CAPACITOR
Figure 65 shows a dielectricinsulated parallelplate capacitor with the spacing s exaggerated for clarity. We assume that s is small, compared to the linear extent of the plates, in order that we may neglect the fringing field at the edges. Then P is uniform, V • P = 0, and p = 0, which is consistent with the fact that p, = 0. The surfacecharge densities + a on the lower plate and —a on the upper plate produce a uniform electric field directed upward. The polarization in the dielectric gives a bound surface charge density —a on the lower surface of the dielectric and + a on the upper surface. These bound charges produce a uniform electric field directed downward that cancels part of the field of the charges situated on the plates. Since the net field between the plates must remain equal to V/s, the free charge densities +a and — a on the plates must be larger than when the dielectric is absent. The presence of the dielectric thus has the effect of increasing the charges on the plates for a given value of V, and hence of increasing the capacitance. b
s
s
h
b
f
f
T o calculate the capacitance, we apply Gauss's law for the displacement D to a cylinder as in Fig. 65. Then the only flux of D through the Gaussian surface is through the top, and D is numerically equal to a . Then. E = o /e e , the potential difference between the plates is a s/e e , and the capacitance f
f
C =
r
f
r
0
0
=
e e A/s, r
0
(627)
167
+
+
+
U
+
+
+
tp
+
+
Figure 65 Dielectricinsulated parallelplate capacitor. The small rectangle is the crosssection of an imaginary cylinder used for calculating D.
where A is the area of one plate. The capacitance is therefore increased by a factor of e through the presence of the dielectric. The measurement of the capacitance of a suitable capacitor with and without a dielectric provides a convenient method for measuring a relative permittivity e . r
r
POISSON'S
AND LAPLACE'S
EQUATIONS
In homogeneous, linear, and isotropic dielectrics, D is proportional to E, and e is independent of the coordinates. Then, from Eqs. 612 and 624, r
V•E =
= p /e e . f
r
(628)
0
Also, since E is —VI , 7
rV
= 
p
 ± ± l ± e
o
= 
P
/
/
M
o
.
(629)
This is Poissoris equation for dielectrics. The expressions involving e are valid only in linear and isotropic dielectrics. Laplace's equation is again. r
V K=0, 2
when both p and p are zero. s
b
(630)
168
Figure 66 (a) Bar electret polarized uniformly parallel to its axis, (b) The E field of the bar electret is the same as that of a pair of circular plates carrying uniform surface charge densities of opposite polarities, (c) Lines of E (black). Lines of D are shown in white inside the electret; outside, they follow the lines of E.
6.11 Summary
6.10
169
ELECTRETS An electret is the electric equivalent of a permanent magnet. In most dielectrics the polarization disappears immediately when the electric field is removed, but some dielectrics retain their polarization for a very long time. Some polymers have extrapolated lifetimes of a few hundred years at room temperature. O n e way of charging a dielectric is to place it in a strong electric field at high temperature. The b o u n d charge density on the surfaces then builds u p slowly as the molecules orient themselves. Free charges are also deposited on the surfaces when sparking occurs between the electrodes and the dielectric. The free and the b o u n d charges have opposite signs. The sample is then cooled to room temperature without removing the electric field. The examples in Sees. 6.2.1 and 6.3.1 apply to a sheet electret with zero free charge density. Figure 66 shows the field of a uniformly polarized bar electret.
6.11
SUMMARY When a dielectric material is placed in an electric field, positive and negative charges within the molecules are displaced, one with respect to the other, and the material becomes polarized. The induced dipole moment per unit volume P is called the electric polarization. This produces real accumulations of charge that we can use to calculate V and E both inside and outside the dielectric: Pb
(62)
V • P,
where p is the volume density and o is the surface density of bound charge. The unit vector is normal to the surface of the dielectric and points oufward. Gauss's law can be expressed in a form that is valid for dielectrics: b
h
V•E
Pf
+
e
Pb
(612)
0
This is one of Maxwell's four fundamental equations of electromagnetism.
170
Dielectrics:
I
In linear and isotropic dielectrics, the dipole moment per molecule p is proportional to E, and P = e
0
Z
,E,
(618)
where % is a dimensionless constant called the electric The electric displacement e
susceptibility.
D = e E + P.
(615)
0
= e d + yjE
= e e E = eE,
0
where e is the relative permittivity
0
and e is the permittivity.
r
(619)
r
Also,
V • D = py,
(616)
f D d a = [pfdx.
(617)
or
Inside a dielectric, Pf
+
Pb =
rV*r.
( ' ^ 6
2
while, at the interface between a dielectric and a conductor, o
f
+ o = o /e . h
f
(626)
r
Poisson's equation for dielectrics is
R

V
=
_Pl±I> e
m
_
p//€f60)
.
(5 29)
o
and Laplace's equation is again V V 2
when p and p are both zero. r
h
=
0,
(650)
Problems
171
As usual, expressions involving e are valid only in linear and isotropic media. An electret is a piece of dielectric material that is permanently polarized. r
PROBLEMS 61E
THE DIPOLE MOMENT p A sample of diamond has a density of 3.5 x 10 kilograms per cubic meter and a polarization of 1 0 " coulomb per square meter. a) Compute the average dipole moment per atom. b) Find the average separation between centers of positive and negative charge. Carbon has a nucleus with a charge + 6 e , surrounded by 6 electrons. 3
7
62E
THE
VOLUME AND SURFACE BOUND CHARGE DENSITIES p AND a Consider a block of dielectric with bound charge densities p and
b
b
b
where the x is the volume of the dielectric and S is its surface. In other words, the total net bound charge is zero. 63
BOUND CHARGE DENSITY AT AN INTERFACE Show that the bound charge density at the interface between two dielectrics 1 and 2 that is crossed by an electric field is (P — P ) • n. The polarization in 1 is P, and is directed into the interface; the polarization in 2 is P and points away from the interface. The unit vector n is normal to the interface and points in the direction from 1 to 2. x
2
2
64E
65
COAXIAL LINE Show that V • E = 0 in the dielectric of a coaxial line. Hint: Apply the divergence theorem to a portion of the dielectric. COAXIAL LINE The capacitance per unit length C of a dielectricinsulated coaxial line is equal to that for an airinsulated line as in Prob. 46, multiplied by e : r
C =
2ne e r
0
In (R /R,) 2
172
Table 62
Diameter B & S Gauge
in Millimeters
20
0.812
22
0.644
24
0.511
26
0.405
28
0.321
30
0.255
32
0.202
34
0.160
For a certain application, one requires a coaxial cable whose outside diameter must not exceed about 10 millimeters. Then R < 5 millimeters. Its capacitance per unit length must be as low as possible. Then R must be 5 millimeters, and R, as small as possible. The cable will be operated at up to 500 volts. The insulating material will be Teflon. Teflon has a relative permittivity of 2.1. It can operate reliably in such a cable at electric field intensities as high as 5 x 10 volts per meter. (The maximum electric field intensity before breakdown is called the dielectric strength of a dielectric. In a uniform field the dielectric strength is larger for thinner specimens. A film of Teflon 0.1 millimeter thick has a dielectric strength of 10 volts per meter.) a) What is the minimum allowable diameter for the inner conductor? Remember that you can solve a transcendental equation by trial and error. b) Table 62 shows the diameters of some commonly available copper wires. Finer wires would be impractical because they would tend to break at the connectors fixed to the ends of the cable. What gauge number do you suggest? c) What will be the capacitance per meter with the wire size you have chosen? 2
2
6
8
COAXIAL
LINE
A coaxial line is composed of an internal conductor of radius R, and an external conductor of radius R , separated by a dielectric. The dielectric should normally 2
Problems
173
fill all the space between the conductors. However, due to an error, the outer radius of the dielectric is R < R , leaving an air space. 2
a) Assuming that the dielectric is well centered, what is the capacitance when Rj = 1 mm. R = 5 mm. R = 4.5 mm, and e = 2.56 (polystyrene)'? 2
r
Solution: The outer surface of the dielectric is an equipotential surface. Thus, we can imagine that it is covered with a thin conducting layer. This does not disturb the field. Then we may consider that we have two cylindrical capacitors, one inside the other. Between P , and R we have a capacitance per unit length of 2ne e r
MR
0
R) t
from Prob. 65. Between R and R . the capacitance per unit length is 2
ln(P /P)' 2
These two capacitors are in series, and the capacitance per unit length between the inner and the outer conductors is given by j_
=
C
ln(P/P.) 2ne e r
1
2ne C =

+
ln(P /P) 2
2ne
0
0
! P , ) ++ llnn((PP ,/ PP) InfPv/P!) ) I..
(2)
^
(3)
2
0
.
—ln(P/P,) + ln(P /P) 2
=
2n x 8.85 x 1 0 "
1 2
, , = 80.2 picofarads/meter.
(4)
In4.5 + In — 2.56 0.9 b) Calculate the percent change in capacitance due to the air film. Solution:
Without the air film, the capacitance would be 2n x 2.56 x 8.85 x 1 0 " C =
1 2
= 88.4 picofarads/meter. In 5
The air film therefore decreases the capacitance by about 10%.
(5)
174
67
Dielectrics:
I
CHARGED WIRE EMBEDDED IN DIELECTRIC: THE FREE AND BOUND CHARGES A conducting wire carrying a charge /. per unit length is embedded along the axis of a circular cylinder of dielectric. The radius of the wire is a; the radius of the cylinder is b. a) Show that the bound charge on the outer surface of the dielectric is equal to the bound charge on the inner surface, except for sign. b) Show that the net charge along the axis is /./e per unit length. r
68
PARALLELPLATE CAPACITOR The space between the plates of a parallelplate capacitor with a plate separation .s and a surface area A is partially filled with a dielectric plate of thickness t < s and area A. a) Show that C = b) Call C the capacitance for t = 0. Draw a curve of C/C as a function of t s for e = 3. 0
0
69
r
SPHERE OF DIELECTRIC WITH A POINT CHARGE AT ITS CENTER A sphere of dielectric of radius R contains a point charge Q at its center. Set R = 20 millimeters, Q = 1 0 " coulomb, e = 3. a) Draw graphs of D, E, V as functions of the distance r from the center, out to r = 100 millimeters. Note the discontinuity in E at the surface. b) Now calculate a at the surface from the value of the polarization P just inside the surface. c) Now use Gauss's law to explain the discontinuity in E. 9
r
b
610
CHARGED DIELECTRIC SPHERE A dielectric sphere of radius R contains a uniform density of free charge Show that the potential at the center is (2e + r
MEASURING
SURFACE
CHARGE
p. f
l)p R
2
f
DENSITIES
ON
DIELECTRICS
Figure 67a shows a schematic diagram of an instrument that has been used to measure charge densities at the surface of dielectrics. See also Prob. 179, The probe P is supported a few millimeters above the surface of the sample and can be displaced horizontally and vertically along three orthogonal directions.
175
Figure 67 (a) Instrument for measuring surface charge densities on dielectrics. A conducting probe P is held close to the surface of the sample S. The probe is connected to an electrometer E through a coaxial cable CC. (h) Equivalent circuit: C\ is the capacitance between the top surface of the sample and ground, C is that between the probe and the sample surface, and C is the capacitance of the coaxial cable, plus the input capacitance of the electrometer. See Prob. 611. 2
3
As we shall see, the surface charge on the sample induces a voltage on P. This voltage is read on the electrometer. An electrometer is a voltmeter that has an extremely high resistance. For example, one type has a resistance of 1 0 ohms. For comparison, common digital voltmeters have resistances of the order of 10 or 10 ohms. A good analog voltmeter draws about 100 microamperes. If it is rated at 10 volts, its resistance is 10 10" = 10 ohms. An electrometer draws essentially zero current and is therefore suitable for measuring the voltage on a small capacitor, which is what we have here. Figure 67b shows the equivalent circuit: we have, effectively, three capacitors connected in series, and the electrometer serves to measure the voltage V. 1 4
6
7
4
a) Find the surface charge density a in terms of C
lt
5
C , C , V. 2
3
Solution: Let the area of the probe be A. Then the charge we are concerned with, at the surface of the dielectric, is Aa. We neglect the fringing field at the edge of the probe. Part of the field of Aa goes to the probe, and part goes to ground. Let Aa = Q + Q\
where Q = Ae E , 0
Q' = Ae^o^i,
2
(D
as in Fig. 67b. The figure also shows the charges induced on the conductors. Since the probe and its lead to the electrometer are exceedingly well insulated, the net charge induced on them is zero. Thus V = Q C , and we must find a, C being known. N o w we have two expressions for the voltage V between the probe and ground: 3
3
v=Q=9L.Q.
c
3
c,
,2, c
2
176
Dielectrics:
I
So
e' = e ^ + ^ .
(3)
Substituting in Eq. 1,
Aa =
61 1 +
£ i +  i ). C, C
Since 2 = C K , 3
a =
C, + C , +
C,C,\ K C,
(5)
A
b) Express
r
Solution: d
= e.eo/4/f!,
C
= e /l/t
2
0
(6)
2
and =
612E
VARIABLE
CAPACITOR
i
^ r,
UTILIZING
+
C3
+
€^r
(7)
2 ) i /
A PRINTEDCIRCUIT
BOARD
A printedcircuit board is a sheet of plastic, about one millimeter thick, one side of which is covered with a thin coating of copper. Part of the copper can be etched away, leaving conducting paths that serve to interconnect resistors, capacitors, and so forth. The terminals of these components are soldered directly to the copper. Figure 68 shows a variable capacitor in which a sliding conducting plate lies under a printedcircuit board that has been etched to give a prescribed variation of capacitance with position z. At the position z. the capacitance is
t Jo where t is the thickness of the plastic sheet of relative permittivity e . Set e = 3, t = 1 millimeter, and find y(z) giving the following values of C, with C expressed in farads and x in meters: a) l ( T z , b) K T z . r
r
9
8
2
177
Q
Figure 68 Variable capacitor made with a printedcircuit board PCB and a conducting plate P. The capacitance between the terminals depends on the position of the plate P and on the shape of the copper foil F. See Prob. 612.
•y
In both cases, draw a curve of y(z) from z = 0 to z = 100 millimeters, showing the unetched region on the circuit board. 613E
EQUIPOTENTIAL SURFACES A liquid dielectric is situated in an electric field. A thin conducting sheet carrying zero net charge is introduced into the dielectric so as to coincide with an equipotential surface. a) Is the electric field disturbed? b) What is the value of a on the surfaces of the sheet at a point where the electric displacement is D] s
614E
615
NONHOMOGENEOUS DIELECTRICS Show that a nonhomogeneous dielectric can have a volume density of bound charge in the absence of a free charge density. FIELD
OF A SHEET OF ELECTRONS TRAPPED IN LUCITE When a block of insulating material such as Lucite is bombarded with highenergy electrons, the electrons penetrate into the material and remain trapped inside. In one particular instance, a 0.1 microampere beam bombarded an area of 25 square centimeters of Lucite (e = 3.2) for f second, and essentially all the electrons were trapped about 6 millimeters below the surface in a region about 2 millimeters thick. The block was 12 millimeters thick. In the following calculation, neglect edge effects and assume a uniform density for the trapped electrons. Assume also that both faces of the Lucite are in contact with grounded conducting plates. Show that a) in the region where the electrons are trapped. p, = —2.000 x 1 0 " coulomb/meter . p = 1.375 x 1 0 " coulomb/meter ; b) in the neutral region, D„ =  2 . 0 0 0 x 1 0 " coulomb/meter , P„ = —1.375 x 1 0 " coulomb meter . E„ =  7 . 0 6 2 x 10 volts/meter, V„ = 7.062 x 10 .v  4,237 volts; c) at the surfaces. a = — 1.375 x 1 0 " coulomb/meter ; r
2
3
2
3
h
5
2
5
2
5
5
?
b
2
178
Dielectrics: I d) in the charged region, dD
c
~dx~
 2 . 0 0 0 x 10
2
coulomb/meter . 3
 2 . 0 0 0 x 10" .v coulomb/meter . 2
2
 7 . 0 6 2 x 10 volts meter , 8
E = c
2
7.068 x 10 x volts/meter, 8
dV = 7.062 x 10 volts/meter . die 2
c
8
2
2
V = 3.531 x 1 0 x  3884 volts. 8
2
c
e) Draw curves of D, E, V as functions of the distance x to the midplane, from x = —6 to x = 6 millimeters. f) Show that the stored energy is 1.88 x 1 0 joule. g) Is there any danger that the block will explode?  4
616
SHEET
ELECTRET A sheet electret is polarized in the direction normal to its surface. Draw a figure showing the polarization P. the surface charge densities +a and 
h
fc
617
RELATION BETWEEN R AND C FOR ANY PAIR OF ELECTRODES When a parallelplate capacitor is filled with a dielectric whose relative permittivity is e , its capacitance is C. If the dielectric is slightly conducting, its resistance is P. Show that RC = e e /a, where o is the conductivity. This rule applies to any pair of electrodes submerged in a medium, as long as the conductivity of the electrodes is much larger than that of the medium. r
r
0
CHAPTER
7
DIELECTRICS:
II
Continuity Conditions at an Interface, and Forces, Displacement Current
7.1 7.1.1 7.1.2 7.1.3 7.1.4 7.1.5
Energy
Density
CONTINUITY CONDITIONS AT THE INTERFACE BETWEEN TWO MEDIA The Potential V The Normal Component of D The Tangential Component of E Bending of Lines of Force Examples: Point Charge Near a Dielectric, Sphere of Dielectric in an Electric Field
7.2
POTENTIAL ENERGY
7.3
FORCES ON CONDUCTORS IN THE PRESENCE OF DIELECTRICS Example: ParallelPlate Capacitor Immersed in a Liquid Dielectric
7.3.1
ENERGY DENSITY
OF A CHARGE
DISTRIBUTION,
7.4 7.4.1
ELECTRIC FORCES ON Example: Curve Tracer
7.5
THE POLARIZATION
7.6 7.6.1
THE DISPLACEMENT CURRENT DENSITY cD[dt Example: DielectricInsulated ParallelPlate Capacitor
7.7
FREQUENCY AND TEMPERATURE DEPENDENCE, ANISOTROPY Examples: Water, Sodium Chloride, Nitrobenzene, Compounds of Titanium
7.7.1 7.8
SUMMARY PROBLEMS
DIELECTRICS
CURRENT
DENSITY
dP/dt
This chapter will complete our study of dielectrics. We shall first discuss the important continuity conditions that apply at the interface between two media. These conditions concern the potential V, the tangential component of E, and the normal component of D. Then we must return to the energy stored in an electrostatic field. This stored energy will give us the forces exerted on conductors immersed in nonconducting liquids. It will also give us the forces acting within the dielectrics themselves. If the electric field is a function of the time, then the motion of charge within the molecules of the dielectric gives a polarization current. The displacement current is equal to the polarization current plus another current that exists whenever the electric field varies with time, even in a vacuum.
7.1
CONTINUITY CONDITIONS A T THE INTERFACE BETWEEN TWO MEDIA The quantities V, E, and D must satisfy certain boundary conditions.
7.1.1
THE
POTENTIAL
V
At the boundary between two media, V must be continuous, for a discontinuity would imply an infinitely large electric field intensity, which is physically impossible.
7. / Continuity Conditions at the Interface Between Two Media
181
The potential is normally set equal to zero at infinity if the charge distribution is of finite extent. T h e potential is constant throughout any conductor, as long as the electric charges are at rest.
7.1.2
THE NORMAL
COMPONENT
OF D
Consider a short Gaussian cylinder drawn about a boundary surface, as in Fig. 71. The end faces of the cylinder are parallel to the boundary and arbitrarily close to it. The boundary carries a free surface charge density a . If the area S is small, D and a do not vary significantly over it and, according to Gauss's law, the flux of D emerging from the flat cylinder is equal to the charge enclosed: f
s
(D
nl

D„ )S 2
= OjS.
(71)
The only flux of D is through the end faces, since the area of the cylindrical surface is arbitrarily small. Thus Dm  Dm = <>{•
Figure 71 Gaussian cylinder on the interface between two media 1 and 2. The difference D — D between the normal components of D is equal to the surface density of free charge a . nl
n2
f
(72)
182
Dielectrics:
II
At the boundary between two dielectric media the free surface charge density o is generally zero and then D„ is continuous across the boundary. On the other hand, if the boundary is between a conductor and a dielectric, and if the electric field is constant, D = 0 in the conductor and D„ = a in the dielectric, a being the free charge density on the surface of the conductor. f
f
f
7.1.3
THE
TANGENTIAL
COMPONENT
OF E
This boundary condition follows from the fact that the line integral of E • dl around any closed path is zero for electrostatic fields. Consider the path shown in Fig. 72, with two sides parallel to the boundary, of length L, and arbitrarily close to it. The other two sides are perpendicular to the boundary. In calculating this line integral, only the first two sides of the path are important, since the lengths of the other two approach zero. If the path is small enough, E, does not vary significantly over it and E L
E, L
n
= 0,
2
(73)
or E
n
= E.
(74)
t2
The tangential component of E is therefore continuous across the boundary.
Figure 72 Closed path of integration crossing the interface between two media 1 and 2. Whatever be the surface charge density a the tangential components of E on either side of the interface are equal; £ = £ 2 f
( 1
(
7.1 Continuity
Conditions at the Interface
Between Two Media
183
If the boundary lies between a dielectric and a conductor, then E = 0 in the conductor and E, = 0 in both media. With static charges, E is therefore normal to the surface of a conductor.
BENDING
OF LINES
OF
FORCE
It follows from the boundary conditions that the D and E vectors change direction at the boundary between two dielectrics. In Fig. 73, using Eq. 72 with a* — 0. D cos (f = D cos 0 , :
2
(75)
2
or e
r l
e
£ 0
i cos 0, = e
r 2
e £ 0
cos 0
2
2
(76)
and, from Eq. 74, Ei sin 0
l
= E sin d . 2
(77)
2
Figure 73 Lines of D or of E crossing the interface between two media 1 and 2. The lines change direction in such a way that e tan 8 = e tan 0,. r l
2
r2
184
Dielectrics:
II
Then, from the last two equations, tan 0!
In
tan 0 ,
(78)
The larger angle from the normal is in the medium with the larger relative permittivity. 7.1.5
EXAMPLES: DIELECTRIC
POINT CHARGE IN AN ELECTRIC
NEAR A DIELECTRIC, FIELD
SPHERE
OF
Figure 74 shows two examples of the bending of lines of force at the boundary between two dielectrics. The lines of D are broken but continuous, since lines of D
1
(a)
Figure 74 (a) Lines of D and equipotentials for a point charge in air near a dielectric. They are not shown near the charge because they get too close together.
185
Figure 74 (cont.) (b) Lines of D and equipotentials near a dielectric sphere situated in air in a uniform electric field. In both cases, the lines of D are indicated by arrows, and the equipotential surfaces are generated by rotating the figures around the horizontal axis.
terminate only on free charges. Some lines of E either originate or terminate at the interface, according as to whether a is positive or negative. In Fig. 74b, the lines of D crowd into the sphere, showing that D is larger inside than outside. However, the equipotentials spread out inside, and E is weaker inside than outside. The density of the lines of E is lower inside than outside, since some of the lines of E coming from outside terminate at the surface. Note that, in the case of Fig. 74b, the field inside the sphere is uniform. Note also that the field outside is hardly disturbed at distances larger than one radius from the surface. b
186
7.2
Dielectrics:
II
POTENTIAL ENERGY OF A CHARGE DISTRIBUTION, ENERGY DENSITY To calculate the potential energy associated with an electric field in dielectric material, we refer to the parallelplate capacitor as in Sec. 4.3. We now have a dielectricfilled capacitor as in Fig. 65. If we add charges dQ until the top plate carries a charge Q and is at a potential V, we must expend an energy
QV =  — ~QV = ^ = = ,i Q 2
V
2 C
2 \A
2^
2
r
le e A r
— ,
(79)
e e /F 0
sA,
(710)
0
= \DEsA,
(711)
where Z) is the electric displacement (Sec. 6.5), E is the electric field intensity, and sA is the volume of the dielectric. The energy density is now DE/2 or, if the dielectric is linear, as is usually the case, € e E /2. The potential energy is then given by the integral of e e E /2 evaluated over the volume occupied by the field, 2
r
0
2
r
0
2
7.3
FORCES ON CONDUCTORS IN THE PRESENCE OF DIELECTRICS The forces between conductors in the presence of dielectrics can be calculated by the method of virtual work that we used in Sec. 4.5. When the conductors are immersed in a liquid dielectric, the forces are always found to be smaller than those in air by the factor e if the charges are the same in both cases. They are larger than in air by the factor e if the electric fields (and hence the voltages) are the same. r
r
The case of solid dielectrics will be illustrated in Probs. 79 and 710.
7.4 Electric Forces on Dielectrics
EXAMPLE: PARALLELPLATE IN A LIQUID DIELECTRIC
CAPACITOR
187
IMMERSED
We found in Sec. 4.5.2 that the force between the plates of an airinsulated parallelplate capacitor is e E A/2, or (a /2e )A, where a is the surface charge density and A is the area of one plate. We can perform a similar calculation for a parallelplate capacitor immersed in a liquid dielectric. The force per unit area is equal to the energy density (Sec. 4.5). Then the mechanical force holding the plates apart is 2
2
0
0
e,f
F,„ =
(712)
E A. 2
For a given electric field strength E, or for a given voltage difference between the plates, the force is larger than in air. Also. F
m
=
D 2e e r
A = 0
2
2e e r
1
Q
e
2e„A
r
a
A,
(713)
0
2
(714)
and, for given charges +Q and — Q on the plates, the force is smaller than in air.
7.4
ELECTRIC
FORCES ON
DIELECTRICS
When a piece of dielectric is placed in an electric field, its dipoles are subjected to electric forces and torques. If the field is uniform, and if the molecules have a permanent dipole moment p. each dipole experiences a torque. T = p x E,
(715)
that tends to align it with the field, but the net force is zero. If the field is nonuniform, then the forces on the two charges of a dipole are unequal, and there is also a net force. Let us consider a dielectric whose molecules do not have a permanent dipole moment. We can find a general expression for the electric force per unit volume in a nonuniform field by considering the dielectricfilled cylindrical capacitor of Fig. 75. The electric field is radial, and its magnitude varies as \jr. The net electric force on a dipole is inward, because the inward force on — Q is
188
Figure 75 Dielectricfilled cylindrical capacitor.
larger than the outward force on + Q. This net inward force is f=
QE  QE r
r + dr
=
dE
(716)
Q ^ s ,
dE
(717)
where p = Qs is the dipole moment. Therefore, if there are N dipoles per unit volume, the electric force per unit volume is dE F
>
=
Hr
Np
(e, 
dE =
(718)
*>
P
clE
= (€ r
1 £)^(e e E /2). 2
r
0
d l)e (E /2), 2
0
(719)
(720)
7.4 Electric Forces on Dielectrics
189
M o r e generally, the electric force per unit volume is proportional to the gradient of the energy density:
(721)
The dielectric tends t o move to where the field is strongest. In the above expression, E is the electric field intensity inside the dielectric, of course.
7.4.1
I
EXAMPLE:
CURVE
TRACER
A curve tracer is an electromechanical device that draws curves; the device is under the control of a computer. The curves are drawn on a sheet of paper that must stay rigidly fixed to the platen. O n e way of holding the paper is shown in Fig. 76. Parallel wires are embedded in the plastic platen about two millimeters apart and charged to plus and minus 300 volts. This gives a highly inhomogeneous field at the surface, with lines of force converging toward the charged wires. Thus V E inside the paper has a vertical component directed into the platen. 2
Platen Figure 76 Curve tracer. The electric field produced by the charged wires embedded in the platen attracts the electrically neutral sheet of paper toward the platen.
190
7.5
Dielectrics:
II
THE POLARIZA
TION CURRENT
DENSITY
dP/dt
When a dielectric is placed in an electric field that is a function of time, the motion of the bound charges gives a polarization current. We can find the polarization current density J as follows. For any volume i bounded by a surface S, the rate at which bound charge flows out through S must be equal to the rate of decrease of the bound charge within S (Sec. 5.1.1): h
£ J „ d a = jp„dx.
(722)
Using the divergence theorem on the left and substituting the value of p on the right from Eq. 62,
h
."ID
"
f V • J„ dx = 
Jr
f V • P dx = f V • — dx.
( Ji
Jr
(723)
(
(!
We have put the c/ct under the integral sign because it is immaterial whether the derivative with respect to the time is calculated first or last. Since x is any volume, the integrands must be equal and the polarization current density is (724) We could have added a constant of integration independent of the coordinates, but such a constant would be of n o interest. We have used the partial derivative with respect to t because P can be a function both of the time and of the space coordinates x, y. z.
7.6
THE DISPLACEMENT
CURRENT
DENSITY
cD/ct
The displacement current density is ,1) d 8 cP — =  ( e E + P) =  e E + — , dt dt ct ct 0
0
(725)
where dP/dt is the polarization current density of the previous section. Note the first term: it can exist even in a vacuum.
7.7 Frequency anil Temperature Dependence, Anisotropy
7.6.1
EXAMPLE:
DIELECTRICINSULATED
PARALLELPLATE
191
CAPACITOR
Let us calculate the displacement current in a dielectricinsulated parallelplate capacitor. Consider a current / flowing through a capacitor as in Fig. 77. We assume that positive charge flows from left to right, that Q = 0 at t = 0, and that edge effects are negligible. On the lefthand side of the capacitor. Q increases and dQ/dt = I. O n the right, a charge of equal magnitude but opposite sign builds up and, again, dQ/dt = /. Now Q = Aa. D = a = Q A. where A is the area of one plate, and a is the surface charge density. The displacement current in the capacitor is dD
dQ
~dt
dt
(726)
I.
The displacement current A(dD/dt) is composed of two parts, first d D A—\— dt
A±(e E)
w
0
d (a\ dt
1 dQ
(727)
dt
and the polarization current dP A— dt
d =A(D dt
Q 1 = dQ/dt
\\da
dQ dt '
(728)
Q I = dQ/dt
Figure 77 Current flowing through a capacitor.
7.7
FREQ UENC Y AND TEMPERA TV RE DEPENDENCE, ANISOTROPY There are three basic polarization processes, (a) In induced or
electronic
polarization, the center of negative charge in a molecule is displaced, relative to the center of positive charge, when an external field is applied, (b) In orientational
polarization, molecules with a permanent
dipole
moment
tend to be aligned by an external field, the magnitude of the susceptibility
192
Dielectrics:
II
being inversely proportional to the temperature, (c) Finally, ionic polarization occurs in ionic crystals: ions of one sign may move, with respect to ions of the other sign, when an external field is applied. For a given magnitude of E. both the magnitude and the phase of p are functions of the frequency, first because of the various polarization processes that come into play as the frequency changes, and also because of the existence of resonances. Thus, since e is a function of the frequency. e is strictly definable only for a pure sinusoidal wave. See Table 61. r
r
In many substances the relative permittivity decreases by a large factor as the temperature is lowered through the freezing point. Anisotropy is another departure from the ideal dielectric behavior. Crystalline solids commonly have different dielectric properties in different crystal directions because the charges that constitute the atoms of the crystal are able to move more easily in some directions than in others. The polarization P is then parallel to E only when E is along certain preferred directions.
7.7.1
EXAMPLES: COMPOUNDS
WATER, SODIUM OF TITANIUM
CHLORIDE,
NITROBENZENE,
Water has a relative permittivity of 81 in an electrostatic field, and of about 1.8 at optical frequencies. The large static value is attributable to the orientation of the permanent dipole moments, but the rotational inertia of the molecules is much too large for any significant response at optical frequencies. Similarly, the relative permittivity of sodium chloride is 5.6 in an electrostatic field and 2.3 at optical frequencies. The larger static value is attributed to ionic motion, which again is impossible at high frequencies. In nitrobenzene, e falls from about 35 to about 3 in changing from the liquid to the solid state at 279 kelvins. In the solid state, the permanent dipoles of the nitrobenzene molecules are fixed rigidly in the crystal lattice and cannot rotate under the influence of an external field. Ceramic capacitors utilize various compounds of titanium as dielectrics. These compounds are used because of their large relative permittivities ranging up to 10.000. However, e varies with the temperature, with the applied voltage, and with the operating frequency. r
r
7.8
SUMMARY At the boundary between two media, both V and the tangential component of E are continuous, but the difference between the normal components of D is equal to the surface density of free charge a . f
Problems
193
The potential energy stored in an electric field can be calculated from the energy density e e £ , ' 2 . Hence the force per unit area exerted on a conductor submerged in a liquid dielectric is* e e £ / 2 . The force per unit volume on a dielectric is 2
r
0
2
r
1  ^
0
V(e e £ /2).
(72/)
2
r
0
The displacement current density at a point is (
D
d
r
i dP
(725)
where dP/dt is the polarization current density. Because of the nature of polarization processes, e is a function of frequency. For most commonly used dielectrics, however, e changes very little, even when the frequency changes by many orders of magnitude (Table 61). In anisotropic media. P is parallel to E only when E is along certain preferred directions. r
r
PROBLEMS 7lE
CONTINUITY CONDITIONS AT AN INTERFACE Discuss the continuity conditions at the surface of the dielectric cylinder of Prob. 67.
72E
CONTINUITY CONDITIONS AT AN INTERFACE Discuss the continuity conditions at the surface of the dielectric sphere of Prob. 69.
73E
ENERGY STORAGE IN CAPACITORS A onemicrofarad capacitor is charged to a potential of one kilovolt. How high could you lift a onekilogram mass with the stored energy, if you could achieve 100°,, efficiency'?
74E
ENERGY STORAGE IN CAPACITORS The maximum stored energy per cubic meter in a capacitor is e e a /2, where a is the dielectric strength of the dielectric. The dielectric strength is the maximum electric field intensity before rupture. See Prob. 65. It is suggested that a small vehicle could be propelled by an electric motor fed by charged capacitors. Comment on this suggestion, assuming that the only problem is one of energy storage. 2
r
0
194
Dielectrics:
II
A good dielectric to use would be Mylar, which has a dielectric strength of 1.5 x 10 volts per meter and a relative permittivity of 3.2. Note that the use of a dielectric increases the capacitance by e and also permits the use of a higher voltage. The dielectric strength of air is only 3 x 10 volts per meter. s
r
6
E, D. a, AND
IF, FOR THE THREE
CAPACITORS
OF FIG. 78
Figure 78 shows three capacitors. In all three the electrodes have an area S and are separated by a distance .s. In A, the dielectric is air. In B, the dielectric has a relative permittivity e . In C, we have the same dielectric, plus a thin film of air. r
a) Find E. D, the surface charge density a on the electrodes, and the energy density W for A and B. Express your results for B in terms of those found for A. t
Solution:
For capacitor A, £ =F/s,
(1)
A
Z) = e £ A
0
a = D A
W
1A
A
A
= e V/s,
(2)
0
= V/s,
= \toEl
(3)
eo
=
{
e V*/s . 2
0
Figure 78 Three capacitors: A. without dielectric; B, with dielectric; C, with dielectric and an air film. See Prob. 75.
(4)
Problems
195
For capacitor B. £„ = Vs = £ .
.
A
D
= e e £
K
r
= e e V/s
B
r
= eD .
0
r
(6)
A
= D = e e F x = ea .
ffB
W
0
(5)
B
r
= i e e E
1B
r
0
r
(7)
A
= 1 e,e K /x
2
2
0
0
2
= e,W . 1A
(8)
b) Find £, D, H , for the dielectric in C, and
Solution:
Since the air film in capacitor C is thin, we may set £„
= V/s = £ .
(9)
A
Then D, a on the lefthand electrode, and W are the same as for capacitor B: l
D
a
= e C .
(10)
= e ff ,
(11)
= e,W .
(12)
r
a
A
r
Wia
A
lA
c) Find £, D, W, for the air film of capacitor C. and a on the righthand electrode. Express again your results in terms of the corresponding quantities found for capacitor A. Solution: The electric displacement D in the air film is the same as in the dielectric, since there are no free charges between the plates. Then Da C
= M>A
(13)
ffc = Da, = e D r
Eca = DcJ*o
A
= e ff , r
= t,D /e A
= e,£ ,
0
(15)
A
WJo = I e o £ i = J e 6 £ 0
(14)
A
2
2
= e rV . 2
r
1A
(16)
BOUND SURFACE CHARGE DENSITY Find the bound surface charge densities on the dielectric of capacitor B of Prob. 75 in terms of V/s.
796
77E
Dielectrics:
II
EXAMPLE OF A LARGE ELECTRIC FORCE One author claims that he can attain fields of 4 x 10 volts per meter over a 2.5 millimeter gap in purified nitrobenzene (« = 35), and that the resulting electric force per unit area on the electrodes is then more than two atmospheres. Is the force really that large? One atmosphere equals about 10 pascals. 7
r
5
78
PERPETUALMOTION MACHINE In Prob. 75 we showed that, for the capacitor C, the energy density in the dielectric and in the air film are. respectively,
ee
V — 2 s~
r
—
2
0
ee —
V
2
and
0
s
2
2
.
r
Now this is disturbing because it appears to mean that the net force on the capacitor is not zero. The forces on the capacitor plates are
ee V
6e V
2
r
2
0
r2
s
2
5
2
0
and
2
= S. s 2
Is there a force on the dielectric? Well, the E inside is uniform and V E = 0, which means that the force density is zero. So it seems that the net force on the capacitor is 2
ee r
V
2
0
(CrDfjS. With Mylar (e = 3.2). s = 0.1 millimeter. S = 1 meter , and V = 1 kilovolt. the force is about 3 x 10" newtons! O n e could therefore propel a large vehicle indefinitely with a set of capacitors like this, fed by a small battery supplying a voltage V at zero current! Where have we erred? Him: (i)The values given above for the forces on the electrodes are correct, (ii) Inside the dielectric, there is no force, (iii) The forces on the faces of the dielectric are not zero, however. Remember that Coulomb's law applies to any net accumulation of charge, irrespective of the presence of matter. 2
r
1
79
SELFCLAMPING CAPACITOR A certain capacitor is formed of two aluminum plates of area A, separated by a sheet of dielectric of thickness /. and connected to a source of voltage V. For various reasons, it is impossible to apply a force of more than a few tens of newtons to press the two plates together mechanically. Since neither the plates nor the dielectric are perfectly flat, the plates are spaced by a distance t, plus a good fraction of a millimeter. This is highly objectionable because the capacitance must be as large as possible. You are asked to investigate whether the electrostatic force of attraction on the plates might not be sufficiently large to reduce the air film thickness to a negligible value.
Problems
197
Set A = 4.38 x 1 0 " square meter, t = 0.762 millimeter. e = 3.0. V = 60 kilovolts. See also the next problem. 2
r
710
ELECTROSTATIC CLAMPS Electrostatic clamps are sometimes used for holding work pieces while they are being machined. They utilize an insulated conducting plate charged to several thousand volts and covered with a thin insulating sheet. The work piece is placed on the sheet and grounded. One particular type operates at 3000 volts and is advertised as having a holding power of 2 x 10 pascals. The insulator is Mylar (e = 3.2). a) Suppose first that there is a film of air on either side of the Mylar. Use the results of Prob. 75 to show that the Mylar film is 15 micrometers thick. b) Now calculate the electric field intensity in the air film. You should find over 6 x 10 volts per meter, which is over 200 times the dielectric strength of air (Prob. 74). Sparking could occur in the air film, and the dielectric could deteriorate. Mylar operates satisfactorily under these conditions, however. c) Now assume that the air film is replaced by a film of transformer oil with a high dielectric strength and a permittivity that is not much different from that of the Mylar. What is the thickness of the Mylar now? 5
P
s
711
CALCULATING AN ELECTRIC FORCE BY THE METHOD OF VIRTUAL WORK Figure 79 shows a pair of parallel conducting plates immersed in a dielectric. The lower plate is fixed in position and grounded; the upper one can slide horizontally and is maintained at a fixed voltage V. Use the method of virtual work (Sec. 4.5) to calculate the horizontal force on the upper plate when V = 1000 volts, s = 1 millimeter. / = 100 millimeters. e = 3.0. r
Figure 79 Pair of conducting plates immersed in a dielectric. The lower one is fixed, while the upper one can slide horizontally. It is possible to find the horizontal force on the upper plate by the method of virtual work. See Prob. 711.
198
712
Dielectrics:
II
ELECTRIC FORCE Show that V £ is in the direction of VE. 2
713
CALCULATING AN ELECTRIC FORCE BY THE METHOD OF VIRTUAL WORK Use the method of virtual work (Sec. 4.5) to calculate the force on the dielectric sheet in Fig. 710 when F = 1000 volts, s = 1 millimeter, / = 100 millimeters, e = 3.0. r
Figure 710 Sheet of dielectric partly inserted between a pair of conducting plates. The force on the dielectric can again be calculated from the principle of virtual work. See Prob. 713. HIGHVOLTAGE
TRANSMISSION
LINES
Losses on highvoltage transmission lines are higher than normal during foggy or rainy weather. One reason for this is that water collects on the wires and forms droplets. Since a droplet carries a charge of the same sign as that of the wire, electrostatic repulsion forces the droplet to elongate and form a sharp point where the electric field intensity can be high enough to ionize the air. The result is a corona discharge in the surrounding air. in which the ions are accelerated by the electric field and collide with neutral molecules, forming more ions and heating the air. Corona discharges are objectionable because they dissipate energy, mostly by heating the air. Over long distances, they can cause heavy losses. They also cause radio and television interference. Droplets collect on the wire in drifting by. They are also attracted by the nonuniform electric field. You are asked to evaluate the importance of the nonuniform electric field. T o do this, assume that the air is stagnant and calculate within what distance of a wire the electrostatic force will be larger than the gravitational force. Clearly, if that distance is much larger than the conductor radius, then electrostatic attraction is important. The transmission line under consideration consists of a pair of conductors 11.7 millimeters in diameter, separated by a distance of 2 meters and operating at 100 kilovolts ( ± 50 kilovolts). You can find an approximate value for the field at a distance r near one wire as follows. If /. is the charge per meter.
2;ie
n
Problems
199
The expression on the right is the field near a single isolated wire, which is a good approximation as long as c is much less than the distance to the other wire. N o w the capacitance per meter for this line is approximately ne /5^ and 0
/. = (ne
5)10 coulombs meter. 5
0
The expression for the force per unit volume given in Sec. 7.4 involves e and the value of E inside a water droplet. For water, at low frequencies, e = 81 (Table 61). T o estimate the £ inside a droplet, one may use the formula for a dielectric sphere in a uniform field.' This is not a bad approximation, because even a raindrop is small, compared to the conductor radius, so that £ does not change much over one drop diameter. Then r
r
e + 2/
2ne r
r
0
Solution: Let us first calculate £ inside a water droplet situated at a distance c from the center of one of the wires: (7re /5)10
5
0
= 361.4/r volts/meter.
2ne r
11 + 2)
0
(1)
The electrostatic force per cubic meter on a droplet is
F
=
1
1\ d
1
81 dr
2
—
x 81 x 8.85 x 1 ( T
,,
x
1 2
.d[i\
4.625 x 1CT — \—A =  9 . 2 5 0 x 1 0 dr \r 5
 5
2
(361.4
V ,1
12)
r newtons/meter . 3
(3)
y
The negative sign indicates that the force is attractive. Now the gravitational force per cubic meter is 1000 g newtons. so F
9.250 x 1 0 " =
100%
5
9800c
5
•
(4
3
This ratio is equal to unity for c = 2.113 millimeters, which is smaller than the radius of the wire. The electrostatic force of attraction on a droplet is therefore negligible.
Electromagnetic * Electromagnetic
+
Fields and Wares, Eq. B55. Fields and Waves, Eq. 4174.
200
715
Dielectrics:
II
ELECTRIC FORCE ON A DIELECTRIC a) Calculate the electric force per cubic meter on the dielectric of a coaxial cable whose inner conductor has a radius of 1 millimeter and whose outer conductor has an inner radius of 5 millimeters. The dielectric has a relative permittivity of 2.5. The outer conductor is grounded, and the inner conductor is maintained at 25 kilovolts. b) Show that the electric force near the inner conductor is about 300 times larger than the gravitational force if the dielectric has the density of water, namely 10 kilograms per cubic meter. 3
716
DISPLACEMENT AND POLARIZATION CURRENTS A capacitor with plates of area A, separated by a dielectric of thickness s and relative permittivity e , is connected to a voltage source V through a resistance R. Calculate the displacement and the polarization currents as functions of,the time. r
7/7
DIRECT ENERGY CONVERSION It is possible to convert thermal energy into electrical energy with the circuit of Fig. 711. With X connected to Z , the battery B charges the barium titanate capacitor C. Then X is disconnected from Z and the capacitor is heated. The relative permittivity of the titanate decreases and the voltage on C increases. Then X is connected to Y, and C recharges the battery B, while supplying energy to R. The capacitor is then cooled and the cycle repeated. Such devices have been built to feed electronic circuits on board satellites. The capacitors are in thermal contact with panels on the outer surface of the satellite that are heated periodically by the sun as the satellite rotates about its own axis in space. Let V be the potential supplied by the battery; Q the capacitance at the temperature T : C and V the capacitance and voltage at T ; \Y the thermal energy required to heat the capacitor; and W the electric energy fed to R. Calculate the efficiency WJW when V = 700 volts, V = 3500 volts. e = 8000. e = 1600, area of one capacitor electrode 1 square meter, thickness of dielectric 0.2 millimeter, specific heat of barium titanate 2.9 x 10 joules per cubic meter per degree, and T — T, = 30 C. B
l
2
2
2
lh
e
lh
B
2
r l
r2
6
2
O
>
R
Figure 711 Circuit for transforming thermal energy into electric energy on board satellites. See Prob. 717.
B
CHAPTER
8
MAGNETIC
FIELDS:
I
The Magnetic
Induction B and the Vector Potential
8.1 8.1.1 8.1.2
THE MAGNETIC INDUCTION B Example: Long Straight Wire Carrying a Current Example: Circular Loop, Magnetic Dipole Moment m
8.2
THE DIVERGENCE
OF B
8.3
MAGNETIC
POLES
8.4 8.4.1
THE VECTOR POTENTIAL Example: Long Straight Wire
A
8.5
THE VECTOR POTENTIAL POTENTIAL V
A AND
8.6
THE LINE INTEGRAL OF THE OVER A CLOSED CURVE
8.7
SUMMARY PROBLEMS
MONO
THE
VECTOR
A
SCALAR POTENTIAL
A
The next eight chapters will deal with the magnetic fields of electric currents and of magnetized matter. We shall start here by studying the magnetic induction B and the vector potential A. These two quantities correspond, respectively, to the electric field intensity E and to the potential V. F o r the moment, we consider only constant conduction currents and nonmagnetic materials.
THE MAGNETIC
INDUCTION
B
Figure 81 shows a circuit carrying a current I. We define the induction at the point P as follows:
magnetic
(81) where the integration is carried out over the closed circuit. As usual, the unit vector I ! points from the source to the point of observation: it points from the element d l to the point P. Magnetic inductions are expressed in teslas. The constant p is defined as follows: 
0
fi
0
= 4n x 10
7
tesla meter/ampere.
(82)
and is called the permeability of free space. If the current I is distributed in space with a current density J amperes per square meter, then / becomes J da and must be put under the integral
203
sign. T h e n J da dl c a n b e w r i t t e n as J dx', w h e r e dx' is a n e l e m e n t of v o l u m e , and
B
^
f
,
^
,
(83)
a s in F i g . 82. T h e i n t e g r a t i o n is c a r r i e d o u t o v e r t h e v o l u m e x o c c u p i e d b y the currents. W e a s s u m e t h a t J is n o t a f u n c t i o n of t i m e a n d t h a t t h e r e a r e n o m a g n e t i c m a t e r i a l s in t h e field.
Figure 82 The magnetic induction dB due to an element J dx' of a volume distribution of current.
204
Magnetic Fields: I
As in electrostatics, we can describe a magnetic field by drawing lines of B that are everywhere tangent to B. Similarly, it is convenient to use the concept of flux, the flux of the magnetic induction B through a surface S, defined as the normal component of B integrated over S: O =
J.
B • da.
(84)
The flux d> is expressed in webers. Thus the tesla is one weber per square meter.
8.1.1
EXAMPLE:
LONG STRAIGHT
WIRE
CARRYING
A
CURRENT
An element dl of a long straight wire carrying a current /, as in Fig. 83, produces a magnetic induction p I dl sin 0 — q>i, r 0
dB = 47i
(85)
where (p is the unit vector pointing in the azimuthal direction. The positive directions for
Figure 83 The magnetic induction d B produced by an element / dl of the current / in a long straight wire.
205
Figure 84 Lines of B in a plane perpendicular to a long straight wire carrying a current /. The density of the lines is inversely proportional to the distance to the wire. Lines close to the wire are not shown.
Expressing dl. sin 0. and r in terms of i. and p.
B =  —
cos x d% ip, =
I
4np J~
(86)
Inp
nl2
The magnitude of B thus falls off inversely as the first power of the distance from an infinitely long wire. T h e lines of B are concentric circles lying in a plane perpendicular to the wire, as in Fig. 84.
EXAMPLE:
CIRCULAR
LOOP. MAGNETIC
DIPOLE
MOMENT
m
A circular loop of radius a carries a current /, as in Fig. 85. An element / dl of current produces a d B having a component dB. on the axis, as indicated in the figure. By symmetry, the total B is along the axis, and dB. =— .cost). 4n r pI 0
B. =
2toj
4n
p lcr 0
cos 0 =
r
(87)
j  ^ .
(88)
r
2
Therefore, on the axis, the magnetic induction is maximum at the center of the ring and drops off as r for z » a . Figure 86 shows lines of B in a plane containing the axis of the loop. 3
2
2
206
Figure 85 The magnetic induction d B produced by an element / dl at a point on the axis of circular current loop of radius a. The projection of dB on the axis is dB..
y
\
Figure 86 Lines of B in a plane containing the axis of the loop of Fig. 85. The direction of the current in the loop is shown by the dot and the cross.
8.2 The Divergence o / B
207
Far from the loop, the field is the same as that of an electric dipole (Sec. 2.5.1), except that the factor 1 /4ne is replaced by p /4n, and that the electric dipole moment p is replaced by the magnetic dipole moment m. The magnetic dipole moment m is a vector whose magnitude is na I and that is normal to the plane of the loop, in the direction given by the righthand screw rule with respect to the current /. 0
0
2
THE DIVERGENCE
OF B
Consider a current element / dl and a volume tk containing a point P as in Fig. 87. T h e current element produces at P a magnetic induction j U
0
dB = ^ 4n
/ dl
x
, r
2
i!
•'.
(89)
As we saw in Sec. 8.1.1, the lines of B due to the element I dl are circles situated in planes perpendicular to a line through dl and centered on it as
Figure 87 The current element / d l produces a magnetic induction B. The net outward flux of B through the surface of the element of volume dx is zero.
208
Magnetic Fields: I
in Fig. 87. F r o m the figure, it is obvious that the net outward flux of the B due to I dl through the surface of the volume dx is zero. Now, any volume can be subdivided into volume elements of the same kind as dx. Therefore, for any volume t bounded by a surface S,
j;
(810)
B • da = 0.
This equation is true of all magnetic fields. Then, using the divergence theorem (Sec. 1.10), V • B dx = 0.
(811)
V • B = 0.
(812)
So
The derivatives contained in the operator V are with respect to the field point where the magnetic induction is B. This is another of Maxwell's equations.
MAGNETIC
MO NO POLES
The statement that V • B is equal to zero implies that magnetic fields are due solely to electric currents and that magnetic "charges" do not exist. Otherwise, one would have the magnetic equivalent of Eq. 612, and the divergence of B would be proportional to the magnetic "charge" density. Elementary magnetic "charges" are called magnetic monopoles. Their existence was postulated by Dirac in 1931, but they have never been observed to date. The theoretical value of the magnetic monopole is2h/e, or 8.271 17 x 1 0 " weber, where h is Planck's constant (6.625 6 x 1 0 " joule second), and e is the charge of the electron, 1.602 1 x 1 0 " coulomb. The surface integral of B • da over a closed surface is equal to the enclosed magnetic charge. Problems 89 and 126 concern two methods that have been used for detecting monopoles in bulk matter. See also Prob. 195. 1 5
3 4
1 9
8.4 The Vector Potential A
THE VECTOR POTENTIAL
209
A
Since V • B = 0, it is reasonable to assume that there exists a vector A such that / B = V x A,
(813)
because the divergence of a curl is identically equal to zero (see Prob. 129). The derivatives in the del operator are evaluated at the field point. We can find an integral for A, starting from Eq. 83,
B = £° f J x %th'. 4tt J*' r
(814)
2
From Probs. 113 and 127,
J x
r
i r
2
f
= v f  ) x J
= V x
J
  V x J .
(815)
Now, the last term on the right is zero, because J is a function of x', y', z', while V contains derivatives with respect to X, y, z. Then
r
47T J ' R
4n J ' r R
(816)
and A
Mo fr dx'. J 4n
(817)
If the current is limited to a conducting wire,
N o t e that we can add to this integral any vector whose curl is zero without affecting the value of B in any way. N o t e also that B depends on the space derivatives of A, and not on A itself. The value of B at a given point can thus be calculated from A, only if A is known in the region around the point considered.
210
Magnetic Fields: I
EXAMPLE:
LONG
STRAIGHT
WIRE
In Sec. 8.1.1 we calculated B for a long straight wire carrying a current /. starting from the definition of B. We can also calculate B starting from (II
HqI
(819)
An
One can see immediately that A is parallel to the wire, since the dl's are all along the wire. We first calculate A for a current of finite length 2L. Referring to Fig. 88. Hoi
r pi L
' An
Jo
^in[z Ikd
dl {p
>
+ (
(820)
+ /2)>2 2
6
.iUL
In p
(p
p
271
(821)
r? y .
+
j P
«
2
(822)
(823)
L ). 2
For L * oo, A tends to infinity logarithmically. We can still try to calculate B, however, because a function can be infinite and still have finite derivatives. For example, if y = x, y » x as x > x . but dy dx remains equal to 1. To calculate B for p « L , we use the fundamental definition of the curl given in Eq. 177, using the rectangular path shown in Fig. 88 for the line integral: 2
2
B = lim
1
ApAr
(824)
(j) A • dl.
as ApA: tends to zero. Since A is parallel to the wire, B = lim
1 ApAr
(825)
[Aip)  A(p + Ap)] A r
— hm — I In 2tt Af. p
In
2/.
(826)
p + Ap
p + Ap
In
(827)
Ap
— hm — In Ap
In
1 +
(828)
Now In (1 + x) =^ x when x is small and. finally, B is equal to p I/2np, 0
as previously.
211
Figure 88 An element / dl of a current / in a long straight wire produces an element of vector potential dA at the point P. The small rectangle is the path of integration used for calculating B.
THE VECTOR POTENTIAL POTENTIAL V
A AND THE
SCALAR
The vector potential A of magnetic fields is in many respects similar to the scalar potential (or, simply, the potential) V of electric fields (Sec. 2.5). Let us return briefly to Chapters 2 and 3. a) With electric fields, the important quantity, in practice, is the intensity E: it is E that gives the stored energy and the forces on charged bodies and on polarized dielectrics. It is also E that causes breakdown in dielectrics. b) The quantity V is related to E through E = — VF. So one can add to V any quantity that is independent of the coordinates without affecting E in any way. Also, the value of E at a point can be calculated only if V is known in the region around that point. c) Finally, the integral for F(Eq. 218) is simpler to evaluate than that for E (Eq. 26). So, in general, it is easier to find E by first calculating V, and then V F , than to calculate E directly. However, if the geometry of the charge distribution is particularly simple, one can use Gauss's law (Sec. 3.2). Then E is easy to calculate directly and there is no need to go through V.
212
Magnetic Fields: I
Now let us see how A compares with V. a) The quantity that corresponds to E is the magnetic induction B: it is B that gives the stored energy and the magnetic forces (Chapter 13). b) The vector potential A is related to B only through B = V x A. One can add to A any quantity whose curl is zero without changing B, and one can deduce from A the value of B at a point only if one knows the value of A in the region around that point. c) The integral for A (Eq. 818) is also simpler to calculate than that for B (Eq. 81). However, with simple current distributions, it is much easier to use Ampere's circuital law (Sec. 9.1) to find B than to calculate A first. So there is a great similarity between V and A. Later on, in Sec. 11.5, we shall find an equation that relates E to both V and A, in timedependent fields.
8.6
THE LINE INTEGRAL OF THE VECTOR POTENTIAL A OVER A CLOSED CURVE Using Stokes's theorem (Sec. 1.13), A • dl = f (V x A) • da,
(829)
where S is any surface bounded by the closed curve C. But, from Eq. 813, the curl of A is B. Then (j), A • dl = J B • da = cp: s
(830)
the line integral of A • dl over a closed curve C is equal to the magnetic flux through C.
8.7
SUMMARY The magnetic induction B due to a circuit C carrying a current / is defined as follows:
Problems
213
Magnetic inductions are measured in teslas, and u is denned to be exactly An x 1 0 " tesla meter per ampere. The magnetic flux through a surface S is 0
7
(D = J B  d a . s
(84)
Magnetic flux is measured in webers, and a tesla is one weber per square meter. The divereence of B is zero: V • B = 0,
(812)
since the magnetic flux through a closed surface is always identically equal to zero, if magnetic monopoles do not exist. This is one of Maxwell's equations. It follows from this that B = V x A, where A is the vector
(813)
potential
The line integral of A • dl over any closed curve C is equal to the magnetic flux through any surface S bounded by C: (j). A • dl = (D.
(830)
PROBLEMS 81E
MAGNETIC FIELD ON THE AXIS OF A CIRCULAR LOOP Plot a curve of B as a function of z on the axis of a circular loop of 100 turns having a mean radius of 100 millimeters and carrying a current of one ampere.
82
SQUARE CURRENT LOOP Compute the magnetic induction B at the center of a square current loop of side a carrying a current /.
214
83
Magnetic Fields: I
MAGNETIC FIELD OF A CHARGED ROTATING DISK An insulating disk of radius R carries a uniform free surface charge density a on one face. It rotates about its axis at an angular velocity to. a) What is the magnitude of the electric field intensity E, close to the disk? b) Show that the surface current density a at the radius r is wra. c) Show that, at the center of the charged surface, 1 B = 
n wRo. 0
d) Calculate E and B for R = 0.1 meter, a = 1 0 and OJ = 1000 radians per second. 84
 6
coulomb per square meter,
SUN
SPOTS See Prob. 83. The Zeeman effect observed in the spectra of sunspots reveals the existence of magnetic inductions as large as 0.4 tesla. Let us assume that the magnetic field is due to a disk of electrons 10 meters in radius rotating at an angular velocity of 3 x 10" radian per second. The thickness of the disk is small compared to its radius. a) Show that the density of electrons required to achieve a B of 0.4 tesla is about 1 0 per square meter. b) Show that the current is about 3 x 1 0 amperes. c) In view of the enormous size of the Coulomb forces, such charge densities are clearly impossible. Then how could such currents exist? 1
7
2
1 9
12
85E
HELMHOLTZ COILS One often wishes to obtain a uniform magnetic field over an appreciable volume. In such cases one uses a pair of Helmholtz coils as in Fig. 89. Show that the field on the axis of symmetry, at the midpoint between the two coils is (0.8)
3
2
p NI/a, 0
where TV is the number of turns in each coil. If you have the courage to calculate the field as a function of z, and its derivatives, you will find that the first, second, and third derivatives are all zero at z = 0! Figure 810 shows B. as a function of z for values of/ up to 0.16a. Roughly speaking, B. is uniform, within 10%, inside a sphere of radius 0.1a.
When a gas is subjected to a strong magnetic field, its spectral lines are split into several components. The splitting is a measure of the magnetic induction.
f
215
Figure 89 Pair of Helmholtz coils. When the spacing between the coils is equal to one radius a, the magnetic field near the center is remarkably uniform. See Prob. 85.
0.2a
I 0.2a
Figure 810 Axial component of B near the center of a pair of Helmholtz coils as in Fig. 89, as a function of z and for various values of r.
HELMHOLTZ COILS You are asked to design a pair of Helmholtz coils (Prob. 85) that will cancel the earth's magnetic field, within 10 percent, over a spherical volume having a radius of 100 millimeters. The magnetic induction of the earth in the laboratory is 5 x 1 0 " tesla and forms an angle of 70 degrees with the horizontal. The laboratory is situated in the Northern Hemisphere. See the footnote to Prob. 811. a) How must the coils be oriented, and in what direction must the current flow? b) Specify the coil diameter and spacing. c) What is the total number of ampereturns required? d) The laboratory has a good supply of N o . 18 enameled copper wire on hand. 5
216
Magnetic Fields: I
This wire has a crosssection of 0.823 square millimeter and a resistance of 21.7 ohms per kilometer. An adjustable power supply is also available. It can supply from 0 to 50 volts at a maximum of two amperes. How many turns should each coil have? Specify the operating current and voltage. Will it be necessary to cool the coils? If so, what do you suggest? 87
LINEAR DISPLACEMENT TRANSDUCER Draw a curve of B as a function of z on the axis of a pair of Helmholtz coils (Prob. 85) with the currents flowing in opposite directions and with the coils separated by a distance 2a. instead of a. for values of z ranging from  a to + a. Note the linearity of the curve over most of this region. One could build a linear displacement transducer for measuring the position of an object by fixing to it a Hall probe (Sec. 10.8.1) and having it move in the field of such a pair of opposing Helmholtz coils. z
88E
THE SPACE DERIVATIVES OF B IN A STATIC FIELD Consider the field of a circular loop, as in Sec. 8.1.2. O n the right and somewhat above the axis, the lines of B flare out, so B. decreases with z and dBJvz is negative. At the same point, B increases with y, so dB /dy is positive. Similarly, cBJdx is positive. a) Why is this, mathematically? How are these derivatives related? b) Compare these derivative for other points on Fig. 86. y
89E
y
MAGNETIC MONOPOLES It is predicted theoretically that a magnetic charge Q* situated in a magnetic field would be subjected to a force Q*B/u . Calculate the energy acquired by a magnetic monopole in a field of 10 teslas over a distance of 0.16 meter. Express your answer in gigaelectronvolts (10 electronvolts). In one experiment for detecting magnetic monopoles, various samples, such as deepsea sediments, were subjected to such a field, as in Fig. 811. None were found. 0
9
Figure 811 Apparatus used in an unsuccessful attempt to observe magnetic monopoles. The device is inserted inside a coil producing an axial B of 10 teslas. A slurry of deepsea sediment flows into the chamber on the right at / and out at O. It was hoped that magnetic monopoles in the sample would be accelerated in the magnetic field and would leave tracks in the photographic plate P. See Prob. 89.
Problems
810
217
MAGNETIC FIELD OF A CHARGED ROTATING SPHERE A conducting sphere of radius R is charged to a potential V and spun about a diameter at an angular velocity w as in Fig. 812. a) Show that the surface charge density a is e V/R. b) Now show that the surface current density is 0
a = e o)V sin 0 = M sin 0, 0
where M is e coV, and 0 is the polar angle. c) Show that the magnetic induction at the center is 0
B = ( 2 / 3 ) u o j F = (2/3)/i„M. 6 o /
0
It turns out that B is uniform inside the sphere. d) What would be the value of B for a sphere 0.1 meter in radius, charged to 10 kilovolts, and spinning at 1 0 revolutions per minute? e) Show that the dipole moment is 4
m = (4/3)7tPv M. 3
f) What is the dipole moment for the above sphere? g) What current flowing through a loop 0.1 meter in diameter would give the same dipole moment?
+
+
+
+ +
r
+
+ +
+ Figure 812 Charged sphere spinning about a diameter. See Probs. 810 and 811.
THE EARTH'S
MAGNETIC
FIELD
The origin of the earth's magnetic field is still largely unknown. According to one model, the earth would carry a surface charge, producing an azimuthal current because of the rotation, and hence the magnetic field, as in Prob. 810.
218
Figure 813 Typical line of B for the earth's magnetic field. The shape of the field inside the earth is unknown With the model discussed in Prob. 811, the B inside would be uniform, and the lines of B would be straight, as in the figure.
This model is attractive, at first sight, because it gives a magnetic field that has the correct configuration, or nearly so, except that one has to disregard the fact that the magnetic and geographic poles do not coincide. As we shall see, this model requires an impossibly large surface charge density a. a) What must be the sign of
+
Solution: Figure 813 shows the earth with its North and South geographic poles, a line of B, and the direction of the angular velocity a). The current must be in the direction shown, and hence a must be negative. b) With this current distribution, it can be shown that is uniform. Use the results of Prob. 810 to deduce the value of M vertical component of B at the poles is 6.2 x 1 0 " tesla. At B is the same, immediately above and immediately below 5
Solution:
the B inside the earth from the fact that the the poles, the value of the surface.
Since 2
B = n M 0
= 6.2 x 1 0 " ,
(1)
5
1.5 M = — 6.2 x 10 = 74 amperes/meter. /'o c) Calculate the magnetic dipole moment of the earth. The earth has a radius of 6.4 x 10 meters.
(2)
6
Solution:
Again from Prob. 810, m = (4/3)7tR M,
(3)
3
= (4/3)ti(6.4 x 10 ) 74 = 8.1 x 1 0 ampere meters . (4) Remember that the magnetic pole that is situated in the Northern Hemisphere is a South" pole: the " N o r t h " pole of a magnetic needle points North. 6
3
2 2
2
Problems
219
d Calculate the surface charge densit) required to give the surface currenl density calculated above. If there were such a surface charge, what would be the value of the vertical electric field intensity? There does exist a negative charge at the surface of the earth. It gives an £ of about 100 volts per meter. See Prob. 42. Also, the maximum electric field intensity that can be sustained in air at normal temperature and pressure is 3 x 10 volts per meter. 6
Solution:
At the equator. I) = 0 7.
= av = M = 74.
(5)
where v is the tangential velocity. Thus
a =
74
= 0.16 coulomb meter . 2
,2;r x 6.4 x 10" )/(24 x 60 x 60)
(6)
This surface charge density would give an electric field intensity
£ =
a
0.16 8.85 x 10 
from Sec. 3.2, which is absurdly large.
1
2
= 1.8 x 1 0
1 0
volts meter.
(7)
CHAPTER
9
MAGNETIC Ampere's
FIELDS:
Circuital
9.1 9.1.1 9.1.2 9.1.3 9.1.4 9.1.5
AMPERE'S Example: Example: Example: Example: Example:
9.2
THE CURL
9.3
SUMMARY
II
Law
CIRCUITAL LAW Long Cylindrical Conductor Toroidal Coil Long Solenoid Refraction of Lines of H at a Current Short Solenoid
PROBLEMS
OF THE MAGNETIC
Sheet
INDUCTION
B
This chapter is devoted to Ampere's circuital law and to the curl of B. Ampere's law is used to calculate magnetic inductions in much the same way as Gauss's law is used to calculate electric field intensities. The value of V x B will follow immediately from Ampere's law by using Stokes's theorem. We still limit ourselves to constant conduction currents and to nonmagnetic media.
AMPERE'S
CIRCUITAL
LA W
We have seen in Sec. 8.1.1 that the magnetic induction vector B near a long straight wire is azimuthal and that its magnitude is p. I/2np, where p is the distance from the wire to the point considered. Thus, over a circle of radius p centered on the wire as in Fig. 91, 0
(91)
This is a general result and, for any closed curve C, (92)
where / is the current enclosed by C. The positive direction for the integration is related to the positive direction for / by the righthand screw rule as in Fig. 91.
222
Figure 91 Positive direction for the integration path around a current /.
F o r a volume distribution of current, (f)
c
B • dl = /t
0
J,
J • da,
(93)
where J is the current density through any surface S bounded by the curve C as in Fig. 92a. This is Ampere's circuital law. In many cases the same current crosses the surface bounded by the curve C several times. With a solenoid, for example, C could follow the axis and return outside the solenoid, as in Fig. 92b. The total current crossing the surface is then the current in each turn multiplied by the number of turns, or the number of ampereturns.
c
(a) Figure 92 (a) Ampere's circuital law states that the line integral of B • dl around C is equal to p times the current through any surface bounded by C. (b) Path of integration C for a solenoid. In this case the line integral of B • dl is &p I. 0
0
9.1 Ampere's Circuital Law
223
The circuital law can be used to calculate B when its magnitude is the same, all along the path of integration. This law is thus somewhat similar to Gauss's law, which is used to compute E, when E is uniform over a surface.
EXAMPLE:
LONG CYLINDRICAL
CONDUCTOR
The long cylindrical conductor of Fig. 93 carries a current / uniformly distributed over its crosssection with a density J =
(94)
I/nR . 2
Outside the conductor, B is azimuthal and independent of (p, so that, according to the circuital law, B =
(95)
p I/2np. 0
Inside the conductor, for a circular path of radius p, PaJitp
1
B
2np
p Jp 0
p Ip 0
2nR
2
(96)
The magnetic induction B therefore increases linearly with p inside the conductor. Outside the conductor, B decreases as 1/p. The curve of B as a function of p is shown in Fig. 94.
Figure 93 Long cylindrical conductor carrying a current /.
224
Magnetic Fields: II
'•0
2.0 3.0 4.0 p (millimeters)
5.0
6.0
Figure 94 The magnetic induction B as a function of radius for a wire of 1 millimeter radius carrying a current of 1 ampere.
EXAMPLE:
TOROIDAL
COIL
A closewound toroidal coil of square crosssection, as in Fig. 95, carries a current /. Along path a, the line integral of B is zero, since there is no current linking this path. Then the azimuthal B is zero in this region. The same applies to c and to any similar path outside the toroid. Then the azimuthal B is zero everywhere outside. Inside, along path b, 2npB = p NI, 0
(97)
where N is the total number of turns, and B = u.N I 2np.
(98)
There exist nonazimuthal components of the magnetic induction outside the toroid. For a path such as d in Fig. 95, the area bounded by the path is crossed
Figure 95 Toroidal coil. The broken lines show paths of integration.
9.1 Ampere's Circuital Law
225
once by the current in the toroidal winding and, at a distance large compared to the outer radius of the toroid, the magnetic induction is that of a single turn along the mean radius. Although the magnetic induction B outside the toroid is essentially zero, the vector potential A is not. This will be evident if one remembers that A is a constant times the integral of / dl/r, where r is the distance between the element dl and the point where A is calculated (Sec. 8.4). For example, at a point close to the winding. A is due mostly to the nearby turns, it is parallel to the current, and it has approximately the same value inside and outside. The vector potential A can therefore exist in a region where there is no B field. This simply means that we can have at the same time A # 0 and V x A = 0. For example, A = ki, where k is a constant, satisfies this condition. We are already familiar with a similar situation in electrostatics: the electric potential can have any uniform value in a region where E = — VV = 0.
9.1.3
EXAMPLE:
LONG
SOLENOID
For the long solenoid of Fig. 96. we select a region remote from the ends, so that end effects will be negligible; also, we assume that the pitch of the winding is small. Let us choose cylindrical coordinates with the raxis coinciding with the axis of symmetry of the solenoid, as in Fig. 97. 1. We first note that B has the following characteristics, both inside and outside the solenoid. a) By symmetry, all three components B , B_, B are functions neither of cp nor of z. b) Moreover, B = 0 for the following reason. Consider an axial cylinder of length / and radius p. either larger than the solenoid radius, or smaller, as in Fig. 96. The integral of B • da over its surface is simply InplB^ since the integrals over the two end faces cancel. But, according to Eq. 810, the integral of B • da over any closed surface is zero. Then B , = 0. c) Consider a rectangular path a inside the solenoid as in Fig. 96. By Ampere's circuital law, the line integral of B • dl around the path is zero. Since B = 0, as we have just found, the line integrals on the vertical sides must cancel, which means that, inside the solenoid, B. is also independent of p. The same applies to a similar path entirely outside the solenoid. 2. Outside the solenoid. a) We have shown above that, outside the solenoid. B . is independent of all three coordinates p.
v
p
(
p
B
v
= pl 0
2np.
(99)
Figure 97 Components of B at P. in cylindrical coordinates.
9.1 Ampere's Circuital Law227
3. Inside the solenoid. a) B = 0 inside because the line integral of B • dl over a circle of radius p, say the top edge of the small cylinder shown in Fig. 96, is 2itpB^, and this must be zero according to Eq. 92 because there is no current enclosed by the path. b) Considering now path c in Fig. 96, and remembering that B , = 0 both inside and outside, and that B. = 0 outside, we see that B.S = p N'IS, where N' is the number of turns per meter. Therefore v
f
0
B = ix N'I. r
(910)
0
The magnetic induction inside a long solenoid in the region remote from the ends is therefore axial, uniform, and equal to p times the number of ampereturns per meter NT. It is much larger than the azimuthal B outside, as long as (l/N')/2np « 1, or as long as the pitch of the winding 1/JV', divided by its circumference 2np, is very small. It is obvious, from Fig. 910, that B. outside a finite solenoid is not zero. This outside B can be larger or smaller than the outside B^, depending on the geometry of the solenoid, but both are much smaller than B. inside. 0
:
EXAMPLE:
REFRACTION
OF LINES
OF B AT A CURRENT
SHEET
Imagine a thin conducting sheet carrying a current density of a amperes per meter, as in Fig. 98. We can understand the refraction of lines of B at a current sheet by proceeding as in Sec. 7.1. Since the divergence of B is zero, the normal component
4
B
Figure 98 Conducting sheet carrying a current density of a amperes per meter. Since V • B = 0, the normal component of B is the same on both sides of the sheet. According to the circuital law, however, the tangential component is not conserved and a line of B is deflected in the direction shown.
228
Magnetic Fields: II
of B is conserved: B„i = B, .
(911)
l2
Also, if we apply Ampere's circuital law to a path of length L that is perpendicular to the sheet as in the figure. fl„LB, L 2
= ,, s
(912)
= B„  /,„*.
(913)
0
B,
2
The line of force is therefore rotated in the clockwise direction for an observer looking in the direction of the vector a. We could have arrived at this result in another way. The magnetic induction B is due to the current sheet itself and to other currents flowing elsewhere. According to Ampere's circuital law. the current sheet produces, just below itself in the figure, a B that is directed to the left and whose magnitude is p / 2 (Prob. 93). Similarly, the B just above the sheet is directed to the right and has the same magnitude. If we add this field to that of the other currents, we see that the tangential components of B must differ as above. a
0
EXAMPLE:
SHORT
SOLENOID
We can calculate B on the axis of the short solenoid of Fig. 99 by summing the contributions of the individual turns, using Eq. 88. If the length of the solenoid is / and if its radius is a. the magnetic induction at the center is Po r+i/2 f + i2 j a NT dz _ Po 2 J " " (a + z Y 2
(914)
=
2
z
1
u a N'I 2
0
a\a
2
+
= HcjN'I sin 8„
z)
2 l!2
(915)
as in Fig. 99. We have assumed that the solenoid is close wound. For a long solenoid. 0 * it/2, and B > fi NT, as in Sec. 9.1.3. At one end. again on the axis, m
0
B = u NT 0
sin 0J2.
(916)
The magnetic induction thus decreases at both ends of the solenoid, and this is of course due to the fact that the lines of B flare out as in Fig. 910. Upon crossing the winding, the radial component of B remains unchanged, but the axial component changes both its magnitude and its sign. For example.
229
Figure 910 Lines of B for a solenoid whose length is twice its diameter.
230
1.5
1.0
0.5
0
0.5
1.0
1.5
Figure 911 The magnetic induction B as a function of z on the axis of a solenoid with a length equal to 5 times its diameter. Note how sharply the field drops off at the ends.
the axial component in the upper lefthand side of the solenoid in Fig. 910 changes from, say, —0.9p N'I to +0.\p N'I, since the surface current density /. is N'l (Sec. 9.1.4). Figure 911 shows B as a function of z for a solenoid with / = 10a. See Prob. 95. 0
o
THE CURL OF THE MAGNETIC
INDUCTION
B
Applying Stokes's theorem to Eq. 93, we find that (917) for any b o u n d e d surface S. Then V x B = HqJ.
(918)
This is another of Maxwell's equations, but not in its final form yet, because we are still limited to static fields, and we have not yet considered magnetic materials.
Problems
9.3
231
SUMMARY Ampere's circuital law states that the lirie integral of B • dl over a closed curve C is equal to the current flowing through any surface S bounded by C: (j) B • dl = fi c
0
j
s
J • da.
(93)
U p o n applying Stokes's theorem to this equation, one finds that V x B = /< J.
(975)
0
PROBLEMS 91E
DEFINITION OF u Show that the permeability of free space fi can be defined as follows: If an infinitely long solenoid carries a current density of one ampere per meter, then the magnetic induction in teslas inside the solenoid is numerically equal to / i . 0
0
0
92
93E
MAGNETIC FIELD OF A CURRENTCARRYING TUBE A length of tubing carries a current / in the longitudinal direction. a) What is the value of B outside? b) How is A oriented outside? c) What is the value of B inside? d) Show that A is uniform inside. MAGNETIC FIELD CLOSE TO A CURRENT SHEET A conducting sheet carries a current density of a amperes per meter. Show that, very close to the sheet, the magnetic induction B due to the current in the sheet is u. x/2 in the direction perpendicular to the current and parallel to the sheet. 0
94E
VAN DE GRAAFF HIGHVOLTAGE GENERATOR In a Van de Graaff highvoltage generator, a charged insulating belt is used to transport electric charges to the highvoltage electrode. a) Calculate the current carried by a belt 0.5 meter wide driven by a pulley 0.1 meter in diameter and rotating at 60 revolutions per second, if the electric field intensity at the surface of the belt is 2 kilovolts per millimeter. b) Calculate the magnetic induction close to the surface of the belt, neglecting edge effects. See Prob. 93.
95
232
Magnetic Fields: II
FIELD
ON THE AXIS OF A SHORT SOLENOID A short solenoid carries a current I and has N' turns per meter. Show that, at any point on the axis.
B =  /J /N'(COS
+
0
cos
a )i 2
where a and a are the angles subtended at the point by a radius R at either end of the solenoid. For example, if the coil has a length 2L, and if the point is situated at a distance x from the center. l
2
L  x 1
96
[R
2
1
L + x
cos a 2
+ (L — x ) ] ' ' 2
2
[R
2
+ (L + x ) ] '
2
2
1
2
FIELD
AT THE CENTER OF A COIL A coil has an inner radius R an outer radius R , and a length L. a) Show that the magnetic induction at the center, when the coil carries a current 7, is 1?
B =
2
ti nIL 0
2
In
a + (i
2
+
/f)"
2
r—T7T,
1 + (1 + P )
where
and n is the turn density, or the number of wires per square meter, b) Show that the length of the wire is \ = Vn = 2nn(y.
2
 l)/?7? , 3
where V is the volume occupied by the wire. 97
CURRENT DISTRIBUTION GIVING A UNIFORM B A long straight conductor has a circular crosssection of radius R and carries a current 7. Inside the conductor, there is a cylindrical hole of radius a whose axis is parallel to the axis of the conductor and at a distance b from it. Show that the magnetic induction inside the hole is uniform and equal to
2n(R
2
See also the next problem.

a )' 2
Problems
233
Hint: Use a uniform current distribution throughout the circle of radius R, plus a current in the opposite direction in the hole. 98D
1
SADDLE COILS Figure 912 shows the construction of a pair of saddle coils. The windings occupy the two outer regions of a pair of overlapping circles. a) In what general direction is the field oriented in the region between the conductors? b) It is said that the magnetic induction is uniform in the hollow. Is this true? What is the value of B? Assume that the length of the coil is infinite. You can solve this one by straight integration if you wish, but there is a much easier way. See the hint for Prob. 97. Such coils are used for magnetohydrodynamic generators. See Prob. 1016.
Figure 912 (a) Pair of saddle coils for producing a transverse magnetic field inside a tubular enclosure. Only one turn of each coil is shown, (b) Crosssection of the windings. See Prob. 98.
99D
TOROIDAL COIL A toroidal coil of N turns has a major radius R and a minor radius r. a) Calculate the magnetic flux by integrating B over a crosssection. b) At what radius does B have its mean value? f
The letter D indicates that the problem is relatively difficult.
CHAPTER
10
MAGNETIC
FIELDS:
Transformation
III
of Electric and Magnetic
Fields
10.1 10.1.1
THE LORENTZ FORCE Example: The CrossedField
10.2
EQUIVALENCE
10.3
REFERENCE FRAMES, AND RELATIVITY
10.4
THE GALILEAN
10.5
THE LORENTZ
10.6 10.6.1
IN VARIANCE OF THE VELOCITY OF LIGHT c Example: The Velocity of Light is Unaffected by the Earth's Orbital Velocity
10.7 10.7.1
INVARIANCE OF ELECTRIC Example: The Electron Charge
10.8
TRANSFORMATION OF ELECTRIC MAGNETIC FIELDS
10.8.1
Example:
10.9
PROBLEMS
Spectrometer
OF E AND v x B OBSERVERS, TRANSFORMATION TRANSFORMATION
The Hall Effect
SUMMARY
Mass
CHARGE AND
In this chapter, we shall start with the longknown fact that a moving charged particle is deflected by a magnetic field. F o r example, the electron beam in a television tube is deflected both horizontally and vertically by the magnetic fields of two sets of coils carrying rapidly varying currents. N o w why should a magnetic field exert a force on an electrically charged particle? The explanation is that the particle "sees" an electric field. In other words, the electrons in the television tube "see" an electric field when they pass between the deflecting coils. Let us consider a more general case. Suppose one has some configuration of electric charges and electric currents that produce an electric field intensity E , and a magnetic induction in a laboratory. Both E and B, are functions of the coordinates .v, v. z. N o w imagine an observer moving at some constant velocity v with respect to the laboratory. This observer is equipped with appropriate instruments for measuring the electric field intensity and the magnetic induction. At every point, he will find a field E , B that is different from E , , B (except if v. E , . B, are all parallel). This chapter concerns the equations that are used to find E . B . given E,. B ^ and inversely. Whenever one expresses E , Bj in terms of E . B , or inversely, one transforms the electromagnetic field from one reference frame to another. Most of the problems at the end of this chapter concern magnetic forces exerted on particles; macroscopic manifestations of magnetic forces will be dealt with in Chapters 13 and 15. t
2
2
x
2
t
2
2
2
236
10.1
Magnetic Fields: 111
THE LORENTZ
FORCE
It is observed experimentally that a charge Q moving at a velocity v in a region where the magnetic induction is B, is subjected to a
magnetic
force F = Qy x B.
(101)
Since this force is perpendicular to v, F • v = 0 and the power supplied to" the particle is zero. T h e Q\ x B force therefore changes the direction of v without changing its magnitude. M o r e generally, if there is also an electric field E, F = <2(E + v x B).
This is the Lorentz
10.1.1
EXAMPLE:
(102)
force.
THE CROSSEDFIELD
MASS
SPECTROMETER
The crossedfield mass spectrometer is illustrated in Fig. 101. The positive ions of mass in and charge Q have a velocity v given by
 me = QV. 2
(103)
In the region of the deflecting plates, they are submitted to an upward force QE and to a downward force QiB. F h e net force is zero for ions having a velocity v = E/B,
(104)
or a mass in = 2QVB E . 2
2
(105)
Fhe mass spectrum is obtained by observing the collector current I as a function of E, for a constant B. This type of mass spectrometer has a poor resolution, mostly because the ion beam is defocused in the fringing electric a n d magnetic fields. It is nonetheless a simple and useful device if one is concerned only with the very lightest elements.
237
9 " A
0 Figure 101 Crossedfield mass spectrometer. Positive ions produced in a source S maintained at a potential V are focused into an ion beam b. In passing through the superposed electric and magnetic fields, the beam splits vertically into its various components. Ions of the proper velocity are undeflected and are collected at C. The spectrometer is enclosed in a vacuum vessel A evacuated by a pump P.
10.2
EQUIVALENCE
OF E AND v x B
The Lorentz force of Eq. 102 is intriguing. Why should v x B have the same effect as the electric field E? Clearly, from Eq. 102, the particle cannot tell whether it "sees" an E or a v x B term. This is illustrated in the above example. It is also illustrated in Prob. 106, where two types of mass spectrometer are compared. The ion trajectory is circular in both types, but the deflecting force is QE in one, and Q\ x B in the other. Thus v x B is, somehow, an electric field intensity. As we shall see in the next sections, the explanation is provided by the theory of relativity. It has to do with reference frames.
10.3
REFERENCE FRAMES, AND RELA TIVITY
OBSERVERS,
By observer, we mean either a h u m a n being equipped with proper instruments, or some device that can make measurements, take photographs, and so on, either automatically or under remote control.
238
Magnetic Fields: III
An observer takes his measurements with respect to his reference frame. For example, one normally measures a magnetic induction with an instrument that is at rest with respect to the earth. This particular frame is usually called the laboratory reference frame. Special relativity is concerned with the observations made by two observers, one of whom has a constant velocity with respect to the other.
10.4
THE GALILEAN
TRANSFORMATION
Imagine two reference frames Si and S as in Fig. 102, with S moving at a velocity vi with respect to S , . N o w imagine some event, say a nuclear reaction, occurring at a certain instant at a certain point in space. An observer in S notes the coordinates of the event as x y z t. An observer in S notes, likewise, x , .v , z , t, and 2
2
{
l s
2
2
u
u
2
2
x, = x
2
+ vt,
(106)
y, = y ,
(io7)
i = i .
(108)
2
2
Figure 102 Two Cartesian coordinate systems, one moving at a velocity v\ with respect to the other in the positive direction of the common xaxis. The two systems overlap when the origins 0 and 0 coincide. We shall always refer to these two coordinate systems whenever we discuss relativistic effects. 1
2
10.5 The Lorentz Transformation
239
We have set t = 0 at the instant when the two reference frames overlap. These equations constitute the Galilean transformation. The inverse transformation, giving the coordinates x , y , ?i terms of x y z is given by interchanging the subscripts 1 and 2, and changing the sign of v. m
2
it
10.5
u
2
u
THE LORENTZ
TRANSFORMA
TION
Although the Galilean transformation is adequate for everyday phenomena, it is only approximate, and the correct equations are those of the Lorentz transformation: x , = y(x
2
+ vt ),
(109)
2
J?I = yi,
(1010)
*i = z ,
(1011)
ti = y ( * 2 + 4 x A
(1012)
2
where 1
(1013)
[1  {v/cfY ' 11
and c is the velocity of light in a vacuum, 2.997 924 58 x 1 0 meters per second. The inverse relationship, giving the coordinates in S in terms of those in S , is again obtained by interchanging the subscripts 1 and 2, and changing the sign of v. N o t e that, in general, t / t . In other words, the observer on Sj does not agree with the observer on S as to the time of occurrence of a given event. s
l
2
1
2
2
The Lorentz transformation forms the basis of special relativity. O n e can deduce from it equations of transformation for the length of an object.
240
Magnetic Fields: III
the duration of an event, a velocity, an acceleration, a force, a mass, and so o n /
10.6
IN VARIANCE
OF THE VELOCITY
OF LIGHT c
Experiments designed to measure the velocity of light c in a vacuum, with respect to reference frames moving at various velocities, always give the same value. Thus c is said to be
10.6.1
invariant.
EXAMPLE: THE VELOCITY OF LIGHT THE EARTH'S ORBITAL VELOCITY
IS UNAFFECTED
BY
The orbital velocity of the earth around the sun is 3 x 10 meters per second. This is two orders of magnitude larger than the tangential velocity due to the rotation of the earth about its own axis. At noontime, the orbital velocity is westward while, at midnight, it is eastward. If one measures c in the eastwest direction in the laboratory, first at noontime and then at midnight, one finds precisely the same value, within the experimental error. 4
10.7
IN VA RIA NCE OF ELECTRIC
CHA RGE
Electric charge is also invariant. This is again an experimental fact.
10.7.1
EXAMPLE:
THE ELECTRON
CHARGE
In the Millikan oildrop experiment, one measures the velocity of an electrically charged microscopic oil drop in an electric field. The velocity is of the order of millimeters per minute and is a measure of the electric charge on the drop. The charge carried by a drop is always a multiple of the electron charge, 1.602 x 10 coulomb. If now one measures in the laboratory frame the charge on an electron emerging from an accelerator with a velocity approaching the velocity of light, by deflecting it in a known magnetic field, one finds again 1.602 x 1 0 coulomb. _ 1g
 1 9
Electromagnetic
Fields and Waves, Chapter 5.
10.8 Transformation
10.8
of Electric and Magnetic Fields
241
TRANSFORMA TION OF ELECTRIC AND MAGNETIC FIELDS The expression for the Lorentz force in Eq. 102 gives us a clue as to how to transform a magnetic field. In this expression, F, (9, E, v, B are all measured in the same reference frame S,, which is normally the laboratory frame: F , = Q{E + v x B,)
(1014)
l
N o w call S the reference frame of the particle, moving at the velocity v at a certain instant. What is the field in S ? Let us assume for the moment that v « c , where c is the velocity of light. In that case, the force is the same in both reference frames. In S . the particle velocity is zero and there can be no magnetic force. Then the force in S must be QE , where 2
2
2
2
2
2
2
E , = E , + v x Bp
(1015)
So the magnetic field B, in frame 1 becomes an electric field v x B, in frame 2. The above equation is valid only for r « c . When v approaches c, it is shown in relativity theory that the forces in the two reference frames are different, a n d the correct value of E is as follows: 2
2
1
2
E21 = 7 ( E , i + v x B,). E
2  I
=E
1 H
,
(1016) (1017)
where the subscripts refer to the components that are either perpendicular or parallel to the velocity vi of frame S with respect to S,, as in Fig. 103; 7 is as defined in Eq. 1013. The magnetic field in S is given by 2
2
x E^. B , = B[.
(1019)
2
Equations 1016 to 1019 are the equations * Electromagnetic
(1018)
of transformation
Fields and Waves, pages 262 and 288.
for E and B.
242
Magnetic Fields: III
Thus, given a field E B
2
B
1 (
(
in frame S
u
one can calculate the field E , 2
in frame S . The inverse transformation is obtained, as usual, by inter2
changing the subscripts 1 and 2, and changing the sign of v. When v
2
« c , y % 1 and 2
E
2
= Ei + v x B,,
(1020)
1 B
2
= B ,  ^ v x E , c
(1021)
N o t e that the expression for the Lorentz force given in Eq. 102 remains true, even if v x c.
10.8.1
EXAMPLE:
THE HALL
EFFECT
Semiconductors contain either one or both of two types of mobile charges, namely conduction electrons and holes (Sec. 5.1). When a current flows through a bar of semiconductor in the presence of a transverse magnetic field B, as in Fig. 104, the mobile charges drift, not only in the direction of the applied electric field E but also in a direction perpendicular to both the applied electric and magnetic fields, because of the Qy x B force. This gives rise to a voltage difference V between the upper and lower electrodes. The drift is similar to the motion of the ions in the massspectrometer of Sec. 10.1.1.
243
(c)
(d) ntype material
ptype material
Figure 104 Hall effect in semiconductors. In ptype materials conduction is due to the drift of positive charges (holes) and the Hall voltage is as shown in (a). In Htype materials conduction is due to the conduction electrons and the Hall voltage has the opposite polarity as in (b). Ordinary conductors such as copper behave as in (b). Figure (c) shows the two opposing transverse forces QE and Qv x B on a positive charge drifting along the axis of the bar at a velocity v. Figure (d) shows how these forces are reversed in «type material. y
If the voltmeter V draws a negligible current, the plates charge up until their field E is sufficient to stop the transverse drift. This transverse electric field is called the Hall field. We shall first calculate the magnitude and direction of the Hall field E by using the Lorentz force. Then, as an exercise, we shall find the field E , B in the reference frame of the charge carriers, to arrive again at E . We assume that the material is zitype as in Fig. 104b and 104d: the charge carriers are negative. y
y
2
2
y
a) The field E must be such that Eq. 104 is satisfied. Then its magnitude is vB. Since v x B is in the positive direction of the yaxis (Fig. 104d), E must point in the negative direction and y
y
vB. as in Figs. 104b and 104d.
(1022)
244
Magnetic Fields: III
b) In the laboratory frame S j , the electric field intensity is E \ + E j, and the magnetic induction is Bk. We have omitted the subscripts 1 for simplicity. To find the field in the reference frame S of the moving charges, we use Eqs. 1020 and 1021. For ntype material, the velocity is — ri, where v is a positive quantity. Then x
2
E
2
= £. i + £ j  ri x B k .
(10231
= £ i + (£,. + vB)i,
(1024)
v
r
v
B
= B k + (l/c )ri x ( £ , i + £ j ) ,
(1025)
= [B + (vE c )]k.
(1026)
2
2
v
2
t
The force in frame 2 is QE , and since it has no v component, £ = —iB as previously. If there are n electrons per cubic meter, each carrying a charge e. 2
v
I = abJ = ab(nev).
(1027)
vBb = Bl/nea.
(1028)
Note that charge carriers of either sign are swept down by the magnetic field. The Hall effect is commonly used for measuring magnetic inductions and for various other purposes, some of which are described in Probs. 87, 1015, and 1712.
10.9
SUMMARY A charge Q moving at a velocity y in superposed electric and magnetic fields E and B is submitted t o the Lorentz
force
F = Q(E + v x B),
(102)
where F, E, v, and B are all measured with respect to the same reference frame S,. T o find the field perceived by the charge Q in its own reference frame S
2
moving at the velocity t i with respect to S , one must use special relativity. t
10.9 Summary
which is based on the Lorentz x
transformation: = y(x
l
V l
245
+ MX
2
(109)
2
= y,
(1010)
2
(1011)
h=y[t
+ ^x )
2
2
(1112)
t
with y
(1013)
2T1/2'
[1 
(v/cf\
and c equal to the velocity of light in a vacuum, 2.997 924 58 x 10 meters per second. It is possible to deduce from this set of equations other transformation equations for a mass, a velocity, etc. A given electric charge Q and the velocity of light c always have the same value, whatever the velocity of the reference frame with respect to which they are measured. 8
F o r a given field E ^ I*! in frame S,, the field in S , moving at a velocity v with respect to S , is given by 2
l
E
2 ±
= y(E
E
2 N
= Ej,
1±
+ v x B,),
B 2 1 = ? ( b
(1017)
U
 ^ X E A
Ban = Bj . If r
2
(1016)
(1018)
(1019)
« c , then y = 1 and 2
E
2
= E , + v x B,,
B
2
= B i  
T
(1020)
y x E . 1
(1021)
246
Magnetic Fields: III
PROBLEMS I01E
THE CYCLOTRON FREQUENCY A particle of mass in. charge Q. and velocity v. describes a circle of radius R in a plane perpendicular to the direction of a uniform magnetic field B. a) Show that BQt = mv
R.
2
b) Show that the angular velocity co = BQ/m. The frequency BQ 2nm is called the cyclotron frequency because it is the frequency at which an ion circulates in a cyclotron. Note how a) is independent of the velocity r of the particle. This is not strictly true, however, because m is itself a function of the velocity:
m m
0
(r /r )]" '
~ [1 
2
2
2
where c is the velocity of light in a vacuum and m is the rest mass. In practice. a> may be considered to be independent of c. for c « c . The mass m is 10 percent larger than m when v is about 0.4c. The above three equations are valid at all velocities. c) Calculate the cyclotron frequency for a deuteron in a field of one tesla. for r « c. 0
2
2
0
2
2
I02E
MOTION OF A CHARGED PARTICLE IN A UNIFORM B A charged particle moves in a region where E = 0 and B is uniform. Show that the particle describes either a circle or a helix.
103E
MAGNETIC MIRRORS Figure 105a shows a crosssection of a solenoid having a uniform turn density, except near the ends where extra turns are added to obtain a higher magnetic induction than near the center. Show qualitatively that, if the axial velocity is not too large, a charged particle that spirals around the axis in a vacuum inside the solenoid will be reflected back when it reaches the higher magnetic field. The regions of higher magnetic field are called magnetic mirrors. Magnetic mirrors are used for confining hightemperature plasmas. (A plasma is a highly ionized gas. Its net charge density is approximately zero.) In the Van Allen radiation hells, charged particles are trapped in the earth's magnetic field, oscillating north and south along the magnetic field lines. They are reflected near the North and South magnetic poles, where the lines crowd together, forming magnetic mirrors, as in Fig. 105b.
247
Solar wind
Figure 105 (a) Crosssection of a solenoid with magnetic mirrors at the ends. {b) Gray lines: magnetic field of the earth. The earth acts as a magnetic dipole. Black lines: magnetic field near the earth, as observed by means of satellites. The solar wind* is composed of protons and electrons evaporated from the surface of the sun. Since the interplanetary pressure is low, there are few collisions and the conductivity is high. In moving through the magnetic field of the earth, the charge particles are subjected to a v x B electric field, and currents flow according to Lenz's law. The black lines show the net field.
104E
HIGHENERGY ELECTRONS IN THE CRAB NEBULA Some astrophysicists believe that there exists, within the Crab Nebula, electrons with energies of about 2 x 1 0 electronvolts spiralling in a magnetic field of 2 x 1 0 ~ tesla. See Prob. 101. a) Calculate the energy W of such an electron in joules. 1 4
Electromagnetic
Fields and Waves, p. 502.
8
248
Magnetic Fields: III
b) Calculate its mass from the relation W
=
Hie . 2
where c is the velocity of light in a vacuum, 3 x 10 meters per second. How does this mass compare with that of a lowvelocity (c « c ) electron, which is9.1 x 1 0 " kilogram? c) Calculate the radius of its orbit, setting v = c and neglecting the velocity component parallel to B. d) Calculate the number of days required to complete one turn. 8
2
2
3 1
105E
MAGNETIC FOCUSING There exist many devices that utilize fine beams of charged particles. The cathoderay tube that is used in television receivers and in oscilloscopes is the bestknown example. The electron microscope is another example. In these devices the particle beam is focused and deflected in much the same way as a light beam in an optical instrument. Beams of charged particles can be focused and deflected by properly shaped electric fields. See, for example, Probs. 211 and 212. The electron beam in a TV tube is focused with electric fields and deflected with magnetic fields. Beams can be deflected along a circular path in a magnetic field as in Prob. 101. Let us see how they can be focused by a magnetic field. Figure 106 shows an electron gun situated inside a long solenoid. The electrons that emerge from the hole in the anode have a small transverse velocity component and, if there is no current in the solenoid, they spread out as in the figure. Let us see what happens when we turn on the magnetic field.
Figure 106 Focusing of an electron beam in a uniform magnetic field: F is a heated filament, A is the anode, and the electron beam is focused at P. The accelerating voltage is F. A filament supply of a few volts is connected between the top two terminals. When the solenoid is not energized, the beam diverges as shown by the dashed lines. See Prob. 105.
Problems
Let the velocity components at the hole be v and c . with r x
1
1
eV =  m ( r + 2
v
2
249
» v . Then 2
,
17) %  mo*.
If v = 0. the electron continues parallel to the axis. If v =F 0, it follows a helical path at an angular velocity cu as in Prob. 101. After one complete turn, it has returned to the axis. So. if B is adjusted correctly, all the electrons will converge at the same point P, as in the figure, and form an image of the hole in the anode. a) Show that this will occur if y
y
L b) Calculate the number of ampere turns per meter IN' in the solenoid, for an accelerating voltage V of 10 kilovolts and a distance L of 0.5 meter. In actual practice, magnetic focusing is achieved with short coils, and not with solenoids. I06E
DEMPSTER MASSSPECTROMETER Figure 210 shows an electrostatic velocity analyser. It was shown in Prob. 211 that a particle of charge Q. mass m, and velocity v passing through the first slit with the proper orientation will reach the detector if v =
QER/m.
Figure 107 shows a Dempster massspectrometer. Here again, the ion describes a circle, except that the centripetal force is now Qy x B. instead of QE. The magnetic field is provided by an electromagnet. a) Show that m =
QR B /2V. 2
2
b) In one particular experiment, a massspectrometer of this latter type was used for the hydrogen ions H^.H^.Hi, with/? = 60.0 millimeters and V = 1000 volts.
Figure 107 Crosssection of a mass spectrometer of the Dempster type. Ions produced at the sourse S are collected at C. The ions are accelerated through a difference of potential V and describe a semicircle of radius R. See Prob. 106.
250
Magnetic Fields: III
The W, ion is a proton. H% is composed of two protons and one electron, and H^ is composed of three protons and two electrons. Find the values of B for these three ions. c) Draw a curve of B as a function of in under the above conditions, from hydrogen to uranium. Would you use this type of spectrometer for heavy ions? Why? +
107
MASS
SPECTROMETER Figure 108 shows a mass spectrometer that separates ions, both according to their velocies and according to their chargetomass ratios. Show that an ion of charge Q, mass in. and velocity v is collected at the point
Figure 108 Mass spectrometer. Ions injected at A follow helical paths in the E and B fields that are both parallel to the yaxis. Each square is a separate collector. See Prob. 107.
I08E
HIGHTEMPERATURE PLASMAS One method of injecting and trapping ions in a hightemperature plasma is illustrated in Fig. 109. Molecular ions of deuterium, D j (two deuterons plus one electron, the deuteron being composed of one proton and one neutron), are injected into a magnetic field and are dissociated into pairs of D ions (deuterons) and electrons +
251
D
+
in a highintensity arc. The radius of the trajectory is reduced and the ions are trapped. See Prob. 101. a) Calculate the radius of curvature P for D J ions having a kinetic energy of 600 kiloelectronvolts in a B of 1.00 tesla. b) Calculate R for the D ions produced in the arc. The D ions have one half the kinetic energy of the D ions. +
+
2
109
HIGHTEMPERATURE PLASMAS The type of discharge shown in Fig. 1010 has been used to produce hightemperature plasmas. The discharge has the shape of a cylindrical shell and is situated in the space between two conducting coaxial cylinders that act as return paths for the current. a) Is there a magnetic field outside the outer cylinder? Inside the inner cylinder? Why? b) Draw a figure showing lines of B. c) Consider the upper part of the discharge. Suppose a positive ion, moving toward the right, has an upward component of velocity. How is its trajectory affected by the magnetic field after it has emerged from the discharge? d) What happens to a positive ion that tends to leave the upper part of the discharge by moving downward?
G
p
Figure 1010 Section through a device that has been utilized for producing a hightemperature plasma P. The copper enclosure, represented by heavy lines, is separated from the plasma by a pair of glass cylinders G. See Prob. 109.
252
Magnetic
Fields: III
e) Electrons move toward the left. What happens to them when they leave the discharge, by moving either upward or downward? 10lOE
IONBEAM DIVERGENCE Let us calculate the forces acting on a particle in a beam of positively charged particles having a velocity v and carrying charges Q. We assume that there are no externally applied electric or magnetic fields. We shall consider positive particles, but it will be a simple matter to apply our results to negative ones. First, there is an outward E, as in Fig. 1011. This is simply a case of electrostatic repulsion. Also, the beam current produces an azimuthal B. and v x B points inward, as in the figure. Thus, there is an outward electric force QE and an inward magnetic force Q\ x B. Does the beam converge or diverge? Experimentally, electron and ion beams always diverge, when left to themselves, but the divergence is slight when the particle velocity r approaches the velocity of light c. We consider a particle situated at the edge of a beam of radius R. The current is / and the particle velocity is v. Calculate a) the charge /. per meter of beam. b  the outward electric force QE. c) the inward magnetic force QvB. d) the net force. You should find that the net force points outward and is proportional to 1 — e / ( r , or to 1  (v/c) (see Eq. 2024). Then the net force tends to zero as r » c. If the particles are negative. E is inward instead of outward, but QE is again outward. Also. B points in the opposite direction and Q\ x B points again inward. In practice, a vacuum is never perfect. Let us say the ions are positive. If their energy is of the order of tens of electronvolts or more, they ionize the residual gas. forming lowenergy positive ions and lowenergy electrons. These positive ions drift 2
0
2
0
Figure 1011 Portion of a positiveion beam and its electric and magnetic fields. See Prob. 1010.
Problems
253
away from the positive beam. The lowenergy electrons, however, remain trapped in the beam and neutralize part of its space charge, thereby reducing E. The magnetic force then tends to pinch the beam. This phenomenon is called the pinch effect. It is also called gas focusing. With gas focusing, some of particles in the beam are scattered away, in colliding with the gas molecules. 1011
ION
THRUSTER Figure 1012 shows an ion thruster that utilizes the magnetic force. Other types of thruster are described in Probs. 214, 215. and 53. An arc A ionizes the gas, which enters from the left, and the ions are blown into the crossed electric and magnetic fields. If the current flowing between the electrodes C and D is /. then the thrust F is Bis. which is the force exerted on the gas ions. We assume that B and s are uniform throughout the thrust chamber, and we neglect the fringing fields. If HI' is the mass of gas flowing per unit time, and if v is the exhaust velocity with respect to the vehicle, then F is m'v, and the kinetic power communicated to the gas is 1
P
,
=  Hi'r
2
2
1
1
= Fv
= 
2
2
Blsr.
If the efficiency is defined as
PG +
where P
D
P
D
is the power dissipated as heat between C and D. show that 1 ~
1
1 2HI~ ~
+
r~
aBx
Figure 1012 Ion thruster. See Prob. 1011.
1 +
1 2~T ~
(TBV
IE'
1+ — Br
254
1012E
1013E
Magnetic
Fields: III
GAMMA For what value of v is the value of y one percent larger than unity? REFERENCE FRAMES Consider two reference frames, 1 and 2, as in Fig. 102, where frame 2 has a velocity v = c/2 with respect to 1. An event is found to occur at .v, = \\ = r , = 1 meter. t = 1 second. Find the coordinates .v , y .  . h1
2
1014
2
2
REFERENCE FRAMES See the preceding problem. If the event occurs at x = y Find v,, Vi, z t. 2
u
1015
= r , = 1 meter, t
2
2
= 1 second.
x
HALL
EFFECT Let us investigate more closely the Hall effect described in Sec. 10.8.1. We assume again that the current is carried by electrons of charge —e. Their effective mass is m*. The effective mass takes into account the periodic forces exerted on the electrons, as they travel through the crystal lattice. As a rule the effective mass is smaller than the mass of an isolated electron. The electrons are subjected to a force F =
e(E
+ v x B).
where E has two components, the applied field E and the Hall field E . The average drift velocity is x
y
F
v =
e
where . // is their mobility. The mobility of a particle in a medium is its average drift velocity, divided by the electric field applied to it. The law F = m*a applies only between collisions with the crystal lattice. a) Show that v = J/(E x
X
v, = .M{E
y
r
=
+
v B).

v B),
y
x
= 0.
b) If i! is the number of conduction electrons per cubic meter, the current density is J = —nev.
Problems
255
Show that J
E = ne.M —
//E.Ji A ,
x
x
T
r
E, + M E B J,. = nejf— i— 1 + .//'tiX
ll J
y
= 0, as in Sec. 10.8.1, .//E B, X
V =
b
a
J/VJ3.
Note that the Hall voltage V is proportional to the product of the applied voltage V and B. This fact makes the Hall effect useful for multiplying one variable by another. See Prob. 1712. When connected in this way, the Hall element has four terminals and is called a Hall generator, or a Hall probe. As a rule, rctype materials are used because of their relatively high mobility. However, the Hall effect exists in all conducting bodies. One important advantage of Hall elements is that they can be made small. For certain applications they are incorporated into integrated circuits. c) Calculate F for b = 1 millimeter, a = 5 millimeters. .// = 1 meters squared per volt second (indium antimonide), V = 1 volt, B = 1 0 tesla. d) Show that, i f £ , = 0, y
x
v
 4
x
AP — = Pn
, , .// B , 2
2
0, and AP is the increase in resistance upon application of the magnetic field. Fhe Hall field E is made equal to zero by making c small, say a few micrometers, and plating conducting strips parallel to t h e v axis, as in Fig. 1013. The element then has only two terminals and is called a magnetoresistor. Magnetoresistors are used for measuring magnetic inductions. See Probs. 153 and 155. y
Figure 1013 Magnetoresistor. See Prob. 1015.
256
Magnetic Fields: III
MAGNETOHYDRODYNAMIC
GENERATOR
Figure 1014 shows schematically the principle of operation of a magnetohyclrodynamic, or MHD generator. The function of an M H D generator is to transform the kinetic energy of a hot gas directly into electric energy. A very hot gas is injected on the left at a high velocity v. The gas is made conducting by injecting a salt such as K C 0 that ionizes readily at high temperature, forming positive ions and free electrons. Conductivities of the order of 100 S i e m e n s per meter are thus achieved, at temperatures of about 3000 kelvins. (The conductivity of copper is 5.8 x 10 Siemens per meter). Of course r « c . The ions and the electrons are deflected in the magnetic field. With the B shown, the positive ions are deflected downward and the electrons upward. The resulting current flows through a load resistance R. giving a voltage difference V. and hence an electric field E between the plates. One such M H D generator is planned to have a power output in the 500megawatt range. It would use a B of several teslas over a region 20 meters long and 3 meters in diameter. 2
7
3
2
2
Figure 1014 Schematic diagram of a magnetohydrodynamic generator. The kinetic energy of a very hot gas. injected on the left at a velocity v, is transformed directly into electric energy. The magnetic field B is that of a pair of coils, outside the chamber. See Prob. 98. The moving ions are deflected either up or down, depending on their sign. See Prob. 1016. a) Let us suppose that E, B. and the velocity v are uniform inside the chamber. These are rather crude assumptions. In particular, v is not uniform because (a) the ions and electrons have a vertical component of velocity, and (b) the vertical velocities give Q\ x B forces that point backward, for both types of particle; these braking forces slow down the horizontal drift of the charged particles. It is through this deceleration that part of the initial kinetic energy of the gas is transformed into electric energy.
Problems
257
The electrodes each have a surface area A and are separated by a distance b. Show that
I
=
V ' R + R,
where R, is the internal resistance of the generator, and V is the value of V when R is infinite. Thevenin's theorem (Sec. 5.12) therefore applies here. Note that, because v « c , the current density is the same in the reference frame of the laboratory as it is in the reference frame of the moving gas. 2
2
Solution: We have two reference frames, that of the laboratory and that of the moving gas. In the reference frame of the gas,
E = E +
v x
2
B.
(1)
We omit the subscript 1 for the quantities measured in the reference frame of the laboratory, for simplicity. Thus
J
2
= E = a(E + v x B), ff
j = j
(2)
2
2
= (vB a
 E) = a (vB 
I = JA = a A \vB  — b
b
b
aA
(3)
(4)
,
V
vBb rtAR
H,
R + R,
(5)
and V = vBb.
R = b:aA. t
(6)
Note that V is v x B/?. and that R is the resistance of a conductor of conductivity CT, length h. and crosssection A. i
b) Find an expression for the efficiency ^
power dissipated in R total power dissipated'
as a function of /. that does not involve the resistance R.
258
oAiB
I
I
(JAIB
Figure 1015 (a) Efficiency as a function of load current for the magnetohydrodynamic generator of Fig. 1014. (£>) Output voltage as a function of load current.
What is the efficiency at / = 0? What is the value of the current when the efficiency is zero? Sketch a curve of S as a function of /. Solution: IR
R
2
rR
+ rR,
R +
(8)
R,
R
R + (b/aA)
oAR + b
(9)
Since oAvBb
(10)
oAR + b" oAR aAvBb
_
RI
~ v~Bb'
V  Rjl
vBb  (b/aA)I
I
vBb
vBb
oAvB
(12)
The efficiency is equal to unity when / = 0. The efficiency is zero when I = aAvB. In that case R = 0. from Eq. 4. the output voltage V is zero, and / = V'/Ri, from Eq. 5. Figure 1015a shows & as a function of /. c) Find an expression for V as a function of / that does not involve R. What is the value of V when the current is zero? What is the value of the current when the voltage is zero? Sketch a curve of V as a function of /.
Problems
259
Solution:
V = IR=
vBb 
7  vBb ( 1
— ).
When the current is zero (R > oo), the output voltage is vBb. The ions then flow through the generator horizontally, undeflected. See Sec. 10.1.1. When the voltage is zero, R = 0 and / is equal to aAvB as above. Figure 1015b shows V as a function of I.
10I7E
ELECTROMAGNETIC FLOWMETERS Electromagnetic flowmeters operate as follows. A fluid, which must be at least slightly conducting (blood, for example), flows in a nonconducting tube between the poles of a magnet. The ions in the fluid are then subjected to magnetic forces Q\ x B in the direction perpendicular to the velocity v of the fluid and to the magnetic induction B. Electrodes placed on either side of the tube and in contact with the fluid thus acquire charges of opposite signs; the resulting voltage difference is a measure of v. Note that, in the electromagnetic flowmeter, ions of both signs have the same velocity v and the polarity of the electrodes is always the same for a given direction of flow and for a given direction of B. Compare with the Hall effect discussed in Sec. 10.8.1. Faraday attempted to measure the velocity of the Fhames river in this way in 1832. The magnetic field was of course that of the earth. Consider an idealized case where the tube has a rectangular crosssection aft, with side b parallel to B, and where the velocity is the same throughout the crosssection. What is the voltage between the electrodes? In fact, the velocity is maximum on the axis and zero at the inner surface of the tube. Moreover, the nonuniform v x B field causes currents to flow within the fluid. It turns out, curiously enough, that, for a tube of circular crosssection and radius a, the voltage between diametrically opposite electrodes is 2avB, where v is the average velocity, which is just what one would expect if the velocity were uniform.
CHAPTER
11
MAGNETIC
FIELDS:
The Faraday Induction
IV Law
111 11.1.1
THE INTEGRAL OF E • dl Example: The Expanding Loop
11.2 11.2.1
THE FA RADA Y INDUCTION LA W Example: Loop Rotating in a Magnetic Field
11.3
LENTS
11.4
THE FARADA Y INDUCTION DIFFERENTIAL FORM
11.5
THE ELECTRIC FIELD INTENSITY E IN TERMS OF V AND A Example: The Electromotance Induced in a Loop by a Pair of Long Parallel Wires Carrying a Variable Current
11.5.1 11.6
LAW
SUMMARY PROBLEMS
LA W IN
In this chapter we shall consider the line integral of E • dl around a closed circuit. This will lead us to still another of Maxwell's equations. In Chapter 2 we saw that, in electrostatic fields, the line integral of E • dl around a closed circuit C is zero. This is not a general rule. If the circuit C, or part of it, moves in a magnetic field, then one must take into account the v x B term of Chapter 10. This will give us the Faraday induction law, according to which the above integral is equal to minus the time derivative of the enclosed magnetic flux. What if, in a given reference frame, we have a changing magnetic field? Then electric field intensity is equal, not to — W as in Chapter 2, but rather to — V I  cA/dt, where A is the vector potential of Sec. 8.4, and Faraday's induction law again applies. 7
THE INTEGRAL
OF E • dl
We have seen in Sec. 2.5 that an electrostatic field is conservative, or that (111) Thus the work performed by an electrostatic field is zero when a charge moves around a closed path. However, there are many cases where one has a closed, conducting circuit, without sources, with part, or all of the circuit moving in a magnetic field. The conduction electrons in the moving conductors then feel an electric
262
Magnetic Fields: IV
field v x B, as in Sec. 10.1. In such cases, the above integral, evaluated over the circuit, is not zero. The v x B field in a conductor moving in a magnetic field is called the induced electric field intensity and the integral of E • dl a r o u n d a closed circuit is called the induced
EXAMPLE:
electromotance.
THE EXPANDING
LOOP
The expanding loop of Fig. 111 has one side that can slide to the right at a velocity v in a region of uniform B. Both v and B are measured in the laboratory reference frame S . In the moving wire, there is an electric field E , = v x B. The line integral of E • dl around the circuit is thus evaluated partly in S and partly in S  In the counterclockwise direction. x
t
2
(112) = 0 + f (v x B) • dl, Jda
(113)
= vwB.
(114)
This is the induced electromotance. It acts in the counterclockwise direction in the figure. If, at a certain instant, the loop has a resistance R , then the current is LWB/R.
We have assumed that R is large so as to make the magnetic field of the current flowing around the loop negligible compared to B .
Figure 111 A conducting wire ad slides at a velocity c on a pair of conducting rails in a region of uniform magnetic induction B. The magnetic force on the electrons in the wire produces a current / in the circuit.
11.2 The Faraday Induction Law
11.2
THE FARAD A Y INDUCTION
263
LA W
The righthand side of Eq. 114 is the area swept by the wire per unit time, multiplied by B, or d8>/dt, where O is the magnetic flux linking the circuit. N o w it is the custom to choose the positive direction for
(H5)
This is the Faraday induction law. the electromotance induced in a circuit is equal to minus the rate of change of the magnetic flux linking the circuit. The path of integration may be chosen at will and need not lie in conducting material. If there are no sources in a circuit, the current is equal to the induced electromotance divided by the resistance of the circuit, exactly as if the electromotance were replaced by a battery of the same voltage and polarity. The induced electromotance is expressed in volts, and it adds algebraically to the voltages of the other sources that may be present in the circuit. If one has a rigid loop and a timedependent B. Eq. 115 still applies. If one has a rigid coil and a timedependent B, then one must add the fluxes through each turn. The sum of these fluxes is called the flux linkage. We shall use the symbol A for flux linkage, keeping O for the surface integral of B • da over a simple loop. For example, if a coil has N turns with all the turns close together as in Fig. 112, A is N times the
(116)
Thus, for circuits other than simple loops, Faraday's induction law becomes
(117)
Flux linkage is expressed in weberturns.
264
Figure 112 Coil with several turns, all close together.
Equation
117 should be used with care. It is not as general as it is
commonly said to be. It is correct (a) if the circuit is rigid and the magnetic induction B is timedependent, or (b) if all or part of the circuit moves in such a way as to produce
an induced
electric
field
intensity
v x B. See
P r o b . 114. As we shall see in Sec. 11.5. the E in the first case is — cA/dt.
EXAMPLE:
LOOP ROTATING
IN A MAGNETIC
FIELD
Figure 113 shows a loop rotating in a constant and uniform magnetic field at an angular velocity to. We shall calculate the induced electromotance using the v x B field in the wire, and then verify the Faraday induction law. This will illustrate the principle of operation of an electric generator. The induced electromotance is 1118) where the integral is evaluated around the loop. Along the top and bottom sides, v x B is perpendicular to dl and the term under the integral is zero. Setting 0 = o>t. and remembering that v = w(a!2). the electromotance induced in the vertical sides is /
= 2aj(a/2)(sin (ot)bB — ouibB sin cot.
(119)
This electromotance is counterclockwise in the figure. It is positive because it is in the positive direction around the loop, with respect to the direction of B, according to the righthand screw rule. Now
(1110)
265
Kigure 113 Loop rotating in a constant and uniform magnetic field B. The vector ri! is normal to the loop.
and
" ~dt'
(1111)
as predicted by the Faraday induction law. The electromotance is zero when the plane of the loop is perpendicular to B. since v along the vertical sides is then parallel to B, and v x B is zero all around the loop. Note that the voltage 1 of Eq. 119 is a sinusoidal function of (. Such voltages are said to be alternating. Chapters 16 to 18 are devoted to alternating voltages and currents.
11.3
LENZ'S
LA W
If t h e flux l i n k a g e A i n c r e a s e s , dA/dt
is p o s i t i v e , a n d t h e e l e c t r o m o t a n c e is
n e g a t i v e , t h a t is, t h e i n d u c e d e l e c t r o m o t a n c e is in t h e n e g a t i v e d i r e c t i o n . O n t h e o t h e r h a n d , if A d e c r e a s e s . dA/dt is n e g a t i v e , a n d t h e e l e c t r o m o t a n c e is in t h e p o s i t i v e d i r e c t i o n .
266
Magnetic Fields: IV
The direction of the induced current is always such that it produces a magnetic field that opposes, to a greater or lesser extent, the change in flux, depending on the resistance in the circuit. Thus, if A increases, the induced current produces an opposing flux. If A decreases, the induced current produces an aiding flux. This is Lenz's law. The currents induced by changing magnetic fields in conductors other than wires are called eddy currents.
11.4
THE FARAD A Y INDUCTION IN DIFFERENTIAL FORM
LA W
The Faraday law stated in Eqs. 115 and 117 gives the electromotance induced in a complete circuit when the flux linkage is a function of the time. It applies to both fixed and deformable paths in both constant and timedependent magnetic fields, with the restriction stated at the end of Sec. 11.2. We shall now find an important equation that concerns the value of the E induced at a given point by a timedependent B. If, in a given reference frame, we have a timedependent B, then we can state the Faraday induction law in differential form as follows. Using Stokes's theorem, Eq. 115 becomes J
s
( V x E , . d a =  ^ =  i j
s
. B . d a ,
,1112,
where S is any surface bounded by the integration path. If the path is fixed in space, we may interchange the order of differentiation and integration on the righthand side and J >
x E)da=
 J
s
^ d a .
(1113)
We have used the partial derivative of B because we now require the rate of change of B with time at a fixed point. Since the above equation is valid for arbitrary surfaces, the integrands must be equal at every point, and
11.5 The Electric Field Intensity E in Terms of V and A
267
This is a general law for stationary media. We have here another of the four Maxwell equations. We have already found two others, namely Eqs. 612 and 812:
V•E
Pf + Pb
V • B = 0.
(1115)
Equation 1114 is a differential equation that relates the space derivatives of E at a particular point to the time rate of change of B at the same point. Both E and B are measured in the same reference frame. The equation does not give the value of E, unless it can be integrated.
11.5
THE ELECTRIC FIELD INTENSITY IN TERMS OF V AND A
E
Equation 1114 will give us an important expression for E in terms of the potentials V and A. Since B = V x A,
(1116)
V x E = ——(V x A) = —V x ~ , dt ct
(1117)
or E + — ]
The term between parentheses must be equal to a quantity whose curl is zero, namely a gradient. Then we can set
E = \V
3A  —.
(1119)
For steady currents, A is a constant and this equation reduces to Eq. 213. The quantity V is therefore the electric potential of Sec. 2.5, which is also known as the scalar potential.
268
Magnetic Fields: IV
Equation 1119 is a general expression for E. It states that an electric field intensity can arise both from accumulations of charge, through the — V I term, and from changing magnetic fields, through the —dA/dt 7
term.
All three quantities, E, A, and V, are measured in the same reference frame. Thus, if we have a field E , , Bj in reference frame 1, the electric field in frame 2, moving at a constant velocity v with respect to 1, is given by Eqs. 1016 and 1017, or by Eq. 1020. However, in any given reference frame, E is given by Eq. 1119 and B is equal to the curl of A as in Eq. 813.
11.5.1
EXAMPLE: THE ELECTROMOTANCE INDUCED IN A LOOP BY A PAIR OF LONG PARALLEL WIRES CARRYING A VARIABLE CURRENT A pair of parallel wires, as in Fig. 114, carries equal currents / in opposite directions, and / increases at the rate dl/dt. We shall first calculate the induced electromotance from Eq. 115 and then from Eq. 1119. a) From Sec. 9.1.1, the current / in wire a produces a magnetic induction B„ = n I/2np , 0
a
(1120)
and a similar relation exists for wire b. The flux through the loop in the direction shown in Fig. 114 is thus
(1121)
r {r + w) a
From Eq. 115, the electromotance induced in the clockwise direction is
( f
J
E
(1122)
b
. d l =  ^ l n
r
(1123)
2n dt
Thus / ' flows in the counterclockwise direction. This is in agreement with Lenz's law: the current / ' produces a magnetic field that opposes the increase in d>. b) Let us now use Eq. 1119 to calculate this same electromotance from the time derivative of the vector potential A, V being equal to zero in this case. From Sec.
269
A
h
I
8A
A
I
P,
dt
Pb
Figure 114 Pair of parallel wires carrying equal currents / in opposite directions in the plane of a closed rectangular loop of wire. When / increases, the induced electromotance gives rise to a current T in the direction shown. The vector potential A and the induced electric field —dA/dt are shown on the horizontal sides of the loop. The induced current / ' flows in the counterclockwise direction because — dA/dt is larger on the lower wire than on the upper wire.
8.4.1. A is parallel to the wires and, if we choose the leftward direction as positive. (I 124)
(1125)
along the lower and upper sides of the loop, respectively. Thus
E
(1126)
(1127)
and, in the clockwise direction. (1128)
as previously.
270
Magnetic Fields: IV
We have disregarded the field —dA/dt along the lefthand and righthand sides, because it is perpendicular to the wires. To find the electromotance induced in the loop by a changing current in a single conductor, we set r » oo, and then h
(j)Edl = ^ ~ l n ( — — V J 2n dt \r + w)
(1129)
a
11.6
SUMMARY The Faraday induction law can be stated as follows: For a circuit C linked by magnetic flux O,
Edl
<7
The integral on the left is called the induced electromotance. The positive directions chosen for
V x E =
Hi"
(1114)
This is another of Maxwell's equations. Both E and B are measured in the same reference frame. The electric field intensity E can be due to accumulations of charge, or to changing magnetic fields, or both. In general,
E =
—VF
7 A "fit"'
where E, F, and A are all measured in the same reference frame.
(1119)
Problems
271
PROBLEMS U1E
BOAT
TESTING TANK A carriage runs on rails on either side of a long tank of water equipped for testing boat models. The rails are 3.0 meters apart and the carriage has a maximum speed of 20 meters per second. a) Calculate the maximum voltage between the rails if the vertical component of the earth's magnetic field is 2.0 x 1 0 " tesla. b) What would be the voltage if the tank were situated at the magnetic equator? 5
112
113E
EXPANDING LOOP A conducting bar slides at a constant velocity v along conducting rails separated by a distance s in a region of uniform magnetic induction B perpendicular to the plane of the rails. A resistance R is connected between the rails. The resistance of the rest of the circuit is negligible. a) Calculate the current / flowing in the circuit. b) How much power is required to move the bar? c) How does this power compare with the power loss in the resistance R! INDUCED CURRENTS A bar magnet is pulled through a conducting ring at a constant velocity as in Fig. 115. Sketch curves of (a) the magnetic flux
Figure 115 See Prob. 113.
114
INDUCED CURRENTS Figure 1 l6a shows a conducting disk rotating in front of a bar magnet, a) In what direction does the current flow in the wire, clockwise or counterclockwise? Why?
272
Figure 116 (a) Conducting disk rotating near the North pole of a bar magnet. The wire is connected to the disk through sliding contacts. (£>) Solenoid wound on a bar magnet. One end of the wire is connected to a sliding contact that can move along the length of the solenoid. b) What happens if both the polarity of the magnet and the direction of rotation are reversed? c) Show that there is no current generated with the setup of Fig. 11 6b, despite the fact that the magnetic flux linking the circuit is not constant. USE
INDUCED
ELECTROMOTANCE
A loop of wire is situated in a timedependent magnetic field with B = 1.00 x 1 0 " cos (27t x 60)f 2
perpendicular to the plane of the loop. Calculate the induced electromotance in a 100turn square loop 100 millimeters on the side. 116
ELECTROMAGNETIC PROSPECTION In electromagnetic prospection, a coil carrying an alternating current causes induced currents to flow in conducting ore bodies, and the alternating magnetic field of these induced currents is detected by means of a second coil placed some distance away from the first one, as in Fig. i l  7 . Let us consider a simple example of induced currents. A thin conducting disk of thickness /?, radius R, and conductivity a is placed in a uniform alternating magnetic field B = B cos cot 0
parallel to the axis of the disk as in Fig. 118. a) Find the induced current density J as a function of the radius. Assume that the conductivity a is small enough to render the magnetic field of the induced current negligible, compared to the applied B. Choose the positive directions for B and for J in the directions shown in the figure. b) Sketch curves of B and J as functions of the time. Explain why B and J are so related.
273
Figure 117 Electromagnetic prospection. Coil A, connected to a source of alternating current, produces an alternating magnetic field. Coil B is held perpendicular to A and is sensitive only to the magnetic field due to the currents induced by A in the ore body C. The figure is not drawn to scale; the coil diameters are of the order of one meter or less.
Figure 118 See Prob. 116.
117E
INDUCTION HEATING It is common practice to heat conductors by subjecting them to an alternating magnetic field. The changing flux induces eddy currents that heat the conductor by the Joule effect (Sec. 5.2.1). The method is called induction heating. It is used extensively for melting metals and for hardening and forging steel. The powers used range from watts to megawatts, and the frequencies from 60 hertz to several hundred kilohertz.
Induction heating has the advantage of convenience and of not contaminating the metal with combustion gases. It even permits heating a conductor enclosed in a vacuum enclosure. Induction heating has another major advantage. At high frequencies, currents flow near the surface of a conductor. This is the skin effect. See Probs. 1617 and 1814. So, by choosing the frequency correctly, one can apply a brief heat treatment down to a known depth. This is particularly important because there are numerous purposes for which one requires steel parts with a hard skin and a soft core: the hard skin resists abrasion and the soft core reduces breakage. Plowshares are heattreated in this way. Induction furnaces are used for melting metals. They consist of large crucibles with capacities ranging up to 30 tons, thermally insulated and surrounded by currentcarrying coils. Operation is usually started with part of the load already molten. Let us consider Fig. 1 l9a where a rod of radius a. length L. and conductivity a is placed inside a solenoid having N' turns per meter and carrying an alternating current / = I cos 0)t. 0
As usual, we neglect end effects. We also assume that the frequency is low enough to avoid the complications due to the skin effect. This assumption is well satisfied at 60 hertz with a rod of graphite
Problems
275
(a = 1.0 x 10 S i e m e n s per meter) having a radius of 60 millimeters. In other words, the magnetic induction of the induced currents is negligible. We also assume that the conductor is nonmagnetic. . This is admittedly a highly simplified illustration of induction heating. Moreover, the power dissipated in the graphite is only a few watts, which is absurdly small for such a large piece. This should nonetheless be a useful exercise on induced currents. Consider a ring of radius /•. thickness dr. and length L inside the conductor, as in Fig. 11 9b. a) Show that the electromotance induced in the ring is ?
u. nr coNT 2
0
0
sin cot.
b) Show that, for a current flowing in the azimuthal direction, the ring has a resistance R =
InrioLdr.
c) Show that the average power dissipated in the ring is ^ (/( o;N7 ) 7T(tL/2
0
3
0
dr.
d) Show that the total average power dissipated in the cylinder is 1 , — (H o)NT )naLa 16 0
0
, .
e) Calculate the power dissipated in the graphite rod, setting L = 1 meter, 7 = 20 amperes, N' = 5000 turns per meter. 0
USD
INDUCED ELECTROMOTANCE A magnetic field is described by
. 2ny
B = £ 0 sin —^ (sin wt)i. A
In this field a square loop of side A/4 lies in the i .plane with its sides parallel to the y and raxes. The loop moves at a constant velocity vj. Calculate the electromotance induced i n the loop as a function of the time i f the trailing edge of the loop is at y = 0 at t = 0. Set co = 2 m l A.
BETATRON Figure 1110 shows the principle of operation of the betatron. The betatron produces electrons having energies of the order of millions of electronvolts. Electrons are held in a circular orbit in a vacuum chamber by a magnetic field B. The electrons
276
/J
i i m i L L J J i n j ^ ^ 11! 1111111! 1111LLL
Figure 1110 Crosssection through the central region of a betatron. The electrons describe a circular orbit in a plane perpendicular to the paper at 0—0, in the field of an electromagnet and inside an evacuated ceramic torus T. Only the pole pieces P of the electromagnet are shown. The core is laminated. The pole pieces are shaped so that the average B inside the orbit is equal to twice that at the orbit. The electromagnet is operated on alternating current, and the electrons are accelerated only during that part of the cycle when both B and dB/dt have the proper signs. See Prob. 119 and Fig. 1111.
Figure 1111 Circular electron orbit in the betatron. It is assumed that the magnetic flux is in the direction shown and that it increases. Here. e =  1.60 x 1 0 " coulomb. 1 9
Problems
277
are accelerated by increasing the magnetic flux linking the orbit. Betatrons serve as sources of xrays. There are very few that are still in use today. Show that the average magnetic induction over the plane of the orbit must be twice the magnetic induction at the orbit, if the orbit radius is to remain fixed when both B and the electron energy increase. Solution: The various vectors we are concerned with are shown in Fig. 1111. In the reference frame of the laboratory, an electron is subjected to the Lorentz force e(E + v x B).
Fhe electric field E is due to the timedependent magnetic field and is — dA/dt. We do not know the value of A. but we do know, from Faraday's induction law, that 1
(1)
We assume that the magnetic flux
e d
(mv) = eE=
(2)
while, in the radial direction, ,m
2
r = Bev.
(3)
In these two equations, all the quantities are positive except e. Equating the two values of mv obtained from Eqs. 2 and 3, mv =
e 2nr
(4) (5)
2
where B is the magnetic induction at the radius r of the electron orbit. Now, by definition,
(6)
where B is the average magnetic induction within the circle of radius r. Then, comparing Eqs. 5 and 6, B is equal to 2B.
278
1110
Magnetic Fields: IV
THE TOLMAN AND BARNETT EFFECTS If a conductor is given an acceleration a, the conduction electrons of mass m and charge — e are subjected to inertia forces —ma. Show that, if E' is the equivalent total electric field intensity and if B is the magnetic induction in the conductor,
This effect was predicted by Maxwell, and was observed for the first time in 1916 by Tolman and collaborators. The inverse effect, namely the acceleration of a body carrying a variable current, was also predicted by Maxwell, and was first observed by Barnett and others in 1930. 1111E
ELECTRIC CONDUITS The U.S. National Electrical Code rules that both conductors of a circuit operating on alternating current, if enclosed in a metallic conduit, must be run in the same conduit. Let us suppose that we have a single conductor carrying an alternating current and enclosed within a conducting tube. Show that both A and dA/dt are longitudinal in the tube. So, with a single wire, a longitudinal current is induced in the tube. This causes a needless power loss, and may even cause sparking at faulty joints. With two conductors carrying equal and opposite currents, both A and cXIct are essentially zero in the conduit.
1112E
THE POTENTIALS V AND A We have seen that E =
 VI  , \ \ ft.
B = V x A.
Show that neither E nor B arc affected if the potentials V and A are replaced by V + (G dt
and
A  VG.
where G is any function of v, r. z, / whose second derivatives exist and are continuous.
CHAPTER MAGNETIC
12 FIELDS:
V
Mutual Inductance M and SelfInductance
L
12.1 12.1.1
MUTUAL INDUCTANCE M Example: Mutual Inductance Between Two Coaxial
12.2 12.2.1
INDUCED ELECTROMOTANCE IN TERMS OF MUTUAL INDUCTANCE Example: The Vector Potential of the Inner Solenoid
12.3 12.3.1 12.3.2
SELFINDUCTANCE L Example: Long Solenoid Example: Toroidal Coil
12.4
COEFFICIENT
12.5 12.5.1 12.5.2 12.5.3
INDUCTORS CONNECTED Zero Mutual Inductance NonZero Mutual Inductance Example: Coaxial Solenoids
IN
12.6 12.6.1 12.6.2 12.6.3
INDUCTORS CONNECTED Zero Mutual Inductance NonZero Mutual Inductance Example: Coaxial Solenoids
IN
12.7 12.7.1 12.7.2
TRANSIENTS IN RL CIRCUITS Example: Inductor L Connected to a Voltage Source V Example: Horizontal Beam Deflection in the CathodeRay Tube of a Television Receiver
12.8
SUMMARY
OF COUPLING
Solenoids
k SERIES
PARALLEL
0
PROBLEMS
In this chapter we shall see how one utilizes the Faraday induction law to calculate induced voltages and currents in electric circuits. In Chapter 5 we ascribed to each branch of a circuit a certain resistance R. If the current flowing in a branch produces an appreciable magnetic field, then the branch also possesses a selfinductance L. If the magnetic field of one branch produces a flux linkage in another branch, then there exists a mutual inductance M between the two branches. Thus, the magnetic properties of the branches are characterized by the two quantities L and M. We shall return to mutual inductance later on, in Chapter 18, which deals with power transfer and transformers.
12.1
MUTUAL
M
INDUCTANCE
To calculate the electromotance induced in one circuit when the current changes in another circuit, it is convenient to express the flux linkage in the first one in terms of the current in the second and of a geometrical factor involving both circuits. Consider the two circuits of Fig. 121. The current /„ in circuit a produces in b a flux linkage A that is proportional to /„: ah
A
fli
= M I. ah
a
(121)
Similarly, if there is a current I through circuit b, the magnetic flux of b linking a is b
Ka
=
M I„. ha
(122)
281
Figure 121 Two circuits a and b. The flux
Since both circuits can be of any shape, one naturally does not expect to find a general relationship between M and M . O n the contrary, it can be shown that M = M f The factor of proportionality M = M = M is called the mutual inductance between the two circuits. Mutual inductance depends solely on the geometry of the two closed circuits and on the position and orientation of one with respect to the other. When multiplied by the current in one circuit, M gives the flux linkage in the other. A similar situation exists in relation with the capacitance between two conductors. The capacitance C depends solely on the geometry of the two conductors and on the position and orientation of one with respect to the other. If one of the conductors is grounded, then the charge induced on it is equal to CV, where V is the voltage applied to the other. Since the mutual inductance is the flux linkage in one circuit per unit of current in the other, inductance is measured in weberturns per ampere, or in henrys. The mutual inductance between two circuits is one henry when a current of one ampere in one of the circuits produces a flux linkage of one weberturn in the other. ab
ab
* Electromagnetic
ha
Fields ami Waves, p. 343.
ba
ab
ba
282
Magnetic Fields: V
The sign of M is chosen as follows. The mutual inductance between two circuits a and b is positive if a current in the positive direction in a produces in b a flux that is in the same direction as one due to a positive current in b. This is illustrated in Fig. 121. The positive directions for the currents are chosen arbitrarily. A pair of coils designed so as to possess a mutual inductance is called a mutual inductor. Mutual inductors are also called transformers.
1.1
EXAMPLE: COAXIAL
MUTUAL INDUCTANCE SOLENOIDS
BETWEEN
TWO
Let us calculate the mutual inductance between two coaxial solenoids as in Fig. 122. We assume that both windings are long, compared to their common diameter 2R. that they have the same number of turns per meter N', and that they are wound in the same direction, as in the figure. We set /„ > l . Let us assume a current /„ in coil a. Then the flux of coil a linking each turn of coil b is b
(123)
2
ah
0
a
and the mutual inductance is M = \JI
B
= N QJI b
= i nR N'N .
(124)
2
a
t 0
h
= ti nR N N /l . 2
0
a
h
a
(125)
I. I,
Figure 122 Coaxial solenoids. The two radii are taken to be approximately equal.
12.2 Induced Electromotance
in Terms of Mutual Inductance
283
We can also calculate the mutual inductance by assuming a current I in coil b. Then b
u nR NT
(126)
2
ba
0
b
This flux links only l N' = N turns of coil a. since B falls rapidly to zero beyond the end of a long solenoid, as we saw in Sec. 9.1.5. Then the mutual inductance is h
M = AJI„
b
= N„JI = n nR N'N 2
b
=
b
0
p nR N N /l , 2
0
a
b
a
(127)
as previously.
12.2
INDUCED ELECTROMOTANCE OF MUTUAL INDUCTANCE
IN
TERMS
From Sec. 11.2, the electromotance induced in circuit b by a change in /„ in Fig. 121 is Edl
=
d A , , h
dt
=
M—. dt
(128)
In this case, E is —dA/dt, where A is the value of the vector potential of the current I at a point on circuit b (Sec. 11.5). Similarly, the electromotance induced in circuit a by a change in I is a
b
Edl
dA
bl
dt
M
dh dt
(129)
These equations are convenient for computing the induced electromotance, since they involve only the mutual inductance and dl/dt, both of which can be measured. We now have a second definition of the henry: the mutual inductance between two circuits is one henry if a current changing at the rate of one ampere per second in one circuit induces an electromotance of one volt in the other.
12.2.1
EXAMPLE:
THE
VECTOR
POTENTIAL
OF THE INNER
SOLENOID
It is paradoxical that a varying current in the inner solenoid should induce an electromotance in the outer one, since we have shown (Sec. 9.1.3) that the magnetic
284
Magnetic Fields: V
induction outside a long solenoid is zero. T h e explanation is that the induced electric field intensity at any given point is equal to the negative time derivative of the vector potential at that point (Eq. 1119) and that the vector potential A does not vanish outside an infinite solenoid, despite the fact that B = V x A does. See also Sec. 9.1.2. We can actually calculate the vector potential just outside the inner solenoid from the mutual inductance. If we consider the direction of the current /„ in the inner solenoid as positive, then the electromotance induced in the outer one is  M i l J i t , or dl, at
ioJt* JVW»£ 2
and the induced electric field intensity — cA/ct is 2nRN
h
PA dt
times smaller:
HoRN' dl, = — . 2 dt
(1210)
Integrating, we have the vector potential at any point close to the solenoid:
A = ^ L
(1211)
in the azimuthal direction.
12.3
SELFINDUCTANCE
L
A single circuit carrying a current / is of course linked by its own flux, as in Fig. 123. The flux linkage A is proportional to the current: A = LI, where L is called the selfinductance,
or simply the inductance
(1212) of the circuit.
Selfinductance, like mutual inductance, depends solely on the geometry and is measured in henrys. It is always positive. A circuit that is designed so as to have a selfinductance is called an inductor. An inductor has a selfinductance of one henry if a current of one ampere produces a flux linkage of one weberturn.
285
Figure 123 Isolated circuit carrying a current / and its flux linkage A.
A change in the current flowing through a circuit produces within the circuit itself an induced electromotance
dt
dt
(1213)
The induced electromotance tends to oppose the change in current, according to Lenz's law, and adds to whatever other voltages are present. If a varying current flows through an inductance L , the voltage across it is L dljdt, as in Fig. 124. An inductor therefore has a selfinductance of one henry if the current flowing through it changes at the rate of one ampere per second when the voltage difference between its terminals is one volt.
V
I
L
O
Figure 124 Idealized inductor with zero resistance. The voltage V is L dljdt.
±r
12.3.1
EXAMPLE:
LONG
SOLENOID
It was shown in Sec. 9.1.3 that the magnetic induction inside a long solenoid, neglecting end effects, is uniform, and that B = HoN'I,
(1214)
where AT is the number of turns per meter. Thus
R\
n
(1215)
286
10
5
Rll Figure 125 Factor K for calculating the inductance of a short solenoid. where N is the total number of turns. / is the length of the solenoid, and R is its radius. Then , A N
0
1
(1216)
The selfinductance of a long solenoid is thus proportional to the square of the number of turns and to its crosssection, and inversely proportional to its length. The inductance of a short solenoid is smaller by a factor K, which is a function of R/l, as in Fig. 125.
EXAMPLE:
TOROIDAL
COIL
A toroidal coil of N turns is wound on a form of nonmagnetic material having a square crosssection, as in Fig. 126. According to the circuital law, or from Eq. 98. the magnetic induction in the azimuthal direction, at a radius p inside the toroid. is
B = poNl/litp.
(1217)
287
Figure 126 Toroidal coil of square crosssection and mean radius R.
Thus the flux linkage is rR W/2 w dp RR ++ WI:
pNI 2
0
jRw/2
2n pNw 2
L =
2R + w
w In
0
2R  w 2R + w
.
2R  iv
(121S
(1219)
(1220)
The selfinductance of a toroidal coil is proportional to the square of the number of turns, like that of a long solenoid.
12.4
COEFFICIENT
OF COUPLING
k
In Sec. 12.1.1 we found that the mutual inductance between the two solenoids of Fig. 122 is M = n 7iR
N N /l .
2
a
0
b
a
(1221)
Now, from Eq. 1216, L„ =
n nR N Jl ,
(1222)
L
n nR N /l ,
(1223)
h
=
2
2
0
2
0
2
a
b
288
Magnetic Fields: V
and
iKU) . m
= (j)
M
(1224)
The mutual inductance is thus proportional to the geometrical mean of the selfinductances. More generally, M = k(L L ) ,
(1225)
112
a
b
where k is the coefficient of coupling, which can be either positive or negative. Moreover, fe < 1.
(1226)
< 1, by hypothesis. For example, in the present case, k = (l /l ) When \k\ % 1, the two circuits are said to be tightly coupled; they are loosely coupled when \k\ « 1. in
h
12.5
12.5.1
INDUCTORS
ZERO
CONNECTED
MUTUAL
IN
a
SERIES
INDUCTANCE
Consider two inductors connected in series as in Fig. 127a or 127b. Their mutual inductance is approximately zero. Then
v=L
4 4 > > T' +L
={L
+L
)d
t

and the effective inductance is simply the sum of the inductances. The same rule applies to any number of inductors connected in series, as long as the mutual inductances are all zero.
289
Tt
lb)
(a)
Figure 127 (a) Two coils connected in series. They are assumed to be far from each other, and their mutual inductance is approximately zero, (b) Two coils connected in series, one perpendicular to the other. T h e flux linkage and the mutual inductance are again approximately zero.
12.5.2
NONZERO
MUTUAL
INDUCTANCE
If the coupling coefficient between the two inductors connected in series is not zero, as in Fig. 1 2  8 , then the voltage on coil 1 is Li times
dl/dt, plus
M times dl/dt .in coil 2 , a n d the voltage on coil 2 is given by a similar expression. T h u s
V = ( L , + M)~
=
(L, + L
2
+ (L
2
+ M ) ^ ,
+ 2M) — .
(1228)
(1229)
M
Figure 128 Two coils connected in series with a nonzero mutual inductance M.
290
Magnetic Fields: V
The effective inductance is thus L = Lj + L
+ 2M.
2
(1230)
Remember that M can be either positive or negative (Sec. 12.1).
EXAMPLE:
COAXIAL
SOLENOIDS
Figure 129a shows the coaxial solenoids of Fig. 122 connected in series, with the lefthand wires connected together. Let us calculate the inductance between a and b. Suppose the current flows from terminal A to terminal B. Then the flux of a points left, while that of b points right. So the flux of b partly cancels that of a, and L
a
+ L
b
(1231)
2M,
where L , L , M are given in Eqs. 1222 to 1224 and M is a positive quantity, here. If the solenoids are connected as in Fig. 129b, a
h
L
a
Figure 129 The coaxial solenoids of Fig. 122, connected in series, (a) with M negative, and (b) with M positive.
+ L„ + 2M.
(1232)
B
A
12.6 Inductors Connected in Parallel
12.6
INDUCTORS
CONNECTED
IN
291
PARALLEL •
12.6.1
ZERO
MUTUAL
INDUCTANCE
We have just seen that two inductors connected in series are treated like resistors in series (Sec. 5.4) when their mutual inductance is zero. One might guess that inductors in parallel are calculated like resistors in parallel (Sec. 5.5) when M is zero. As we shall see, this is correct. We now have two inductors connected in parallel as in Fig. 1210, with zero M. Then
The total current I is I + I . If L is the selfinductance that is equivalent to L and L in parallel, t
l
2
2
(1234)
(1235)
(1236)
(1237)
and
L =
as we guessed at the beginning.
LL X
L,
2
+
L
2
(1238)
292
I
t Tl
Figure 1210 Two coils connected in parallel with approximately zero mutual inductance.
12.6.2
NONZERO
MUTUAL
INDUCTANCE
Let us now consider Fig. 1211. which shows two coupled inductors, connected in parallel. If L is the equivalent inductance,
V
=
L
dl
(1239)
Jt'
dt
(1240)
dt
Also, for the lefthand coil,
dt
dt
dt
dt
(1241)
and, for the righthand coil,
(1242)
293
1
A
A/
O U
2
Figure 1211 Two coils connected in parallel with a nonzero mutual inductance M. Here, M is positive: a positive (downward) current in L, produces a downward B in L , while a positive (downward) current in L also produces a downward B in L , . 2
2
F r o m the last two equations,
(L {
M) ±
= ( L
d
2
 M )
d
^ .
(1243)
Then, from Eqs. 1240 and 1241,
L
2
L
=

MJ
dt
\
1
L

2
MJ
dt
and L,(L
— M) + M ( L . 
2
(L
L L X

2
M ) + (Lj 
2
L, + L
2
M 
M) M)
2
2M'
When M = 0, we revert to the previous case.
(1246)
294
12.6.3
Magnetic Fields: V
EXAMPLE:
COAXIAL
SOLENOIDS
Figure 1212 shows again the coaxial solenoids of Fig. 122, this time connected in parallel. With the connections shown in Fig. 1212a, M is positive and the inductance between terminals A and B is M
2
L =
L
2M
+ L„
a
(1247)
with the L's and the M on the right as in Eqs. 1222 to 1224. If the connections to one of the solenoids are interchanged as in Fig. 1212b, L =
LL a
L+ a
b
 M L+ b
2M'
(1248)
Here again, M is the positive quantity calculated as in Eq. 1224
Figure 1212 The pair of coaxial solenoids of Fig. 122, connected in parallel, (a) with M positive, (b) with M negative.
12.7
TRANSIENTS
IN RL
CIRCUITS
When a circuit comprising resistances and inductances is disturbed in some way, whether by changing the applied voltages or currents, or by modifying
12.7 Transients in RL Circuits
295
the circuit, the currents take some time to adjust themselves to their new steadystate values. We discussed a similar phenomenon in relation with RC circuits in Sec. 5.14. EXAMPLE:
INDUCTOR
L CONNECTED
TO A VOLTAGE
SOURCE
V
0
Figure 1213 shows an inductor L connected to a voltage source F . The inductor and its connecting wires are made of a single continuous wire of conductivity a and crosssection a. a) The current is constant. If the wire is in air (e « 1), its surface charge density is e £ „ , where E„ is the normal component of E just outside the wire. This follows from Gauss's law (Sec. 3.2) because £„ = 0 inside the wire, as we shall show below. Similarly, if the wire is submerged in a dielectric, the surface charge density is e e £ „ . At any given point on the surface of the wire, the field £„ is proportional to F . It depends on the configuration of the circuit, as well as on the position and configuration of the neighboring bodies. Inside the wire, the volume charge density is zero, as we saw in Sec. 5.1.2. Also, 0
r
0
r
0
0
E =
 V F = J/ff.
(1249)
Thus £ and VF, like J, are the same everywhere inside the wire. The lines of force inside the wire are parallel and, if the length of the wire is /, /VK = /  = I— = IR = V . a ao 0
T h u s VF inside t h e w i r e is simply
1,, /.
V
+ + +++
Figure 1213 Inductor £ connected to a voltage source.
(1250)
296
t (seconds)
Figure 1214 Current through the inductor of Fig. 1213 as a function of the time. It is assumed that / = 0 at r = 0 and that RjL = 1.
b) The current builds up from zero to its steady state value V /R. We assume that the time the field takes to propagate from the source to the inductor is negligible. We also neglect the stray capacitance between the turns of the inductor and between the wire and ground. Since the wire has both a resistance R and an inductance L, Kirchoff's voltage law (Sec. 5.7) gives us that 0
dl L  + RI= dt
(1251)
V. 0
We saw in Sec 12.3 that the voltage across a pure inductance is L dljdt. From Sec. 5.13, the solution of this equation is
R
Ke
(1252)
where K is a constant of integration. By hypothesis, / = 0 at / = 0, and thus K  V jR. Then 0
(1253)
The current 1 increases with time as in Fig. 1214. N o t e the analogy with Fig. 523. The current density J is again the same throughout the wire since, by hypothesis, there is zero capacitance, and hence no accumulation of charge at the surface of the wire. Thus
J
I
E =  = — a aa
R
I.
(1254)
12.7 Transients in RL Circuits
297
and £ is uniform throughout the wire. It builds up gradually, like /. Also. RI = El.
(1255)
From Eq. 1119,
RI = j)^\V
 ^Y«U,
(1256)
VFdl ^(j)Adl,
(1257)
where the integration is evaluated along the wire, in the direction of E, clockwise in Fig. 1213. Now, from Eq. 854, the line integral of A • dl around a simple circuit is equal to the enclosed flux. More generally, the integral is equal to the flux linkage:
Adl = A
(1258)
and dA
r
RI =  c i v K  d l
,
(1259)
dt
J
r
= d)VFdl
J
L dl
at
.
(1260)
from Eq. 1212. Comparing with Eq. 1251, we find that )VFdl.
(1261)
This integral is therefore a constant. The negative sign comes from the fact that the integral is the voltage of the lefthand terminal with respect to the righthand one, or — F . Inside the wires that lead from the source to the coil, the magnetic field is weak and VF is approximately equal to the £ of Eq. 1254, with / a function of the time as in Eq. 1253. Inside the coiled part of the wire, A is not negligible. The — V F field points in the same direction as the current, while —dA/dt points in the opposite direction. At t = 0, when the source is connected, A = 0 but I^A/f'fj is large, / = 0, and the two terms on the right in Eqs. 1256, 1257, 1259, 1260 cancel. At the beginning, most of the voltage d r o p appears across the coil while, under steadystate conditions, the potential d r o p is uniformly distributed along the wire. 0
298
Magnetic Fields: V
EXAMPLE: HORIZONTAL BEAM DEFLECTION IN THE CATHODERAY TUBE OF A TELEVISION RECEIVER The electron beam sweeps horizontal lines across the screen, one below the other, until the whole screen has been covered. The process then repeats itself. The image is obtained by modulating the beam intensity. The beam is deflected by the magnetic fields of two pairs of coils, one for the horizontal motion, and one for the vertical. The beam sweeps horizontally at a constant velocity, and returns at a much higher velocity. T o sweep horizontally at a constant velocity, the current through the pair of horizontal deflection coils must increase linearly with time. This is achieved by applying a constant voltage V to the coils, for then 1
dl L — dt
12.8
V I « — t. L
(1262)
SUMMARY The mutual inductance M between two circuits is equal to the magnetic flux linking one circuit per unit current flowing in the other: 9
Kb = M I .
(121)
ah a
The mutual inductance depends solely on the geometry and on the relative positions and orientations of the two circuits. It has the same value, whichever circuit produces the magnetic flux linking the other. Mutual inductance is measured in henrys. Mutual inductance can be either positive or negative. The mutual inductance between two circuits a and b is positive if a current in the positive direction in a produces in b a flux that is in the same direction as that produced by a positive current in b. The electromotance induced in circuit b by a change in current in circuit a is
Edl
The selfinductance
=  M ^ .
dt
(128)
L of a circuit is its flux linkage per unit current.
Problems
299
Thus
Edl =  L ~ . dt
(1213)
The voltage across an inductance L carrying a current I is L dl/dt. The coefficient of coupling k between two circuits is defined by M = k(L L„) ' , 1 2
a
(1225)
and can have values ranging from —1 to + 1 . It is zero if none of the flux of one circuit links the other. The effective inductance of two inductors connected in series is L = L
l
+ L
2
+ 2M,
(1230)
where M is the mutual inductance, which can be either positive or negative. The effective inductance of two inductors connected in parallel is
L = J ^ U L ^ . L + L  2M l
{l2
46)
2
When a voltage is applied t o an inductive circuit, the currents take some time to adjust to their steadystate values.
PROBLEMS 121E
Starting from Eq. 125, show that p is expressed in henrys per meter.
122E
MUTUAL INDUCTANCE A long straight wire is situated on the axis of a toroidal coil of N turns as in Fig. 1215. Show that
0
u bN* ( b\ M = — In 1 +  . 2n \ a 0
300
Figure 1215 See Prob. 122,
123E
MUTUAL INDUCTANCE a) Show that the mutual inductance between two coaxial coils, separated by a distance z as in Fig. 1216, one of radius a and N turns, and the other of radius b « a and N turns is a
h
nu N N a b 2
0
2(a
a
2
2
b
+ z ) ' ' 2
3
2
We shall use this result in Prob. 147. You can calculate this in one of two ways. You can either assume a current in coil a and calculate the flux through coil b, or assume a current in b and calculate the flux through a. From Sec. 12.1, M is the same, either way. Now, if the current is in a, you know how to calculate the field at b, from Sec. 8.1.2, since b « a. If the current is in b, then the calculation is vastly more difficult because you have to calculate B away from the axis. So you should calculate the magnetic flux through coil b due to a current through coil a. b) How does the mutual inductance vary when the small coil is rotated around the vertical axis?
Figure 1216 See Prob. 123.
N turns a
Problems
301
c) What if one rotated the large coil around a vertical axis? Would the mutual inductance vary in the same way? •
I24E
A OUTSIDE A SOLENOID Show that the vector potential at a distance r from the axis outside an infinite solenoid of radius R and N' turns per meter carrying a current / is A =
n N'R II2r. 2
0
The vector potential is azimuthal. I25E
A INSIDE A SOLENOID Show that the vector potential at a distance r from the axis, inside an infinite solenoid of radius R and N' turns per meter carrying a current / is A =
n Nir/2. 0
Here also, the vector potential is azimuthal. 126
MAGNETIC MONOPOLES Magnetic monopoles are the magnetic equivalent of electric charges. See Sec. 8.3. Figure 1217 shows schematically one device that has been used in the search for magnetic monopoles in bulk matter. At the beginning of the experiment. SW is closed and there is zero current through the iVturn superconducting coil C. A sample of matter SA, with a mass of the order of one kilogram, is made to follow a closed path P traversing C. Imagine that the sample contains one monopole of charge e* weber with radial lines of B. As it approaches a given turn of C. the flux through that turn is e*/2 weber
i
i O
,«.
p
SA Figure 1217 Magnetic monopole detector. The sample SA is placed on a conveyor and follows path P with the switch SW closed. After hundreds of passes, the switch is opened and the voltage pulse across R is observed on an oscilloscope. The coil C is superconducting. See Prob. 126.
302
Magnetic Fields: V
to the right, and, a moment later, it is e*/2 weber to the left. Thus, on emerging at the righthand end of C, the flux linkage through C due to the monopole has increased by Ne*. However, the resistance of the coil is zero and the induced current keeps the flux linkage zero. Calculate the current per monopole after 100 passes through a coil of 1200 turns with a selfinductance of 75 millihenrys. The current is detected by opening the switch SW and observing the voltage pulse on R. All the measurements to date have given null results.
ROGOWSKI
COIL
Figure 1218 shows a Rogowski coil surrounding a currentcarrying wire. The coil has N turns, and r » r'. Rogowski coils are used for measuring large fluctuating currents, particularly large current pulses. a) If the value of R is chosen large enough, V is unaffected by that part of the circuit that is to the right of the vertical dotted line. Show that Nr'
2
dl
2Vdl'
(1)
Show that, if C is now chosen large enough to make V « V, then Nr'
2
V
=
Po  2r
I RC
Figure 1218 Rogowski coil for measuring a heavy current pulse in the axial wire. With the RC circuit shown, the voltage V is proportional to I. See Prob. 127.
12)
Problems
303
See Prob. 530. The product P C must be large compared to l/f, where / is the repetition frequency, dl/dt being of the order of If. The current I need not be sinusoidal. Solution:
Under those conditions, V is the induced electromotance: d (
dcp V
=
N—
=
dt
_ ^ r '
2
N—
dt
V
u I\
,
T t r '
0
2
—
2nr)
,
3
d l
2r
dt'
n Nr' /2r.
Thus the mutual inductance is Also,
2
0
dQ V = PR + V % PR = — R, dt dV * P C — .
(5)
(6)
Comparing now Eqs. 4 and 6,
F « u '°
.
0
2r
7)
PC
We have set V equal to zero when / is zero. For example, the current / could be in the form of short pulses and, between pulses, both / and V would be zero. b) Show that, if the capacitor is shorted and if R is made small so that V is decreased by a factor of, say, 20 or more, then F * I/N. The selfinductance of the coil is L « p N r' /2r. 2
(8)
2
0
Solution: The induced electromotance is equal to the sum of the voltage drops across the inductance L of the coil and the resistance P : u Nr' ™2r
dl dP  = L — + RI'. dt dt
2
0
r
(9)
If P were very large, then RI' would be nearly equal to the induced electromotance. O n the other hand, if R is small, we may set
L
' E dt
v
^ l 2r
d
l dt
u o ,
304
Magnetic Fields: V
Substituting now the value of L, Li N r' — 2r 2
dl'
2
0
dt
n Nr' x— 2r
2
0
dl dt
.
(11)
1
I' x I/N.
128E
'
(12)
INDUCED CURRENTS An azimuthal current is induced in a conducting tube of length /, average radius a, and thickness b, with b « a. a, Show that its resistance and inductance are given by R = 2na/(jbl,
L =
[i na j\. 2
0
b) Calculate L, R, L/R for a copper tube (a = 5.8 x 1 0 Siemens per meter), one meter long, 10 millimeters in diameter, and with a wall thickness of 1 millimeter. 7
129
COAXIAL LINES Coaxial lines, as in Fig. 1219, are widely used for the interconnection of electronic equipment and for longdistance telephony. They can be used with either direct^ or alternating currents, up to very high frequencies, where the wavelength c/f is of the same order of magnitude as the diameter of the line. At these frequencies, of the order of 1 0 hertz, the field inside the line becomes much more complicated than that shown in the figure and quite unmanageable. There is zero electric field outside the line. Since the outer conductor carries the same current as the inner one, there is also zero magnetic field, from Ampere's circuital law (Sec. 9.1). A coaxial line acts as a cylindrical capacitor. We found its capacitance per meter in Prob. 65: 1 0
_ ~ In where R and R x
2
2ne e r
0
(Rt/Rj
are the radii of the inner and outer conductors, respectively.
/
Figure 1219 Coaxial line. See Prob. 129.
Problems
305
A coaxial line is also inductive, since there is an azimuthal magnetic field in the annular space between the two conductors. Show that the selfinductance per meter is.
1210
COAXIAL LINES It is known that highfrequency currents do not penetrate into a conductor as do lowfrequency currents. This is called the skin effect. See Prob. 117. Would you expect the selfinductance of a coaxial line to increase or to decrease with increasing frequency?
1211
LONG SOLENOID WITH CENTER TAP Consider a long solenoid with connections at both ends, A and C, and with a center tap B. The two halves are in series, and thus Lc A
= L
AB
+ L
BC
+ 2M,
where M is their mutual inductance. Now. if one uses Sec. 12.3.1 to calculate the three L's, one finds that M = 0. This is absurd because the coefficient of coupling is surely not zero. Can you explain this strange result?
1212
TRANSIENT
IN RLC
CIRCUIT
The switch in the circuit of Fig. 1220 is initially open, and then closed at t = 0.
C Figure 1220 See Prob. 1212.
Draw a curve ofQ as a function off for V = 100 volts, L = 1 henry, C = 1 microfarad, R = 8000 ohms, with t varying from 0 to 20 milliseconds.
306
Magnetic Fields: V
Solution:
From Kirchoff's voltage law. RI + L
or, since / =
dl dt
Q h— C
1.
(1)
dQjdt, dQ Q dQ L ~ + R± + ~ = V. dt dt C 2
(2)
1
The particular solution of this equation is Q = CV = 1CT ,
(3)
4
and the complementary function is the solution of dQ dQ Q L ~ + R — + — dt dt C 2
0.
2
(4)
Let us set 15)
Q = Ae"' Then, substituting in Eq. 4 and dividing by Q, Ln
2
1 + Rn + C
(6)
0.
R
R
2L
4L
1
2
2
~ LC
(7)
Note that there are two values of n, say n , and n . Then 2
Q = Ae"" + Be" '.
(S)
2
where A and B are unknown constants of integration. Here R/2L = 4000. the bracket of Eq. 7 is 1.5 x 10 , and the general solution is 7
Q = e  4 0 0 0 1 1 ^ 3 8 7 3 ! + Be" = Ae~ ' 121
+ Be '
+ 10"
+ 10"
lsli
(9) (10)
Since (2 = 0 a t r = 0. A + B =  1 0 " . Also, 4
dQ dt
\21Ae'
7873Be"
(ID
307
1.5fx 10"
4
0
10
20 ms
f
Figure 1221 The charge on the capacitor of Fig. 1220, as a function of the time for various values of R. The time is expressed in milliseconds. When the resistance is small, the charge overshoots and oscillates about its asymptotic value.
and I = 0 at r = 0, so that 127/4 + 78735 = 0, A = 1.016 x 10" . 4
Q = 1.016 x 1 0 " e " 4
(12)
B = 1.640 x 1 0 " .
(13)
6
1 2 7
' + 1.640 x l O " ? " 6
7 8 7 3
' + 10" . 4
(14)
As a check, we can show that L dljdt = 100 at f = 0. At that instant, 7 = 0 and the voltage drop on R is zero. Figure 1221 shows Q as a function of t for R = 500, 2000, and 8000 ohms. With R = 2000 ohms, the bracket is zero. The righthand side of Eq. 8 then becomes (A + Bt) exp nt. With R = 500 ohms, the bracket is negative and we again have another type of solution. 1213
VOLTAGE SURGES ON INDUCTORS If the current through a large inductor is interrupted suddenly, dljdt is large and transient voltages appear across the switch and across the coil. The voltages can
308
I
Figure 1222 Inductor L and its resistance R fed by a source V through a switch represented here by a resistor R . See Prob. 1213.
a
L
9
s
be large enough to damage both. Care must therefore be taken to reduce the current slowly. This is particularly important when using large electromagnets. N o problem arises, however, on connecting the source to the inductor. See Sec. 12.7.1. Figure 1222 shows the inductor L and its resistance R fed by a source V. T h e resistance R plays the role of the switch. At first. R is zero and / = VjR. Then R is quickly increased to some value much larger than R. Set L = 10 henrys. R = 10 ohms. V = 100 volts, dljdt =  1 0 amperes per second. a) Show that the voltages across the switch and across the inductor both rise to about 10 kilovolts. Of course the insulation of the switch and the insulation between turns on the inductor break down much before the voltage can rise to such a high value. b) It is possible to suppress the transient by using a diode. A diode is a twoterminal device that has either a very low or a very high resistance, depending on the polarity of the applied voltage, as in Fig. 1223. In other words, a diode will pass a current in only one direction. Of course, if the inverse voltage is sufficiently high, current will pass in the forbidden direction and the diode will be damaged. s
s
s
3
Figure 1223 The symbol for a diode is a triangle and a bar. In (a), the diode has a low resistance and can usually be considered as a short circuit. In (/>), the resistance is high and is usually considered to be infinite. The curve of / as a function of the applied voltage is nonlinear, even for forward voltages.
j • +
*
^1 UT)
*
_
*
' = 0 . I ^1 (/?)
Problems
309
A given type of diode can be used up to a rated maximum current, and up to a rated maximum inverse voltage. You are given a diode that is rated at 20 amperes and 200 volts peak inverse voltage. How should it be used? What is the current through the diode as a function of the time if the switch is opened at f = 0? What happens to the energy stored in the magnetic field of the inductor after the switch is opened? 1214D
TRANSIENT IN RLC CIRCUIT One wishes to charge a capacitor as quickly as possible. The source supplies an opencircuit voltage V and has an internal resistance R (Sec. 5.12), as in Fig. 1224a. It is suggested that the rate of charge could be increased by adding an inductor with L = R C/4 in series with the capacitor as in Fig. 1224b. Draw curves of the voltage V across the capacitor as a function of time for F = 100 volts, R = 100 ohms, C = 1 microfarad, with and without the inductor. Is the suggestion valid? s
2
s
Oi
Oi
3 c
V
S
 ±
(b) Figure 1224 Capacitor C charged by a source F having an internal resistance R. The inductor L in {b) has been added in the hope that the capacitor will charge more rapidly. See Prob. 1214. s
I2I5D
TRANSIENT IN RLC CIRCUIT In the circuit of Fig. 1225 the switch is initially open. The capacitor is charged to the voltage V and no current is drawn from the source. Set V = 100 volts, L = 1 henry, R = 10 ohms, C = 1 microfarad, a) The switch is closed. Show that / = 10(1 
e'
310
L
R
Figure 1225 See Prob. 1215.
b) After a time t » L!R, the switch is opened. Show that, from that time on, Qx
 1 0 ~ V ' cos 10 r + 1 ( T . 5
3
I * O . l e " ' sin 10 f. 5
3
4
CHAPTER MAGNETIC Magnetic
13 FIELDS:
VI
Forces
13.1.1
FORCE ON A WIRE CARRYING MAGNETIC FIELD Example: The Hodoscope
13.2
MAGNETIC
13.3
MAGNETIC
13.4 13.4.1
MAGNETIC ENERGY Example: The Long Solenoid
13.5
MAGNETIC ENERGY IN TERMS AND OF THE INDUCTANCE L Example: Coaxial Solenoids
13.1
13.5.1
A CURRENT
IN A
PRESSURE ENERGY
DENSITY
OF THE
CURRENT
13.6.1
MAGNETIC FORCE BETWEEN TWO ELECTRIC CURRENTS Example: The Force Between Two Infinitely Long Parallel
13.7
MAGNETIC
13.8
SUMMARY
13.6
PROBLEMS
FORCES
WITHIN
AN ISOLATED
I
Wires
CIRCUIT
In Chapters 4 and 7, we studied the energy stored in an electric field, and the resulting electric forces. Then, in Chapter 10, we studied the magnetic forces on individual charged particles. N o w we must study magnetic energy and macroscopic magnetic forces. Magnetic forces are, in practice, so much larger than electric forces that they lend themselves to many more applications. For example, electric motors almost invariably use magnetic forces!* We continue to assume that the materials in the field are nonmagnetic. We shall deal with the magnetic forces exerted by electromagnets in Chapter 15.
13.1
FORCE ON A WIRE CARRYING IN A MAGNETIC FIELD
A
CURRENT
Consider a wire of crosssectional area a, carrying a current / as in Fig. 131. There are n conduction electrons per unit volume, each one carrying a charge — e at a velocity v. If the wire is situated in a magnetic field B, due to currents flowing elsewhere, the magnetic force on one electron is — ev x B (Sec. 10.1) and,
Although electrostatic motors are exceedingly rare, they have been studied extensively. See, for example, A. D. Moore, Editor, Electrostatics and Its Applications, John Wiley, New York, 1973.
f
313
Figure 131 Wire of crosssection a carrying a current /. The charges — e move at a velocity v. The magnetic field B is due to currents flowing elsewhere.
for a length dl containing na dl conduction electrons, dF = nadl(ev
x B).
(131)
N o w the charge passing through a given crosssection per second is the charge on the carriers contained in a length v of the wire and I =
— nae\.
(132)
Then dF = Idl
x B = / d l x B,
(133)
where dl is in the direction of the current. For a straight length / of wire carrying a current / in a uniform magnetic field B. F = II x B
(134)
Figure 132 shows the lines of B near a currentcarrying wire, perpendicular to a uniform magnetic field. The force on a length / is then simply IIB, in the direction shown. As for electric fields, it is useful to imagine that lines ofB really exist, are under tension, and repel each other laterally. This simple model gives the correct direction for magnetic forces, as can be seen from Fig. 132.
314
(a)
(b)
Figure 132 Magnetic field near a currentcarrying wire situated in a uniform magnetic field, (a) Lines of B for a current perpendicular to the paper, (b) Lines of B for a uniform field parallel to the paper, (c) Superposition of fields in (a) and (b) and the resulting magnetic force F. The wire carries a current of 10 amperes, the uniform field has a B of 2 x 1 0 ~ tesla, and the point where the lines of B are broken is at 10 millimeters from the center of the wire. 4
EXAMPLE:
THE
HODOSCOPE
The hodoscope is a device that simulates the trajectory of a charged particle in a magnetic field. The principle involved is simple: if the charged particle, of mass m, charge Q, and velocity v, is replaced by a light wire fixed at the two ends of the trajectory and carrying a current I, the wire will follow the trajectory if mv/Q =
T/L
(135)
315
Figure 133 (a) Charge Q moving at a velocity v in a magnetic field B. The radius of curvature of the trajectory is R . (b) Light wire carrying a current / in the opposite direction in the same magnetic field. The tension in the wire is T, and its radius of curvature is R . It is shown that the wire has the same radius of curvature as the trajectory if mv/Q is equal to T/I. The angle dQ is infinitesimally small. t
w
where T is the tension in the wire. This statement is by no means obvious, but we shall demonstrate its validity. The advantage of the socalled floating wire lies in the fact that it is much easier to experiment with a wire than with an ion beam. Let us consider a region where the ion beam is perpendicular to B as in Fig. 133a. Then the charge Q is subjected to a force QvB. Thus QvB = mv /R„
(136)
2
where R, is the radius of curvature of the trajectory, and R, = mv/BQ.
(137)
In Fig. 133b we have replaced the charged particles by a light wire carrying a current / flowing in the opposite direction. If the element dl is in equilibrium, the outward force BI dl is compensated by the inward component of the tension forces T: Bl dl = IT sin (dd/2) = IT dO/2 = T dl/R . w
(138)
Fhe radius of curvature of the wire is thus R
w
= T/BI,
(139)
and the two radii of curvature will be the same if mv/Q
=
T/l.
(1310)
316
Magnetic
Fields: VI
For example, if we have a proton (in = 1.7 x 1 0 " kilogram) beam with a kinetic energy of 10 electronvolts (10 x 1.6 x 10" joule), then T/I is about 0.15 and. if / = 1 ampere, T is 0.15 newton. Note that the particle will be deflected downward if the magnetic force is downward, but the wire will curve downward if the force is upward. The magnetic forces must therefore be directed in opposite directions. If the magnetic field is not uniform, and if the beam is not perpendicular to B, then the wire does not always follow the trajectory. For example, magnetic fields are often used both to deflect and focus an ion beam. In such cases the focusing forces on the beam can become defocusing forces on the wire, which is then deflected away from the trajectory. 2 7
6
13.2
MAGNETIC
6
1 9
PRESSURE
In Chapter 10 we considered the magnetic forces acting on individual particles, and we have just seen how one can calculate the force on a currentcarrying wire. If, now, we have a current sheet, it is appropriate to think in terms of magnetic pressure. Let us imagine a flat current sheet carrying a amperes per meter and situated in a uniform tangential magnetic field B/2 due to currents flowing elsewhere, as in Fig. 134a, with a normal to B. We adjust a until it produces an aiding field B/2 on one side and an opposing field B/2 on the other side, as in Fig. 134b. Then, from P r o b . 93, a = B'p and the force per unit area, or the pressure, is ot. times the field B 2 due to the currents flowing elsewhere, or B /2p . So, whenever one has a current flowing through a conducting sheet, with a magnetic induction B on one side and zero magnetic induction on the other side, the sheet is subjected to a pressure B /2p . This pressure is exerted in the same direction as if the magnetic field were replaced by a compressed gas. We have arrived at this result for a flat current sheet, but the same applies to any current sheet when B = 0 on only one side. N o w we saw in Sec. 13.1 that, qualitatively, one can explain magnetic forces by assuming that the lines of B (a, repell each other laterally, and (b) are under tension. The magnetic pressure we have here is clearly ascribable to the lateral repulsion, since the lines are roughly parallel to the current sheet. 0
2
0
2
0
317
Figure 134 (a) A current sheet carries a current density of a amperes per meter of its width and is situated in a uniform magnetic field B/2 due to currents flowing elsewhere, (b) The total magnetic field is that of (a), plus that of the current sheet. By adjusting a so that it just cancels the magnetic field on the lefthand side of the sheet, we have a total field B on the right.
Since the lines are under tension, there should also be a force of attraction in the direction of B. This is correct, as we shall see later on in Sec. 15.3. In fact, along the lines of B, the force per unit area is again B /2p . 2
0
13.3
MAGNETIC
ENERGY
DENSITY
In Sec. 4.5 we found that, in an electrostatic field, the a conductor and the energy density in the field are N o w we have just found that the magnetic pressure therefore expect the energy density in a magnetic field This turns out to be c o r r e c t / Figure 135 shows the magnetic pressure, or density, B /2ti , as a function of B. 2
0
+
Electromagnetic Fields and Waves, p. 367.
force per unit area on e E /2. both equal to is B /2fi . One might to be equal to B /2[i . 2
0
2
0
2
0
the magnetic energy
318
10
1
l
l./"T
10"* 1 0 "
3
10
10 p (pascals) 10
8
'TO
4
10
2
1
1
10"
1
6
1
1
10"'
2
1
1
1
1
10 10 B (teslas)
1 2
1
10
3
Figure 135 Magnetic pressure B /2fi as a function of B. The magnetic pressure is equal to the magnetic energy density. Atmospheric pressure at sea level is about 10 pascals. 2
0
5
13.4
MAGNETIC
ENERGY
Since the magnetic energy density is B /2u , 2
0
the energy stored in a magnetic
field is
$ B ch,
(1311)
2
y
Ho
2
where the integral is evaluated over all space.
13.4.1
EXAMPLE:
THE LONG
SOLENOID
The magnetic energy stored in the field of a long solenoid of Af turns, radius R, and length /, carrying a current / is
inR l). 2
2/i
(1312)
2
(1313)
21 The magnetic pressure is 1 [ i NlV t 0
pN
2
0
(1314)
13.5 Magnetic Energy in Terms of Current 1 and Inductance L
13.5
319
MAGNETIC ENERGY IN TERMS OF THE CURRENT I AND OF THE INDUCTANCE L If one compares Eq. 1313 for the magnetic energy stored in the field of a solenoid with Eq. 1216 for the inductance of a solenoid, one finds that
M ^ W i
L
/
(1315)
2
This is a general result: the magnetic energy stored in the field of a circuit of inductance L carrying a current I is \LI . You will recall from Sec. 4.3 that the energy stored in the electric field of a capacitor of capacitance C charged to a voltage V is \CV . If one has two circuits with inductances L and L , and having a mutual inductance M, then the magnetic energy is 2
2
a
W
= ~ LJ
m
+ ~ LI
2
+
2
b
a
fc
(1316)
MlJ . h
This is shown in Prob. 1316. 13.5.1
EXAMPLE:
COAXIAL
SOLENOIDS
We can easily verify the above equation for coaxial solenoids. We first evaluate the righthand side. From Eqs. 1216 and 127,
 LJ
+  L„I
2
2
1 fi N nR 2
2
0
2
+
MIJ„
1 n N nR
2
2
0
ti nR
b
0
fN I 2
2
\ l„
NI
2 2
+ —\
r
• l„
H N N nR
2
0
b
'
a
2
b
(1317)
2N N I f a
b a
(131J
l
a
To calculate W , we use the fact that the energy density is B /2[i . l we have a total B of 2
m
0
b
>A
'B
Over the length
320
Magnetic Fields: VI
giving a magnetic energy 1
(NJ
2
N„I \
2
L
>b J
a
2po
\
b
Over the length /„ — l , we have only the field of coil a and a magnetic energy b
2
v
/.
'
Then the magnetic energy stored in the field of the coaxial solenoids is
II,"
ii 7iR 0
2
2
(N I l 2
V
Njll
2
a
b
\l
N.NJJt
lb
Nil
2
/.
I.
N ! !, 1
W
2
.
(13
m
which reduces to the expression on the righthand side of Eq. 131S
13.6
MAGNETIC ELECTRIC
FORCE BETWEEN CURRENTS
TWO
It is well k n o w n t h a t c i r c u i t s c a r r y i n g e l e c t r i c c u r r e n t s e x e r t forces o n e a c h o t h e r . In F i g . 136, t h e force e x e r t e d by c i r c u i t a on c i r c u i t b is o b t a i n e d b y i n t e g r a t i n g Eq. 133 o v e r c i r c u i t b:
Kb = I (f d\ x B b
w h e r e B is defined a s in Eq. 81.
b
h
(1320) a
13.7 Magnetic Forces Within an Isolated Circuit
EXAMPLE: THE FORCE BETWEEN LONG PARALLEL WIRES
TWO
321
INFINITELY
The force between two infinitely long parallel wires carrying currents /„ and I , separated by a distance p, as in Fig. 137, is calculated as follows. The force acting on an element l d l is b
h
b
x BJ,
(1321)
I _dl ii lJ2np, 0
(1322)
p IJ /2np.
(1323)
d F = /„(dl
dF =
t
b
b
and the force per unit length is dF — = dU
0
b
The force is attractive if the currents are in the same direction; and it is repulsive if they are in opposite directions.
Figure 137 Two long parallel wires carrying currents in the same direction. The element of force d F acting on element dl is in the direction shown. fc
13.7
MAGNETIC FORCES WITHIN AN ISOLATED CIRCUIT In an isolated circuit the current flows in its own magnetic field and is therefore subjected to a force. This is illustrated in Fig. 138: the interaction of
322
x R
S
v W — f — o — b — f t i)
•i
i
4 Figure 138 Schematic diagram of a rail gun. The battery A charges, through a resistance R, the capacitor C, which can be made to discharge through the line by closing switch S. The role of the capacitor is to store electric charge and to supply a very large current to the loop for a very short time. If side D is allowed to move, it moves to the right under the action of the magnetic force F.
the current / with its own magnetic induction B produces a force that tends to extend the circuit to the right. Magnetic forces within an isolated circuit are usually calculated by postulating a small displacement and using the principle of conservation of energy. See Prob. 1325.
13.8
SUMMARY The force on an element of wire of length dl carrying a current I in a region where the magnetic induction is B, is dF = / dl
x
B.
U33)
If a conducting sheet situated in a magnetic field carries a current that cancels the magnetic field on one side, then it is subjected to a magnetic pressure B /2p , where B is the magnetic induction on the other side. This expression also gives the energy density in the magnetic field. The magnetic energy stored in a magnetic field is obtained by integrating the energy density: 2
0
(1311)
Problems
323
If the field is due to an inductance L, then W
= (1 2)LI
m
(1315)
2
or, if we have two inductances L and L with a mutual inductance M. then a
W
= (l/2)L„/
m
The magnetic force
h
2 a
+ ll 21/.,,/,; + M / / . a
(1316)
0
exerted by a circuit a on a circuit b is
F
flft
= h§ A\ h
x B„.
h
(7520)
M agnetic forces also exist within an isolated circuit. They are calculated by postulating a small displacement and using the principle of the conservation of energy.
PROBLEMS 131E
MAGNETIC FORCE Calculate the magnetic force on an arc 50 millimeters long carrying a current of 400 amperes in a direction perpendicular to a uniform B of 5 x 1 0 " tesla. Highcurrent circuit breakers often comprise coils that generate a magnetic field to blow out the arc that forms when the contacts open. 2
132E
MAGNETIC FORCE a) Find the current density necessary to float a copper wire in the earth's magnetic field at the equator. Assume a field of 1 0 ~ tesla. The density of copper is 8.9 x 10 kilograms per cubic meter. b) Will the wire become hot? The conductivity of copper is 5.80 x 10 Siemens per meter. c) In what direction must the current flow? See Prob. 811. d) What would happen if the experiment were performed at one of the magnetic poles? 4
3
7
133E
MAGNETIC FORCE Calculate the force due to the earth's magnetic field on a horizontal wire 100 meters long carrying a current of 50 amperes due north. Set B = 5 x 1 0 " tesla, pointing downward at an angle of 70 degrees with the horizontal. See Prob. 811. 5
324
Magnetic Fields: VI
134E
MAGNETIC FORCE Show that the total force on a closed circuit carrying a current / in a uniform magnetic field is zero.
135E
ELECTROMAGNETIC PUMPS The conduction current density in a liquid metal is
J = ff(E + v x f
B),
where a is the conductivity, E is the electric field intensity, v is the velocity of the fluid, and B is the magnetic induction, all quantities measured in the frame of reference of the laboratory. Show that the magnetic force per unit volume of fluid is
ff(E + v x
B)
x
B.
For example, in crossed electric and magnetic fields, a conducting fluid is pushed in the direction of E x B. The field v x B then opposes E. This is the principle of operation of electromagnetic pumps. 136E
HOMOPOLAR GENERATOR AND HOMOPOLAR MOTOR Figure 139 shows a homopolar generator. It consists in a conducting disk rotating in an axial magnetic field. In one particular case, B = 1 tesla, R = 0.5 meter, and the angular velocity is 3000 revolutions per minute. Calculate the output voltage. Homopolar generators are inherently highcurrent, lowvoltage devices. Homopolar generators using superconducting coils and with power outputs of several megawatts are used for the purification of metals by electrolysis. If a voltage is applied to the output terminals of a homopolar generator, it becomes a homopolar motor. Homopolar motors can provide large torques. They are suitable, in particular, for ship propulsion.
Figure 139 Homopolar generator. See Prob. 136.
Problems
137
325
HOMOPOLAR MOTOR See Prob. 136. N o w consider the device illustrated in Fig. 1310. Will the wheel turn when the switch is closed? If so, in which direction and why?*
Figure 1310 Spiral coil. A voltage is applied between the axis and the periphery. Contact to the periphery is made with a carbon brush like that used on motors. Does the disk turn? See Prob. 137. The disk could be made with a printedcircuit board.
138
MAGNETIC PRESSURE So as to clarify the concept of magnetic pressure, let us consider, first a single solenoid, and then a pair of coaxial solenoids. a) Figure 13lla shows a longitudinal section through a solenoid. The small element identified by vertical bars (1) produces its own magnetic field, both outside and inside the solenoid, and (2) is situated in the field produced by the rest of the winding.
fTYTYTVTYTYn TYTYTYri rTYTYlYTYTYTYT) UA1ALAJAIALJ J A I A I A J J U A l A J A l A i A i A L '
(MXKDQXDQXKD0(Tj(IXJXD(D(]XD
00©00©0©©0©0©0©00 fTYTYTYTYTYT) UAIAlAiAJAIAlAiAJTUAlAiAiAlAlAiAi/
(a) Figure 1311 (a) Longitudinal section through a long solenoid. In Prob. 138, we consider the fields close to the small element identified by vertical bars, (b) Section through a pair of coaxial solenoids.
f
See the American Journal of Physics, Volume 38, 1970, p. 1273.
326
Magnetic Fields: VI
Show the first field by means of solid arrows, and the second one by means of dashed arrows. We are of course thinking here of the fields infinitely close to an infinitely thin winding. Now can you explain why the magnetic pressure is B 2/< ? ^ — b) Figure 131 lb shows a pair of coaxial solenoids carrying equal currents in opposite directions. In this example we have three fields. Show, on either side of the element, (a) solid arrows for the field of the element, (b) dashed arrows for the field of the rest of the inner solenoid, (c) wavy arrows for the field of the outer solenoid. What is the magnetic pressure on the inner solenoid? 2
0
139E
MAGNETIC PRESSURE a) Show that the magnetic pressure is about 4 B atmospheres. One atmosphere is about 10 pascals. b) Draw a loglog plot of the electric force per unit area je E . in pascals, as a function of E. The maximum electric field intensity that can be maintained in air at normal temperature and pressure is 3 x 10 volts per meter. c) Discuss the similarities and differences between the magnetic pressure and the electric force per unit area. 2
5
2
0
6
1310E
MAGNETIC PRESSURE It is possible to attain very high pressures by discharging a large capacitor through a hollow wire. a) Show that, for a thin tube of radius R carrying a current /. the inward magnetic pressure is p,„ = tt I
Sn R .
2
2
0
2
b) Calculate the pressure for a current of 30 kiloamperes in a tube one millimeter in diameter. One atmosphere is approximately 10 pascals. 5
131 IE
MAGNETIC PRESSURE Magnetic fields are used for performing various mechanical tasks that require a high power level for a very short time. For example, if a light aluminum tube is inserted axially in a solenoid, and if the solenoid, is suddenly connected to a charged capacitor, the induced electromotance /
Problems
327
a) Calculate the pressure on the tube for B = 1 tesla. bl What would the pressure be if the conducting tube were parallel to the axis, but some distance away? '312E
ENERGY STORAGE Compare the energies per unit volume in (a) a magnetic field of 1.0 tesla and (b) an electrostatic field of 10 volts per meter. 6
'313E
MAGNETIC PRESSURE Imagine that the current in a long solenoid is maintained constant, while the magnetic pressure increases the radius from R to R + dR. Then the magnetic energy mcreases by
a) Show that the mechanical work done by the magnetic pressure p„, is 2nRlp„, dR. b) Show that the mcrease in magnetic energy is equal to the mechanical work done. c) During the expansion, the magnetic induction inside the solenoid remains constant because B depends only on the number of turns per meter and on the current. So, as the radius increases, the flux linkage also increases. Show that the extra energy supplied by the source during the expansion, l(N (!<£>), is just twice the mechanical work done. We conclude that one half of the energy supplied by the source serves to perform mechanical work, and the other half serves to increase the magnetic energy. See Prob. 413. 1314
FLUX
COMPRESSION Flux compression is one method of obtaining a magnetic field with a very large B. For example, imagine a light conducting tube situated inside a solenoid producing a steady B . The annular space between the tube and the solenoid is filled with an explosive, and the solenoid is reinforced externally. If the tube is now imploded, it will be crushed, an azimuthal current will flow, and the internal magnetic pressure B /2u will build up until it is close to the external gas pressure. a) Show that, if the radius of the tube shrinks very rapidly. 0
2
0
B *
B (R 'Rf. 0
0!
where 6 is the magnetic induction when the radius has been reduced to P. b) Show that the surface current density in the tube must be about 10 amperes per meter to achieve a field of 10 teslas. c) If the initial B is 10 teslas. and if the tube has initially a diameter of 100 millimeters, what should be the value of B when the tube is compressed to a diameter of about 10 millimeters? 9
3
328
Magnetic Fields: VI
d) Calculate the resulting increase in magnetic energy. Assume that the cylinder is 200 millimeters long and that the current in the solenoid is constant. Neglect end effects. Flux compression can also be achieved by means of a conducting piston shot axially into a solenoid. y If the radius of the solenoid is R . and if the radius of the piston is K, the magnetic induction in the annular region between the piston and the solenoid, becomes 0
S
B%
0
(R/R r 0
1315E
PULSED MAGNETIC FIELDS Extremely high magnetic fields can be obtained by discharging a capacitor through a lowinductance coil. The capacitor leads must of course have a low inductance. Such fast capacitors cost approximately two dollars per joule of storedenergy capacity. a) Estimate the cost of a capacitor that could store an energy equal to that of a 100tesla magnetic field occupying a volume of one liter. b) Estimate the cost of the electricity required to charge the capacitors. c) Calculate the magnetic pressure in atmospheres, at 100 teslas. One atmosphere is about 10 pascals. 5
1316E
MAGNETIC ENERGY It was stated in Sec. 13.5 that, if we have two inductances L and L , carrying currents I and I , a
a
b
b
W
m
= LJ l
+\L ll
2
+
b
a
MIJ . b
You can prove this quite easily. Let coil a produce flux linkages A
aa
in coil a
and
A
and
A
ab
in coil b.
Similarly, let coil h produce flux linkages bb
in coil b.
a) Find W in terms of the currents and of the flux linkages. m
1317E
b) Verify the above equation. ENERGY STORAGE a) Show that the magnetic energy stored in an isolated circuit carrying a current / and with a flux linkage A is W
m
= (l/2)/A.
b) Calculate W for a long solenoid using this formula. m
Problems
329
1318
ENERGY STORAGE a) A constantvoltage source set at a voltage V is connected to an inductance L of negligible resistance at t = 0. Find, as functions of the time, (i) the current /. (ii) the energy supplied by the source, and (iii) the energy stored in the magnetic field. b) A constantcurrent source set at a current / is connected to a capacitor C at r = 0. Find, as functions of the time, (i) the voltage V on the capacitor, (ii) the energy supplied by the source, and (iii) the energy stored in the electric field.
1319
ENERGY STORAGE Electrical utilities must be able to meet peak power demands. It is therefore desirable to store energy during periods of low demand and feed it back into the grid when needed. O n e way of storing energy is to pump water into an elevated reservoir. It has been suggested that large amounts of energy could also be stored in the magnetic fields of superconducting coils. a) Compare the energy densities in (i) an electric field of 1 0 volts per meter in Mylar, for which e = 3.2. and (ii) a magnetic field of 8 teslas. Fhese fields are about as high as present technology will permit. F h e magnetic energy density is larger by two orders of magnitude. b) It is suggested that 10 gigawatthours be stored in a solenoid with a lengthtodiameter ratio of 20. The solenoid would be made in four parts, joined end to end to form a quasitoroidal coil located in a tunnel excavated in solid rock. Disregard the toroidal shape in your calculation. The magnetic induction would be 8 teslas. (i) Calculate the total length and the diameter, (ii) Calculate the magnetic pressure in atmospheres, (iii) Calculate the surface area that would require cryogenic thermal insulation. 8
r
1320
SUPERCONDUCTING POWER TRANSMISSION LINE A superconducting power transmission line has been proposed that would have the following characteristics. It would carry 1 0 " watts at 200 kilovolts DC over a distance of 1 0 kilometers. The conductors would have a circular crosssection of 5 square centimeters and would be held 5 centimeters apart, center to center. a) Calculate the magnetic force per meter. It so happens that the force of attraction is the same as if the conductors were replaced by fine wires, 5 centimeters apart. b) Calculate B / 2 p midway between the two conductors. c) Calculate the stored energy and its cost at 0.5 cent per kilowatthour. The selfinductance of such a line is 0.66 microhenry per meter. 3
2
0
1321
ELECTRIC MOTORS AND MOVINGCOIL METERS Electric motors, as well as movingcoil voltmeters and ammeters, utilize the torque exerted on a currentcarrying coil situated in a magnetic field. Consider an Nturn rectangular coil situated in a uniform magnetic field as in Fig. 1312. a) Calculate the forces on sides 1, 2, 3, 4. b) Show that the torque T = NIBab sin 0.
330
Figure 1312 Rectangular coil whose normal n forms an angle 0 with a uniform magnetic field. See Prob. 1321.
In an electric motor, we have essentially a number of such coils, offset by, sa; 15 degrees, around the axis, and connected to a set of contacts called a commutator fixed to the axis. Connection to the coils is made by carbon brushes that ride on th commutator. The coils and their commutator form the armature of the motor. Th magnetic field is supplied by the stator coils. The magnetic flux is carried by laminatci iron, both in the armature and in the stator yoke. Moving coil voltmeters and ammeters use a uniform radial magnetic field as ii Fig. 1313. The torque is then simply NIBab and is independent of 0. A light spira spring exerts a restoring torque that is proportional to 0. giving a deflection 0 propor tional to /.
Figure 1313 Moving coil and magnetic field in a voltmeter or ammeter. The magnetic field is supplied by a permanent magnet. The parts marked / are made of iron.
Problems
331
1322
MAGNETIC TORQUE Consider a square singleturn coil of side a, carrying a current /. a) Show that it tends to orient itself in a magnetic field in such a way that the total magnetic flux linking the coil is maximum. b) Show that the torque exerted on the coil is m x B, where m is the magnetic moment of the coil (Sec. 8.1.2), and B is the magnetic induction when the current in the coil is zero. This result is independent of the shape of the coil.
1323
ATTITUDE CONTROL FOR SATELLITES See the previous problem. Many satellites require attitude control to keep them properly oriented. For example, communication satellites must keep their antenna systems on target. Attitude control requires a method for exerting appropriate torques as they are required. Attitude control can be achieved to a certain extent by means of coils whose magnetic fields interact with that of the earth. a) Show that the torque exerted by such a coil is NIBA
sin 0.
where N is the number of turns, / is the current. B is the magnetic induction due to the earth, A is the area of the coil, and 0 is the angle between the earth's magnetic field and the normal to the coil. b) Calculate the number of ampereturns required for a coil wound around the outside surface of a satellite whose diameter is 1.14 meters. The torque required at 0 = 5 degrees is 1 ( F newtonmeter. and the magnetic induction at an altitude of 700 kilometers over the equator, where the orbit of the satellite will be situated, is 4.0 x 1 0 " tesla. 3
5
1324
MECHANICAL FORCES ON AN ISOLATED CIRCUIT Show that, if the geometry of an isolated active circuit is altered, the energy supplied by the source is equal to twice the increase in magnetic energy. Thus the mechanical work performed is equal to the increase in magnetic energy. Assume that the current is kept constant. It follows that, on this assumption, the force on an element of an active circuit is equal to the rate of change of magnetic energy. See Probs. 1313 and 1325.
RAIL
GUN. OR PLASMA
GUN
Calculate the force on the movable link D in the circuit of Fig. 138. Instead of being a metallic rod. the link can also be an electric arc. The device then accelerates blobs of plasma and is called a plasma gun. Statellite thrusters can be made in this way. See Probs. 214. 215. 53, and 1011.
332
Magnetic Fields: VI
Solution: We cannot solve this problem by integrating the magnetic force J x B over the volume of the link, since neither J nor B are known, or even easily calculated. Instead, we shall find the force by investigating the magnetic and mechanical energies involved. If we set x = 0 at the initial position of the link, then the inductance L in t h l T ' circuit is L + L'x, where L'x is the inductance associated with the magnetic flux linking the rails. L' being the selfinductance per meter. During the discharge of the capacitor C we can neglect the battery and the resistance R. since the current flowing in that part of the circuit is negligible. Let the resistance on the righthand side of the circuit be only that of the arc. R'. At any instant, the power supplied by the capacitor serves to (a) increase the magnetic energy \LI . (b) increase the kinetic energy, and (c) dissipate energy in the resistance R'. Thus 0
2
(1) where F is the driving force and v is the velocity of the link. Note that both L and / are functions of the time. Now what is the value of VI The voltage V (a) increases the flux linkage in the inductance and (b) gives a voltage drop IR' on the resistor R': (LI) +
IR\
(2)
where LI is the flux linkage. Combining Eqs. 1 and 2.
(3)
(4)
2
(5)
I1 LI . 2
2
(6)
CHAPTER
14
MAGNETIC
FIELDS:
Magnetic
VII
Materials
14.1
THE
14.2
THE EQUIVALENT
SURFACE
CURRENT
DENSITY
et
14.3
THE EQUIVALENT
VOLUME
CURRENT
DENSITY
J,
14.4 14.4.1
CALCULATION OF MAGNETIC FIELDS IN MAGNETIZED MATERIAL Example: Uniformly Magnetized Bar Magnet
14.5
THE MAGNETIC
14.6 14.6.1
AMPERE'S CIRCUITAL LAW Example: Straight Wire Carrying a Current I and in Magnetic Material
14.7
MAGNETIC SUSCEPTIBILITY AND RELATIVE PERMEABILITY
14.8 14.9
MAGNETIZATION
FIELD
THE MAGNETIZATION
INTENSITY
CURVE
HYSTERESIS
14.9.1
Example:
14.10
BOUNDARY
14.11
M
Transformer
SUMMARY PROBLEMS
Iron
CONDITIONS
e
(
ORIGINATING
H Embedded
/„„ PERMEABILITY p r
p.
Thus far we have studied only those magnetic fields that are attributable to the motion of free charges. Now, on the atomic scale, all bodies contain spinning electrons that move around in orbits, and these electrons also produce magnetic fields. O u r purpose in this chapter is to express the magnetic fields of these atomic currents in macroscopic terms. Magnetic materials are similar to dielectrics in that individual charges or systems of charges can possess magnetic moments (Sec. 8.1.2), and these moments, when properly oriented, produce a resultant magnetic moment in a macroscopic body. Such a body is then said to be magnetized. In most atoms the magnetic moments associated with the orbital and spinning motions of the electrons cancel. If the cancellation is not complete, the material is said to be paramagnetic. When such a substance is placed in a magnetic field, its atoms are subjected to a torque that tends to align them with the field, but thermal agitation tends to destroy the alignment. This phenomenon is analogous to the alignment of polar molecules in dielectrics. In diamagnetic materials, the elementary moments are not permanent but are induced according to the Faraday induction law (Sec. 11.2). All materials are diamagnetic, but orientational magnetization may predominate. Magnetic devices use ferromagnetic materials, such as iron, in which the magnetization can be orders of magnitude larger than that of either para or diamagnetic substances. This large magnetization results from electron spin and is associated with g r o u p phenomena in which all the elementary moments in a small region, known as a domain, are aligned. The magnetization of one domain may be oriented at r a n d o m with respect to that of a neighboring domain. There is one important difference between dielectric and magnetic materials. In most dielectrics D is proportional to E and the medium is said
14.2 The Equivalent Surface Current Density a,,
335
to be linear. Ferromagnetic materials are not only highly nonlinear, but their behavior also depends on their previous history. The calculation of the fields associated with magnetic materials is therefore largely empirical. This will be the subject of the next chapter.
14.1
THE MAGNETIZA
TION M
The magnetization M is the magnetic moment per unit volume at a given point. If m is the average magnetic dipole moment per atom, and if N is the number of atoms per unit volume, M = A/m,
(141)
if the individual dipole moments in the element of volume considered are all aligned in the same direction. The magnetization M is measured in amperes per meter, and it corresponds to the polarization P in dielectrics (Sec. 6.1).
14.2
THE EQUIVALENT
SURFACE
CURRENT
DENSITY
a
e
Figure 141 shows a cylinder of material divided into square cells of crosssectional area a , where a » 0. Imagine that each cell carries a clockwise surface current of M amperes per meter, as in the figure. Then each onemeter length of cell has a magnetic moment of a M ampere meter squared (Sec. 8.1.2) and the magnetic moment per cubic meter is M amperes per meter in the direction shown in the figure. Now the current in one cell is canceled by the currents in the adjoining cells, except at the periphery of the material. Thus, if the material has a magnetization M, the equivalent current density ct at the surface is M. More generally, the equivalent, or Amperian, surface current density is 2
2
e
a, = M x n,,
(142)
where n, is a unit vector normal to the surface and pointing outward. These equivalent currents do not dissipate energy because they do not involve electron drift and scattering processes like those associated with conduction currents.
336
Figure 141 Ampere's model for the equivalent current in a cylinder of magnetized material.
14.3
THE EQUIVALENT
VOLUME CURRENT
DENSITY J
e
Let us now suppose that the magnetization M is a function of the xcoordinate as in Fig. 142, with M = px,
(143)
where p is a positive constant. Since M is expressed in amperes per meter, p is expressed in amperes per square meter. At the surface of the material, the equivalent current density « is again given by Eq. 142, with M equal to the local value of the magnetization Thus a is a function of x, as in the figure. Inside the material, currents parallel to the xaxis cancel. Currents parallel to the yaxis do not cancel. Along any of the vertical planes inside, the lefthand side of a given cell carries an upward current of e
e
while the righthand side carries a downward current of
337
y
X
Figure 142 Magnetized cylinder in which M is a function of the xcoordinate: M increases linearly with x as in Eq. 143.
Along any vertical plane, we therefore have a net downward current of pa amperes. N o t e that this downward current is independent of x. This is simply because we have assumed that M is directly proportional to x. So, along every vertical division between cells, we have a net downward current of pa amperes. The net downward current density is thus p, since we have pa amperes for every interval a. In other words, we have an equivalent current density J
e
=
pi
(144)
We can now show that J is equal to V x M in the following way. Along the curve C of Fig. 142, M • dl is zero over the lefthand end, because M is perpendicular to dl. The same applies to the righthand end. Thus e
(145)
=
pa.
(146)
where the line integral is evaluated in the direction shown in the figure.
338
Magnetic Fields: VII
Since the area of the loop is a x 1, the ycomponent of the curl of M is — p, from Sec. 1.12, and
j
= (v x rviy.
e
(147^
M o r e generally, the equivalent volume current density is
J = V x M.
(148)
e
14.4
CALCULA TION OF MAGNETIC ORIGINATING IN MAGNETIZED
FIELDS MATERIAL
Let us first recall the reasoning we followed in Sec. 6.3 to find the electric field due to polarized dielectric material. We found that the charge displacement that occurs on polarization results in the accumulation of net densities of bound charge, a on the surfaces and p inside. We then argued that any net charge density gives rise to an electric field, exactly as if it were in a vacuum. As a consequence, E can be calculated at any point in space from a and p . b
b
h
b
We now have an entirely analogous situation in the magnetic fields originating in magnetized materials. The magnetization gives equivalent currents, with densities a , on the surfaces and J inside. These currents produce magnetic fields, both inside and outside the material, exactly as if they were situated in a vacuum. L
e
In principle, one can calculate B at any point in space from the equivalent current densities a and J . In practice, the vector M is an unknown function of the coordinates inside the material, and one can d o little more than guess its value. We shall return to this subject in the next chapter. e
EXAMPLE:
UNIFORMLY
e
MAGNETIZED
BAR
MAGNET
The uniformly magnetized bar magnet of Fig. 141 is a good example to use at this point. We have shown in Sec. 14.2 that the cylindrical surface carries an equivalent current density x that is equal to M. Inside the magnet, V x M = 0, and the volume current density J is zero. Over the end faces. M is parallel to the normal unit vector n , , and a, is zero. e
f
14.5 The Magnetic Field Intensity H
339
The B field of a cylinder uniformly magnetized in the direction of its axis of symmetry is therefore identical to that of a solenoid of the same dimensions with JV' turns per meter and carrying a current / such that IN' = M, as in Fig. 910. In a real bar magnet the magnetic moments of the individual atoms tend to align themselves with the B field, so that the magnetization M is weaker near the ends. The end faces also carry Amperian currents, since M x n is zero only on the axis. The net result is that there are two "poles," one at each end of the magnet, from which lines of B radiate in all directions outside the magnet. The poles are most conspicuous if the bar magnet is long and thin.
14.5
THE MAGNETIC
FIELD INTENSITY
H
We found in Sec. 9.2 that, for a steady current density of free charge for nonmagnetic materials, V x B = /< J/
and
(149)
0
Now we have just seen that magnetized material can be replaced by its equivalent currents for calculating B. Consequently, if we have magnetized material as well as steady currents, V x B = n {i Q
f
+ j ), e
(1410,
where J is the equivalent volume current density. This equation is of course valid only at points where the derivatives of B with respect to the coordinates x, y, z exist. It is therefore not applicable at the surface of magnetic materials and we disregard a ,. e
(
Substituting V x M for J , t>
V x B = p. (J 0
f
+ \
x M),
V x ^— — mJ = J . f
(1411,
(1412,
The vector between parentheses, whose curl is equal to the free current density at the point, is the magnetic field intensity H = —  M, Ih
(1413,
and is expressed in amperes per meter. Note that H and M are expressed in the same units.
340
Magnetic
Fields:
VII
Thus B = n (H 0
+ M).
(1414)
This equation is to be compared with Eq. 615, which applies to dielectricsrf E = — (D e
P).
(1415)
0
14.6
AMPERE'S
CIRCUITAL
LA W
Rewriting now Eq. 1412, we have that, for steady currents, either inside or outside magnetic material, V x H = J.
(1416)
f
Integrating over a surface S, J" (V x H ) d a = J J /  d a s
(1417)
s
or, using Stoke's theorem on the lefthand side, ( f H •«« = / „
(1418)
c
where C is the curve bounding the surface S, and I is the current of free charges linking the curve C. Note that I does not include the equivalent currents. The term on the left is called the magnetomotance. A magnetomotance is expressed in amperes, or in ampereturns. This is a more general form of Ampere's circuital law (Sec. 9.1), in that it can be used to calculate H even in the presence of magnetic materials. It is rigorously valid, however, only for steady currents; we shall deal with variable currents in Chapter 19. r
f
14.6.1
EXAMPLE: EMBEDDED
STRAIGHT WIRE IN MAGNETIC
CARRYING MATERIAL
A CURRENT
I
AND
Figure 143 shows a straight wire carrying a current / and embedded in magnetic material. The magnetomotance around the path shown circling the current I is equal to /, and H = I/2nr, exactly as if the wire were situated in a vacuum.
341
Figure 143 Straight wire carrying a current / and embedded in magnetic material.
If the material is isotropic, and if it has the shape of a circular cylinder with / along its axis, M is azimuthal, like H .
14.7
MAGNETIC SUSCEPTIBILITY x , PERMEABILITY AND RELATIVE PERMEABILITY \i m
n,
r
As for dielectrics (Sec. 6.6), it is convenient to define a magnetic ~/ such that
susceptibility
m
H.
(1419)
B = p ( H + M),
(1420)
M =
Z m
Now, since 0
then B = jU (l + z J H = p p H 0
0
r
= pU,
(1421)
where PR
=
1 + Xm,
/' =
MO/V
The quantity p is the permeability and p is the relative permeability. X and p are dimensionless quantities. r
m
r
(1422)
Both
342
Magnetic Fields:
VII
Equation 1420 is general, but Eqs. 1421 and 1422 are based on the assumption that the material is both isotropic and linear, in other words, that M is proportional to H and in the same direction. This assumption is unfortunately not valid in ferromagnetic materials, as we shall see in next section. The magnetic susceptibility of paramagnetic substances is smaller than unity by several orders of magnitude and is proportional to the inverse of the absolute temperature. In diamagnetic materials, the magnetization is in the direction opposite to the external field; the relative permeability is less than unity and is independent of the temperature. If orientational magnetization predominates, the resultant relative permeability is greater than unity.
14.8
THE MAGNETIZA
TION CUR VE
O n e can measure B for various values of H with a Rowland ring whose minor radius is much smaller than its major radius, as in Fig. 144. The function of winding a, which has N turns and carries a current /„, is to produce a known magnetic field intensity a
H = NJJlnr
Figure 144 Rowland ring for the determination of B as a function of H in a ferromagnetic substance.
(1423)
343
H (ampereturns meter) Figure 145 Magnetization curves for various magnetic materials: a, Permalloy; b, annealed pure iron: c, ductile cast iron; d, Alnico 5. The numbers shown are the relative permeabilities B//i H. 0
in the sample. The magnetic induction is B = cp/S,
(1424)
where S is the crosssectional area of the core, and O is the magnetic flux. O n e can measure changes in 0 , and hence changes in B, by changing / and integrating the electromotance induced in winding b as in Prob. 1416. Both B and H are therefore easily measurable. If we start with an unmagnetized sample and increase the current in coil a, the magnetic induction also increases, but irregularly, as in Fig. 145. All ferromagnetic substances have such Sshaped magnetization curves. N o t e the large variations in u, for a given material. Whenever one uses p in relation with ferromagnetic materials, one refers to the ratio B/u H, for specific values of B or of H. A characteristic feature of the magnetization curve is the saturation induction beyond which M increases no further. Maximum alignment of the domains is then achieved, after which dB = fi dH. The saturation induction lies in the range from 1 to 2 teslas, depending on the material. r
0
0
344
Figure 146 Magnetization curve ab and hysteresis loop bcdefgb.
14.9
HYSTERESIS Consider now Fig. 146. Curve ab is the magnetization curve. Once point b has been reached, if the current in winding a of Fig. 144 is reduced to zero, B decreases along be. The magnitude of the magnetic induction at c is the remanence or the retentivity for the particular sample of material. If the current is then reversed in direction and increased. B reaches a point d where it is reduced to zero. The magnitude of H at this point is known as the coercive force. On further increasing the current in the same direction a point e, symmetrical to point b, is reached. If the current is now reduced, reversed, and increased, the point b is again reached. The closed curve bcdefgb is known as a hysteresis loop. If, at any point, the current is varied in a smaller cycle, a small hysteresis loop is described. Energy is required to describe a hysteresis cycle. This can be shown by considering Fig. 147. When the current in winding a of Fig. 144 is increasing, the electromotance induced in the winding opposes the increase
345
Figure 147 The shaded area gives the energy per unit volume required to go from « to b on the hysteresis loop. The energy per unit volume required to describe a complete hysteresis loop is equal to the area enclosed by the loop.
in current, according to Lenz's law (Sec. 11.3), and the extra power spent by the source is
where S is the crosssectional area of the ring, N is the number of turns in winding a. and B is the average magnetic induction in the core. Also,
where / is the mean circumference of the ring, T = SI is its volume, and IN/I = H, as in Eq. 1423. T h u s W,
= T [ HdB b
(1427)
346
Magnetic Fields:
VII
is the energy supplied by the source in going from the point g to the point b in Fig. 147. This integral corresponds to the shaded area in the figure and is equal to the energy supplied per unit volume of the magnetic core. When the current is in the same direction but is decreasing, the polarivf of the induced electromotance is reversed, according to Lenz's law, with the result that the energy W
2
= t P H dB
(1428)
is returned to the source. Finally, the energy supplied by the source during one cycle is W=tj)HdB,
(1429)
where the integral is evaluated around the hysteresis loop. The area of the hysteresis loop in teslaampere turns per meter or in weberampere turns per cubic meter is therefore the number of joules dissipated per cubic meter and per cycle in the core. Hysteresis losses can be minimized by selecting a material with a narrow hysteresis loop.
14.9.1
EXAMPLE:
TRANSFORMER
IRON
For the transformer iron of Fig. 147. the area enclosed by the hysteresis loop is approximately 150 weberampere turns per cubic meter or 150 joules per cubic meter and per cycle, or 1.1 watts per kilogram at 60 hertz. This iron is suitable for use in a power transformer, but other alloys are available with lower hysteresis losses.
14.10
BO UNDA R Y
CONDITIONS
Let us examine the continuity conditions that B and H must obey at the interface between two media. We shall proceed as in Sec. 7.1. Figure 148a shows a short cylindrical volume whose top and bottom faces are parallel and infinitely close to the interface. Since there is zero net flux through the cylindrical surface (Sec. 8.2), the flux through the top face must equal that through the bottom, and B„i = B . n2
(1430)
The normal component of B is therefore continuous across the interface.
347
(b) Figure 148 (a) Gaussian surface at the interface between two media, (b) Closed path crossing the interface.
Consider now Fig. 148b. The closed path has two sides parallel to the interface and close to it. From the circuital law (Sec. 14.6), H d l = 7,
(1431)
where / is the conduction current linking the path. If the two sides parallel to the interface are infinitely close to it, 7 is equal to zero, H
n
= H, t2
and the tangential component of H is continuous across an interface.
(1432)
348
Magnetic Fields:
VII
If we can set B equal to uH in both media, the relative permeabilities being those that correspond to the actual values of H. then uuH ri
l}
cos 0
l
= p, fj, H
l
r2
0
cos 9 ,
2
2
(1433/f^
from the continuity of the normal component of B, and H sin 0j = H x
sin 0 ,
2
2
(1434)
from the continuity of the tangential component of H. Then tan 6j
n
tan 0
Uri
2
ri
(1435)
The lines of B, or of H. are farther away from the normal in the medium having the larger permeability.
14.11
SUMMARY The magnetization
M is the magnetic moment per unit volume.
One can calculate the magnetic field of a piece of magnetic material by assuming an equivalent surface current density a, = M x n,.
(142)
where n, is a unit vector normal to the surface and pointing outward, and an equivalent volume current density J = V x M. e
(148)
The magnetic field intensity is
H = —  M,
(1413)
l<0
and V x H = J . F
(1416)
Problems
349
This is Ampere's circuital law in terms of H. In integral form,
ic
(1418)
H • dl = I
where I is the current of free charges linking the curve C. If a magnetic material is linear and isotropic, then f
15 = /'o(l + Zm)H = MortH =
pH.
(1421)
where y is the magnetic susceptibility, p is the relative permeability, and p is the permeability. Ferromagnetic materials are highly nonlinear, and often anisotropic. The relation between B and H depends on the previous history of the field, and is represented by the hysteresis loop. The area of a hysteresis loop gives the energy dissipated in the material per cubic meter and per cycle. At the boundary between two materials, the tangential component of H and the normal component of B are continuous. m
r
PROBLEMS 141E
MAGNETIC FIELD OF THE EARTH A sphere of radius R is uniformly magnetized in the direction parallel to a diameter. The external field of such a sphere is closely similar to that of the earth. See Prob. 811. Show that the equivalent surface current density is the same as if the sphere carried a uniform surface charge density a and rotated at an angular velocity co such
142E
EQUIVALENT CURRENTS An iron torus whose major radius is much larger than its minor radius is magnetized in the azimuthal direction with M uniform. What can you say about a,'? t
143E
EQUIVALENT CURRENTS A long tube is uniformly magnetized in the direction parallel to its axis. What is the value of B inside the tube?
144
DIELECTRICS AND MAGNETIC MATERIALS COMPARED a) Imagine a parallelplate capacitor whose plates are charged and insulated. How are E and D affected by the introduction of a dielectric between the plates?
350
Magnetic Fields:
VII
b) Show a polar molecule of the dielectric oriented in the field. c) Is its energy maximum or minimum? d) Now imagine a long solenoid carrying a fixed current. How are B and H affected by the introduction of a cylinder of ferromagnetic material inside the solenoid? / e) Show a small current loop representing the magnetic dipole moment of an atom, and show the direction of the current on the loop. f) Is its energy maximum or minimum? 145
I46E
MAGNETIC TORQUE Show that the torque exerted on a permanent magnet of dipole moment m situated in a magnetic field B is m x B. See Prob. 1322. MEASUREMENT OF M The following method has been used to measure the magnetization M of a small sphere of material, induced by an applied uniform magnetic field B as in Fig. 149. In the figure. HG is a Hall generator (Sec. 10.8.1) with its current flowing parallel to B . Since Hall generators are sensitive only to magnetic fields perpendicular to their current flow, this one measures only the dipole field originating in the small sphere. 0
0
Figure 149 Setup for measuring the magnetization M induced in a small spherical sample of material by a uniform field B . The Hall generator HG is oriented so as to be sensitive to the vcomponent of the field originating in the sphere, and not to B„. See Prob. 146. 0
Problems
351
If the magnetization is Af, the magnetic moment m of the sphere is (4/3)7rP M, and it turns out that the xcomponent of the dipole field of the sample at HG is 3
3p m sin 0 cos 0 0
An
r
3
Thus, at a given angle 0, this field decreases as 1/r . At what angle 0 will the measured field be largest? 3
147
MICROMETEORITE DETECTOR Figure 1410 shows an instrument that has been devised to detect ferromagnetic micrometeorites falling to the ground. The instrument detects particles as small as 10 micrometers in diameter and measures their magnetic moment, which is a measure of their size and composition. A particle of radius b is magnetized in the field of the solenoid and acquires a magnetic moment in. It is then equivalent to a small coil having a magnetic moment m. The mutual inductance between the particle and the coil C of mean radius a = m. shown in the figure is then the M of Prob. 123. with nb N I 2
h h
I
Figure 1410 Section through a micrometeorite detector. Air is sucked through a funnel F. and a particle p is magnetized in the field B of a solenoid S. In passing through the tube, the particle induces a voltage in coil C. See Prob. 147.
352
Magnetic Fields:
VII
a) Show that the voltage induced in the coil is 3Li N a
2
0
a
where r is the velocity of the particle. b) Draw a curve of z(a + r ) as a function of z for a = 10 millimeters. The coordinate z should vary from —50 to + 5 0 millimeters. Since the velocity is constant, z is proportional to the time t. 2
148E
2
 5
2
MECHANICAL DISPLACEMENT TRANSDUCER One often wishes to obtain a voltage that is proportional to the displacement of an object from a known reference position. If there are no magnetic materials nearby, one can fix to the object a small permanent magnet and measure its field with a Hall generator, as in Fig. 1411. The Hall element is sensitive only to the vcomponent of the magnetic field and. from Prob. 146. 3// '"
xz
0
*
=
~4V(x + _  ) ' 2
2
52
a) Draw a curve of xz/(x + r ) ' as a function of z for v = 100 millimeters and for z = —50 to + 5 0 millimeters. b) Over what range does B deviate by less than 5% from the tangent to the curve at z = 0? The linear region is longer for larger values of x. but B decreases rapidly with x. 2
2
2
x
x
x
Figure 1411 Hall generator HG used as a mechanical displacement transducer. The moving part P slides along the zaxis. The Hall generator is oriented so as to monitor the xcomponent of the magnetic field of the small permanent magnet M fixed to P. See Prob. 148.
149
HG
MAGNETIZED DISK A thin disk of iron of radius a and thickness t is magnetized in the direction parallel to its axis. C 'alculate 11 and B on the axis, both inside and outside the iron.
Problems
I4I0E
353
TOROIDAL COIL WITH MAGNETIC CORE Compare the fields inside two similar toroidal coils, both carrying a current /, one with a nonmagnetic core and the other with a magnetic core. How are the equivalent currents on the magnetic core oriented with respect to those in the coil?
1411
EQUIVALENT CURRENTS A long wire of radius a carries a current / and is situated on the axis of a long hollow iron cylinder of inner radius b and outer radius c. a) Compute the flux of B inside a section of the cylinder / meters long. b) Find the equivalent current density on the inner and outer iron surfaces, and find the direction of these equivalent currents relative to the current in the wire. c) Find B at distances r > c from the wire. How would this value be affected if the iron cylinder were removed?
1412E
THE DIVERGENCE OF H We have seen in Sec. 8.2 that V • B is zero. This equation is valid even in nonlinear, nonhomogeneous, and nonisotropic media. Set B = u. u M. Under what conditions is V • H # 0? r
w
1413E
1414
n
THE MAGNETIZATION CURVE Use Fig. 145 to find the relative permeability of annealed pure iron at a magnetic induction of one tesla. ROWLAND RING A ring of ductile castiron (Fig. 144) has a major radius of 200 millimeters and a minor radius of 10 millimeters. A 500turn toroidal coil is wound over it. a) What is the value of B inside the ring when the current through the coil is 2.4 amperes? Use Fig. 145. b) A 10turn coil is wound over the first one. Calculate the voltage induced in it if the current in the large coil suddenly increases by a small amount at the rate of 10 amperes per second. Assume that u remains constant. r
1415E
1416
THE WEBER AMPERETURN Show that one weber ampereturn is one joule.
ROWLAND
RING
Figure 1412 shows how one can measure B as a function of H to obtain the magnetization and hysteresis curves of a ring of magnetic material. Winding a is fed by an adjustable power supply. As in Sec. 14.8. H =
NJJlnr.
Fhe voltage across winding b is integrated with the circuit of Prob. 532.
354
The crosssection of the ring is S and winding b has N turns. Show that b
RCV/N S
B =
b
if. at the beginning of the experiment. B = V = 0. The integrating circuit draws essentially zero current. Solution: The voltage across winding h is
dB
~dt
~dt'
=
(1)
since there is zero current in b, and hence no voltage d r o p caused by the resistance and inductance of the winding. Then, from Prob. 532, disregarding the negative sign. 1 v
=
p
dB
RC J° " Tt N
s
=
N
"
and B =
1417
TRANSFORMER HUM Why does a transformer hum?
RCV/N S. b
SB/RC
 2 )
355
2
2\ 40
I
I
20 0 20 H (ampereturns,meter)
I
I
40
Figure 1413 Hysteresis loop for the alloy Deltamax. See Prob. 1418.
1418E
POWER LOSS DUE TO HYSTERESIS Figure 1413 shows the hysteresis loop for a nickeliron alloy called Deltamax. What is the approximate value of the power dissipation per cubic meter and per cycle when the material is driven to saturation both ways?
1419
THE FLUXGATE MAGNETOMETER AND THE PEAKING STRIP A wire is would around a strip of Deltamax (Fig. 1413). forming a solenoid. A few turns of wire are then wound over the solenoid, and connected to an oscilloscope. An alternating current is applied to the solenoid, driving the Deltamax to saturation in both directions. a) What is the shape of the waveform observed on the oscilloscope? b) What happens to the pattern on the oscilloscope if a steady magnetic field is now applied parallel to the strip? It can be shown that, upon application of the steady field, the output voltage contains a component at double the frequency of the applied alternating current and proportional to the magnitude of the steady field. This is the principle of operation of the fluxgate magnetometer. Fluxgate magnetometers are often used on board satellites for measuring magnetic fields in outer space. They are useful from about 1 0 to 1CT tesla. _ 1 0
356
Magnetic Fields:
VII
The peaking strip also utilizes a solenoid wound on a material with a square hysteresis loop like Deltamax (Fig. 1413), and a secondary winding. It is used differently, however. The peaking strip is placed inside a solenoid whose current is adjusted so as to cancel the ambient field, making the pattern on the oscilloscope symmetrical. The peaking strip is used to measure magnetic inductions in the range of about 1 ( T to 1 0 " tesla. 6
2
CHAPTER MAGNETIC Magnetic
15 FIELDS:
VIII
Circuits
15.1
MAGNETIC
15.2 15.2.1
MAGNETIC CIRCUIT WITH Example: Electromagnet
15.3
MAGNETIC FORCES ELECTROMAGNETS
15.4
SUMMARY PROBLEMS
CIRCUITS AN AIR
EXERTED
BY
GAP
In general, it is not possible to calculate magnetic fields accurately, when magnetic materials are involved. There are several reasons for this, (a) The relation between H and B for ferromagnetic materials is nonlinear. It even depends on the previous history of the material, as we saw in Sec. 14.9. Also, ferromagnetic materials are often nonisotropic. (b) The iron cores that are used to confine and guide the magnetic flux are often quite inefficient. As we shall see, a large part of the flux can be situated outside the core, (c) Permanent magnets are not as simple as one would like them to be: their magnetization M is nonuniform and depends on the presence of neighboring magnetic materials. It is nonetheless necessary to be able to make approximate calculations. The calculation serves to design a model on which magnetic fields or magnetic forces can be measured, and which can then be modified to give the final design.
15.1
MAGNETIC
CIRCUITS
Figure 151 shows a ferromagnetic core around which is wound a short coil of N turns carrying a current /. We wish to calculate the magnetic flux O through the core. In the absence of ferromagnetic material, the lines of B are a s shown in the figure. At first sight, one expects the B inside the core to be much larger close to the winding than on the opposite side. This is not the case, however, and B is of the same order of magnitude at all points within the ferromagnetic material.
359
Figure 151 Ferromagnetic toroid with concentrated winding. The lines offeree shown are in the plane of the toroid and are similar to those of Fig. 86. They apply only when there is no iron present.
This can be understood as follows. The magnetic induction of the current / magnetizes the core in the region near the coil, and this magnetization gives equivalent currents that both increase B and extend it along the core. This further increases and extends the magnetization, and hence B, until the lines of B extend all around the core. Of course some of the lines of B escape into the air and then return to the core to pass again through the coil. This constitutes the leakage flux that may, or may not, be negligible. For example, if the toroid is made up of a long thin wire, most of the flux leaks across from one side of the ring to the other and the flux at P is negligible compared to that near the coil. Let us assume that the crosssection of the toroid is large enough to render the leakage flux negligible. Then, applying Ampere's circuital law, Eq. 1418, to a circular path of radius r going all around inside the toroid, (151) InrB
r
(152)
o
NI,
B
H fi NI/2nr.
fi ft
r
0
(153)
360
Magnetic Fields:
VIII
Taking R to be the radius corresponding to the average value of B, and R to be the minor radius of the toroid. {
2
n n nRlNII2nR r
0
(154)
v
The flux through the core is therefore the same as if we had a toroidal coil (Sec. 9.1.2) of the same size and if the number of ampereturns were increased by a factor of a . In other words, for each ampereturn in the coil there are u — 1 ampereturns in the core. The amplification can be as high as 10 . This equation shows that the magnetic flux is given by the magnetomotance NI multiplied by the factor r
5
r
UrUonRl/lnRu which is called the permeance of the magnetic circuit. The inverse of the permeance is called the reluctance. Thus ^ /V/ NI (1) = — = St InRJp^nRV
(155) ' '
and the reluctance is
9t =
InRJurHouRl.
(156)
Reluctance is expressed in henrys" The analogy with Ohm's law is obvious: if an electromotance V were induced in the core, the current would be
r (157)
InRxlanRj
where 2nR\lanR\ is the resistance of the core. Thus the corresponding quantities in electric and magnetic circuits are as follows: Current / Current density J Conductivity a
Magnetic flux (P Magnetic induction B Permeability p = p p r
0
15.2 Magnetic Circuit with an Air Gap
Electromotance i Electric field intensity E Conductance G = l/R Resistance R
361
Magnetomotance NI Magnetic field intensity H Permeance \f. A Reluctance 88.
r
J
There is one important difference between electric and magnetic circuits: the magnetic flux cannot be made to follow a magnetic circuit in the manner that an electric current follows a conducting path. Indeed, a magnetic circuit behaves much as an electric circuit would if it were submerged in t a p water: part of the current would flow through the components, and the rest would flow through the water. If a magnetic circuit is not properly designed, the leakage flux can easily be an order of magnitude larger than that flowing around the circuit.
15.2
MAGNETIC
CIRCUIT
WITH AN AIR GAP
Figure 152 shows a circuit with an air gap whose crosssection is different from that of the softiron yoke. Each winding provides NI/2 ampereturns. We wish to calculate the magnetic induction in the air gap. We assume that the leakage flux is negligible. As we shall see, this assumption will result in quite a large error. Applying Ampere's circuital law to the circuit, we see that NI = Hik + H l , g g
(158)
where the subscript i refers to the iron yoke and g to the air g a p ; /, and l are the path lengths. The path length /, in the iron can be taken to be the length measured along the center of the crosssection of the yoke. If we neglect leakage flux, the flux of B must be the same over any crosssection of the magnetic circuit, so g
B A = BgAg, i
i
(159)
where A and A are, respectively, the crosssections of the iron yoke and of the air gap. t
g
362
Soft iron yoke
• Ii
•11 Figure 152 Electromagnet. The coils have been cut out to expose the iron yoke.
Combining these two equations.
NI,
(1510)
and the magnetic flux is
NI
I;
(1511)
The magnetic flux is therefore equal to the magnetomotance divided by the sum of the reluctances of the iron and of the air gap. This is a general law: reluctances in series in a magnetic circuit add in the same way as resistances in series in an electric circuit; permeances in parallel add like conductances in parallel.
15.2 Magnetic Circuit with an Air Gap
363
Since we have neglected leakage flux, the above equation can only provide an upper limit for . = IJi.irP.QAj be the reluctance of the iron, and &
N o w let
=
g
l lp A g
0
g
be the reluctance of the air gap. N o t e that
//,/,
Hl
g g
=
.^,
(1512)
=
® 4>.
(1513)
g
If 09 « dig, as is usually the case, since p t
r
»
1, Eqs. 158 and 1511
become NI
% Hl,
*
EXAMPLE:
(1514)
g g
* NIp A /l . 0
g
(1515)
g
ELECTROMAGNET
Let us calculate B and the stored energy in the electromagnet of Fig. 152. If there is a total of 10,000 turns in the two windings, and setting / = 1.00 ampere. A = 10* square millimeters, A = 5 x 1 0 square millimeters, p = 1,000,/, = 900 millimeters, and /,, = 10 millimeters, then x
3
g
r
10
<1>
4
0.9
1516)
10
10 x 4;: x 1 0 " x 1 0 ~ 3
7
2
4TT X 1 0 " x 5 x 1 0 " 7
= 6.0 x 1 0 " weber,
(1517)
3
B = 6.0 x 1 0 " / 5 x 1 0 " = 1.2 teslas. 3
3
3
g
(1518)
The selfinductance is L = NQ/I = 10 x 6.0 x 10~ /1.00 = 60 henrys. 4
3
15191
The length / is measured along the middle of the crosssection of the yoke. The stored energy is f
(1/2)L/ = (1/2) x 60 x 1.00 = 30 joules. 2
(1520)
364
Magnetic Fields:
VIII
In this particular case the leakage flux is 70% of the flux in the gap. In other words, the magnetic induction in the gap is not 1.2 teslas, but only 1.2 1.7 = 0.71 tesla. There exist empirical formulae for estimating leakage flux for simple geometries.
15.3
MA GNETIC FORCES EXER TED BY ELECTROMAGNETS Electromagnets are commonly used to actuate various mechanisms such as switches, valves, and so forth. A switch activated by an electromagnet is called a relay. These electromagnets comprise a coil and an iron core that is made in two parts, one fixed and one movable, called the armature, with an air gap between the two. The coil and its core together are usually termed a solenoid. The magnetic force is attractive. In the simplest cases, one has a flat air gap, perpendicular to B. Then the magnetic force of attraction is B /2p: times the crosssection of the gap (Sec. 13.2) where B is calculated as in Sec. 15.2. If the solenoid is energized with direct current, then B is approximately proportional to the inverse of the gap length, and since the force is proportional to B , the force increases rapidly with decreasing gap length. The air gap is often designed so that the attractive force will vary with the gap length in some prescribed way. The design of the magnetic circuit is largely empirical. 2
0
2
15.4
SUMMARY The concept of magnetic circuit is widely used whenever the magnetic flux is guided mostly through magnetic material. We then have a magnetic equivalent of O h m ' s law: O = NI/M, where NI is the magnetomotance of the magnetic circuit.
(155)
of the energizing coil and J ? is the reluctance
Problems
365
The correspondence between electric and magnetic circuits is as follows: Current / Current density J Conductivity rr Electromotance f~ Electric field intensity E Conductance Resistance R
Magnetic Flux