continuity law [(8.0.1) and (8.0.2)], as well as the associated continuity conditions [(8.0.3) and (8.0.4)], are satisfied by an interior magnetic field intensity that is uniform and an exterior one that is zero.
| (a) | Show that Ampère's differential law and the magnetic flux |
| (b) | What is the interior field? |
| (c) | A is continuous at r = a because otherwise the H field would have a singularity. Determine A. |
d/2 and two in the negative z direction at x = d/2, y
= 0. (With the line charges representing line currents, the
cross-section is the same as shown in Fig. P4.4.3.)
Show that in the limit where r \gg d, A_z = - (
o
id2/4
)(r-2) cos 2
. (Note that distances must be
approximated accurately to order d2.)
that is much less than the inner radius b and outer
radius a. It supports a current density J = J_o i_.
Show that along the z axis,
direction.
With the wire driven by a current source, the resulting current
distribution is a surface current at r = R having the density
K = K_o sin 
i_, where K_o is a given constant.
There are no other currents. Show that at the center of the
coil, H = (2K_o/3)iz.
, where
K_o is again a constant. Find H at the center of the coil.
, where J_o is a given constant. Outside
this region is free space and no other sources of H. Determine
H at the origin.
a. Show that
the lines \Psi = const in the x - y plane are circles.
o H_o is trapped
between perfectly conducting plane parallel plates that extend to
infinity to the left and right. At the origin on the lower plate is a
perfectly conducting half-cylinder of radius R.
| (a) | Show that if s \gg R, then |
| (b) | Show that a plot of H would appear as in the left half
of Fig. 8.4.2 turned on its side.
|
. The magnetic field
intensity is uniform far from the hemisphere. Determine \Psi and
H.
to z = +\infty at (x, y) = (0, -h) is
a wire. The wire is parallel to a perfectly conducting plane at y = 0.
When t = 0, a current step i = I u_-1(t) is applied in the +z
direction to the wire.
| (a) | Show that in the region y < 0, |
| (b) | Show that the surface current density at y = 0 is
K_z = - ih/ (x2 + h2).
|
z directions is shown in Fig. P8.4.6. Surrounded by free space,
a sheet of current has the surface current density K_o iz
uniformly distributed between x = b and x = a. The plane x = 0
is perfectly conducting.
| (a) | Determine \Psi in the region 0 < x. |
| (b) | Find K in the plane x = 0. |
| (a) | Given that the winding carries a current i, show that |
| (b) | Show that the inductance per unit length of the winding is
L = o N2/8.
|
| (a) | What are the continuity |
| (b) | Find \Psi, and hence H, in regions (a) and (b) outside and inside the winding, respectively. |
| (c) | With the understanding that the rotor is
wound using one wire, so that each turn is in series with the next and
a wire carrying the current in the +z direction at returns the
current in the -z direction at -, what is the inductance of the
rotor coil? Why is it independent of the rotor position _o?
|
of r = R, A_z must be continuous there.
| (a) | Show that if H_ is to be finite in the neighborhood
|
| (b) | Show that A is given by
 |
| (c) | The loop designated by C' in Fig. 1.4.2 has a
length l in the z direction, an inner leg at r = 0,
and an outer leg at r = a > R. Use A
to show that the flux linked is
 |
| (a) | Determine A_z in the regions r < b and b < r < a. |
| (b) |
Use A to determine the flux linked by a closed rectangular loop
having length \l in the z direction and each of its four sides in a
plane of constant . Two of the sides are parallel to the z axis,
one at radius r = c and the other at r = 0. The other two, respectively,
join the ends of these segments, running radially from r = 0 to r = c.
|
o H = \mo [H_r(r, z)ir + H_z(r,
z)iz].
That is, the magnetic flux density is axially symmetric and does not
have a component.
| (a) | Show that |
| (b) | Show that the flux passing between contours at r = a and
r = b is
 |
| (a) | derive the vector potential, (20), |
| (b) | Show that the current is as given by (21). |
| (c) | In the limit where b/a \gg 1, show that the response has the dependence on b/a shown in the plot of Fig. 8.6.11. |
| (d) | Show that in the opposite limit, where
b/a \ll 1, the total current in the lower plate (21)
is consistent with a magnetic field intensity between the upper and
lower plates that is uniform (with respect to y) and hence equal to
(\Lambda /bo )ix. Note that
 |
that is small compared to all
dimensions. Passing through these gaps is a magnetic flux (per unit
length in the z direction) \Lambda (t). One method of solution
is suggested by Example 6.6.3.
(b) to the right and left, respectively, of the plane x = 0.
| (a) | Find A in regions (a) and |
| (b) | Sketch H. |
x/a)iz, where i
is the current carried by each wire and |n_o sin ( x/a)| is the
number of turns per unit area. This region is surrounded by perfectly
conducting walls at y = 0 and y = b and at x = -a and x = a.
The length l in the z direction is much greater than either a or
b.
| (a) | Show that |
| (b) | Show that the inductance of the winding is
 |
| (c) | Sketch H. |
| (a) | Determine A. |
| (b) | Assume that all the wires are connected to the wall by a terminating plate at z = l and that each is driven by a current source i(t) in the plane z = 0. Note that it has been assumed that each of these current sources is the same function of time. What is the voltage v(x, y, t) of these sources? |
| (a) | Determine A. |
| (b) | Find the inductance L.
|
A superposition integral for the field intensity rather than the vector potential proves convenient, especially if the fields are determined numerically from given current distributions. This superposition integral is provided by Biot-Savart's law in Sec. 8.3.
The determination of fields from given current distributions is important for at least two reasons. First, by contrast with EQS systems, where the charges seldom have known distributions, the use of conductors wrapped with insulation (wires) makes feasible the imposition of prescribed current distributions. In the absence of magnetizable materials such as iron, electromagnets are an example. The Biot-Savart law is often used to design electromagnets.
A second reason for seeking solutions to the magnetic version of the superposition integral is its role as the particular solution in boundary value problems. Typical of this class of situations is a time varying magnetic field within some region bounded at least in part by a highly conducting material. For reasons cited in Sec. 8.4, currents are induced in the surfaces of ``perfect'' conductors in such a way that the normal magnetic flux density is zero. The homogeneous solutions needed to satisfy boundary conditions are familiar from Chap. 5, because they are solutions to Laplace's equation. The method of images and boundary value point of view needed to solve such problems is the subject of Sec. 8.5.
In many practical configurations, the currents are confined to
relatively thin regions where they can be modeled by surface currents
or line currents. In these situations, where H is irrotational
in the volumes of
interest, it is natural to exploit a scalar potential that is
analogous to the electric potential 
introduced in Chap. 4. This
scalar magnetic potential \Psi, which is the theme of Secs. 8.6 and
8.7, is used extensively in Chap. 9 for systems containing
magnetizable materials.
The vector potential is the natural variable for evaluating the
flux passing through a surface. (We will see in Sec. 8.4 that
the time-rate-of-change of flux determines the terminal voltage of an
inductor.) Integration of (1) over the open surface S of Fig.
8.1.1 gives
ds around the contour enclosing the
surface.
), then
the polar components of the flux density would follow from (1)
using the curl operator in polar coordinates.
The magnetic field intensity for the line current i of Fig. 8.1.3 was determined in Sec. 1.4 using Ampère's integral law. Thus, the magnetic flux density is given by (1.4.10) asThe vector potential is determined by first writing (1) in polar coordinates.
Because there is no r component of the flux density, it follows that A_z is independent of To illustrate the use of the vector potential to evaluate the net flux through a surface, consider the flux through the coil shown in Fig. 1.7.2a. The contour running parallel to the z axis in the z direction is at a radius d from the wire while that returning the current in the -z direction is at radius R. The length in the z direction is l. Thus, (7) gives a net flux of Note that this result is in agreement with (1.7.5), which was obtained by evaluating the specific surface integral.
o H. For the two-dimensional fields pictured by A_z in
Fig. 8.1.2, think of (b) as held fixed and let the line denoted by (a)
trace out a surface of constant A_z. The flux per unit length
passing between the line denoted by (b) and any line in this surface
of constant A_z is the same. Thus, there is no flux density normal
to a surface of constant A_z. That is, lines of constant A_z
are the lines of magnetic flux density.
For the line current of Fig. 8.1.3, the surfaces of constant A_z given by (10) appear in a constant z plane as concentric circles, as shown in Fig. 8.1.4. With arrows added to indicate direction, these represent the flux density. Drawn for equal increments of A_z, the magnetic flux trapped between pairs of lines is the same. Thus, as the distance between lines increases, the flux density decreases.
directions only, the
vector potential again has a single component.
multiplied by the circumferences. The
contour is closed by adjacent oppositely directed segments joining
points (a) and (b) in a plane of constant . Thus, the
contributions to the line integral of (4) from these segments
cancel, even if A had components in the direction of ds on these
segments. Thus, the net flux through the annulus is simply the
axisymmetric stream function \Lambda at (a) relative to that
at (b)
) is called Stokes' stream function.where Lines of flux density are tangential to the axisymmetric surfaces of constant \Lambda. Just as A_z provides a ready visualization of the flux lines in two-dimensions, \Lambda portrays the axisymmetric flux lines.
Example 8.7.1. Dipole Field of a Current Loop
A three-dimensional magnetic dipole field is observed if a small circular current loop sustains a circulating current i. By ``small'' we mean that the radius R is much less than the distance to the point of observation. The configuration, shown in Fig. 8.2.1, has rotational symmetry with respect to the z axis, so the magnetic field intensity is independent of
In the evaluation of (6), the source is confined to the wire which is assumed to have a cross-sectional area, a, of dimensions small compared to the radius R. With da' used to denote an incremental area of the wire perpendicular to the current, the incremental source in (6) can be written as Because the current density is confined to a wire having a cross-section that is much smaller than the radius R, integration over the cross-section of the wire can be taken as a multiplication of the current density by the area with distance |r - r' | essentially held fixed. Thus, (6) becomes where (r2 = x2 + y2 + z2).
Because r \gg R, we can expand the source-observer distance and drop R2 compared to r2 in this expression. Thus,
6 [(a + b)n \approx an + nban-1]and (9) becomes Only the third and fifth terms in the integrand contribute to the integral, which is Here, the expression has been written so as to recognize that the Cartesian unit vectors combine to become the unit vector in the Written in terms of the magnetic dipole moment m, defined as the product of the area enclosed by the loop and the current i, this is the desired dipole vector potential. The magnetic field intensity is obtained by taking the curl of this expression, as called for by (8.1.1) in spherical coordinates.
The lines of magnetic field flux density, and hence H, are tangent to the surfaces of constant \Lambda = 2 o
Two-Dimensional Current and Vector Potential Distributions
A current density that is z directed but independent of z gives rise to a vector potential that is also in the z direction and dependent only on x and y. This is evident from the vector Poisson's equation, which reduces to (4). It also follows from the a two dimensional version of the superposition integral.Observe that the vector potential for a z-directed uniform line current on the z axis is given by (8.1.10). Positioned at an arbitrary source location denoted by r' and illustrated in Fig. 8.2.3, this line current results in the vector potential
Note that, as in the case of the two-dimensional superposition integral for the electric potential, (4.5.20), the position vectors r and r' are two dimensional in the sense that they lie in any x - y plane. The current i is now replaced by the current density at the source coordinate multiplied by the differential area da'. The latter could be dx' , dy', for example. The total vector potential is then the superposition of contributions from all of the current distribution. Thus, the two-dimensional vector potential superposition integral is In essence, this represents the z component of the three dimensional superposition integral, (6), with the integration on z' carried out. Note that the two-dimensional current distribution is automatically solenoidal.
With the identification
/
Thus, the z component of the vector Poisson's equation reduces to Laplace's equation.
Because the fields are two-dimensional, A_z is the only component of A. In terms of A_z, the magnetic flux density can be written in polar coordinates (8.1.9) or Cartesian coordinates (8.1.6)
The Cartesian expression is useful for recognizing the form of A_z as y It follows that as r In terms of the vector potential, the condition that there be no flux density normal to the cylinder surface, where r = R, is met if These boundary conditions are met by taking a linear combination of the solutions to (4) that have the same To match the uniform vector potential at infinity, C = - Remember, the field lines follow the ``lines'' of constant A_z, which are shown in Fig. 8.4.2. The field lines are excluded by the cylinder.
Example 8.7.2. Terminal Voltage Computed From Vector Potential
The perfectly conducting cylinder in an imposed time-varying magnetic field intensity H_o (t) from Example 8.4.1 is shown in Fig. 8.4.5. An N turn coil has been added that has legs of length l in the z direction closed at the ends by segments perpendicular to the z axis. Thus, wires run in the +z direction at (r = a,
The total flux linked is N times the flux linked by one turn. In view of (8.1.7), is therefore given by simply
The specific A_z is given by (10), so (16) becomes Thus, the terminal voltage follows from (14) as The terminal voltage reflects the rate of change of the flux linked by the coil, regardless of the origins of that flux. In this case, it is due to the currents used to excite the electromagnet responsible for H_o and the currents induced in the cylinder to expel the flux from its interior. We consider next cases where the flux is due to current in the coil itself.
Boundary Value Solution By Superposition of Sources
By judicious arrangements of currents, it is possible to obtain solutions to practical boundary value problems. In Chap. 4, superposition of charges was used to solve boundary value problems involving perfect conductors in EQS systems. As we shall now show, a family of solutions for two-dimensional MQS systems can be obtained by superposition of the fields due to current sources.In two-dimensional configurations, any surface of constant A_z can be replaced by the surface of a perfect conductor. Moreover, in the free space region between conductors, A_z satisfies Laplace's equation. Thus, any two-dimensional configuration from Chaps. 4 and 5 can be replaced by one where the potential lines are field lines. The equipotential surfaces of the EQS perfect conductors become the perfectly conducting surfaces of an MQS system with tangential lines of magnetic field intensity.
Illustration. Field Trapped Between Hyperbolic Perfect Conductors
The two-dimensional potential distribution of Example 4.1.1 suggests the vector potential A_z =
This solution is the superposition of the fields of four line currents. Two directed in the +z direction are at infinity in the first and third quadrants while two in the -z direction are in the second and third quadrants.
Example 8.7.4. Field and Inductance of Oppositely Directed Currents in Parallel Perfectly Conducting Cylinders
The cross-section of a pair of parallel perfectly conducting cylinders that extend to
Example 4.6.3 suggests our strategy. Instead of superimposing the potentials With the identification of variables
7 This is one of the few places where there is the possibility of confusingThis expression is identical to that for the antidual EQS configuration, (4.6.18). We can conclude that the line currents should be located at a = (l2 - R2)1/2 and that the constant k used in that deduction (4.6.20) is identified using (26).
Thus, the surfaces of constant A_z are circular cylinders and represent the field lines shown in Fig. 8.4.8.
The inductance per unit length L is now deduced from (27).
In the limit where the conductors represent wires that are thin compared to their spacing, the inductance per unit length of (28) is approximated using (4.6.28).
Once the vector potential has been determined, it is possible to evaluate the distribution of current density on the conductors. Note that the currents tend to concentrate on the inside surfaces of the conductors, where the magnetic field intensity is more intense. We are one step short of a general relationship between the capacitance per unit length and inductance per unit length of a pair of parallel perfect conductors, regardless of the cross-sectional geometry. With
where c is the velocity of light (3.1.16). Note that inductance per unit length of parallel circular conductors given by (28) and the capacitance per unit length for the same conductors under ``open-circuit'' conditions (4.6.27) satisfy the general relation of (30).
Method of Images
In the presence of a planar perfect conductor, the zero normal flux condition can be satisfied by symmetrically mounting source distributions on both sides of the plane. This approach is familiar from Sec. 4.7, where the boundary condition required a plane of symmetry on which the tangential electric field was zero. Here, we require that the field intensity be tangential to the boundary. For two-dimensional configurations, the analogy between the electric potential and A_z makes the image method of Sec. 4.7 directly applicable here. In both cases, the symmetry plane is one of constant potential (The most obvious example is an infinitely long line current at a distance d/2 from a perfectly conducting plane. If Fig. 4.7.1 were a picture of line charges rather than point charges, this would be the antidual situation. The appropriate image is then an oppositely directed line current located at a distance d/2 to the other side of the perfectly conducting plane. By making a pair of symmetrically located line currents the image for this pair of currents, the boundary condition on yet another plane can be satisfied, the analog to the configuration of Fig. 4.7.3. The following demonstration is intended to emphasize that the perfectly conducting symmetry plane carries a surface current that terminates the field in the region of interest. \begindemo2Surface Currents Induced in Ground Plane by Overhead Conductor The metal cylinder mounted over a metal ground-plane shown in Fig. 8.4.10 is familiar from Demonstration 4.7.1. Rather than being insulated from the ground plane and driven by a voltage source, this cylinder is shorted to the ground plane at one end and driven by a current source at the other. The height l is small compared to the length so that the two-dimensional model describes the field distribution in the mid-region.
A probe is used to measure the magnetic flux density tangential to the metal ground-plane. The distribution of this field, and hence of the surface current density in the adjacent metal, can be determined by recognizing that the ground plane boundary condition of no normal flux density is met by symmetrically mounting a distribution of oppositely directed currents below the metal sheet. This is just what was done in determining the fields for the pair of cylindrical conductors, Fig. 8.4.7. Thus, (25) is the image solution for the region x \geq 0. In terms of x and y, The flux density tangential to the ground-plane is Normalized to H_o = i/ To appreciate the physical origins of this distribution, a dc current source is used in place of the ac source. Then, the distribution of current in the sheet is dictated by the rules of steady conduction as enunciated in the first half of Chap. 7. If the sheet is long enough compared to its width, the current is uniformly distributed over the sheet and over the cross-section of the cylinder. By contrast with the high-frequency a-c case where the field is terminated by surface currents in the sheet, the magnetic field now extends below the sheet.
The method of images is not restricted to the two-dimensional
situations where there is a convenient analogy between and
A_z. In the following example, of a three-dimensional field,
the symmetry conditions are viewed without the aid of the vector
potential.
A current loop with time varying current i is mounted a distance h above a perfectly conducting plane, as in Fig. 8.4.11. Its axis is inclined at an angle
To satisfy the boundary condition in the plane of the perfectly conducting sheet, an image loop is mounted as shown in Fig. 8.4.12. For each current segment in the actual loop, there is a segment in the image loop giving rise to an oppositely directed vertical component of H. Thus, the net normal flux density in the plane of the perfect conductor is zero.
The MQS configurations of this section are approached as were the EQS systems of Secs. 5.1 and 5.6. The volume V of interest is bounded by perfect conductors. There can also be a distribution of prescribed current density within V, perhaps by means of windings driven by current sources.As in Secs. 5.1 and 5.6, the magnetic field intensity within V is divided into particular and homogeneous parts.
Although it is often easier to find the particular solution by a more direct method, the given current distribution can be used in the vector potential superposition integral to find a particular solution A_p that satisfies Ampère's law and the flux continuity condition at each point within V. Alternatively, H_p can be found directly from the Biot-Savart law, (8.3.7).
Note that this integration is confined to the volume V, where the distribution of current density is known. Then, the homogeneous solution satisfies the source-free forms of Ampère's law and the flux continuity law.
In this chapter, we have emphasized the vector potential as a way of representing solenoidal fields. In Chap. 5, we used the scalar potential to represent irrotational fields. By design, the homogeneous field is both solenoidal and irrotational, so we now have a choice of representations, at least for the homogeneous part. The vector potential is convenient if the flux is to be found and if the analogies discussed in Sec. 8.4 are to be exploited. However, unless A has only one component, as it fortunately does in many important cases, it only replaces one vector with another in the field description. A scalar potential representation of H preserves the advantage of letting a scalar represent a vector in three dimensional configurations. Although we could now introduce the scalar potential, we will find that it is particularly useful in dealing with magnetization problems. Thus, the magnetic scalar potential will be introduced in Sec. 8.6 and used extensively in Chap. 9 and we proceed here to use A_h to represent the homogeneous solution. The flux continuity law is automatically satisfied by letting and substitution into (3) shows that which of course is the vector Poisson's equation, (8.2.3), with the current density set equal to zero.
Example 8.8.1. Inductive Attenuator
The cross-section of two conducting electrodes that extend to infinity in the
The magnetic fields are two-dimensional and there are no sources in the region of interest. Thus, B can be represented in terms of A_z, which satisfies the z component of (6).
The walls are perfectly conducting in the sense that they are modeled as having no normal B (8.4.1). This means that A_z is constant on these walls (8.4.3). We define A_z to be zero on the vertical and bottom walls. Then, A_z must be equal to \Lambda on the upper electrode so that the flux per unit length in the z direction through the gaps is \Lambda. (Note that the z axis is out of the page, so that according to (8.1.7), \Lambda is positive if directed as shown in the figure.) With the identification of variables the boundary value problem is now formally identical to the EQS capacitive attenuator that was the theme of Sec. 5.5. Thus, it follows from (5.5.9) that The lines of magnetic flux density are the lines of constant A_z. Thus, they are the equipotential ``lines'' of Fig. 5.5.3, shown in Fig. 8.5.1 with arrows added to indicate the field direction. Remember, there is a z-directed surface current density that is proportional to the tangential field intensity. For the flux lines shown, K_z is out of the page in the upper electrode and returned into the page on the side walls and (to an extent determined by b relative to a) on the bottom wall as well. From the cross-sectional view given by Fig. 8.5.1, the provision for the current through the driven plate at the top to recirculate through the side and bottom plates is not shown. The following demonstration emphasises the implied current paths at the ends of the configuration.
Demonstration 8.8.1. Inductive Attenuator
One configuration described by Example 8.5.1 is shown in Fig. 8.5.2. Here, the upper plate is shorted to the adjacent walls at the near end and driven at the far end through a step-down transformer by a 20 kHz oscillator. The driving voltage v(t) at the far end of the upper plate is measured by means of an oscilloscope. The lower plate is shorted to the side-walls at the far end and also connected to these walls at the near end, but in such a way that the induced current i(t) can be measured by means of a current probe.The walls and upper and lower plates are made from brass or copper. To insure that the resistances of the plate terminations are negligible, they are made from heavy copper wire with the connections soldered. (To make it possible to adjust the spacing b, braided wire is used for the shorts on the lower electrode.) If the length w of the plates in the z direction is large compared to a and b, H within the volume follows from (10). The surface current density K_z in the lower plate then follows from evaluation of the tangential H on its surface. In turn, the total current follows from integration of K_z over the width, a, of the plate.
With the objective of relating this current to the driving voltage, note that (8.4.14) so that with the driving voltage a sinusoid of magnitude V, Thus, in terms of the driving voltage, the output current is i_o sin ( t) where it follows from (11) and (13) that
We have found that the output current, normalized to U, has the dependence on spacing between upper and lower plates shown by the insert to Fig. 8.5.2. With the spacing b small compared to a, almost all of the current through the upper plate is returned in the lower one and the field between is essentially uniform. As the spacing b becomes comparable to the distance a between the side walls, most of the current through the upper electrode is returned in these side walls. Thus, for large b/a, the normalized output current of Fig. 8.5.2 reflects the exponential decay in the -y direction of the field.
Value is added to this demonstration if it is compared to its EQS antidual, Demonstration 5.5.1. For the EQS configuration, the lower plate was properly constrained to essentially the same potential as the walls by connecting it to these side walls through a resistance (which was then used to measure the induced current). Up to frequencies above 100 Hz in the EQS case, this resistance could be as high as that of the oscilloscope (say 1 M\Omega) and still constrain the lower plate to essentially the same zero potential as the walls. In the MQS case, we did not use a resistance to connect the lower plate to the side walls (and hence provide a means of measuring the output current) because that resistance would have had to have been extremely low, even at 20 kHz, to prevent flux from leaking through the gaps between the lower plate and the side walls. We used the current probe instead. The effects of finite conductivity in MQS systems are the subject of Chap. 10. In a final example, we exemplify how the particular and homogeneous solutions are combined to satisfy boundary condtions while also illustrating how the inductance of a distributed winding is determined.
Example 8.8.2. Field and Inductance of Distributed Winding Bounded by Perfect Conductor
The cross-section of a distributed winding of radius a is shown in Fig. 8.5.3. It consists of turns carrying current i in the +z direction at a location (r,
The total number of wires N in the left ball of the coil is so that the current density is The windings are very long in the z direction so that effects of the end-turns are ignored and the fields taken as independent of z. The coil is bounded at r = a by a perfect conductor. With the following steps we determine the field distribution throughout the winding and finally its inductance.
The vector potential must satisfy (8.2.4) and is z-independent. In polar coordinates:
First, we look for a particular solution. If it is to take a product form, inspection shows that sin The magnetic flux density normal to the perfectly conducting surface at r = a must be zero, so the total vector potential must be constant there. It follows that, at r = a, the homogeneous solution, A_zh must be the negative of the particular solution, A_zp.
A linear combination of the two solutions to Laplace's equation that have the same The coefficient D must be zero so that the solution is finite at the origin. Then, the coefficient C is adjusted to make (21) satisfy the condition of (20). Hence, the sum of the particular and homogeneous solutions is A graphical representation of what has been accomplished is given by Fig. 8.5.4, where the surfaces of constant A_z (and hence the lines of field intensity) are shown for the particular, homogeneous and total solutions.
Each turn of the coil links a different magnetic flux. Thus, to determine the total flux linked by the distribution of turns, it is necessary to carry out an integration. To do this, first observe that the flux linked by the turns with their right legs within the area rd Here, l is the length of the system in the z direction. The total flux linked by all of the turns is obtained by integrating over all of the turns.
Substitution for A_z from (22) then gives where L will be recognized as the inductance.
Example 8.8.1. Vector Potential of a Line Current
The magnetic field intensity for the line current i of Fig. 8.6.3 was determined in Sec. 1.4 and in Example 8.3.1 using Ampère's integral law. The magnetic flux density is given by (1.4.10) asThe vector potential is determined by expressing curl A in polar coordinates.
Because there is no r component of the flux density, it follows that A_z is independent of To illustrate the use of the vector potential to evaluate the net flux through a surface, consider the flux through the coil shown in Fig. 1.7.2a. The contour running parallel to the z axis in the z direction is at a radius d from the wire while that returning the current in the -z direction is at radius R. The length in the z direction is l. Thus, (5) gives a net flux of Note that this result is in agreement with (1.7.5), which was obtained by evaluating the specific surface integral.
