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Electromagnetic Potentials and Topics for Circuits and Systems RaoCh05v3.qxd 12/18/03 4:08 PM Page 282 CHAPTER 5 Electromagnetic Potentials and Topics for Circuits and Systems In Chapters 2, 3, and 4, we introduced progressively Maxwell’s equations and studied uniform plane waves and associated topics. Two quantities of funda- mental importance, resulting from Maxwell’s equations in differential form, are the electromagnetic potentials: the electric scalar potential and the magnetic vector potential.We introduce these quantities in this chapter and also consider several topics of relevance to circuits and systems. We begin the discussion of topics for circuits and systems with two impor- tant differential equations involving the electric potential and discuss several applications based on the solution of these equations, including the analysis of a p-n junction semiconductor and arrangements involving two parallel conduc- tors. We then introduce an important relationship between the (lumped) circuit parameters, capacitance, conductance, and inductance for infinitely long, paral- lel perfect conductor arrangements, and consider their determination. Next we turn our attention to electric- and magnetic-field systems, that is, systems in which either the electric field or the magnetic field is predominant, leading from quasistatic extensions of the static fields existing in the structures when the frequency of the source driving the structure is zero. The concepts of electric- and magnetic-field systems are important in the study of electro- mechanics. We shall also consider magnetic circuits, an important class of mag- netic field systems, and the topic of electromechanical energy conversion. 5.1 GRADIENT, LAPLACIAN, AND THE POTENTIAL FUNCTIONS # -١؋A = 0. It then fol ١,Magnetic In Example 3.9, we showed that for any vector A B = 0, that # ١,vector lows from Gauss’ law for the magnetic field in differential form potential the magnetic flux density vector B can be expressed as the curl of another 282 RaoCh05v3.qxd 12/18/03 4:08 PM Page 283 5.1 Gradient, Laplacian, and the Potential Functions 283 vector A; that is, (١؋A (5.1 = B The vector A in (5.1) is known as the magnetic vector potential. ١؋E =-0B 0t, Gradient ,Substituting (5.1) into Faraday’s law in differential form > and rearranging, we then obtain 0 ١؋A = 0 + ١؋E 0t 1 2 or 0A (١؋ E + = 0 (5.2 a 0t b If the curl of a vector is equal to the null vector, that vector can be ex- pressed as the gradient of a scalar, since the curl of the gradient of a scalar func- tion is identically equal to the null vector. The gradient of a scalar, say, £, £ del £) is defined in such a manner that the increment d£ in) £١ denoted from a point P to a neighboring point Q is given by (dl (5.3 # £١=£d where dl is the differential length vector from P to Q.Applying Stokes’ theorem and a surface S bounded by closed path C, we then have £١؋١ to the vector # # dl £١ = dS £١؋١ LS 1 2 CC = d£ (5.4) CC = 0 for any single-valued function £. Since (5.4) holds for an arbitrary S, it follows that (0 (5.5=£١؋١ To obtain the expression for the gradient in the Cartesian coordinate sys- tem, we write 0£ 0£ 0£ d£= dx + dy + dz 0x 0y 0z (5.6) 0£ 0£ 0£ # = a + a + a dx a + dy a + dz a a 0x x 0y y 0z z b 1 x y z2 RaoCh05v3.qxd 12/18/03 4:08 PM Page 284 284 Chapter 5 Electromagnetic Potentials and Topics for Circuits and Systems Then comparing with (5.3), we observe that 0£ 0£ 0£ (a + a + a (5.7 =£١ 0x x 0y y 0z z Note that the right side of (5.7) is simply the vector obtained by applying the del operator to the scalar function £. It is for this reason that the gradient of £ is -Expressions for the gradient in cylindrical and spherical coordi .£١ written as nate systems are derived in Appendix B. These are as follows: CYLINDRICAL 0£ 0£ 0£ + 1 + =£١ ar af az (5.8a) 0r r 0f 0z SPHERICAL 0£ 1 0£ 1 0£ (a + au + af (5.8b =£١ 0r r r 0u r sin u 0f Physical To discuss the physical interpretation of the gradient, let us consider a sur- £ £ interpretation face on which is equal to a constant, say,0, and a point P on that surface, as of gradient shown in Fig. 5.1(a). If we now consider another point Q1 on the same surface and an infinitesimal distance away from P, d£ between these two points is zero £ since is constant on the surface. Thus, for the vector dl1 drawn from P to £١ = # £١ Q1, [ ]P dl1 0 and hence [ ]P is perpendicular to dl1. Since this is true for ⌽ ϭ ⌽ 0 ⌽ ϭ ⌽ ϩ ⌽ 0 d ⌽ ϭ ⌽ 0 P dl3 Q an Q3 dl2 dl dl1 Q2 a Q1 P an (a) (b) FIGURE 5.1 For discussing the physical interpretation of the gradient of a scalar function. RaoCh05v3.qxd 12/18/03 4:08 PM Page 285 5.1 Gradient, Laplacian, and the Potential Functions 285 £١ £ Á all points Q1, Q2, Q3, on the constant surface, it follows that [ ]P must be Á normal to all possible infinitesimal length vectors dl1, dl2, dl3, drawn at P and hence is normal to the surface. Denoting an to be the unit normal vector to the surface at P, we then have £١ = £١ [ ]P ƒ ƒ P an (5.9) £ £ Let us now consider two surfaces on which is constant, having values 0 £ + £ £=£ and 0 d , as shown in Fig. 5.1(b). Let P and Q be points on the 0 £=£ + £ and 0 d surfaces, respectively, and dl be the vector drawn from P to Q. Then from (5.3) and (5.9), # £١ =£ d [ ]P dl # £١ = ƒ ƒ P an dl £١ = ƒ ƒ P dl cos a where a is the angle between an at P and dl. Thus, d£ (ƒ = (5.10 £١ ƒ P dl cos a Since dl cos a is the distance between the two surfaces along an and hence is the £١ shortest distance between them, it follows that ƒ ƒ P is the maximum rate of in- crease of £ at the point P. Thus, the gradient of a scalar function £ at a point is a vector having magnitude equal to the maximum rate of increase of £ at that point and is directed along the direction of the maximum rate of increase, which is normal to the constant £ surface passing through that point; that is, d£ (a (5.11 =£١ dn n where dn is a differential length along an. The concept of the gradient of a scalar function we just discussed is often utilized to find a unit vector normal to a given surface. We shall illustrate this by means of an example. Example 5.1 Finding unit vector normal to a surface by using the gradient concept Let us find the unit vector normal to the surface y = x2 at the point (2, 4, 1) by using the concept of the gradient of a scalar. Writing the equation for the surface as x2 - y = 0 we note that the scalar function that is constant on the surface is given by £ x, y, z = x2 - y 1 2 RaoCh05v3.qxd 12/18/03 4:08 PM Page 286 286 Chapter 5 Electromagnetic Potentials and Topics for Circuits and Systems The gradient of the scalar function is then given by x2 - y ١=£١ 1 2 0 x2 - y 0 x2 - y 0 x2 - y = 1 2 a + 1 2 a + 1 2 a 0x x 0y y 0z z = - 2xax ay The value of the gradient at the point (2, 4, 1) is [2 2 a - a ] = 4a - a . Thus, the 1 2 x y 1 x y2 required unit vector is - 4ax ay 4 1 a = Ϯ = Ϯ a - a n - x y ƒ4ax ay ƒ a 117 117 b Electric Returning now to (5.2), we write scalar potential 0A (5.12) £١-= + E 0t where we have chosen the scalar to be -£, the reason for the minus sign to be explained in Section 5.2. Rearranging (5.12), we obtain 0A (5.13) -£١-= E 0t The quantity £ in (5.13) is known as the electric scalar potential. Electro- The electric scalar potential £ and the magnetic vector potential A are magnetic known as the electromagnetic potentials. As we shall show later in this section, potentials the electric scalar potential is related to the source charge density r, whereas the magnetic vector potential is related to the source current density J. For the time-varying case, the two are not independent, since the charge and current densities are related through the continuity equation. For a given J, it is suffi- cient to determine A, since B can be found from (5.1) and then E can be found ,١؋H = J +0D 0t. For static fields, that is by using Ampère’s circuital law > for 0 0t = 0, the two potentials are independent. Equation (5.1) remains unal- > We shall consider the static field .£١-= tered, whereas (5.13) reduces to E case in Section 5.2. To proceed further, we recall that Maxwell’s equations in differential form are given by 0B (١؋E =- (5.14a 0t 0D (١؋H = J + (5.14b 0t (D = r (5.14c # ١ (B = 0 (5.14d # ١ RaoCh05v3.qxd 12/18/03 4:08 PM Page 287 5.1 Gradient, Laplacian, and the Potential Functions 287 From (5.14d), we expressed B in the manner (١؋A (5.15 = B and then from (5.14a), we obtained 0A (5.16) -£١-= E 0t We now substitute (5.16) and (5.15) into (5.14c) and (5.14b), respectively, to obtain 0A r (5.17a) = -£١ - # ١ a 0t b e 0 0A (mJ (5.17b = -£١ - ١؋١؋A - me 0t a 0t b We now define the Laplacian of a scalar quantity £, denoted §2£ (del Laplacian of squared £) as a scalar # (5.18) £١ ١=2£§ In Cartesian coordinates, 0£ 0£ 0£ a + a + a =£١ 0x x 0y y 0z z 0 0 0A Ay Az + + A = x # ١ 0x 0y 0z so that 0 0£ 0 0£ 0 0£ 2 § £= + + 0x a 0x b 0y a 0y b 0z a 0z b or 02£ 02£ 02£ §2£= + + (5.19) 0x2 0y2 0z2 Note that the Laplacian of a scalar is a scalar quantity.
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