The Vector Potential and the Vector Poisson Equation

A general solution to (8.0.2) is

where

is theAvector potential. Just asis the "integral" of the EQS equationE= -gradcurl, so too is (1) the "integral" of (8.0.2). Remember that we could add an arbitrary constant to without affectingE= 0. In the case of the vector potential, we can add the gradient of an arbitrary scalar function toEwithout affectingA. Indeed, becauseHx (\nabla ) = 0, we can replacebyA. The curl ofA^{'}=A+is the same as ofA.A^{'}

We can interpret (1) as the specification of

in terms of the assumedly known physicalAfield. But as pointed out in the introduction, to uniquely specify a vector field, both its curl and divergence must be given. In order to specifyHuniquely, we must also give its divergence. Just what we specify here is a matter of convenience and will vary in accordance with the application. In MQS systems, we shall find it convenient to make the vector potential solenoidalA

Specification of the potential in this way is sometimes called setting the gauge, and with (2) we have established the

Coulomb gauge.We turn now to the evaluation of

, and henceA, from the MQS Ampère's law and magnetic flux continuity law, (8.0.1) and (8.0.2). The latter is automatically satisfied by letting the magnetic flux density be represented in terms of the vector potential, (1). Substituting (1) into Ampère's law (8.0.1) then givesH

The following identity holds.

The reason for defining

as solenoidal was to eliminate theAterm in this expression and to reduce (3) to theAvector Poisson's equation.

The vector Laplacian on the left in this expression is defined

in Cartesian coordinatesas having components that are the scalar Laplacian operating on the respective components of. Thus, (5) is equivalent to three scalar Poisson's equations, one for each Cartesian component of the vector equation. For example, theAzcomponent is

With the identification of

Aand_{z}, this expression becomes the scalar Poisson's equation of Chap. 4, (4.2.2). The integral of this latter equation is the superposition integral, (4.5.3). Thus, identification of variables gives as the integral of (6)_{o}J_{z}/_{o}

and two similar equations for the other two components of

. Reconstructing the vectorAby multiplying (7) byAand adding the correspondingi_{z}xandycomponents, we obtain thesuperposition integral for the vector potential.

Remember,

is the coordinate of the current density source, whiler^{'}is the coordinate of the point at whichris evaluated, the observer coordinate. Given the current density everywhere, this integration provides the vector potential. Hence, in principle, the flux densityAis determined by carrying out the integration and then taking the curl in accordance with (1)._{o}H

The theorem at the end of Sec. 8.0 makes it clear that the solution provided by (8) is indeed unique when the current density is given everywhere.

In order that

xbe a physical flux density,Acannot be an arbitrary vector field. BecauseJ(r)div (curl)of any vector is identically equal to zero, the divergence of the quasistatic Ampère's law, (8.0.1), gives( xand thusH) = 0 = \nablaJ

The current distributions of magnetoquasistatics must be solenoidal.Of course, we know from the discussion of uniqueness given in Sec. 8.0 that (9) does not uniquely specify the current distribution. In an Ohmic conductor, stationary current distributions satisfying (9) were determined in Secs. 7.1-7.5. Thus, any of these distributions can be used in (8). Even under dynamic conditions, (9) remains valid for MQS systems. However, in Secs. 8.4-8.6 and as will be discussed in detail in Chap. 10, if time rates of change become too rapid, Faraday's law demands a rotational electric field which plays a role in determining the distribution of current density. For now, we assume that the current distribution is that for steady Ohmic conduction.

\sectnonumTwo-Dimensional Current and Vector Potential Distributions

Suppose a current distribution

exists through all of space. Then the vector potential isJ=iJ_{z}_{z}(x, y)zdirected, according to (8), and itszcomponent obeys the scalar Poisson equation

But this is formally the same expression, (4.5.3), as that of the scalar potential produced by a charge distribution

(x.^{'}, y^{'})

It was inconvenient to integrate the above equation directly. Instead, we determined the field of a line charge from symmetry and Gauss' law and integrated the resulting expression to obtain the potential (4.5.18)

where

ris the distance from the line charger = (x - xand^{'})^{2}+ (y - y^{'})^{2}ris the reference radius. The scalar potential can thus be evaluated from the two-dimensional integral_{o}

The vector potential of a two-dimensional

z-directed current distribution obeys the same equation and thus has a solution by analogy, after a proper interchange of parameters.

Two important consequences emerge from this derivation.

by a given charge distribution

(x, y), has an MQS analog vector potentialAcaused by a current density_{z}(x, y)Jwith the same spatial distribution as_{z}(x, y)(x, y). The magnetic field follows from (1) and thus

Therefore the lines of magnetic flux density are perpendicular to the gradient of

A. A plot of field lines and equipotential lines of the EQS problem is transformed into a plot of an MQS field problem by interpreting the equipotential lines as the lines of magnetic flux density._{z}Lines of constantAare lines of magnetic flux._{z}

The following illustrates the integration called for in (8). The fields associated with singular current distributions will be used in later sections and chapters.

(a)Every two-dimensional EQS potential (x, y)produced(b)The vector potential of a line current of magnitude ialong thezdirection is given by analogy with (12),

which is consistent with the magnetic field

given by (1.4.10), if one makes use of the curl expression in polar coordinates,H=i_{}(i/2 r)

## Example 8.1.1. Field Associated with a Current Sheet

A

z-directed current density is uniformly distributed over a strip located betweenxand_{2}xas shown in Fig. 8.1.1. The thickness of the sheet, , is very small compared to other dimensions of interest. So, the integration of (14) in the_{1}ydirection amounts to a multiplication of the current density by . The vector potential is therefore determined by completing the integration onx^{'}

where

K._{o}\equiv J_{z}

Figure 8.1.1Cross-section of surfaces of constantAand lines of magnetic flux density for the uniform sheet of current shown._{z}This integral is carried out in Example 4.5.3, where the two dimensional electric potential of a charged strip was determined. Thus, with

, (4.5.24) becomes the desired vector potential._{o}/_{o}_{o}K_{o}

The profiles of surfaces of constant

Aare shown in Fig. 8.1.1. Remember, these are also the lines of magnetic flux density,_{z}._{o}H

## Example 8.1.2. Two-Dimensional Magnetic Dipole Field

A pair of closely spaced conductors carrying oppositely directed currents of magnitude

iis shown in Fig. 8.1.2. The currents extend to+and-infinity in thezdirection, so the resulting fields are two-dimensional and can be represented byA. In polar coordinates, the distance from the right conductor, which is at a distance_{z}dfrom thezaxis, to the observer location is essentiallyr - d cos. TheAfor each wire takes the form of (16), with_{z}rthe distance from the wire to the point of observation. Thus, superposition of the vector potentials due to the two wires gives

In the limit

d \ll r, this expression becomes

Thus, the surfaces of constant

Ahave intersections with planes of constant_{z}zthat are circular, as shown in Fig. 8.1.3. These are also the lines of magnetic flux density, which follow from (17).

Figure 8.1.2A pair of wires having the spacingdcarry the currentiin opposite directions parallel to thezaxis. The two-dimensional dipole field is shown in Fig. 8.1.3.

Figure 8.1.3Cross-sections of surfaces of constantAand hence lines of magnetic flux density for configuration of Fig. 8.1.2._{z}If the line currents are replaced by line charges, the resulting equipotential lines (intersections of the equipotential surfaces with the

x - yplane) coincide with the magnetic field lines shown in Fig. 8.1.3. Thus, the lines of electric field intensity for the electric dual of the magnetic configuration shown in Fig. 8.1.3 originate on the positive line charge on the right and terminate on the negative line charge at the left, following lines that are perpendicular to those shown.