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4.6. DIVERGENCE 85 4.6 Divergence [m0044] In this section, we present the divergence operator, which provides a way to calculate the ﬂux associated with a point in space. First, let us review the concept of ﬂux. The integral of a vector ﬁeld over a surface is a scalar quantity known as ﬂux. Speciﬁcally, the ﬂux F of a vector ﬁeld A(r) over a surface S is Z S A · ds = F (4.93) Note that A could be fairly described as a ﬂux density; i.e., a quantity having units equal to the units of F, but divided by area (i.e., m2). Also worth noting is that the ﬂux of a vector ﬁeld that has unit magnitude and is normal to all points on S is simply the area of S. It is quite useful to identify some electromagnetic quantities as either ﬂuxes or ﬂux densities. Here are two important examples: • The electric ﬂux density D, having units of C/m2, is a description of the electric ﬁeld as a ﬂux density. (See Section 2.4 for more about electric ﬂux density.) The integral of D over a closed surface yields the enclosed charge Qencl, having units of C. This relationship is known as Gauss’ Law: I S D · ds = Qencl (4.94) (See Section 5.5 for more about Gauss’ Law.) • The magnetic ﬂux density B, having units of Wb/m2, is a description of the magnetic ﬁeld as a ﬂux density. (See Section 2.5 for more about magnetic ﬂux density.) The integral of B over a surface (open or closed) yields the magnetic ﬂux Φ, having units of Wb: Z S B · ds = Φ (4.95) This is important because, for example, the time rate of change of Φ is proportional to electric potential. (See Section 8.3 for more about this principle, called Faraday’s Law.) Summarizing: Flux is the scalar quantity obtained by integrating a vector ﬁeld, interpreted in this case as a ﬂux density, over a speciﬁed surface. The concept of ﬂux applies to a surface of ﬁnite size. However, what is frequently of interest is behavior at a single point, as opposed to the sum or average over a region of space. For example, returning to the idea of electric ﬂux density (D), perhaps we are not concerned about the total charge (units of C) enclosed by a surface, but rather the charge density (C/m3) at a point. In this case, we could begin with Equation 4.94 and divide both sides of the equation by the volume V enclosed by S: H S D · ds V = Qencl V (4.96) Now we let V shrink to zero, giving us an expression that must be true at whatever point we decide to converge upon. Taking the limit as V →0: lim V →0 H S D · ds V = lim V →0 Qencl V (4.97) The quantity on the right hand side is by deﬁnition the volume charge density ρv (units of C/m3) at the point at which we converge. The left hand side is the divergence of D, sometimes abbreviated “div D.” Thus, the above equation can be written div D = ρv (4.98) Summarizing: Divergence is the ﬂux per unit volume through an inﬁnitesimally-small closed surface surrounding a point. We will typically not actually want to integrate and take a limit in order to calculate divergence. Fortunately, we do not have to. It turns out that this operation can be expressed as the dot product ∇· D; where, for example, ∇≜ˆx ∂ ∂x + ˆy ∂ ∂y + ˆz ∂ ∂z (4.99)	Electromagnetics_Vol1_Page_100_Chunk1401
86 CHAPTER 4. VECTOR ANALYSIS in the Cartesian coordinate system. This is the same “∇” that appears in the deﬁnition of the gradient operator (Section 4.5) and is same operator that often arises when considering other differential operators. If we expand D in terms of its Cartesian components: D = ˆxDx + ˆyDy + ˆzDz (4.100) Then div D = ∇· D = ∂Dx ∂x + ∂Dy ∂y + ∂Dz ∂z (4.101) This seems to make sense for two reasons. First, it is dimensionally correct. Taking the derivative of a quantity having units of C/m2 with respect to distance yields a quantity having units of C/m3. Second, it makes sense that ﬂux from a point should be related to the sum of the rates of change of the ﬂux density in each basis direction. Summarizing: The divergence of a vector ﬁeld A is ∇· A. Example 4.5. Divergence of a uniform ﬁeld. A ﬁeld A that is constant with respect to position is said to be uniform. A completely general description of such a ﬁeld is A = ˆxAx + ˆyAy + ˆzAz where Ax, Ay, and Az are all constants. We see immediately that the divergence of such a ﬁeld must be zero. That is, ∇· A = 0 because each component of A is constant with respect to position. This also makes sense from the perspective of the “ﬂux through an inﬁnitesimally-small closed surface” interpretation of divergence. If the ﬂux is uniform, the ﬂux into the surface equals the ﬂux out of the surface resulting in a net ﬂux of zero. Example 4.6. Divergence of a linearly-increasing ﬁeld. Consider a ﬁeld A = ˆxA0x where A0 is a constant. The divergence of A is ∇· A = A0. If we interpret A as a ﬂux density, then we have found that the net ﬂux per unit volume is simply the rate at which the ﬂux density is increasing with distance. To compute divergence in other coordinate systems, we merely need to know ∇for those systems. In the cylindrical system: ∇= ˆρ1 ρ ∂ ∂ρρ + ˆφ1 ρ ∂ ∂φρ + ˆz ∂ ∂z (4.102) and in the spherical system: ∇= ˆr 1 r2 ∂ ∂rr2 + ˆθ 1 r sin θ ∂ ∂θ sin θ + ˆφ 1 r sin θ ∂ ∂φ (4.103) Alternatively, one may use the explicit expressions for divergence given in Appendix B.2. Example 4.7. Divergence of a radially-decreasing ﬁeld. Consider a vector ﬁeld that is directed radially outward from a point and which decreases linearly with distance; i.e., A = ˆrA0/r where A0 is a constant. In this case, the divergence is most easily computed in the spherical coordinate system since partial derivatives in all but one direction (r) equal zero. Neglecting terms that include these zero-valued partial derivatives, we ﬁnd: ∇· A = 1 r2 ∂ ∂r r2 A0 r = A0 r2 (4.104) In other words, if we interpret A as a ﬂux density, then the ﬂux per unit volume is decreasing with as the square of distance from the origin. It is useful to know that divergence, like ∇itself, is a linear operator; that is, for any constant scalars a and b and vector ﬁelds A and B: ∇· (aA + bB) = a∇· A + b∇· B (4.105) Additional Reading: • “Divergence” on Wikipedia.	Electromagnetics_Vol1_Page_101_Chunk1402
4.7. DIVERGENCE THEOREM 87 4.7 Divergence Theorem [m0046] The Divergence Theorem relates an integral over a volume to an integral over the surface bounding that volume. This is useful in a number of situations that arise in electromagnetic analysis. In this section, we derive this theorem. Consider a vector ﬁeld A representing a ﬂux density, such as the electric ﬂux density D or magnetic ﬂux density B. The divergence of A is ∇· A = f (4.106) where is f(r) is the ﬂux per unit volume through an inﬁnitesimally-small closed surface surrounding the point at r. Since f is ﬂux per unit volume, we can obtain ﬂux for any larger contiguous volume V by integrating over V; i.e., ﬂux through V = Z V f dv (4.107) In the Cartesian system, V can be interpreted as a three-dimensional grid of inﬁnitesimally-small cubes having side lengths dx, dy, and dz, respectively. Note that the ﬂux out of any face of one of these cubes is equal to the ﬂux into the cube that is adjacent through that face. That is, the portion of the total ﬂux that ﬂows between cubes cancels when added together. In fact, the only ﬂuxes which do not cancel in the integration over V are those corresponding to faces which lie on the bounding surface S, since the integration stops there. Stating this mathematically: Z V f dv = I S A · ds (4.108) Thus, we have converted a volume integral into a surface integral. To obtain the Divergence Theorem, we return to Equation 4.106. Integrating both sides of that equation over V, we obtain Z V (∇· A) dv = Z V f dv (4.109) Now applying Equation 4.108 to the right hand side: Z V (∇· A) dv = I S A · ds (4.110) The Divergence Theorem (Equation 4.110) states that the integral of the divergence of a vector ﬁeld over a volume is equal to the ﬂux of that ﬁeld through the surface bounding that volume. The principal utility of the Divergence Theorem is to convert problems that are deﬁned in terms of quantities known throughout a volume into problems that are deﬁned in terms of quantities known over the bounding surface and vice-versa. Additional Reading: • “Divergence theorem” on Wikipedia.	Electromagnetics_Vol1_Page_102_Chunk1403
88 CHAPTER 4. VECTOR ANALYSIS 4.8 Curl [m0048] Curl is an operation, which when applied to a vector ﬁeld, quantiﬁes the circulation of that ﬁeld. The concept of circulation has several applications in electromagnetics. Two of these applications correspond to directly to Maxwell’s Equations: • The circulation of an electric ﬁeld is proportional to the rate of change of the magnetic ﬁeld. This is a statement of the Maxwell-Faraday Equation (Section 8.8), which includes as a special case Kirchoff’s Voltage Law for electrostatics (Section 5.11). • The circulation of a magnetic ﬁeld is proportional to the source current and the rate of change of the electric ﬁeld. This is a statement of Ampere’s Law (Sections 7.9 and 8.9) Thus, we are motivated to formally deﬁne circulation and then to ﬁgure out how to most conveniently apply the concept in mathematical analysis. The result is the curl operator. So, we begin with the concept of circulation: “Circulation” is the integral of a vector ﬁeld over a closed path. Speciﬁcally, the circulation of the vector ﬁeld A(r) over the closed path C is I C A · dl (4.111) The circulation of a uniform vector ﬁeld is zero for any valid path. For example, the circulation of A = ˆxA0 is zero because non-zero contributions at each point on C cancel out when summed over the closed path. On the other hand, the circulation of A = ˆφA0 over a circular path of constant ρ and z is a non-zero constant, since the non-zero contributions to the integral at each point on the curve are equal and accumulate when summed over the path. z I H c⃝K. Kikkeri CC BY SA 4.0 Figure 4.20: Magnetic ﬁeld intensity due to a current ﬂowing along the z axis. Example 4.8. Circulation of the magnetic ﬁeld intensity surrounding a line current. Consider a current I (units of A) ﬂowing along the z axis in the +z direction, as shown in Figure 4.20. It is known that this current gives rise to a magnetic ﬁeld intensity H = ˆφH0/ρ, where H0 is a constant having units of A since the units of H are A/m. (Feel free to consult Section 7.5 for the details; however, no additional information is needed to follow the example being presented here.) The circulation of H along any circular path of radius a in a plane of constant z is therefore I C H · dl = Z 2π φ=0 ˆφH0 a · ˆφ a dφ = 2πH0 Note that the circulation of H in this case has two remarkable features: (1) It is independent of the radius of the path of integration; and (2) it has units of A, which suggests a current. In fact, it turns out that the circulation of H in this case is equal to the enclosed source current I. Furthermore, it turns out that the circulation of H along any path enclosing the source current is equal to the source current! These ﬁndings are consequences of Ampere’s Law, as noted above. Curl is, in part, an answer to the question of what the circulation at a point in space is. In other words, what	Electromagnetics_Vol1_Page_103_Chunk1404
4.8. CURL 89 is the circulation as C shrinks to it’s smallest possible size. The answer in one sense is zero, since the arclength of C is zero in this limit – there is nothing to integrate over. However, if we ask instead what is the circulation per unit area in the limit, then the result should be the non-trivial value of interest. To express this mathematically, we constrain C to lie in a plane, and deﬁne S to be the open surface bounded by C in this case. Then, we deﬁne the scalar part of the curl of A to be: lim ∆s→0 H C A · dl ∆s (4.112) where ∆s is the area of S, and (important!) we require C and S to lie in the plane that maximizes the above result. Because S and it’s boundary C lie in a plane, it is possible to assign a direction to the result. The chosen direction is the normal ˆn to the plane in which C and S lie. Because there are two normals at each point on a plane, we specify the one that satisﬁes the right hand rule. This rule, applied to the curl, states that the correct normal is the one which points through the plane in the same direction as the ﬁngers of the right hand when the thumb of your right hand is aligned along C in the direction of integration. Why is this the correct orientation of ˆn? See Section 4.9 for the answer to that question. For the purposes of this section, it sufﬁces to consider this to be simply an arbitrary sign convention. Now with the normal vector ˆn unambiguously deﬁned, we can now formally deﬁne the curl operation as follows: curl A ≜lim ∆s→0 ˆn H C A · dl ∆s (4.113) where, once again, ∆s is the area of S, and we select S to lie in the plane that maximizes the magnitude of the above result. Summarizing: The curl operator quantiﬁes the circulation of a vector ﬁeld at a point. The magnitude of the curl of a vector ﬁeld is the circulation, per unit area, at a point and such that the closed path of integration shrinks to enclose zero area while being constrained to lie in the plane that maximizes the magnitude of the result. The direction of the curl is determined by the right-hand rule applied to the associated path of integration. Curl is a very important operator in electromagnetic analysis. However, the deﬁnition (Equation 4.113) is usually quite difﬁcult to apply. Remarkably, however, it turns out that the curl operation can be deﬁned in terms of the ∇operator; that is, the same ∇operator associated with the gradient, divergence, and Laplacian operators. Here is that deﬁnition: curl A ≜∇× A (4.114) For example: In Cartesian coordinates, ∇≜ˆx ∂ ∂x + ˆy ∂ ∂y + ˆz ∂ ∂z (4.115) and A = ˆxAx + ˆyAy + ˆzAz (4.116) so curl can be calculated as follows: ∇× A = ˆx ˆy ˆz ∂ ∂x ∂ ∂y ∂ ∂z Ax Ay Az (4.117) or, evaluating the determinant: ∇× A = ˆx ∂Az ∂y −∂Ay ∂z + ˆy ∂Ax ∂z −∂Az ∂x + ˆz ∂Ay ∂x −∂Ax ∂y (4.118) Expressions for curl in each of the three major coordinate systems is provided in Appendix B.2. It is useful to know is that curl, like ∇itself, is a linear operator; that is, for any constant scalars a and b and vector ﬁelds A and B: ∇× (aA + bB) = a∇× A + b∇× B (4.119) Additional Reading: • “Curl (mathematics)” on Wikipedia.	Electromagnetics_Vol1_Page_104_Chunk1405
90 CHAPTER 4. VECTOR ANALYSIS 4.9 Stokes’ Theorem [m0051] Stokes’ Theorem relates an integral over an open surface to an integral over the curve bounding that surface. This relationship has a number of applications in electromagnetic theory. Here is the theorem: Z S (∇× A) · ds = I C A · dl (4.120) where S is the open surface bounded by the closed path C. The direction of the surface normal ds = ˆnds is related to the direction of integration along C by the right-hand rule, illustrated in Figure 4.21. In this case, the right-hand rule states that the correct normal is the one that points through the surface in the same direction as the ﬁngers of the right hand when the thumb of your right hand is aligned along C in the direction of integration. Stokes’ Theorem is a purely mathematical result and not a principle of electromagnetics per se. The relevance of the theorem to electromagnetic theory is primarily as a tool in the associated mathematical analysis. Usually the theorem is employed to transform a problem expressed in terms of an integration over a surface into an integration over a closed path or vice-versa. For more information on the theorem and its derivation, see “Additional Reading” at the end of this section. c⃝Cronholm144 (modiﬁed) CC BY SA 3.0 Figure 4.21: The relative orientations of the direc- tion of integration C and surface normal ˆn in Stokes’ Theorem. Additional Reading: • “Stokes’ Theorem” on Wikipedia.	Electromagnetics_Vol1_Page_105_Chunk1406
4.10. THE LAPLACIAN OPERATOR 91 4.10 The Laplacian Operator [m0099] The Laplacian ∇2f of a ﬁeld f(r) is the divergence of the gradient of that ﬁeld: ∇2f ≜∇· (∇f) (4.121) Note that the Laplacian is essentially a deﬁnition of the second derivative with respect to the three spatial dimensions. For example, in Cartesian coordinates, ∇2f = ∂2f ∂x2 + ∂2f ∂y2 + ∂2f ∂z2 (4.122) as can be readily veriﬁed by applying the deﬁnitions of gradient and divergence in Cartesian coordinates to Equation 4.121. The Laplacian relates the electric potential (i.e., V , units of V) to electric charge density (i.e., ρv, units of C/m3). This relationship is known as Poisson’s Equation (Section 5.15): ∇2V = −ρv ǫ (4.123) where ǫ is the permittivity of the medium. The fact that V is related to ρv in this way should not be surprising, since electric ﬁeld intensity (E, units of V/m) is proportional to the derivative of V with respect to distance (via the gradient) and ρv is proportional to the derivative of E with respect to distance (via the divergence). The Laplacian operator can also be applied to vector ﬁelds; for example, Equation 4.122 is valid even if the scalar ﬁeld “f” is replaced with a vector ﬁeld. In the Cartesian coordinate system, the Laplacian of the vector ﬁeld A = ˆxAx + ˆyAy + ˆzAz is ∇2A = ˆx∇2Ax + ˆy∇2Ay + ˆz∇2Az (4.124) An important application of the Laplacian operator of vector ﬁelds is the wave equation; e.g., the wave equation for E in a lossless and source-free region is (Section 9.2) ∇2E + β2E = 0 (4.125) where β is the phase propagation constant. It is sometimes useful to know that the Laplacian of a vector ﬁeld can be expressed in terms of the gradient, divergence, and curl as follows: ∇2A = ∇(∇· A) −∇× (∇× A) (4.126) The Laplacian operator in the cylindrical and spherical coordinate systems is given in Appendix B.2. Additional Reading: • “Laplace operator” on Wikipedia. [m0043]	Electromagnetics_Vol1_Page_106_Chunk1407
92 CHAPTER 4. VECTOR ANALYSIS Image Credits Fig. 4.1: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0006 fCartesian.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.2: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0006 fPositionFixed.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.3: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0006 fPositionFree.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.4: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0006 fCartesianBasis.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.5: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0006 fVecCart.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.6: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0006 fRelativePosition.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.7: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0006 fDotProductSpecialCases.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.8: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0006 fDotProduct.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.9: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0006 fCrossProduct.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.11: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0096 fCylindricalCoordinates.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.13: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:Cylindrical Coordinates Example.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.14: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:Cylindrical Coordinates-Area of a Circle.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.15: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0096 CylinderArea.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.16: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:Spherical Coordinate System.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.18: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0097 fArc.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.19: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0097 fSphereArea.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.20: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:Biot Savart Example.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.21: c⃝Cronholm144, https://en.wikipedia.org/wiki/File:Stokes’ Theorem.svg, CC BY SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/). Heavily modiﬁed by author.	Electromagnetics_Vol1_Page_107_Chunk1408
Chapter 5 Electrostatics [m0116] Electrostatics is the theory of the electric ﬁeld in conditions in which its behavior is independent of magnetic ﬁelds, including • The electric ﬁeld associated with ﬁxed distributions of electric charge • Capacitance (the ability of a structure to store energy in an electric ﬁeld) • The energy associated with the electrostatic ﬁeld • Steady current induced in a conducting material in the presence of an electrostatic ﬁeld (essentially, Ohm’s Law) The term “static” refers to the fact that these aspects of electromagnetic theory can be developed by assuming sources are time-invariant; we might say that electrostatics is the study of the electric ﬁeld at DC. However, many aspects of electrostatics are relevant to AC, radio frequency, and higher-frequency applications as well. Additional Reading: • “Electrostatics” on Wikipedia. 5.1 Coulomb’s Law [m0102] Consider two charge-bearing particles in free space, identiﬁed as “particle 1” and “particle 2” in Figure 5.1. Let the charges borne by these particles be q1 and q2, and let R be the distance between them. If the particles bear charges of the same sign (i.e., if q1q2 is positive), then the particles repel; otherwise, they attract. This repulsion or attraction can be quantiﬁed as a force experienced by each particle. Physical observations reveal that the magnitude of the force is proportional to q1q2, and inversely proportional to R2. For particle 2 we ﬁnd: F = ˆRF0 q1q2 R2 (5.1) where ˆR is the unit vector pointing from the particle 1 to the particle 2, and F0 is a constant. The value of F0 must have units of inverse permittivity; i.e., (F/m)−1. This is most easily seen by dimensional analysis of the above relationship, including the suspected factor: C · C F/m · m2 = C · C F · m = C · C C/V · m = C · V m = J m = N where we have used the facts that 1 F = 1 C/V, 1 V = 1 J/C, and 1 N = 1 J/m. This ﬁnding suggests that F0 ∝ǫ−1, where ǫ is the permittivity of the medium in which the particles exist. Observations conﬁrm that the force is in fact inversely proportional to the permittivity, with an additional factor of 1/4π (unitless). Putting this all together we obtain what is commonly known as Coulomb’s Law: F = ˆR q1q2 4πǫR2 (5.2) Electromagnetics Vol 1. c⃝2018 S.W. Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics-vol-1	Electromagnetics_Vol1_Page_108_Chunk1409
94 CHAPTER 5. ELECTROSTATICS F q2 q1 particle Å particle Æ ÇF R c⃝K. Kikkeri CC BY SA 4.0 Figure 5.1: Coulomb’s Law describes the force per- ceived by pairs of charged particles. Subsequently, the force perceived by particle 2 is equal and opposite; i.e., equal to −F. Separately, it is known that F can be described in terms of the electric ﬁeld intensity E1 associated with particle 1: F = q2E1 (5.3) This is essentially the deﬁnition of E1, as explained in Section 2.2. Combining this result with Coulomb’s Law, we obtain a means to directly calculate the ﬁeld associated with the ﬁrst particle in the absence of the second particle: E1 = ˆR q1 4πǫR2 (5.4) where now ˆRR is the vector beginning at the particle 1 and ending at the point to be evaluated. The electric ﬁeld intensity associated with a point charge (Equation 5.4) is (1) directed away from positive charge, (2) proportional to the magni- tude of the charge, (3) inversely proportional to the permittivity of the medium, and (3) inversely proportional to distance squared. We have described this result as originating from Coulomb’s Law, which is based on physical observations. However, the same result may be obtained directly from Maxwell’s Equations using Gauss’ Law (Section 5.5). Example 5.1. Electric ﬁeld of a point charge at the origin. A common starting point in electrostatic analysis is the ﬁeld associated with a particle bearing charge q at the origin of the coordinate system. Because the electric ﬁeld is directed radially away from a positively-charged source particle in all directions, this ﬁeld is most conveniently described in the spherical coordinate system. Thus, ˆR becomes ˆr, R becomes r, and we have E(r) = ˆr q 4πǫr2 (5.5) Here’s a numerical example. What is the electric ﬁeld intensity at a distance 1 µm from a single electron located at the origin, in free space? In this case, q ∼= −1.60 × 10−19 C (don’t forget that minus sign!), ǫ = ǫ0, r = 1 µm, and we ﬁnd: E(r) = −ˆr (1.44 kV/m) (5.6) This is large relative to electric ﬁeld strengths commonly encountered in engineering applications. The strong electric ﬁeld of the electron is not readily apparent because electrons in common materials tend to be accompanied by roughly equal amounts of positive charge, such as the protons of atoms. Sometimes, however, the effect of individual electrons does become signiﬁcant in practical electronics through a phenomenon known as shot noise. Additional Reading: • “Coulomb’s Law” on Wikipedia. • “Shot Noise” on Wikipedia.	Electromagnetics_Vol1_Page_109_Chunk1410
5.2. ELECTRIC FIELD DUE TO POINT CHARGES 95 5.2 Electric Field Due to Point Charges [m0103] The electric ﬁeld intensity associated with a single particle bearing charge q1, located at the origin, is (Section 5.1) E(r) = ˆr q1 4πǫr2 (5.7) If this particle is instead located at some position r1, then the above expression may be written as follows: E(r; r1) = r −r1 \|r −r1\| q1 4πǫ \|r −r1\|2 (5.8) or, combining like terms in the denominator: E(r; r1) = r −r1 \|r −r1\|3 q1 4πǫ (5.9) Now let us consider the ﬁeld due to multiple such particles. Under the usual assumptions about the permittivity of the medium (reminder: Section 2.8), the property of superposition applies. Using this principle, we conclude: The electric ﬁeld resulting from a set of charged particles is equal to the sum of the ﬁelds associ- ated with the individual particles. Stated mathematically: E(r) = N X n=1 E(r; rn) (5.10) where N is the number of particles. Thus, we have E(r) = 1 4πǫ N X n=1 r −rn \|r −rn\|3 qn (5.11) 5.3 Charge Distributions [m0100] In principle, the smallest unit of electric charge that can be isolated is the charge of a single electron, which is ∼= −1.60 × 10−19 C. This is very small, and we rarely deal with electrons one at a time, so it is usually more convenient to describe charge as a quantity that is continuous over some region of space. In particular, it is convenient to describe charge as being distributed in one of three ways: along a curve, over a surface, or within a volume. Line Charge Distribution. Imagine that charge is distributed along a curve C through space. Let ∆q be the total charge along a short segment of the curve, and let ∆l be the length of this segment. The line charge density ρl at any point along the curve is deﬁned as ρl ≜lim ∆l→0 ∆q ∆l = dq dl (5.12) which has units of C/m. We may then deﬁne ρl to be a function of position along the curve, parameterized by l; e.g., ρl(l). Then, the total charge Q along the curve is Q = Z C ρl(l) dl (5.13) which has units of C. In other words, line charge density integrated over length yields total charge. Surface Charge Distribution. Imagine that charge is distributed over a surface. Let ∆q be the total charge on a small patch on this surface, and let ∆s be the area of this patch. The surface charge density ρs at any point on the surface is deﬁned as ρs ≜lim ∆s→0 ∆q ∆s = dq ds (5.14) which has units of C/m2. Let us deﬁne ρs to be a function of position on this surface. Then the total charge over a surface S is Q = Z S ρs ds (5.15) In other words, surface charge density integrated over a surface yields total charge. Volume Charge Distribution. Imagine that charge is distributed over a volume. Let ∆q be the total charge	Electromagnetics_Vol1_Page_110_Chunk1411
96 CHAPTER 5. ELECTROSTATICS in a small cell within this volume, and let ∆v be the volume of this cell. The volume charge density ρv at any point in the volume is deﬁned as ρv ≜lim ∆v→0 ∆q ∆v = dq dv (5.16) which has units of C/m3. Since ρv is a function of position within this volume, the total charge within a volume V is Q = Z V ρv dv (5.17) In other words, volume charge density integrated over a volume yields total charge. 5.4 Electric Field Due to a Continuous Distribution of Charge [m0104] The electric ﬁeld intensity associated with N charged particles is (Section 5.2): E(r) = 1 4πǫ N X n=1 r −rn \|r −rn\|3 qn (5.18) where qn and rn are the charge and position of the nth particle. However, it is common to have a continuous distribution of charge as opposed to a countable number of charged particles. In this section, we extend Equation 5.18 using the concept of continuous distribution of charge (Section 5.3) so that we may address this more general class of problems. Distribution of Charge Along a Curve. Consider a continuous distribution of charge along a curve C. The curve can be divided into short segments of length ∆l. Then, the charge associated with the nth segment, located at rn, is qn = ρl(rn) ∆l (5.19) where ρl is charge density (units of C/m) at rn. Substituting this expression into Equation 5.18, we obtain E(r) = 1 4πǫ N X n=1 r −rn \|r −rn\|3 ρl(rn) ∆l (5.20) Taking the limit as ∆l →0 yields: E(r) = 1 4πǫ Z C r −r′ \|r −r′\|3 ρl(r′) dl (5.21) where r′ represents the varying position along C with integration. The simplest example of a curve is a straight line. It is straightforward to use Equation 5.21 to determine the electric ﬁeld due to a distribution of charge along a straight line. However, it is much easier to analyze that particular distribution using Gauss’ Law, as	Electromagnetics_Vol1_Page_111_Chunk1412
5.4. ELECTRIC FIELD DUE TO A CONTINUOUS DISTRIBUTION OF CHARGE 97 shown in Section 5.6. The following example addresses a charge distribution for which Equation 5.21 is more appropriate. Example 5.2. Electric ﬁeld along the axis of a ring of uniformly-distributed charge. Consider a ring of radius a in the z = 0 plane, centered on the origin, as shown in Figure 5.2. Let the charge density along this ring be uniform and equal to ρl (C/m). Find the electric ﬁeld along the z axis. Solution. The source charge position is given in cylindrical coordinates as r′ = ˆρa (5.22) The position of a ﬁeld point along the z axis is simply r = ˆzz (5.23) Thus, r −r′ = −ˆρa + ˆzz (5.24) and \|r −r′\| = p a2 + z2 (5.25) Equation 5.21 becomes: E(z) = 1 4πǫ Z 2π 0 −ˆρa + ˆzz [a2 + z2]3/2 ρl (a dφ) (5.26) Pulling factors that do not vary with φ out of the integral and factoring into separate integrals for the ˆφ and ˆz components, we obtain: ρl a 4πǫ [a2 + z2]3/2 −a Z 2π 0 ˆρ dφ + ˆzz Z 2π 0 dφ (5.27) The second integral is equal to 2π. The ﬁrst integral is equal to zero. To see this, note that the integral is simply summing values of ˆρ for all possible values of φ. Since ˆρ(φ + π) = −ˆρ(φ), the integrand for any given value of φ is equal and opposite the integrand π radians later. (This is one example of a symmetry argument.) Thus, y x z ρa ÈÉ ÊË r - r Ì point c⃝K. Kikkeri CC BY SA 4.0 Figure 5.2: Calculating the electric ﬁeld along the axis of a ring of charge. we obtain E(z) = ˆzρl a 2ǫ z [a2 + z2]3/2 (5.28) It is a good exercise to conﬁrm that this result is dimensionally correct. It is also recommended to conﬁrm that when z ≫a, the result is approximately the same as that expected from a particle having the same total charge as the ring. Distribution of Charge Over a Surface. Consider a continuous distribution of charge over a surface S. The surface can be divided into small patches having area ∆s. Then, the charge associated with the nth patch, located at rn, is qn = ρs(rn) ∆s (5.29) where ρs is the surface charge density (units of C/m2) at rn. Substituting this expression into Equation 5.18, we obtain E(r) = 1 4πǫ N X n=1 r −rn \|r −rn\|3 ρs(rn) ∆s (5.30) Taking the limit as ∆s →0 yields: E(r) = 1 4πǫ Z S r −r′ \|r −r′\|3 ρs(r′) ds (5.31)	Electromagnetics_Vol1_Page_112_Chunk1413
98 CHAPTER 5. ELECTROSTATICS where r′ represents the varying position over S with integration. Example 5.3. Electric ﬁeld along the axis of a disk of uniformly-distributed charge. Consider a circular disk of radius a in the z = 0 plane, centered on the origin, as shown in Figure 5.3. Let the charge density over this disk be uniform and equal to ρs (C/m2). Find the electric ﬁeld along the z axis. Solution. The source charge position is given in cylindrical coordinates as r′ = ˆρρ (5.32) The position of a ﬁeld point along the z axis is simply r = ˆzz (5.33) Thus, r −r′ = −ˆρρ + ˆzz (5.34) and \|r −r′\| = p ρ2 + z2 (5.35) Equation 5.31 becomes: E(z) = 1 4πǫ Z a ρ=0 Z 2π φ=0 −ˆρρ + ˆzz [ρ2 + z2]3/2 ρs (ρ dρ dφ) (5.36) To solve this integral, ﬁrst rearrange the double integral into a single integral over φ followed by integration over ρ: ρs 4πǫ Z a ρ=0 ρ [ρ2 + z2]3/2 Z 2π φ=0 (−ˆρρ + ˆzz) dφ dρ (5.37) Now we address the integration over φ shown in the square brackets in the above expression: Z 2π φ=0 (−ˆρρ + ˆzz) dφ = −ρ Z 2π φ=0 ˆρdφ+ˆzz Z 2π φ=0 dφ (5.38) The ﬁrst integral on the right is zero for the following reason. As the integral progresses in φ, the vector ˆρ rotates. Because the integration is over a complete revolution (i.e., φ from 0 to 2π), the contribution from each pointing of ˆρ is canceled out by another pointing of ˆρ that is in the opposite direction. Since there is an equal number of these canceling pairs of pointings, the result is zero. Thus: Z 2π φ=0 (−ˆρρ + ˆzz) dφ = 0 + ˆzz Z 2π φ=0 dφ = ˆz2πz (5.39) Substituting this into Expression 5.37 we obtain: ρs 4πǫ Z a ρ=0 ρ [ρ2 + z2]3/2 [ˆz2πz] dρ =ˆzρsz 2ǫ Z a ρ=0 ρ dρ [ρ2 + z2]3/2 (5.40) This integral can be solved using integration by parts and trigonometric substitution. Since the solution is tedious and there is no particular principle of electromagnetics demonstrated by this solution, we shall simply state the result: Z a ρ=0 ρ dρ [ρ2 + z2]3/2 = −1 p ρ2 + z2 a ρ=0 = −1 √ a2 + z2 + 1 \|z\| (5.41) Substituting this result: E(z) = ˆzρsz 2ǫ −1 √ a2 + z2 + 1 \|z\| = ˆzρs 2ǫ −z √ a2 + z2 + z \|z\| = ˆzρs 2ǫ −z √ a2 + z2 + sgn z (5.42) where “sgn” is the “signum” function; i.e., sgn z = +1 for z > 0 and sgn z = −1 for z < 0. Summarizing: E(z) = ˆzρs 2ǫ sgn z − z √ a2 + z2 (5.43) It is a good exercise to conﬁrm that this result is dimensionally correct and yields an electric ﬁeld vector that points in the expected direction and with the expected dependence on a and z.	Electromagnetics_Vol1_Page_113_Chunk1414
5.4. ELECTRIC FIELD DUE TO A CONTINUOUS DISTRIBUTION OF CHARGE 99 y x z ρρ r - rÍ a ÎÏÐ Ñ point c⃝K. Kikkeri CC BY SA 4.0 Figure 5.3: Calculating the electric ﬁeld along the axis of a disk of charge. A special case of the “disk of charge” scenario considered in the preceding example is an inﬁnite sheet of charge. The electric ﬁeld from an inﬁnite sheet of charge is a useful theoretical result. We get the ﬁeld in this case simply by letting a →∞in Equation 5.43, yielding: E(r) = ˆzρs 2ǫsgn z (5.44) Again, it is useful to conﬁrm that this is dimensionally correct: C/m2 divided by F/m yields V/m. Also, note that Equation 5.44 is the electric ﬁeld at any point above or below the charge sheet – not just on z axis. This follows from symmetry. From the perspective of any point in space, the edges of the sheet are the same distance (i.e., inﬁnitely far) away. Distribution of Charge In a Volume. Consider a continuous distribution of charge within a volume V. The volume can be divided into small cells (volume elements) having volume ∆v. Then, the charge associated with the nth cell, located at rn, is qn = ρv(rn) ∆v (5.45) where ρv is volume charge density (units of C/m3) at rn. Substituting this expression into Equation 5.18, we obtain E(r) = 1 4πǫ N X n=1 r −rn \|r −rn\|3 ρv(rn) ∆v (5.46) Taking the limit as ∆v →0 yields: E(r) = 1 4πǫ Z V r −r′ \|r −r′\|3 ρv(r′) dv (5.47) where r′ represents the varying position over V with integration.	Electromagnetics_Vol1_Page_114_Chunk1415
100 CHAPTER 5. ELECTROSTATICS 5.5 Gauss’ Law: Integral Form [m0014] Gauss’ Law is one of the four fundamental laws of classical electromagnetics, collectively known as Maxwell’s Equations. Before diving in, the reader is strongly encouraged to review Section 2.4. In that section, Gauss’ Law emerges from the interpretation of the electric ﬁeld as a ﬂux density. Section 2.4 does not actually identify Gauss’ Law, but here it is: Gauss’ Law (Equation 5.48) states that the ﬂux of the electric ﬁeld through a closed surface is equal to the enclosed charge. Gauss’ Law is expressed mathematically as follows: I S D · ds = Qencl (5.48) where D is the electric ﬂux density ǫE, S is a closed surface with differential surface normal ds, and Qencl is the enclosed charge. We can see the law is dimensionally correct; D has units of C/m2, thus integrating D over a surface gives a quantity with units of C/m2 · m2 = C, which are the units of charge. Gauss’ Law has a number of applications in electromagnetic theory. One of them, as explored below, is as a method to compute the electric ﬁeld in response to a distribution of electric charge. Note that a method to do this, based on Coulomb’s Law, is described in Sections 5.1, 5.2, and 5.4. Gauss’ Law provides an alternative method that is easier or more useful in certain applications. Example 5.4. Electric ﬁeld associated with a charged particle, using Gauss’ Law. In this example, we demonstrate the ability of Gauss’ Law to predict the ﬁeld associated with a charge distribution. Let us do this for the simplest possible charge distribution. A particle of charge q located at the origin, for which we already have the answer (Section 5.1). Solution. Gauss’ Law applies to any surface that encloses the charge, so for simplicity we chose a sphere of radius r centered at the origin. Note that Qencl on the right hand side is just q for any surface having r > 0. Gauss’ Law in this case becomes Z π θ=0 Z 2π φ=0 D ·	Electromagnetics_Vol1_Page_115_Chunk1416
5.6. ELECTRIC FIELD DUE TO AN INFINITE LINE CHARGE USING GAUSS’ LAW 101 Additional Reading: • “Gauss’ Law” on Wikipedia. 5.6 Electric Field Due to an Inﬁnite Line Charge using Gauss’ Law [m0149] Section 5.5 explains one application of Gauss’ Law, which is to ﬁnd the electric ﬁeld due to a charged particle. In this section, we present another application – the electric ﬁeld due to an inﬁnite line of charge. The result serves as a useful “building block” in a number of other problems, including determination of the capacitance of coaxial cable (Section 5.24). Although this problem can be solved using the “direct” approach described in Section 5.4 (and it is an excellent exercise to do so), the Gauss’ Law approach demonstrated here turns out to be relatively simple. Example 5.5. Electric ﬁeld associated with an inﬁnite line charge, using Gauss’ Law. Use Gauss’ Law to determine the electric ﬁeld intensity due to an inﬁnite line of charge along the z axis, having charge density ρl (units of C/m), as shown in Figure 5.4. Solution. Gauss’ Law requires integration over a surface that encloses the charge. So, our ﬁrst problem is to determine a suitable surface. A cylinder of radius a that is concentric with the z axis, as shown in Figure 5.4, is maximally symmetric with the charge distribution and so is likely to yield the simplest possible analysis. At ﬁrst glance, it seems that we may have a problem since the charge extends to inﬁnity in the +z and −z directions, so it’s not clear how to enclose all of the charge. Let’s suppress that concern for a moment and simply choose a cylinder of ﬁnite length l. In principle, we can solve the problem ﬁrst for this cylinder of ﬁnite size, which contains only a fraction of the charge, and then later let l →∞to capture the rest of the charge. (In fact, we’ll ﬁnd when the time comes it will not be necessary to do that, but we shall prepare for it anyway.)	Electromagnetics_Vol1_Page_116_Chunk1417
102 CHAPTER 5. ELECTROSTATICS Here’s Gauss’ Law: I S D · ds = Qencl (5.54) where D is the electric ﬂux density ǫE, S is a closed surface with outward-facing differential surface normal ds, and Qencl is the enclosed charge. The ﬁrst order of business is to constrain the form of D using a symmetry argument, as follows. Consider the ﬁeld of a point charge q at the origin (Section 5.5): D = ˆr q 4πr2 (5.55) We can “assemble” an inﬁnite line of charge by adding particles in pairs. One pair is added at a time, with one particle on the +z axis and the other on the −z axis, with each located an equal distance from the origin. We continue to add particle pairs in this manner until the resulting charge extends continuously to inﬁnity in both directions. The principle of superposition indicates that the resulting ﬁeld will be the sum of the ﬁelds of the particles (Section 5.2). Thus, we see that D cannot have any component in the ˆφ direction because none of the ﬁelds of the constituent particles have a component in that direction. Similarly, we see that the magnitude of D cannot depend on φ because none of the ﬁelds of the constituent particles depends on φ and because the charge distribution is identical (“invariant”) with rotation in φ. Also, note that for any choice of z the distribution of charge above and below that plane of constant z is identical; therefore, D cannot be a function of z and D cannot have any component in the ˆz direction. Therefore, the direction of D must be radially outward; i.e., in the ˆρ direction, as follows: D = ˆρDρ(ρ) (5.56) Next, we observe that Qencl on the right hand side of Equation 5.54 is equal to ρll. Thus, we obtain I S [ˆρDρ(ρ)] · ds = ρll (5.57) The cylinder S consists of a ﬂat top, curved side, and ﬂat bottom. Expanding the above equation to reﬂect this, we obtain ρll = Z top [ˆρDρ(ρ)] · (+ˆzds) + Z side [ˆρDρ(ρ)] · (+ˆρds) + Z bottom [ˆρDρ(ρ)] · (−ˆzds) (5.58) Examination of the dot products indicates that the integrals associated with the top and bottom surfaces must be zero. In other words, the ﬂux through the top and bottom is zero because D is perpendicular to these surfaces. We are left with ρll = Z side [Dρ(ρ)] ds (5.59) The side surface is an open cylinder of radius ρ = a, so Dρ(ρ) = Dρ(a), a constant over this surface. Thus: ρll = Z side [Dρ(a)] ds = [Dρ(a)] Z side ds (5.60) The remaining integral is simply the area of the side surface, which is 2πa · l. Solving for Dρ(a) we obtain Dρ(a) = ρll 2πal = ρl 2πa (5.61) Remarkably, we see Dρ(a) is independent of l, So the concern raised in the beginning of this solution – that we wouldn’t be able to enclose all of the charge – doesn’t matter. Completing the solution, we note the result must be the same for any value of ρ (not just ρ = a), so D = ˆρDρ(ρ) = ˆρ ρl 2πρ (5.62) and since D = ǫE: E = ˆρ ρl 2πǫρ (5.63) This completes the solution. We have found that the electric ﬁeld is directed radially away from	Electromagnetics_Vol1_Page_117_Chunk1418
5.7. GAUSS’ LAW: DIFFERENTIAL FORM 103 z a ρ ρ l Ò Ó Ô c⃝K. Kikkeri CC BY SA 4.0 Figure 5.4: Finding the electric ﬁeld of an inﬁnite line of charge using Gauss’ Law. the line charge, and decreases in magnitude in inverse proportion to distance from the line charge. Suggestion: Check to ensure that this solution is dimensionally correct. 5.7 Gauss’ Law: Differential Form [m0045] The integral form of Gauss’ Law (Section 5.5) is a calculation of enclosed charge Qencl using the surrounding density of electric ﬂux: I S D · ds = Qencl (5.64) where D is electric ﬂux density and S is the enclosing surface. It is also sometimes necessary to do the inverse calculation (i.e., determine electric ﬁeld associated with a charge distribution). This is sometimes possible using Equation 5.64 if the symmetry of the problem permits; see examples in Section 5.5 and 5.6. If the problem does not exhibit the necessary symmetry, then it seems that one must fall back to the family of techniques presented in Section 5.4 requiring direct integration over the charge, which is derived from Coulomb’s Law. However, even the Coulomb’s Law / direct integration approach has a limitation that is very important to recognize: It does not account for the presence of structures that may inﬂuence the electric ﬁeld. For example, the electric ﬁeld due to a charge in free space is different from the electric ﬁeld due to the same charge located near a perfectly-conducting surface. In fact, these approaches do not account for the possibility of any spatial variation in material composition, which rules out their use in many engineering applications. To address this broader scope of problems, we require an alternative form of Gauss’ Law that applies at individual points in space. That is, we require Gauss’ Law expressed in the form of a differential equation, as opposed to an integral equation. This facilitates the use of Gauss’ Law even in problems that do not exhibit sufﬁcient symmetry and that involve material boundaries and spatial variations in material constitutive parameters. Given this differential equation and the boundary conditions imposed by structure and materials, we may then solve for the electric ﬁeld in these more complicated scenarios. In this section, we derive the desired differential form of Gauss’ Law. Elsewhere (in particular, in Section 5.15)	Electromagnetics_Vol1_Page_118_Chunk1419
104 CHAPTER 5. ELECTROSTATICS we use this equation as a tool to ﬁnd electric ﬁelds in problems involving material boundaries. There are in fact two methods to develop the desired differential equation. One method is via the deﬁnition of divergence, whereas the other is via the divergence theorem. Both methods are presented below because each provides a different bit of insight. Let’s explore the ﬁrst method: Derivation via the deﬁnition of divergence. Let the geometrical volume enclosed by S be V, which has volume V (units of m3). Dividing both sides of Equation 5.64 by V and taking the limit as V →0: lim V →0 H S D · ds V = lim V →0 Qencl V (5.65) The quantity on the right hand side is the volume charge density ρv (units of C/m3) at the point at which we converge after letting the volume go to zero. The left hand side is, by deﬁnition, the divergence of D, indicated in mathematical notation as “∇· D” (see Section 4.6). Thus, we have Gauss’ Law in differential form: ∇· D = ρv (5.66) To interpret this equation, recall that divergence is simply the ﬂux (in this case, electric ﬂux) per unit volume. Gauss’ Law in differential form (Equation 5.66) says that the electric ﬂux per unit volume origi- nating from a point in space is equal to the vol- ume charge density at that point. Derivation via the divergence theorem. Equation 5.66 may also be obtained from Equation 5.64 using the Divergence Theorem (Section 4.7), which in the present case may be written: Z V (∇· D) dv = I S D · ds (5.67) From Equation 5.64, we see that the right hand side of the equation may be replaced with the enclosed charge: Z V (∇· D) dv = Qencl (5.68) Furthermore, the enclosed charge can be expressed as an integration of the volume charge density ρv over V: Z V (∇· D) dv = Z V ρvdv (5.69) The above relationship must hold regardless of the speciﬁc location or shape of V. The only way this is possible is if the integrands are equal. Thus, ∇· D = ρv, and we have obtained Equation 5.66. Example 5.6. Determining the charge density at a point, given the associated electric ﬁeld. The electric ﬁeld intensity in free space is E(r) = ˆxAx2 + ˆyBz + ˆzCx2z where A = 3 V/m3, B = 2 V/m2, and C = 1 V/m4. What is the charge density at r = ˆx2 −ˆy2 m? Solution. First, we use D = ǫE to get D. Since the problem is in free space, ǫ = ǫ0. Thus we have that the volume charge density is ρv = ∇· D = ∇· (ǫ0E) = ǫ0∇· E = ǫ0 ∂ ∂x	Electromagnetics_Vol1_Page_119_Chunk1420
5.8. FORCE, ENERGY, AND POTENTIAL DIFFERENCE 105 constrain possible solutions for the electric ﬁeld. For that, we might also need Kirchoff’s Voltage Law; see Section 5.11. Before moving on, it is worth noting that Equation 5.66 can be solved in the special case in which there are no boundary conditions to satisfy; i.e., for charge only, in a uniform and unbounded medium. In fact, no additional electromagnetic theory is required to do this. Here’s the solution: D(r) = 1 4π Z V r −r′ \|r −r′\|3 ρv(r′) dv (5.70) which we recognize as one of the results obtained in Section 5.4 (after dividing both sides by ǫ to get E). It is reasonable to conclude that Gauss’ Law (in either integral or differential form) is fundamental, whereas Coulomb’s Law is merely a consequence of Gauss’ Law. Additional Reading: • “Gauss’ Law” on Wikipedia. • “Partial differential equation” on Wikipedia. • “Boundary value problem” on Wikipedia. 5.8 Force, Energy, and Potential Difference [m0061] The force Fe experienced by a particle at location r bearing charge q in an electric ﬁeld intensity E is (see Sections 2.2 and/or Section 5.1) Fe = qE(r) (5.71) If left alone in free space, this particle would immediately begin to move. The resulting displacement represents a loss of potential energy. This loss can quantiﬁed using the concept of work, W. The incremental work ∆W done by moving the particle a short distance ∆l, over which we assume the change in Fe is negligible, is ∆W ≈−Fe ·ˆl∆l (5.72) where in this case ˆl is the unit vector in the direction of the motion; i.e., the direction of Fe. The minus sign indicates that potential energy of the system consisting of the electric ﬁeld and the particle is being reduced. Like a spring that was previously compressed and is now released, the system is “relaxing.” To conﬁrm that work deﬁned in this way is an expression of energy, consider the units. The product of force (units of N) and distance (units of m) has units of N·m, and 1 N·m is 1 J of energy. Now, what if the motion of the particle is due to factors other than the force associated with the electric ﬁeld? For example, we might consider “resetting” the system to it’s original condition by applying an external force to overcome Fe. Equation 5.72 still represents the change in potential energy of the system, but now ˆl changes sign. The same magnitude of work is done, but now this work is positive. In other words, positive work requires the application of an external force that opposes and overcomes the force associated with the electric ﬁeld, thereby increasing the potential energy of the system. With respect to the analogy of a mechanical spring used above, positive work is achieved by compressing the spring.	Electromagnetics_Vol1_Page_120_Chunk1421
106 CHAPTER 5. ELECTROSTATICS It is also worth noting that the purpose of the dot product in Equation 5.72 is to ensure that only the component of motion parallel to the direction of the electric ﬁeld is included in the energy tally. This is simply because motion in any other direction cannot be due to E, and therefore does not increase or decrease the associated potential energy. We can make the relationship between work and the electric ﬁeld explicit by substituting Equation 5.71 into Equation 5.72, yielding ∆W ≈−qE(r) ·ˆl∆l (5.73) Equation 5.73 gives the work only for a short distance around r. Now let us try to generalize this result. If we wish to know the work done over a larger distance, then we must account for the possibility that E varies along the path taken. To do this, we may sum contributions from points along the path traced out by the particle, i.e., W ≈ N X n=1 ∆W(rn) (5.74) where rn are positions deﬁning the path. Substituting Equation 5.73, we have W ≈−q N X n=1 E(rn) ·ˆl(rn)∆l (5.75) Taking the limit as ∆l →0 we obtain W = −q Z C E(r) ·ˆl(r)dl (5.76) where C is the path (previously, the sequence of rn’s) followed. Now omitting the explicit dependence on r in the integrand for clarity: W = −q Z C E · dl (5.77) where dl = ˆldl as usual. Now, we are able to determine the change in potential energy for a charged particle moving along any path in space, given the electric ﬁeld. At this point, it is convenient to formally deﬁne the electric potential difference V21 between the start point (1) and end point (2) of C. V21 is deﬁned as the work done by traversing C, per unit of charge: V21 ≜W q (5.78) This has units of J/C, which is volts (V). Substituting Equation 5.77, we obtain: V21 = − Z C E · dl (5.79) An advantage of analysis in terms of electrical potential as opposed to energy is that we will no longer have to explicitly state the value of the charge involved. The potential difference V21 between two points in space, given by Equation 5.79, is the change in potential energy of a charged particle divided by the charge of the particle. Potential energy is also commonly known as “voltage” and has units of V. Example 5.7. Potential difference in a uniform electric ﬁeld. Consider an electric ﬁeld E(r) = ˆzE0, which is constant in both magnitude and direction throughout the domain of the problem. The path of interest is a line beginning at ˆzz1 and ending at ˆzz2. What is V21? (It’s worth noting that the answer to this problem is a building block for a vast number of problems in electromagnetic analysis.) Solution. From Equation 5.79 we have V21 = − Z z2 z1 (ˆzE0) · ˆzdz = −E0(z2 −z1) (5.80) Note V21 is simply the electric ﬁeld intensity times the distance between the points. This may seem familiar. For example, compare this to the ﬁndings of the battery-charged capacitor experiment described in Section 2.2. There too we ﬁnd that potential difference equals electric ﬁeld intensity times distance, and the signs agree.	Electromagnetics_Vol1_Page_121_Chunk1422
5.9. INDEPENDENCE OF PATH 107 The solution to the preceding example is simple because the direct path between the two points is parallel to the electric ﬁeld. If the path between the points had been perpendicular to E, then the solution is even easier – V21 is simply zero. In all other cases, V21 is proportional to the component of the direct path between the start and end points that is parallel to E, as determined by the dot product. 5.9 Independence of Path [m0062] In Section 5.8, we found that the potential difference (“voltage”) associated with a path C in an electric ﬁeld intensity E is V21 = − Z C E · dl (5.81) where the curve begins at point 1 and ends at point 2. Let these points be identiﬁed using the position vectors r1 and r2, respectively.1 Then: V21 = − Z r2 r1, along C E · dl (5.82) The associated work done by a particle bearing charge q is W21 = qV21 (5.83) This work represents the change in potential energy of the system consisting of the electric ﬁeld and the charged particle. So, it must also be true that W21 = W2 −W1 (5.84) where W2 and W1 are the potential energies when the particle is at r2 and r1, respectively. It is clear from the above equation that W21 does not depend on C; it depends only on the positions of the start and end points and not on any of the intermediate points along C. That is, V21 = − Z r2 r1 E · dl , independent of C (5.85) Since the result of the integration in Equation 5.85 is independent of the path of integration, any path that begins at r1 and ends at r2 yields the same value of W21 and V21. We refer to this concept as independence of path. The integral of the electric ﬁeld over a path be- tween two points depends only on the locations of the start and end points and is independent of the path taken between those points. 1See Section 4.1 for a refresher on this concept.	Electromagnetics_Vol1_Page_122_Chunk1423
108 CHAPTER 5. ELECTROSTATICS A practical application of this concept is that some paths may be easier to use than others, so there may be an advantage in computing the integral in Equation 5.85 using some path other than the path actually traversed. 5.10 Kirchoff’s Voltage Law for Electrostatics: Integral Form [m0016] As explained in Section 5.9, the electrical potential at point r2 relative to r1 in an electric ﬁeld E (V/m) is V21 = − Z r2 r1 E · dl (5.86) where the path of integration may be any path that begins and ends at the speciﬁed points. Consider what happens if the selected path through space begins and ends at the same point; i.e., r2 = r1. In this case, the path of integration is a closed loop. Since V21 depends only on the positions of the start and end points and because the potential energy at those points is the same, we conclude: I E · dl = 0 (5.87) This principle is known as Kirchoff’s Voltage Law for Electrostatics. Kirchoff’s Voltage Law for Electrostatics (Equa- tion 5.87) states that the integral of the electric ﬁeld over a closed path is zero. It is worth noting that this law is a generalization of a principle of which the reader is likely already aware. In electric circuit theory, the sum of voltages over any closed loop in a circuit is zero. This is also known as Kirchoff’s Voltage Law because it is precisely the same principle. To obtain Equation 5.87 for an electric circuit, simply partition the closed path into branches, with each branch representing one component. Then, the integral of E over each branch is the branch voltage; i.e., units of V/m times units of m yields units of V. Then, the sum of these branch voltages over any closed loop is zero, as dictated by Equation 5.87. Finally, be advised that Equation 5.87 is speciﬁc to electrostatics. In electrostatics, it is assumed that the electric ﬁeld is independent of the magnetic ﬁeld. This is true if the magnetic ﬁeld is either zero or not	Electromagnetics_Vol1_Page_123_Chunk1424
5.11. KIRCHOFF’S VOLTAGE LAW FOR ELECTROSTATICS: DIFFERENTIAL FORM 109 time-varying. If the magnetic ﬁeld is time-varying, then Equation 5.87 must be modiﬁed to account for the effect of the magnetic ﬁeld, which is to make the right hand size potentially different from zero. The generalized version of this expression that correctly accounts for that effect is known as the Maxwell-Faraday Equation (Section 8.8). Additional Reading: • “Maxwell’s Equations” on Wikipedia. • “Kirchoff’s Circuit Laws” on Wikipedia. 5.11 Kirchoff’s Voltage Law for Electrostatics: Differential Form [m0152] The integral form of Kirchoff’s Voltage Law for electrostatics (KVL; Section 5.10) states that an integral of the electric ﬁeld along a closed path is equal to zero: I C E · dl = 0 (5.88) where E is electric ﬁeld intensity and C is the closed curve. In this section, we derive the differential form of this equation. In some applications, this differential equation, combined with boundary conditions imposed by structure and materials (Sections 5.17 and 5.18), can be used to solve for the electric ﬁeld in arbitrarily complicated scenarios. A more immediate reason for considering this differential equation is that we gain a little more insight into the behavior of the electric ﬁeld, disclosed at the end of this section. The equation we seek may be obtained using Stokes’ Theorem (Section 4.9), which in the present case may be written: Z S (∇× E) · ds = I C E · dl (5.89) where S is any surface bounded by C, and ds is the normal to that surface with direction determined by right-hand rule. The integral form of KVL tells us that the right hand side of the above equation is zero, so: Z S (∇× E) · ds = 0 (5.90) The above relationship must hold regardless of the speciﬁc location or shape of S. The only way this is possible for all possible surfaces is if the integrand is zero at every point in space. Thus, we obtain the desired expression: ∇× E = 0 (5.91) Summarizing: The differential form of Kirchoff’s Voltage Law for electrostatics (Equation 5.91) states that the curl of the electrostatic ﬁeld is zero.	Electromagnetics_Vol1_Page_124_Chunk1425
110 CHAPTER 5. ELECTROSTATICS Equation 5.91 is a partial differential equation. As noted above, this equation, combined with the appropriate boundary conditions, can be solved for the electric ﬁeld in arbitrarily-complicated scenarios. Interestingly, it is not the only such equation available for this purpose – Gauss’ Law (Section 5.7) also does this. Thus, we see a system of partial differential equations emerging, and one may correctly infer that that the electric ﬁeld is not necessarily fully constrained by either equation alone. Additional Reading: • “Maxwell’s Equations” on Wikipedia. • “Boundary value problem” on Wikipedia. 5.12 Electric Potential Field Due to Point Charges [m0064] The electric ﬁeld intensity due to a point charge q at the origin is (see Section 5.1 or 5.5) E = ˆr q 4πǫr2 (5.92) In Sections 5.8 and 5.9, it was determined that the potential difference measured from position r1 to position r2 is V21 = − Z r2 r1 E · dl (5.93) This method for calculating potential difference is often a bit awkward. To see why, consider an example from circuit theory, shown in Figure 5.5. In this example, consisting of a single resistor and a ground node, we’ve identiﬁed four quantities: • The resistance R • The current I through the resistor • The node voltage V1, which is the potential difference measured from ground to the left side of the resistor • The node voltage V2, which is the potential difference measured from ground to the right side of the resistor Let’s say we wish to calculate the potential difference V21 across the resistor. There are two ways this can be done: • V21 = −IR • V21 = V2 −V1 The advantage of the second method is that it is not necessary to know I, R, or indeed anything about what is happening between the nodes; it is only necessary to know the node voltages. The point is that it is often convenient to have a common datum – in	Electromagnetics_Vol1_Page_125_Chunk1426
5.12. ELECTRIC POTENTIAL FIELD DUE TO POINT CHARGES 111 R I V1 V Õ Ö × Ø Ù ÚÛ Ü ÝÞ ßà áâ ã äå æ ç è éêë ìí îï c⃝K. Kikkeri CC BY SA 4.0 Figure 5.5: A resistor in a larger circuit, used as an example to demonstrate the concept of node voltages. this example, ground – with respect to which the potential differences at all other locations of interest can be deﬁned. When we have this, calculating potential differences reduced to simply subtracting predetermined node potentials. So, can we establish a datum in general electrostatic problems that works the same way? The answer is yes. The datum is arbitrarily chosen to be a sphere that encompasses the universe; i.e., a sphere with radius →∞. Employing this choice of datum, we can use Equation 5.93 to deﬁne V (r), the potential at point r, as follows: V (r) ≜− Z r ∞ E · dl (5.94) The electrical potential at a point, given by Equa- tion 5.94, is deﬁned as the potential difference measured beginning at a sphere of inﬁnite radius and ending at the point r. The potential obtained in this manner is with respect to the potential in- ﬁnitely far away. In the particular case where E is due to the point charge at the origin: V (r) = − Z r ∞ h ˆr q 4πǫr2 i · dl (5.95) The principle of independence of path (Section 5.9) asserts that the path of integration doesn’t matter as long as the path begins at the datum at inﬁnity and ends at r. So, we should choose the easiest such path. The radial symmetry of the problem indicates that the easiest path will be a line of constant θ and φ, so we choose dl = ˆrdr. Continuing: V (r) = − Z r ∞ h ˆr q 4πǫr2 i · [ˆrdr] (5.96) = −q 4πǫ Z r ∞ 1 r2 dr (5.97) = + q 4πǫ 1 r r ∞ (5.98) so V (r) = + q 4πǫr (5.99) (Suggestion: Conﬁrm that Equation 5.99 is dimensionally correct.) In the context of the circuit theory example above, this is the “node voltage” at r when the datum is deﬁned to be the surface of a sphere at inﬁnity. Subsequently, we may calculate the potential difference from any point r1 to any other point r2 as V21 = V (r2) −V (r1) (5.100) and that will typically be a lot easier than using Equation 5.93. It is not often that one deals with systems consisting of a single charged particle. So, for the above technique to be truly useful, we need a straightforward way to determine the potential ﬁeld V (r) for arbitrary distributions of charge. The ﬁrst step in developing a more general expression is to determine the result for a particle located at a point r′ somewhere other than the origin. Since Equation 5.99 depends only on charge and the distance between the ﬁeld point r and r′, we have V (r; r′) ≜+ q′ 4πǫ \|r −r′\| (5.101) where, for notational consistency, we use the symbol q′ to indicate the charge. Now applying superposition, the potential ﬁeld due to N charges is V (r) = N X n=1 V (r; rn) (5.102) Substituting Equation 5.101 we obtain: V (r) = 1 4πǫ N X n=1 qn \|r −rn\| (5.103)	Electromagnetics_Vol1_Page_126_Chunk1427
112 CHAPTER 5. ELECTROSTATICS Equation 5.103 gives the electric potential at a speciﬁed location due to a ﬁnite number of charged particles. The potential ﬁeld due to continuous distributions of charge is addressed in Section 5.13. 5.13 Electric Potential Field due to a Continuous Distribution of Charge [m0065] The electrostatic potential ﬁeld at r associated with N charged particles is V (r) = 1 4πǫ N X n=1 qn \|r −rn\| (5.104) where qn and rn are the charge and position of the nth particle. However, it is more common to have a continuous distribution of charge as opposed to a countable number of charged particles. We now consider how to compute V (r) three types of these commonly-encountered distributions. Before beginning, it’s worth noting that the methods will be essentially the same, from a mathematical viewpoint, as those developed in Section 5.4; therefore, a review of that section may be helpful before attempting this section. Continuous Distribution of Charge Along a Curve. Consider a continuous distribution of charge along a curve C. The curve can be divided into short segments of length ∆l. Then, the charge associated with the nth segment, located at rn, is qn = ρl(rn) ∆l (5.105) where ρl is the line charge density (units of C/m) at rn. Substituting this expression into Equation 5.104, we obtain V(r) = 1 4πǫ N X n=1 ρl(rn) \|r −rn\|∆l (5.106) Taking the limit as ∆l →0 yields: V (r) = 1 4πǫ Z C ρl(l) \|r −r′\|dl (5.107) where r′ represents the varying position along C with integration along the length l. Continuous Distribution of Charge Over a Surface. Consider a continuous distribution of charge	Electromagnetics_Vol1_Page_127_Chunk1428
5.14. ELECTRIC FIELD AS THE GRADIENT OF POTENTIAL 113 over a surface S. The surface can be divided into small patches having area ∆s. Then, the charge associated with the nth patch, located at rn, is qn = ρs(rn) ∆s (5.108) where ρs is surface charge density (units of C/m2) at rn. Substituting this expression into Equation 5.104, we obtain V (r) = 1 4πǫ N X n=1 ρs(rn) \|r −rn\| ∆s (5.109) Taking the limit as ∆s →0 yields: V (r) = 1 4πǫ Z S ρs(r′) \|r −r′\| ds (5.110) where r′ represents the varying position over S with integration. Continuous Distribution of Charge in a Volume. Consider a continuous distribution of charge within a volume V. The volume can be divided into small cells (volume elements) having area ∆v. Then, the charge associated with the nth cell, located at rn, is qn = ρv(rn) ∆v (5.111) where ρv is the volume charge density (units of C/m3) at rn. Substituting this expression into Equation 5.104, we obtain V (r) = 1 4πǫ N X n=1 ρv(rn) \|r −rn\| ∆v (5.112) Taking the limit as ∆v →0 yields: V (r) = 1 4πǫ Z V ρv(r′) \|r −r′\| dv (5.113) where r′ represents the varying position over V with integration. 5.14 Electric Field as the Gradient of Potential [m0063] In Section 5.8, it was determined that the electrical potential difference V21 measured over a path C is given by V21 = − Z C E(r) · dl (5.114) where E(r) is the electric ﬁeld intensity at each point r along C. In Section 5.12, we deﬁned the scalar electric potential ﬁeld V (r) as the electric potential difference at r relative to a datum at inﬁnity. In this section, we address the “inverse problem” – namely, how to calculate E(r) given V (r). Speciﬁcally, we are interested in a direct “point-wise” mathematical transform from one to the other. Since Equation 5.114 is in the form of an integral, it should not come as a surprise that the desired expression will be in the form of a differential equation. We begin by identifying the contribution of an inﬁnitesimal length of the integral to the total integral in Equation 5.114. At point r, this is dV = −E(r) · dl (5.115) Although we can proceed using any coordinate system, the following derivation is particularly simple in Cartesian coordinates. In Cartesian coordinates, dl = ˆxdx + ˆydy + ˆzdz (5.116) We also note that for any scalar function of position, including V (r), it is true that dV = ∂V ∂x dx + ∂V ∂y dy + ∂V ∂z dz (5.117) Note the above relationship is not speciﬁc to electromagnetics; it is simply mathematics. Also note that dx = dl · ˆx and so on for dy and dz. Making these substitutions into the above equation, we obtain: dV = ∂V ∂x (dl · ˆx) + ∂V ∂y (dl · ˆy) + ∂V ∂z (dl · ˆz) (5.118) This equation may be rearranged as follows: dV = ˆx ∂ ∂x + ˆy ∂ ∂y + ˆz ∂ ∂z V · dl (5.119)	Electromagnetics_Vol1_Page_128_Chunk1429
114 CHAPTER 5. ELECTROSTATICS Comparing the above equation to Equation 5.115, we ﬁnd: E(r) = − ˆx ∂ ∂x + ˆy ∂ ∂y + ˆz ∂ ∂z V (5.120) Note that the quantity in square brackets is the gradient operator “∇” (Section 4.5). Thus, we may write E = −∇V (5.121) which is the relationship we seek. The electric ﬁeld intensity at a point is the gra- dient of the electric potential at that point after a change of sign (Equation 5.121). Using Equation 5.121, we can immediately ﬁnd the electric ﬁeld at any point r if we can describe V as a function of r. Furthermore, this relationship between V and E has a useful physical interpretation. Recall that the gradient of a scalar ﬁeld is a vector that points in the direction in which that ﬁeld increases most quickly. Therefore: The electric ﬁeld points in the direction in which the electric potential most rapidly decreases. This result should not come as a complete surprise; for example, the reader should already be aware that the electric ﬁeld points away from regions of net positive charge and toward regions of net negative charge (Sections 2.2 and/or 5.1). What is new here is that both the magnitude and direction of the electric ﬁeld may be determined given only the potential ﬁeld, without having to consider the charge that is the physical source of the electrostatic ﬁeld. Example 5.8. Electric ﬁeld of a charged particle, beginning with the potential ﬁeld. In this example, we determine the electric ﬁeld of a particle bearing charge q located at the origin. This may be done in a “direct” fashion using Coulomb’s Law (Section 5.1). However, here we have the opportunity to ﬁnd the electric ﬁeld using a different method. In Section 5.12 we found the scalar potential for this source was: V (r) = q 4πǫr (5.122) So, we may obtain the electric ﬁeld using Equation 5.121: E = −∇V = −∇ q 4πǫr (5.123) Here V (r) is expressed in spherical coordinates, so we have (Section B.2): E = − ˆr ∂ ∂r + ˆθ1 r ∂ ∂θ + ˆφ 1 r sin θ ∂ ∂φ q 4πǫr (5.124) In this case, V (r) does not vary with φ or θ, so the second and third terms of the gradient are zero. This leaves E = −ˆr ∂ ∂r q 4πǫr = −ˆr q 4πǫ ∂ ∂r 1 r = −ˆr q 4πǫ −1 r2 (5.125) So we ﬁnd E = +ˆr q 4πǫr2 (5.126) as was determined in Section 5.1.	Electromagnetics_Vol1_Page_129_Chunk1430
5.15. POISSON’S AND LAPLACE’S EQUATIONS 115 5.15 Poisson’s and Laplace’s Equations [m0067] The electric scalar potential ﬁeld V (r), deﬁned in Section 5.12, is useful for a number of reasons including the ability to conveniently compute potential differences (i.e., V21 = V (r2) −V (r1)) and the ability to conveniently determine the electric ﬁeld by taking the gradient (i.e., E = −∇V ). One way to obtain V (r) is by integration over the source charge distribution, as described in Section 5.13. This method is awkward in the presence of material interfaces, which impose boundary conditions on the solutions that must be satisﬁed simultaneously. For example, the electric potential on a perfectly conducting surface is constant2 – a constraint which is not taken into account in any of the expressions in Section 5.13. In this section, we develop an alternative approach to calculating V (r) that accommodates these boundary conditions, and thereby facilitates the analysis of the scalar potential ﬁeld in the vicinity of structures and spatially-varying material properties. This alternative approach is based on Poisson’s Equation, which we now derive. We begin with the differential form of Gauss’ Law (Section 5.7): ∇· D = ρv (5.127) Using the relationship D = ǫE (and keeping in mind our standard assumptions about material properties, summarized in Section 2.8) we obtain ∇· E = ρv ǫ (5.128) Next, we apply the relationship (Section 5.14): E = −∇V (5.129) yielding ∇· ∇V = −ρv ǫ (5.130) This is Poisson’s Equation, but it is not in the form in which it is commonly employed. To obtain the 2This fact is probably already known to the reader from past study of elementary circuit theory; however, this is established in the context of electromagnetics in Section 5.19. alternative form, consider the operator ∇· ∇in Cartesian coordinates: ∇· ∇= ∂ ∂x ˆx + ∂ ∂y ˆy + ∂ ∂z ˆz · ∂ ∂x ˆx + ∂ ∂y ˆy + ∂ ∂z ˆz = ∂2 ∂x2 + ∂2 ∂y2 + ∂2 ∂z2 = ∇2 (5.131) i.e., the operator ∇· ∇is identically the Laplacian operator ∇2 (Section 4.10). Furthermore, this is true regardless of the coordinate system employed. Thus, we obtain the following form of Poisson’s Equation: ∇2V = −ρv ǫ (5.132) Poisson’s Equation (Equation 5.132) states that the Laplacian of the electric potential ﬁeld is equal to the volume charge density divided by the permittivity, with a change of sign. Note that Poisson’s Equation is a partial differential equation, and therefore can be solved using well-known techniques already established for such equations. In fact, Poisson’s Equation is an inhomogeneous differential equation, with the inhomogeneous part −ρv/ǫ representing the source of the ﬁeld. In the presence of material structure, we identify the relevant boundary conditions at the interfaces between materials, and the task of ﬁnding V (r) is reduced to the purely mathematical task of solving the associated boundary value problem (see “Additional Reading” at the end of this section). This approach is particularly effective when one of the materials is a perfect conductor or can be modeled as such a material. This is because – as noted at the beginning of this section – the electric potential at all points on the surface of a perfect conductor must be equal, resulting in a particularly simple boundary condition. In many other applications, the charge responsible for the electric ﬁeld lies outside the domain of the problem; i.e., we have non-zero electric ﬁeld (hence, potentially non-zero electric potential) in a region that is free of charge. In this case, Poisson’s Equation simpliﬁes to Laplace’s Equation: ∇2V = 0 (source-free region) (5.133)	Electromagnetics_Vol1_Page_130_Chunk1431
116 CHAPTER 5. ELECTROSTATICS Laplace’s Equation (Equation 5.133) states that the Laplacian of the electric potential ﬁeld is zero in a source-free region. Like Poisson’s Equation, Laplace’s Equation, combined with the relevant boundary conditions, can be used to solve for V (r), but only in regions that contain no charge. Additional Reading: • “Poisson’s equation” on Wikipedia. • “Boundary value problem” on Wikipedia. • “Laplace’s equation” on Wikipedia. 5.16 Potential Field Within a Parallel Plate Capacitor [m0068] This section presents a simple example that demonstrates the use of Laplace’s Equation (Section 5.15) to determine the potential ﬁeld in a source free region. The example, shown in Figure 5.6, pertains to an important structure in electromagnetic theory – the parallel plate capacitor. Here we are concerned only with the potential ﬁeld V (r) between the plates of the capacitor; you do not need to be familiar with capacitance or capacitors to follow this section (although you’re welcome to look ahead to Section 5.22 for a preview, if desired). What is recommended before beginning is a review of the battery-charged capacitor experiment discussed in Section 2.2. In this section you’ll see a rigorous derivation of what we ﬁgured out in an informal way in that section. The parallel-plate capacitor in Figure 5.6 consists of two perfectly-conducting circular disks separated by a distance d by a spacer material having permittivity ǫ. There is no charge present in the spacer material, so Laplace’s Equation applies. That equation is (Section 5.15): ∇2V = 0 (source-free region) (5.134) Let VC be the potential difference between the plates, which would also be the potential difference across the terminals of the capacitor. The radius a of the plates is larger than d by enough that we may neglect what is going on at at the edges of the plates – more on this will be said as we work the problem. Under this assumption, what is the electric potential ﬁeld V (r) between the plates? This problem has cylindrical symmetry, so it makes sense to continue to use cylindrical coordinates with the z axis being perpendicular to the plates. Equation 5.134 in cylindrical coordinates is: 1 ρ ∂ ∂ρ ρ ∂ ∂ρ + 1 ρ2 ∂2 ∂φ2 + ∂2 ∂z2 V = 0 (5.135) or perhaps a little more clearly written as follows: 1 ρ ∂ ∂ρ ρ∂V ∂ρ + 1 ρ2 ∂2V ∂φ2 + ∂2V ∂z2 = 0 (5.136)	Electromagnetics_Vol1_Page_131_Chunk1432
5.16. POTENTIAL FIELD WITHIN A PARALLEL PLATE CAPACITOR 117 d z ρ a V-+VC V- Figure 5.6: A parallel plate capacitor, as a demonstra- tion of the use of Laplace’s Equation. Since the problem has radial symmetry, ∂V/∂φ = 0. Since d ≪a, we expect the ﬁelds to be approximately constant with ρ until we get close to the edge of the plates. Therefore, we assume ∂V/∂ρ is negligible and can be taken to be zero. Thus, we are left with ∂2V ∂z2 ≈0 for ρ ≪a (5.137) The general solution to Equation 5.137 is obtained simply by integrating both sides twice, yielding V (z) = c1z + c2 (5.138) where c1 and c2 are constants that must be consistent with the boundary conditions. Thus, we must develop appropriate boundary conditions. Let the node voltage at the negative (z = 0) terminal be V−. Then the voltage at the positive (z = +d) terminal is V−+ VC. Therefore: V (z = 0) = V− (5.139) V (z = +d) = V−+ VC (5.140) These are the relevant boundary conditions. Substituting V (z = 0) = V−into Equation 5.138 yields c2 = V−. Substituting V (z = +d) = V−+ VC into Equation 5.138 yields c1 = VC/d. Thus, the answer to the problem is V (z) ≈VC d z + V−for ρ ≪a (5.141) Note that the above result is dimensionally correct and conﬁrms that the potential deep inside a “thin” parallel plate capacitor changes linearly with distance between the plates. Further, you should ﬁnd that application of the equation E = −∇V (Section 5.14) to the solution above yields the expected result for the electric ﬁeld intensity: E ≈−ˆzVC/d. This is precisely the result that we arrived at (without the aid of Laplace’s Equation) in Section 2.2. A reasonable question to ask at this point would be, what about the potential ﬁeld close to the edge of the plates, or, for that matter, beyond the plates? The ﬁeld in this region is referred to as a fringing ﬁeld. For the fringing ﬁeld, ∂V/∂ρ is no longer negligible and must be taken into account. In addition, it is necessary to modify the boundary conditions to account for the outside surfaces of the plates (that is, the sides of the plates that face away from the dielectric) and to account for the effect of the boundary between the spacer material and free space. These issues make the problem much more difﬁcult. When an accurate calculation of a fringing ﬁeld is necessary, it is common to resort to a numerical solution of Laplace’s Equation. Fortunately, accurate calculation of fringing ﬁelds is usually not required in practical engineering applications.	Electromagnetics_Vol1_Page_132_Chunk1433
118 CHAPTER 5. ELECTROSTATICS 5.17 Boundary Conditions on the Electric Field Intensity (E) [m0020] In homogeneous media, electromagnetic quantities vary smoothly and continuously. At an interface between dissimilar media, however, it is possible for electromagnetic quantities to be discontinuous. These discontinuities can be described mathematically as boundary conditions and used to to constrain solutions for the associated electromagnetic quantities. In this section, we derive boundary conditions on the electric ﬁeld intensity E. To begin, consider a region consisting of only two media that meet at an interface deﬁned by the mathematical surface S, as shown in Figure 5.7. If either one of the materials is a perfect electrical conductor (PEC), then S is an equipotential surface; i.e., the electric potential V is constant everywhere on S. Since E is proportional to the spatial rate of change of potential (recall E = −∇V ; Section 5.14), we ﬁnd: The component of E that is tangent to a perfectly- conducting surface is zero. This is sometimes expressed informally as follows: Etan = 0 on PEC surface (5.142) where “Etan” is understood to be the component of E that is tangent to S. Since the tangential component PEC E E E E (equipotential) c⃝K. Kikkeri CC BY SA 4.0 Figure 5.7: At the surface of a perfectly-conducting region, E may be perpendicular to the surface (two leftmost possibilities), but may not exhibit a compo- nent that is tangent to the surface (two rightmost pos- sibilities). B A w t c⃝K. Kikkeri CC BY SA 4.0 Figure 5.8: Use of KVL to determine the boundary condition on E. of E on the surface of a perfect conductor is zero, the electric ﬁeld at the surface must be oriented entirely in the direction perpendicular to the surface, as shown in Figure 5.7. The following equation expresses precisely the same idea, but includes the calculation of the tangential component as part of the statement: E × ˆn = 0 (on PEC surface) (5.143) where ˆn is either normal (i.e., unit vector perpendicular to the surface) to each point on S. This expression works because the cross product of any two vectors is perpendicular to either vector (Section 4.1), and any vector which is perpendicular to ˆn is tangent to S. We now determine a more general boundary condition that applies even when neither of the media bordering S is a perfect conductor. The desired boundary condition can be obtained directly from Kirchoff’s Voltage Law (KVL; Section 5.10): I C E · dl = 0 (5.144) Let the closed path of integration take the form of a rectangle centered on S, as shown in Figure 5.8. Let the sides A, B, C, and D be perpendicular or parallel to the surface, respectively. Let the length of the perpendicular sides be w, and let the length of the	Electromagnetics_Vol1_Page_133_Chunk1434
5.17. BOUNDARY CONDITIONS ON THE ELECTRIC FIELD INTENSITY (E) 119 parallel sides be l. From KVL we have I C E · dl = Z A E · dl + Z B E · dl + Z C E · dl + Z D E · dl = 0 (5.145) Now, let us reduce w and l together while (1) maintaining a constant ratio w/l ≪1 and (2) keeping C centered on S. In this process, the contributions from the B and D segments become equal in magnitude but opposite in sign; i.e., Z B E · dl + Z D E · dl →0 (5.146) This leaves I C E · dl → Z A E · dl + Z C E · dl →0 (5.147) Let us deﬁne the unit vector ˆt (“tangent”) as shown in Figure 5.8. When the lengths of sides A and C become sufﬁciently small, we can write the above expression as follows: E1 · ˆt∆l −E2 · ˆt∆l →0 (5.148) where E1 and E2 are the ﬁelds evaluated on the two sides of the boundary and ∆l →0 is the length of sides A and C while this is happening. Note that the only way Equation 5.148 can be true is if the tangential components of E1 and E2 are equal. In other words: The tangential component of E must be continu- ous across an interface between dissimilar media. Note that this is a generalization of the result we obtained earlier for the case in which one of the media was a PEC – in that case, the tangent component of E on the other side of the interface must be zero because it is zero in the PEC medium. As before, we can express this idea in compact mathematical notation. Using the same idea used to obtain Equation 5.143, we have found E1 × ˆn = E2 × ˆn on S (5.149) or, as it is more commonly written: ˆn × (E1 −E2) = 0 on S (5.150) We conclude this section with a note about the broader applicability of this boundary condition: Equation 5.150 is the boundary condition that ap- plies to E for both the electrostatic and the gen- eral (time-varying) case. Although a complete explanation is not possible without the use of the Maxwell-Faraday Equation (Section 8.8), the reason why this boundary condition applies in the time-varying case can be disclosed here. In the presence of time-varying magnetic ﬁelds, the right-hand side of Equation 5.144 may become non-zero and is proportional to the area deﬁned by the closed loop. However, the above derivation requires the area of this loop to approach zero, in which case the possible difference from Equation 5.144 also converges to zero. Therefore, the boundary condition expressed in Equation 5.150 applies generally.	Electromagnetics_Vol1_Page_134_Chunk1435
120 CHAPTER 5. ELECTROSTATICS 5.18 Boundary Conditions on the Electric Flux Density (D) [m0021] In this section, we derive boundary conditions on the electric ﬂux density D. The considerations are quite similar to those encountered in the development of boundary conditions on the electric ﬁeld intensity (E) in Section 5.17, so the reader may ﬁnd it useful to review that section before attempting this section. This section also assumes familiarity with the concepts of electric ﬂux, electric ﬂux density, and Gauss’ Law; for a refresher, Sections 2.4 and 5.5 are suggested. To begin, consider a region at which two otherwise-homogeneous media meet at an interface deﬁned by the mathematical surface S, as shown in Figure 5.9. Let one of these regions be a perfect electrical conductor (PEC). In Section 5.17, we established that the tangential component of the electric ﬁeld must be zero, and therefore, the electric ﬁeld is directed entirely in the direction perpendicular to the surface. We further know that the electric ﬁeld within the conductor is identically zero. Therefore, D at any point on S is entirely in the direction perpendicular to the surface and pointing into the non-conducting medium. However, it is also possible to determine the magnitude of D. We shall demonstrate in this section that At the surface of a perfect conductor, the magni- tude of D is equal to the surface charge density ρs (units of C/m2) at that point. PEC D=nρs c⃝K. Kikkeri CC BY SA 4.0 Figure 5.9: The component of D that is perpendicular to a perfectly-conducting surface is equal to the charge density on the surface. region 1 region 2 n a h h c⃝K. Kikkeri CC BY SA 4.0 Figure 5.10: Use of Gauss’ Law to determine the boundary condition on D. The following equation expresses precisely the same idea, but includes the calculation of the perpendicular component as part of the statement: D · ˆn = ρs (on PEC surface) (5.151) where ˆn is the normal to S pointing into the non-conducting region. (Note that the orientation of ˆn is now important; we have assumed ˆn points into region 1, and we must now stick with this choice.) Before proceeding with the derivation, it may be useful to note that this result is not surprising. The very deﬁnition of electric ﬂux (Section 2.4) indicates that D should correspond in the same way to a surface charge density. However, we can show this rigorously, and in the process we can generalize this result to the more-general case in which neither of the two materials are PEC. The desired more-general boundary condition may be obtained from the integral form of Gauss’ Law (Section 5.5), as illustrated in Figure 5.10. Let the surface of integration S′ take the form of closed cylinder centered at a point on the interface and for which the ﬂat ends are parallel to the surface and perpendicular to ˆn. Let the radius of this cylinder be a, and let the length of the cylinder be 2h. From Gauss’ Law we have I S′ D · ds = Z top D · ds + Z side D · ds + Z bottom D · ds = Qencl (5.152)	Electromagnetics_Vol1_Page_135_Chunk1436
5.18. BOUNDARY CONDITIONS ON THE ELECTRIC FLUX DENSITY (D) 121 where the “top” and “bottom” are in Regions 1 and 2, respectively, and Qencl is the charge enclosed by S′. Now let us reduce h and a together while (1) maintaining a constant ratio h/a ≪1 and (2) keeping S′ centered on S. Because h ≪a, the area of the side can be made negligible relative to the area of the top and bottom. Then as h →0 we are left with Z top D · ds + Z bottom D · ds →Qencl (5.153) As the area of the top and bottom sides become inﬁnitesimal, the variation in D over these areas becomes negligible. Now we have simply: D1 · ˆn∆A + D2 · (−ˆn) ∆A →Qencl (5.154) where D1 and D2 are the electric ﬂux density vectors in medium 1 and medium 2, respectively, and ∆A is the area of the top and bottom sides. The above expression can be rewritten ˆn · (D1 −D2) →Qencl ∆A (5.155) Note that the left side of the equation must represent a actual, physical surface charge; this is apparent from dimensional analysis and the fact that h is now inﬁnitesimally small. Therefore: ˆn · (D1 −D2) = ρs (5.156) where, as noted above, ˆn points into region 1. Summarizing: Any discontinuity in the normal component of the electric ﬂux density across the boundary be- tween two material regions is equal to the surface charge. Now let us verify that this is consistent with our preliminary ﬁnding, in which Region 2 was a PEC. In that case D2 = 0, so we see that Equation 5.151 is satisﬁed, as expected. If neither Region 1 nor Region 2 is PEC and there is no surface charge on the interface, then we ﬁnd ˆn · (D1 −D2) = 0; i.e., In the absence of surface charge, the normal com- ponent of the electric ﬂux density must be contin- uous across the boundary. Finally, we note that since D = ǫE, Equation 5.156 implies the following boundary condition on E: ˆn · (ǫ1E1 −ǫ2E2) = ρs (5.157) where ǫ1 and ǫ2 are the permittivities in Regions 1 and 2, respectively. The above equation illustrates one reason why we sometimes prefer the “ﬂux” interpretation of the electric ﬁeld to the “ﬁeld intensity” interpretation of the electric ﬁeld.	Electromagnetics_Vol1_Page_136_Chunk1437
122 CHAPTER 5. ELECTROSTATICS 5.19 Charge and Electric Field for a Perfectly Conducting Region [m0025] In this section, we consider the behavior of charge and the electric ﬁeld in the vicinity of a perfect electrical conductor (PEC). First, note that the electric ﬁeld – both the electric ﬁeld intensity E and electric ﬂux density D – throughout a PEC region is zero. This is because the electrical potential throughout a PEC region must be constant. (This idea is explored further in Section 6.3.) Recall that the electric ﬁeld is proportional to the spatial rate of change of electrical potential (i.e., E = −∇V ; Section 5.14). Thus, the electric ﬁeld must be zero throughout a PEC region. Second, the electric ﬁeld is oriented directly away from (i.e., perpendicular to) the PEC surface, and the magnitude of D is equal to the surface charge density ρs (C/m2) (Section 5.18). Now we address the question of charge distribution; i.e., the location and density of charge. Consider the scenario shown in Figure 5.11. Here, a ﬂat slab of PEC material is embedded in dielectric material.3 The 3For the purposes of this section, it sufﬁces to interpret “dielec- tric” as a “nonconducting and well characterized entirely in terms of c⃝K. Kikkeri CC BY SA 4.0 Figure 5.11: An inﬁnite ﬂat slab of PEC in the pres- ence of an applied electric ﬁeld. thickness of the slab is ﬁnite, whereas the length and width of the slab is inﬁnite. The region above the slab is deﬁned as Region 1 and has permittivity ǫ1. The region below the slab is deﬁned as Region 2 and has permittivity ǫ2. Electric ﬁelds E1 and E2 are present in Regions 1 and 2, respectively, as shown in Figure 5.11. To begin, let us assume that these ﬁelds are the result of some external stimulus that results in the direction of these ﬁelds being generally upward, as shown in Figure 5.11. Now, what do we know about E1 and E2? First, both ﬁelds must satisfy the relevant boundary conditions. That is, the component of E1 that is tangent to the upper PEC surface is zero, so that E1 is directed entirely in a direction perpendicular to the surface. Similarly, the component of E2 that is tangent to the lower PEC surface is zero, so that E2 is directed entirely in a direction perpendicular to the surface. At this point we have not determined the magnitudes or signs of E1 and E2; we have established only that there are no non-zero components tangential to (i.e., parallel to) the PEC surfaces. Next, recall that the electric ﬁeld must be zero within the slab. This means that there must be zero net charge within the slab, since any other distribution of charge will result in a non-zero electric ﬁeld, and subsequently a potential difference between locations within the slab. Therefore: There can be no static charge within a PEC. It follows that Charge associated with a PEC lies entirely on the surface. Outside the slab, the boundary conditions on D1 and D2 in the dielectric regions require these ﬁelds to be non-zero when the surface charge density on the PEC is non-zero. The surface charge supports the discontinuity in the normal component of the electric ﬁelds. Speciﬁcally, D1 and D2 have the same magnitude \|ρs\| because the surface charge densities on both sides of the slab have equal magnitude. However, the electric ﬁeld intensity E1 = D1/ǫ1, whereas E2 = D2/ǫ2; i.e., these are different. That its permittivity.” For more, see Section 5.20.	Electromagnetics_Vol1_Page_137_Chunk1438
5.20. DIELECTRIC MEDIA 123 − −−−−−−−− + ++ + + + + + −−− − + + + + + − − −−−− − + + + + ++ + + by C. Burks (modiﬁed) Figure 5.12: Electric ﬁeld lines due to a point charge in the vicinity of PEC regions (shaded) of various shapes. is, the electric ﬁeld intensities are unequal unless the permittivities in each dielectric region are equal. Finally, let us consider the structure of the electric ﬁeld in more general cases. Figure 5.12 shows ﬁeld lines in a homogeneous dielectric material in which a point charge and PEC regions of various shapes are embedded. Note that electric ﬁeld lines now bend in the dielectric so as to satisfy the requirement that the tangential component of the electric ﬁeld be zero on PEC surfaces. Also note that the charge distribution arranges itself on the PEC surfaces so as to maintain zero electric ﬁeld and constant potential within the cube. 5.20 Dielectric Media [m0107] Dielectric is particular category of materials that exhibit low conductivity4 because their constituent molecules remain intact when exposed to an electric ﬁeld, as opposed to shedding electrons as is the case in good conductors. Subsequently, dielectrics do not effectively pass current, and are therefore considered “good insulators” as well as “poor conductors.” An important application of dielectrics in electrical engineering is as a spacer material in printed circuit boards (PCBs), coaxial cables, and capacitors. Examples of dielectrics include air, glass, teﬂon, and ﬁberglass epoxy (the material used in common “FR4” printed circuit boards). These and other dielectrics are listed along with values of their constitutive parameters in Section A.1. The electromagnetic properties of dielectric materials are quantiﬁed primarily by relative permittivity ǫr (Section 2.3), which ranges from very close to 1 upward to roughly 50, and is less than 6 or so for most commonly-encountered materials having low moisture content. The permeability of dielectric materials is approximately equal to the free-space value (i.e., µ ≈µ0); therefore, these materials are said to be “non-magnetic.” Additional Reading: • “Dielectric” on Wikipedia. 4See Section 2.8 for a refresher on parameters describing prop- erties of materials.	Electromagnetics_Vol1_Page_138_Chunk1439
124 CHAPTER 5. ELECTROSTATICS 5.21 Dielectric Breakdown [m0109] The permittivity of an ideal dielectric is independent of the magnitude of an applied electric ﬁeld; the material is said to be “linear.”5 However, all practical dielectrics fail in this respect with sufﬁciently strong electric ﬁeld. Typically, the failure is abrupt and is observed as a sudden, dramatic increase in conductivity, signaling that electrons are being successfully dislodged from their host molecules. The threshold value of the electric ﬁeld intensity at which this occurs is known as the dielectric strength, and the sudden change in behavior observed in the presence of an electric ﬁeld greater than this threshold value is known as dielectric breakdown. Dielectric strength varies from about 3 MV/m for air to about 200 MV/m in mica (a dielectric commonly used in capacitors). Dielectric breakdown is typically accompanied by “arcing,” which is a sudden ﬂow of current associated with the breakdown. A well known example of this phenomenon is lightning, which occurs when charge is exchanged between sky and ground when air (a dielectric) exhibits breakdown. Dielectric breakdown in solids typically damages the material. Additional Reading: • “Electrical Breakdown” on Wikipedia. 5See Section 2.8 for a review of this concept. 5.22 Capacitance [m0112] When separate regions of positive and negative charge exist in proximity, Coulomb forces (Section 5.1) will attempt to decrease the separation between the charges. As noted in Section 5.8, this can be interpreted as a tendency of a system to reduce its potential energy. If the charges are ﬁxed in place, then the potential energy remains constant. This potential energy is proportional to the Coulomb force. Referring back to Section 5.1, the Coulomb force is: • Proportional to quantity of positive charge squared • Inversely proportional to the separation between the charges squared • Inversely proportional to the permittivity of the material separating the charges Therefore, the potential energy of the system is likewise dependent on charge, separation, and permittivity. Furthermore, we see that the ability of a system to store energy in this manner depends on the geometry of the charge distribution and the permittivity of the intervening material. Now recall that the electric ﬁeld intensity E is essentially deﬁned in terms of the Coulomb force; i.e., F = qE (Section 2.2). So, rather than thinking of the potential energy of the system as being associated with the Coulomb force, it is equally valid to think of the potential energy as being stored in the electric ﬁeld associated with the charge distribution. It follows from the previous paragraph that the energy stored in the electric ﬁeld depends on the geometry of the charge distribution and the permittivity of the intervening media. This relationship is what we mean by capacitance. We summarize as follows: Capacitance is the ability of a structure to store energy in an electric ﬁeld. and	Electromagnetics_Vol1_Page_139_Chunk1440
5.22. CAPACITANCE 125 The capacitance of a structure depends on its ge- ometry and the permittivity of the medium sepa- rating regions of positive and negative charge. Note that capacitance does not depend on charge, which we view as either a stimulus or response from this point of view. The corresponding response or stimulus, respectively, is the potential associated with this charge. This leads to the following deﬁnition: C ≜Q+ V (5.158) where Q+ (units of C) is the total positive charge, V (units of V) is the potential associated with this charge (deﬁned such that it is positive), and C (units of F) is the associated capacitance. So: In practice, capacitance is deﬁned as the ratio of charge present on one conductor of a two- conductor system to the potential difference be- tween the conductors (Equation 5.158). In other words, a structure is said to have greater capacitance if it stores more charge – and therefore stores more energy – in response to a given potential difference. Figure 5.13 shows the relevant features of this deﬁnition. Here, a battery imposes the potential difference V between two regions of perfectly-conducting material. Q+ is the total charge on the surface of the PEC region attached to the positive terminal of the battery. An equal amount of negative charge appears on the surface of the PEC region attached to the negative terminal of the battery (Section 5.19). This charge distribution gives rise to an electric ﬁeld. Assuming the two PEC regions are ﬁxed in place, Q+ will increase linearly with increasing V , at a rate determined by the capacitance C of the structure. A capacitor is a device that is designed to exhibit a speciﬁed capacitance. We can now make the connection to the concept of the capacitor as it appears in elementary circuit theory. In circuit theory, the behavior of devices is characterized in terms of terminal voltage VT in response to terminal current IT , and vice versa. First, note that current does not ð - ñ ò ó ô õ ö ÷ - - - - - E V øùúû ü ýþÿ c g	Electromagnetics_Vol1_Page_140_Chunk1441
126 CHAPTER 5. ELECTROSTATICS Finally, solving for IT we obtain the differential form of this relationship: IT (t) = C d dtVT (t) (5.162) Additional Reading: • “Capacitance” on Wikipedia. • “Capacitor” on Wikipedia. 5.23 The Thin Parallel Plate Capacitor [m0070] Let us now determine the capacitance of a common type of capacitor known as the thin parallel plate capacitor, shown in Figure 5.14. This capacitor consists of two ﬂat plates, each having area A, separated by distance d. To facilitate discussion, let us place the origin of the coordinate system at the center of the lower plate, with the +z axis directed toward the upper plate such that the upper plate lies in the z = +d plane. Below we shall ﬁnd the capacitance by assuming a particular charge on one plate, using the boundary condition on the electric ﬂux density D to relate this charge density to the internal electric ﬁeld, and then integrating over the electric ﬁeld between the plates to obtain the potential difference. Then, capacitance is the ratio of the assumed charge to the resulting potential difference. The principal difﬁculty in this approach is ﬁnding the electric ﬁeld. To appreciate the problem, ﬁrst consider that if the area of the plates was inﬁnite, then the electric ﬁeld would be very simple; it would begin at the positively-charged plate and extend in a perpendicular direction toward the negatively-charged plate (Section 5.19). Furthermore, the ﬁeld would be constant everywhere between the plates. This much is apparent from symmetry alone. However, when the plate area is ﬁnite, then we expect a fringing ﬁeld to emerge. “Fringing ﬁeld” is simply a term applied to the non-uniform ﬁeld that appears near the edge of the plates. The ﬁeld is non-uniform in this region d z ρ Figure 5.14: Thin parallel plate capacitor.	Electromagnetics_Vol1_Page_141_Chunk1442
5.23. THE THIN PARALLEL PLATE CAPACITOR 127 because the boundary conditions on the outside (outward-facing) surfaces of the plates have a signiﬁcant effect in this region. In the central region of the capacitor, however, the ﬁeld is not much different from the ﬁeld that exists in the case of inﬁnite plate area. In any parallel plate capacitor having ﬁnite plate area, some fraction of the energy will be stored by the approximately uniform ﬁeld of the central region, and the rest will be stored in the fringing ﬁeld. We can make the latter negligible relative to the former by making the capacitor very “thin,” in the sense that the smallest identiﬁable dimension of the plate is much greater than d. Under this condition, we may obtain a good approximation of the capacitance by simply neglecting the fringing ﬁeld, since an insigniﬁcant fraction of the energy is stored there. Imposing the “thin” condition leads to three additional simpliﬁcations. First, the surface charge distribution may be assumed to be approximately uniform over the plate, which greatly simpliﬁes the analysis. Second, the shape of the plates becomes irrelevant; they might be circular, square, triangular, etc. When computing capacitance in the “thin” case, only the plate area A is important. Third, the thickness of each of the plates becomes irrelevant. We are now ready to determine the capacitance of the thin parallel plate capacitor. Here are the steps: 1. Assume a total positive charge Q+ on the upper plate. 2. Invoking the “thin” condition, we assume the charge density on the plates is uniform. Thus, the surface charge density on bottom side of the upper plate is ρs,+ = Q+/A (C/m2). 3. From the boundary condition on the bottom surface of the upper plate, D on this surface is −ˆzρs,+. 4. The total charge on the lower plate, Q−, must be equal and opposite the total charge on the upper plate; i.e, Q−= −Q+. Similarly, the surface charge density on the upper surface of the lower plate, ρs,−, must be −ρs,+. 5. From the boundary condition on the top surface of the lower plate (Section 5.18), D on this surface is +ˆzρs,−. Since +ˆzρs,−= −ˆzρs,+, D on the facing sides of the plates is equal. 6. Again invoking the “thin” condition, we assume D between the plates has approximately the same structure as we would see if the plate area was inﬁnite. Therefore, we are justiﬁed in assuming D ≈−ˆzρs,+ everywhere between the plates. (You might also see that this is self-evident from the deﬁnition of D as the ﬂux density of electric charge (Section 2.4).) 7. With an expression for the electric ﬁeld in hand, we may now compute the potential difference V between the plates as follows (Section 5.8): V = − Z C E · dl = − Z d 0 1 ǫ D · (ˆzdz) = − Z d 0 −ˆzρs,+ ǫ · (ˆzdz) = +ρs,+ d ǫ (5.163) 8. Finally, C = Q+ V = ρs,+ A ρs,+ d/ǫ = ǫA d (5.164) Summarizing: C ≈ǫA d (5.165) The capacitance of a parallel plate capacitor hav- ing plate separation much less than the size of the plate is given by Equation 5.165. This is an approximation because the fringing ﬁeld is ne- glected. It’s worth noting that this is dimensionally correct; i.e., F/m times m2 divided by m yields F. It’s also worth noting the effect of the various parameters: Capacitance increases in proportion to permittiv- ity and plate area and decreases in proportion to distance between the plates.	Electromagnetics_Vol1_Page_142_Chunk1443
128 CHAPTER 5. ELECTROSTATICS Example 5.9. Printed circuit board capacitance. Printed circuit boards commonly include a “ground plane,” which serves as the voltage datum for the board, and at least one “power plane,” which is used to distribute a DC supply voltage (See “Additional Reading” at the end of this section). These planes are separated by a dielectric material, and the resulting structure exhibits capacitance. This capacitance may be viewed as an equivalent discrete capacitor in parallel with the power supply. The value of this equivalent capacitor may be either negligible, signiﬁcant and beneﬁcial, or signiﬁcant and harmful. So, it is useful to know the value of this equivalent capacitor. For a common type of circuit board, the dielectric thickness is about 1.6 mm and the relative permittivity of the material is about 4.5. If the area in common between the ground and power planes is 25 cm2, what is the value of the equivalent capacitor? Solution. From the problem statement, ǫ ∼= 4.5ǫ0, A ∼= 25 cm2 = 2.5 × 10−3 m2, and d ∼= 1.6 mm. Using Equation 5.165, the value of the equivalent capacitor is 62.3 pF. Additional Reading: • “Printed circuit board” on Wikipedia. 5.24 Capacitance of a Coaxial Structure [m0113] Let us now determine the capacitance of coaxially-arranged conductors, shown in Figure 5.15. Among other applications, this information is useful in the analysis of voltage and current waves on coaxial transmission line, as addressed in Sections 3.4 and 3.10. For our present purposes, we may model the structure as consisting of two concentric perfectly-conducting cylinders of radii a and b, separated by an ideal dielectric having permittivity ǫs. We place the +z axis along the common axis of the concentric cylinders so that the cylinders may be described as constant-coordinate surfaces ρ = a and ρ = b. In this section, we shall ﬁnd the capacitance by assuming a total charge Q+ on the inner conductor and integrating over the associated electric ﬁeld to obtain the voltage between the conductors. Then, capacitance is computed as the ratio of the assumed charge to the resulting potential difference. This strategy is the same as that employed in Section 5.23 for the parallel plate capacitor, so it may be useful to review that section before attempting this derivation. The ﬁrst step is to ﬁnd the electric ﬁeld inside the structure. This is relatively simple if we assume that b - + a z l V c⃝K. Kikkeri CC BY SA 4.0 Figure 5.15: Determining the capacitance of a coaxial structure.	Electromagnetics_Vol1_Page_143_Chunk1444
5.24. CAPACITANCE OF A COAXIAL STRUCTURE 129 the structure has inﬁnite length (i.e., l →∞), since then there are no fringing ﬁelds and the internal ﬁeld will be utterly constant with respect to z. In the central region of a ﬁnite-length capacitor, however, the ﬁeld is not much different from the ﬁeld that exists in the case of inﬁnite length, and if the energy storage in fringing ﬁelds is negligible compared to the energy storage in this central region then there is no harm in assuming the internal ﬁeld is constant with z. Alternatively, we may think of the length l as pertaining to one short section of a much longer structure and thereby obtain the capacitance per length as opposed to the total capacitance. Note that the latter is exactly what we need for the transmission line lumped-element equivalent circuit model (Section 3.4). To determine the capacitance, we invoke the deﬁnition (Section 5.22): C ≜Q+ V (5.166) where Q+ is the charge on the positively-charged conductor and V is the potential measured from the negative conductor to the positive conductor. The charge on the inner conductor is uniformly-distributed with density ρl = Q+ l (5.167) which has units of C/m. Now we will determine the electric ﬁeld intensity E, integrate E over a path between conductors to get V , and then apply Equation 5.166 to obtain the capacitance. The electric ﬁeld intensity for this scenario was determined in Section 5.6, “Electric Field Due to an Inﬁnite Line Charge using Gauss’ Law,” where we found E = ˆρ ρl 2πǫsρ (5.168) The reader should note that in that section we were considering merely a line of charge; not a coaxial structure. So, on what basis do we claim the ﬁeld is the same? This is a consequence of Gauss’ Law (Section 5.5) I S D · ds = Qencl (5.169) which we used in Section 5.6 to ﬁnd the ﬁeld. If in this new problem we specify the same cylindrical surface S with radius ρ < b, then the enclosed charge is the same. Furthermore, the presence of the outer conductor does not change the radial symmetry of the problem, and nothing else remains that can change the outcome. This is worth noting for future reference: The electric ﬁeld inside a coaxial structure com- prised of concentric conductors and having uni- form charge density on the inner conductor is identical to the electric ﬁeld of a line charge in free space having the same charge density. Next, we get V using (Section 5.8) V = − Z C E · dl (5.170) where C is any path from the negatively-charged outer conductor to the positively-charged inner conductor. Since this can be any such path (Section 5.9), we may as well choose the simplest one. This path is the one that traverses a radial of constant φ and z. Thus: V = − Z a ρ=b ˆρ ρl 2πǫsρ · (ˆρdρ) = −ρl 2πǫs Z a ρ=b dρ ρ = + ρl 2πǫs Z b ρ=a dρ ρ = + ρl 2πǫs ln b a (5.171) Wrapping up: C ≜Q+ V = ρll (ρl/2πǫs) ln (b/a) (5.172) Note that factors of ρl in the numerator and denominator cancel out, leaving: C = 2πǫsl ln (b/a) (5.173) Note that this expression is dimensionally correct, having units of F. Also note that the expression depends only on materials (through ǫs) and geometry (through l, a, and b). The expression does not depend on charge or voltage, which would imply non-linear behavior.	Electromagnetics_Vol1_Page_144_Chunk1445
130 CHAPTER 5. ELECTROSTATICS To make the connection back to lumped-element transmission line model parameters (Sections 3.4 and 3.10), we simply divide by l to get the per-unit length parameter: C′ = 2πǫs ln (b/a) (5.174) Example 5.10. Capacitance of RG-59 coaxial cable. RG-59 coaxial cable consists of an inner conductor having radius 0.292 mm, an outer conductor having radius 1.855 mm, and a polyethylene spacing material having relative permittivity 2.25. Estimate the capacitance per length of RG-59. Solution. From the problem statement, a = 0.292 mm, b = 1.855 mm, and ǫs = 2.25ǫ0. Using Equation 5.174 we ﬁnd C′ = 67.7 pF/m. 5.25 Electrostatic Energy [m0114] Consider a structure consisting of two perfect conductors, both ﬁxed in position and separated by an ideal dielectric. This could be a capacitor, or it could be one of a variety of capacitive structures that are not explicitly intended to be a capacitor – for example, a printed circuit board. When a potential difference is applied between the two conducting regions, a positive charge Q+ will appear on the surface of the conductor at the higher potential, and a negative charge Q−= −Q+ will appear on the surface of the conductor at the lower potential (Section 5.19). Assuming the conductors are not free to move, potential energy is stored in the electric ﬁeld associated with the surface charges (Section 5.22). We now ask the question, what is the energy stored in this ﬁeld? The answer to this question has relevance in several engineering applications. For example, when capacitors are used as batteries, it is useful to know to amount of energy that can be stored. Also, any system that includes capacitors or has unintended capacitance is using some fraction of the energy delivered by the power supply to charge the associated structures. In many electronic systems – and in digital systems in particular – capacitances are periodically charged and subsequently discharged at a regular rate. Since power is energy per unit time, this cyclic charging and discharging of capacitors consumes power. Therefore, energy storage in capacitors contributes to the power consumption of modern electronic systems. We’ll delve into that topic in more detail in Example 5.11. Since capacitance C relates the charge Q+ to the potential difference V between the conductors, this is the natural place to start. From the deﬁnition of capacitance (Section 5.22): V = Q+ C (5.175) From Section 5.8, electric potential is deﬁned as the work done (i.e., energy injected) by moving a charged particle, per unit of charge; i.e., V = We q (5.176)	Electromagnetics_Vol1_Page_145_Chunk1446
5.25. ELECTROSTATIC ENERGY 131 where q is the charge borne by the particle and We (units of J) is the work done by moving this particle across the potential difference V . Since we are dealing with charge distributions as opposed to charged particles, it is useful to express this in terms of the contribution ∆We made to We by a small charge ∆q. Letting ∆q approach zero we have dWe = V dq (5.177) Now consider what must happen to transition the system from having zero charge (q = 0) to the fully-charged but static condition (q = Q+). This requires moving the differential amount of charge dq across the potential difference between conductors, beginning with q = 0 and continuing until q = Q+. Therefore, the total amount of work done in this process is: We = Z Q+ q=0 dWe = Z Q+ 0 V dq = Z Q+ 0 q C dq = 1 2 Q2 + C (5.178) Equation 5.178 can be expressed entirely in terms of electrical potential by noting again that C = Q+/V , so We = 1 2CV 2 (5.179) Since there are no other processes to account for the injected energy, the energy stored in the electric ﬁeld is equal to We. Summarizing: The energy stored in the electric ﬁeld of a capac- itor (or a capacitive structure) is given by Equa- tion 5.179. Example 5.11. Why multicore computing is power-neutral. Readers are likely aware that computers increasingly use multicore processors as opposed to single-core processors. For our present purposes, a “core” is deﬁned as the smallest combination of circuitry that performs independent computation. A multicore processor consists of multiple identical cores that run in parallel. Since a multicore processor consists of N identical processors, you might expect power consumption to increase by N relative to a single-core processor. However, this is not the case. To see why, ﬁrst realize that the power consumption of a modern computing core is dominated by the energy required to continuously charge and discharge the multitude of capacitances within the core. From Equation 5.179, the required energy is 1 2C0V 2 0 per clock cycle, where C0 is the sum capacitance (remember, capacitors in parallel add) and V0 is the supply voltage. Power is energy per unit time, so the power consumption for a single core is P0 = 1 2C0V 2 0 f0 (5.180) where f0 is the clock frequency. In a N-core processor, the sum capacitance is increased by N. However, the frequency is decreased by N since the same amount of computation is (nominally) distributed among the N cores. Therefore, the power consumed by an N-core processor is PN = 1 2 (NC0) V 2 0 f0 N = P0 (5.181) In other words, the increase in power associated with replication of hardware is nominally offset by the decrease in power enabled by reducing the clock rate. In yet other words, the total energy of the N-core processor is N times the energy of the single core processor at any given time; however, the multicore processor needs to recharge capacitances 1/N times as often. Before moving on, it should be noted that the usual reason for pursuing a multicore design is to increase the amount of computation that can be done; i.e., to increase the product f0N. Nevertheless, it is extremely helpful that power consumption is proportional to f0 only, and is independent of N. The thin parallel plate capacitor (Section 5.23) is representative of a large number of practical	Electromagnetics_Vol1_Page_146_Chunk1447
132 CHAPTER 5. ELECTROSTATICS applications, so it is instructive to consider the implications of Equation 5.179 for this structure in particular. For the thin parallel plate capacitor, C ≈ǫA d (5.182) where A is the plate area, d is the separation between the plates, and ǫ is the permittivity of the material between the plates. This is an approximation because the fringing ﬁeld is neglected; we shall proceed as if this is an exact expression. Applying Equation 5.179: We = 1 2 ǫA d (Ed)2 (5.183) where E is the magnitude of the electric ﬁeld intensity between the plates. Rearranging factors, we obtain: We = 1 2ǫE2 (Ad) (5.184) Recall that the electric ﬁeld intensity in the thin parallel plate capacitor is approximately uniform. Therefore, the density of energy stored in the capacitor is also approximately uniform. Noting that the product Ad is the volume of the capacitor, we ﬁnd that the energy density is we = We Ad = 1 2ǫE2 (5.185) which has units of energy per unit volume (J/m3). The above expression provides an alternative method to compute the total electrostatic energy. Within a mathematical volume V, the total electrostatic energy is simply the integral of the energy density over V; i.e., We = Z V we dv (5.186) This works even if E and ǫ vary with position. So, even though we arrived at this result using the example of the thin parallel-plate capacitor, our ﬁndings at this point apply generally. Substituting Equation 5.185 we obtain: We = 1 2 Z V ǫE2dv (5.187) Summarizing: The energy stored by the electric ﬁeld present within a volume is given by Equation 5.187. It’s worth noting that this energy increases with the permittivity of the medium, which makes sense since capacitance is proportional to permittivity. [m0034]	Electromagnetics_Vol1_Page_147_Chunk1448
5.25. ELECTROSTATIC ENERGY 133 Image Credits Fig. 5.1: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0102 fCoulombsLaw.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.2: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0104 fRing.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.3: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0104 fDisk.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.4: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:4.5 m0149 commons.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.5: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0064 fResistor.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.7: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0020 fPEC.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.8: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0020 fKVL.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.9: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0021 fPEC.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.10: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0021 fGL.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.11: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0025 fFlatSlab.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.12: by C. Burks, https://commons.wikimedia.org/wiki/File:Electrostatic induction.svg, in public domain. Minor modiﬁcations. Fig. 5.15: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0113 fCapCoax.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).	Electromagnetics_Vol1_Page_148_Chunk1449
Chapter 6 Steady Current and Conductivity 6.1 Convection and Conduction Currents [m0110] In practice, we deal with two physical mechanisms for current: convection and conduction. The distinction between these types of current is important in electromagnetic analysis. Convection current consists of charged particles moving in response to mechanical forces, as opposed to being guided by the electric ﬁeld (Sections 2.2 and/or 5.1). An example of a convection current is a cloud bearing free electrons that moves through the atmosphere driven by wind. Conduction current consists of charged particles moving in response to the electric ﬁeld and not merely being carried by motion of the surrounding material. In some materials, the electric ﬁeld is also able to dislodge weakly-bound electrons from atoms, which then subsequently travel some distance before reassociating with other atoms. For this reason, the individual electrons in a conduction current do not necessarily travel the full distance over which the current is perceived to exist. The distinction between convection and conduction is important because Ohm’s Law (Section 6.3) – which speciﬁes the relationship between electric ﬁeld intensity and current – applies only to conduction current. Additional Reading: • “Electric current” on Wikipedia. Electromagnetics Vol 1. c⃝2018 S.W. Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics-vol-1	Electromagnetics_Vol1_Page_149_Chunk1450
6.2. CURRENT DISTRIBUTIONS 135 6.2 Current Distributions [m0101] In elementary electric circuit theory, current is the rate at which electric charge passes a particular point in a circuit. For example, 1 A is 1 C per second. In this view current is a scalar quantity, and there are only two possible directions because charge is constrained to ﬂow along deﬁned channels. Direction is identiﬁed by the sign of the current with respect to a reference direction, which is in turn deﬁned with respect to a reference voltage polarity. The convention in electrical engineering deﬁnes positive current as the ﬂow of positive charge into the positive voltage terminal of a passive device such as a resistor, capacitor, or inductor. For an active device such as a battery, positive current corresponds to the ﬂow of positive charge out of the positive voltage terminal. However, this kind of thinking only holds up when the systems being considered are well-described as “lumped” devices connected by inﬁnitesimally thin, ﬁlament-like connections representing wires and circuit board traces. Many important problems in electrical engineering concern situations in which the ﬂow of current is not limited in this way. Examples include wire- and pin-type interconnects at radio frequencies, circuit board and enclosure grounding, and physical phenomena such as lightning. To accommodate the more general class of problems, we must deﬁne current as a vector quantity. Furthermore, current in these problems can spread out over surfaces and within volumes, so we must also consider spatial distributions of current. Line Current. As noted above, if a current I is constrained to follow a particular path, then the only other consideration is direction. Thus, a line current is speciﬁed mathematically as ˆlI, where the direction ˆl may vary with position along the path. For example, in a straight wire ˆl is constant, whereas in a coil ˆl varies with position along the coil. Surface Current Distribution. In some cases, current may be distributed over a surface. For example, the radio-frequency current on a wire of radius a made from a metal with sufﬁciently high conductivity can be modeled as a uniform surface current existing on the wire surface. In this case, the current is best described as a surface current density Js, which is the total current I on the wire divided by the circumference 2πa of the wire: Js = ˆu I 2πa (units of A/m) (6.1) where ˆu is the direction of current ﬂow. Volume Current Distribution. Imagine that current is distributed within a volume. Let ˆu∆i be the current passing through a small open planar surface deﬁned within this volume, and let ∆s be the area of this surface. The volume current density J at any point in the volume is deﬁned as J ≜lim ∆s→0 ˆu∆i ∆s = ˆu di ds (units of A/m2) (6.2) In general, J is a function of position within this volume. Subsequently the total current passing through a surface S is I = Z S J · ds (units of A) (6.3) In other words, volume current density integrated over a surface yields total current through that surface. You might recognize this as a calculation of ﬂux, and it is.1 Example 6.1. Current and current density in a wire of circular cross-section. Figure 6.1 shows a straight wire having cross-sectional radius a = 3 mm. A battery is connected across the two ends of the wire resulting in a volume current density J = ˆz8 A/m2, which is uniform throughout the wire. Find the net current I through the wire. Solution. The net current is I = Z S J · ds We choose S to be the cross-section perpendicular to the axis of the wire. Also, we choose ds to point such that I is positive with respect to the sign convention shown in Figure 6.1 , which is the usual choice in electric 1However, it is not common to refer to net current as a ﬂux; go ﬁgure!	Electromagnetics_Vol1_Page_150_Chunk1451
136 CHAPTER 6. STEADY CURRENT AND CONDUCTIVITY J I a z - + ds c⃝K. Kikkeri CC BY SA 4.0 Figure 6.1: Net current I and current density J in a wire of circular cross-section. circuit analysis. With these choices, we have I = Z a ρ=0 Z 2π φ=0	Electromagnetics_Vol1_Page_151_Chunk1452
6.3. CONDUCTIVITY 137 Conductivity σ is expressed in units of S/m, where 1 S = 1 Ω−1. It is important to note that the current being addressed here is conduction current, and not convection current, displacement current, or some other form of current – see Section 6.2 for elaboration. Summarizing: Ohm’s Law for Electromagnetics (Equation 6.4) states that volume density of conduction current (A/m2) equals conductivity (S/m) times electric ﬁeld intensity (V/m). The reader is likely aware that there is also an “Ohm’s Law” in electric circuit theory, which states that current I (units of A) is voltage V (units of V) divided by resistance R (units of Ω); i.e., I = V/R. This is in fact a special case of Equation 6.4; see Section 6.4 for more about this. As mentioned above, σ depends on both the availability and mobility of charge within the material. At the two extremes, we have perfect insulators, for which σ = 0, and perfect conductors, for which σ →∞. Some materials approach these extremes, whereas others fall midway between these conditions. Here are a few classes of materials that are frequently encountered: • A perfect vacuum – “free space” – contains no charge and therefore is a perfect insulator with σ = 0. • Good insulators typically have conductivities ≪10−10 S/m, which is sufﬁciently low that the resulting currents can usually be ignored. The most important example is air, which has a conductivity only slightly greater than that of free space. An important class of good insulators is lossless dielectrics,2 which are well-characterized in terms of permittivity (ǫ) alone, and for which µ = µ0 and σ = 0 may usually be assumed. • Poor insulators have conductivities that are low, but nevertheless sufﬁciently high that the 2See Section 5.20 for a discussion of dielectric materials. resulting currents cannot be ignored. For example, the dielectric material that is used to separate the conductors in a transmission line must be considered a poor insulator as opposed to a good (effectively lossless) insulator in order to characterize loss per length along the transmission line, which can be signiﬁcant.3 These lossy dielectrics are well-characterized in terms of ǫ and σ, and typically µ = µ0 can be assumed. • Semiconductors such as those materials used in integrated circuits have intermediate conductivities, typically in the range 10−4 to 10+1 S/m. • Good conductors are materials with very high conductivities, typically greater than 105 S/m. An important category of good conductors includes metals, with certain metals including alloys of aluminum, copper, and gold reaching conductivities on the order of 108 S/m. In such materials, minuscule electric ﬁelds give rise to large currents. There is no signiﬁcant storage of energy in such materials, and so the concept of permittivity is not relevant for good conductors. The reader should take care to note that terms such as good conductor and poor insulator are qualitative and subject to context. What may be considered a “good insulator” in one application may be considered to be a “poor insulator” in another. One relevant category of material was not included in the above list – namely, perfect conductors. A perfect conductor is a material in which σ →∞. It is tempting to interpret this as meaning that any electric ﬁeld gives rise to inﬁnite current density; however, this is not plausible even in the ideal limit. Instead, this condition is interpreted as meaning that E = J/σ →0 throughout the material; i.e., E is zero independently of any current ﬂow in the material. An important consequence is that the potential ﬁeld V is equal to a constant value throughout the material. (Recall E = −∇V (Section 5.14), so constant V means E = 0.) We refer to the volume occupied by such a material as an equipotential volume. This concept is useful as an approximation of the behavior 3Review Sections 3.4, 3.9, and associated sections for a refresher on this issue.	Electromagnetics_Vol1_Page_152_Chunk1453
138 CHAPTER 6. STEADY CURRENT AND CONDUCTIVITY of good conductors; for example, metals are often modeled as perfectly-conducting equipotential volumes in order to simplify analysis. A perfect conductor is a material for which σ → ∞, E →0, and subsequently V (the electric po- tential) is constant. One ﬁnal note: It is important to remain aware of the assumptions we have made about materials in this book, which are summarized in Section 2.8. In particular, we continue to assume that materials are linear unless otherwise indicated. For example, whereas air is normally considered to be a good insulator and therefore a poor conductor, anyone who has ever witnessed lightning has seen a demonstration that under the right conditions – i.e., sufﬁciently large potential difference between earth and sky – current will readily ﬂow through air. This particular situation is known as dielectric breakdown (see Section 5.21). The non-linearity of materials can become evident before reaching the point of dielectric breakdown, so one must be careful to consider this possibility when dealing with strong electric ﬁelds. Additional Reading: • “Electrical Resistivity and Conductivity” on Wikipedia. • “Ohm’s Law” on Wikipedia. 6.4 Resistance [m0071] The concept of resistance is most likely familiar to readers via Ohm’s Law for Devices; i.e., V = IR where V is the potential difference associated with a current I. This is correct, but it is not the whole story. Let’s begin with a statement of intent: Resistance R (Ω) is a characterization of the con- ductivity of a device (as opposed to a material) in terms of Ohms Law for Devices; i.e., V = IR. Resistance is a property of devices such as resistors, which are intended to provide resistance, as well as being a property of most practical electronic devices, whether it is desired or not. Resistance is a manifestation of the conductivity of the materials comprising the device, which subsequently leads to the “V = IR” relationship. This brings us to a very important point and a common source of confusion. Resistance is not necessarily the real part of impedance. Let’s take a moment to elaborate. Impedance (Z) is deﬁned as the ratio of voltage to current; i.e., V/I; or equivalently in the phasor domain as eV /eI. Most devices – not just devices exhibiting resistance – can be characterized in terms of this ratio. Consider for example the input impedance of a terminated transmission line (Section 3.15). This impedance may have a non-zero real-valued component even when the transmission line and the terminating load are comprised of perfect conductors. Summarizing: Resistance results in a real-valued impedance. However, not all devices exhibiting a real-valued impedance exhibit resistance. Furthermore, the real component of a complex-valued impedance does not necessarily represent resistance. Restating the main point in yet other words: Resistance pertains to limited conductivity, not simply to voltage-current ratio. Also important to realize is that whereas conductivity σ (units of S/m) is a property of materials, resistance depends on both conductivity and the geometry of the	Electromagnetics_Vol1_Page_153_Chunk1454
6.4. RESISTANCE 139 E I a z - + V z=0 z= σ c⃝K. Kikkeri CC BY SA 4.0 Figure 6.2: Analysis of the resistance of straight wire of circular cross-section. device. In this section, we address the question of how the resistance of a device can be is determined. The following example serves this purpose. Figure 6.2 shows a straight wire of length l centered on the z axis, forming a cylinder of material having conductivity σ and cross-sectional radius a. The ends of the cylinder are covered by perfectly-conducting plates to which the terminals are attached. A battery is connected to the terminals, resulting in a uniform internal electric ﬁeld E between the plates. To calculate R for this device, let us ﬁrst calculate V , then I, and ﬁnally R = V/I. First, we compute the potential difference using the following result from Section 5.8: V = − Z C E · dl (6.5) Here we can view V as a “given” (being the voltage of the battery), but wish to evaluate the right hand side so as to learn something about the effect of the conductivity and geometry of the wire. The appropriate choice of C begins at z = 0 and ends at z = l, following the axis of the cylinder. (Remember: C deﬁnes the reference direction for increasing potential, so the resulting potential difference will be the node voltage at the end point minus the node voltage at the start point.) Thus, dl = ˆzdz and we have V = − Z l z=0 E · (ˆzdz) We do not yet know E; however, we know it is constant throughout the device and points in the −ˆz direction since this is the direction of current ﬂow and Ohm’s Law for Electromagnetics (Section 6.3) requires the electric ﬁeld to point in the same direction. Thus, we may write E = −ˆzEz where Ez is a constant. We now ﬁnd: V = − Z l z=0 (−ˆzEz) · (ˆzdz) = +Ez Z l z=0 dz = +Ezl The current I is given by (Section 6.2) I = Z S J · ds We choose the surface S to be the cross-section perpendicular to the axis of the wire. Also, we choose ds to point such that I is positive with respect to the sign convention shown in Figure 6.2. With these choices, we have I = Z a ρ=0 Z 2π φ=0 J · (−ˆz ρ dρ dφ) From Ohm’s Law for Electromagnetics, we have J = σE = −ˆzσEz (6.6) So now I = Z a ρ=0 Z 2π φ=0 (−ˆzσEz) · (−ˆzρ dρ dφ) = σEz Z a ρ=0 Z 2π φ=0 ρ dρ dφ = σEz	Electromagnetics_Vol1_Page_154_Chunk1455
140 CHAPTER 6. STEADY CURRENT AND CONDUCTIVITY i.e., the resistance of a wire having cross-sectional area A – regardless of the shape of the cross-section, is given by the above equation. The resistance of a right cylinder of material, given by Equation 6.7, is proportional to length and inversely proportional to cross-sectional area and conductivity. It is important to remember that Equation 6.7 presumes that the volume current density J is uniform over the cross-section of the wire. This is an excellent approximation for thin wires at “low” frequencies including, of course, DC. At higher frequencies it may not be a good assumption that J is uniformly-distributed over the cross-section of the wire, and at sufﬁciently high frequencies one ﬁnds instead that the current is effectively limited to the exterior surface of the wire. In the “high frequency” case, A in Equation 6.7 is reduced from the physical area to a smaller value corresponding to the reduced area through which most of the current ﬂows. Therefore, R increases with increasing frequency. To quantify the high frequency behavior of R (including determination of what constitutes “high frequency” in this context) one requires concepts beyond the theory of electrostatics, so this is addressed elsewhere. Example 6.2. Resistance of 22 AWG hookup wire. A common type of wire found in DC applications is 22AWG (“American Wire Gauge”; see “Additional Resources” at the end of this section) copper solid-conductor hookup wire. This type of wire has circular cross-section with diameter 0.644 mm. What is the resistance of 3 m of this wire? Assume copper conductivity of 58 MS/m. Solution. From the problem statement, the diameter 2a = 0.644 mm, σ = 58 × 106 S/m, and l = 3 m. The cross-sectional area is A = πa2 ∼= 3.26 × 10−7 m2. Using Equation 6.7 we obtain R = 159 mΩ. Example 6.3. Resistance of steel pipe. A pipe is 3 m long and has inner and outer radii of 5 mm and 7 mm respectively. It is made from steel having conductivity 4 MS/m. What is the DC resistance of this pipe? Solution. We can use Equation 6.7 if we can determine the cross-sectional area A through which the current ﬂows. This area is simply the area deﬁned by the outer radius, πb2, minus the area deﬁned by the inner radius πa2. Thus, A = πb2 −πa2 ∼= 7.54 × 10−5 m2. From the problem statement, we also determine that σ = 4 × 106 S/m and l = 3 m. Using Equation 6.7 we obtain R ∼= 9.95 mΩ. Additional Reading: • “Resistor” on Wikipedia. • “Ohm’s Law” on Wikipedia. • “American wire gauge” on Wikipedia.	Electromagnetics_Vol1_Page_155_Chunk1456
6.5. CONDUCTANCE 141 6.5 Conductance [m0105] Conductance, like resistance (Section 6.4), is a property of devices. Speciﬁcally: Conductance G (Ω−1 or S) is the reciprocal of resistance R. Therefore, conductance depends on both the conductivity of the materials used in the device, as well as the geometry of the device. A natural question to ask is, why do we require the concept of conductance, if it simply the reciprocal of resistance? The short answer is that the concept of conductance is not required; or, rather, we need only resistance or conductance and not both. Nevertheless, the concept appears in engineering analysis for two reasons: • Conductance is sometimes considered to be a more intuitive description of the underlying physics in cases where the applied voltage is considered to be the independent “stimulus” and current is considered to be the response. This is why conductance appears in the lumped element model for transmission lines (Section 3.4), for example. • Characterization in terms of conductance may be preferred when considering the behavior of devices in parallel, since the conductance of a parallel combination is simply the sum of the conductances of the devices. Example 6.4. Conductance of a coaxial structure. Let us now determine the conductance of a structure consisting of coaxially-arranged conductors separated by a lossy dielectric, as shown in Figure 6.3. The conductance per unit length G′ (i.e., S/m) of this structure is of interest in determining the characteristic impedance of coaxial transmission line, as addressed in Sections 3.4 and 3.10. For our present purposes, we may model the structure as two concentric perfectly-conducting cylinders of radii a and b, separated by a lossy dielectric having conductivity σs. We place the +z axis along the common axis of the concentric cylinders so that the cylinders may be described as constant-coordinate surfaces ρ = a and ρ = b. There are at least 2 ways to solve this problem. One method is to follow the procedure that was used to ﬁnd the capacitance of this structure in Section 5.24. Adapting that approach to the present problem, one would assume a potential difference V between the conductors, from that determine the resulting electric ﬁeld intensity E, and then using Ohm’s Law for Electromagnetics (Section 6.3) determine the density J = σsE of the current that leaks directly between conductors. From this, one is able to determine the total leakage current I, and subsequently the conductance G ≜I/V . Although highly recommended as an exercise for the student, in this section we take an alternative approach so as to demonstrate that there are a variety of approaches available for such problems. The method we shall use below is as follows: (1) Assume a leakage current I between the conductors; (2) Determine the associated current density J, which is possible using only geometrical considerations; (3) Determine the associated electric ﬁeld intensity E using J/σs; (4) Integrate E over a path between the conductors to get V . Then, as before, conductance G ≜I/V . The current I is deﬁned as shown in Figure 6.3, with reference direction according to the engineering convention that positive current ﬂows out of the positive terminal of a source. The associated current density must ﬂow in the same direction, and the circular symmetry of the problem therefore constrains J to have the form J = ˆρ I A (6.8) where A is the area through which I ﬂows. In other words, current ﬂows radially outward from the inner conductor to the outer conductor, with	Electromagnetics_Vol1_Page_156_Chunk1457
142 CHAPTER 6. STEADY CURRENT AND CONDUCTIVITY density that diminishes inversely with the area through which the total current ﬂows. (It may be helpful to view J as a ﬂux density and I as a ﬂux, as noted in Section 6.2.) This area is simply circumference 2πρ times length l, so J = ˆρ I 2πρl (6.9) which exhibits the correct units of A/m2. Now from Ohm’s Law for Electromagnetics we ﬁnd the electric ﬁeld within the structure is E = J σs = ˆρ I 2πρlσs (6.10) Next we get V using (Section 5.8) V = − Z C E · dl (6.11) where C is any path from the negatively-charged outer conductor to the positively-charged inner conductor. Since this can be any such path (Section 5.9), we should choose the simplest one. The simplest path is the one that traverses a radial of constant φ and z. Thus: V = − Z a ρ=b ˆρ I 2πρlσs · (ˆρdρ) = − I 2πlσs Z a ρ=b dρ ρ = + I 2πlσs Z b ρ=a dρ ρ = + I 2πlσs ln b a (6.12) Wrapping up: G ≜I V = I (I/2πlσs) ln (b/a) (6.13) Note that factors of I in the numerator and denominator cancel out, leaving: G = 2πlσs ln (b/a) (6.14) b a z ρ l I s J Figure 6.3: Determining the conductance of a struc- ture consisting of coaxially-arranged conductors sepa- rated by a lossy dielectric. Note that Equation 6.14 is dimensionally correct, having units of S = Ω−1. Also note that this is expression depends only on materials (through σs) and geometry (through l, a, and b). Notably it does not depend on current or voltage, which would imply non-linear behavior. To make the connection back to lumped-element transmission line model parameters (Sections 3.4 and 3.10), we simply divide by l to get the per-length parameter: G′ = 2πσs ln (b/a) (6.15) Example 6.5. Conductance of RG-59 coaxial cable. RG-59 coaxial cable consists of an inner conductor having radius 0.292 mm, an outer conductor having radius 1.855 mm, and a polyethylene spacing material exhibiting conductivity of about 5.9 × 10−5 S/m. Estimate the conductance per length of RG-59. Solution. From the problem statement, a = 0.292 mm, b = 1.855 mm, and σs ∼= 5.9 × 10−5 S/m. Using Equation 6.15, we ﬁnd G′ ∼= 200 µS/m.	Electromagnetics_Vol1_Page_157_Chunk1458
6.6. POWER DISSIPATION IN CONDUCTING MEDIA 143 6.6 Power Dissipation in Conducting Media [m0106] The displacement of charge in response to the force exerted by an electric ﬁeld constitutes a reduction in the potential energy of the system (Section 5.8). If the charge is part of a steady current, there must be an associated loss of energy that occurs at a steady rate. Since power is energy per unit time, the loss of energy associated with current is expressible as power dissipation. In this section, we address two questions: (1) How much power is dissipated in this manner, and (2) What happens to the lost energy? First, recall that work is force times distance traversed in response to that force (Section 5.8). Stated mathematically: ∆W = +F · ∆l (6.16) where the vector F is the force (units of N) exerted by the electric ﬁeld, the vector ∆l is the direction and distance (units of m) traversed, and ∆W is the work done (units of J) as a result. Note that a “+” has been explicitly indicated; this is to emphasize the distinction from the work being considered in Section 5.8. In that section, the work “∆W” represented energy from an external source that was being used to increase the potential energy of the system by moving charge “upstream” relative to the electric ﬁeld. Now, ∆W represents this internal energy as it is escaping from the system in the form of kinetic energy; therefore, positive ∆W now means a reduction in potential energy, hence the sign change.4 The associated power ∆P (units of W) is ∆W divided by the time ∆t (units of s) required for the distance ∆l to be traversed: ∆P = ∆W ∆t = F · ∆l ∆t (6.17) Now we’d like to express force in terms of the electric ﬁeld exerting this force. Recall that the force exerted by an electric ﬁeld intensity E (units of V/m) on a 4It could be argued that it is bad form to use the same variable to represent both tallies; nevertheless, it is common practice and so we simply remind the reader that it is important to be aware of the deﬁnitions of variables each time they are (re)introduced. particle bearing charge q (units of C) is qE (Section 2.2). However, we’d like to express this force in terms of a current, as opposed to a charge. An expression in terms of current can be constructed as follows. First, note that the total charge in a small volume “cell” is the volume charge density ρv (units of C/m3) times the volume ∆v of the cell; i.e., q = ρv∆v (Section 5.3). Therefore: F = qE = ρv ∆v E (6.18) and subsequently ∆P = ρv ∆v E · ∆l ∆t = E · ρv ∆l ∆t ∆v (6.19) The quantity in parentheses has units of C/m3 · m · s−1, which is A/m2. Apparently this quantity is the volume current density J, so we have ∆P = E · J ∆v (6.20) In the limit as ∆v →0 we have dP = E · J dv (6.21) and integrating over the volume V of interest we obtain P = Z V dP = Z V E · J dv (6.22) The above expression is commonly known as Joule’s Law. In our situation, it is convenient to use Ohm’s Law for Electromagnetics (J = σE; Section 6.3) to get everything in terms of materials properties (σ), geometry (V), and the electric ﬁeld: P = Z V E · (σE) dv (6.23) which is simply P = Z V σ \|E\|2 dv (6.24) Thus: The power dissipation associated with current is given by Equation 6.24. This power is pro- portional to conductivity and proportional to the electric ﬁeld magnitude squared.	Electromagnetics_Vol1_Page_158_Chunk1459
144 CHAPTER 6. STEADY CURRENT AND CONDUCTIVITY This result facilitates the analysis of power dissipation in materials exhibiting loss; i.e., having ﬁnite conductivity. But what is the power dissipation in a perfectly conducting material? For such a material, σ →∞and E →0 no matter how much current is applied (Section 6.3). In this case, Equation 6.24 is not very helpful. However, as we just noted, being a perfect conductor means E →0 no matter how much current is applied, so from Equation 6.22 we have found that: No power is dissipated in a perfect conductor. When conductivity is ﬁnite, Equation 6.24 serves as a more-general version of a concept from elementary circuit theory, as we shall now demonstrate. Let E = ˆzEz, so \|E\|2 = E2 z. Then Equation 6.24 becomes: P = Z V σE2 z dv = σE2 z Z V dv (6.25) The second integral in Equation 6.25 is a calculation of volume. Let’s assume V is a cylinder aligned along the z axis. The volume of this cylinder is the cross-sectional area A times the length l. Then the above equation becomes: P = σE2 z A l (6.26) For reasons that will become apparent very shortly, let’s reorganize the above expression as follows: P = (σEzA) (Ezl) (6.27) Note that σEz is the current density in A/m2, which when multiplied by A gives the total current. Therefore, the quantity in the ﬁrst set of parentheses is simply the current I. Also note that Ezl is the potential difference over the length l, which is simply the node-to-node voltage V (Section 5.8). Therefore, we have found: P = IV (6.28) as expected from elementary circuit theory. Now, what happens to the energy associated with this dissipation of power? The displacement of charge carriers in the material is limited by the conductivity, which itself is ﬁnite because, simply put, other constituents of the material get in the way. If charge is being displaced as described in this section, then energy is being used to displace the charge-bearing particles. The motion of constituent particles is observed as heat – in fact, this is essentially the deﬁnition of heat. Therefore: The power dissipation associated with the ﬂow of current in any material that is not a perfect con- ductor manifests as heat. This phenomenon is known as joule heating, ohmic heating, and by other names. This conversion of electrical energy to heat is the method of operation for toasters, electric space heaters, and many other devices that generate heat. It is of course also the reason that all practical electronic devices generate heat. Additional Reading: • “Joule Heating” on Wikipedia. [m0057]	Electromagnetics_Vol1_Page_159_Chunk1460
6.6. POWER DISSIPATION IN CONDUCTING MEDIA 145 Image Credits Fig. 6.1: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0101 fMatEx1.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 6.2: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0071 fMatEx1.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).	Electromagnetics_Vol1_Page_160_Chunk1461
Chapter 7 Magnetostatics [m0117] Magnetostatics is the theory of the magnetic ﬁeld in conditions in which its behavior is independent of electric ﬁelds, including • The magnetic ﬁeld associated with various spatial distributions of steady current • The energy associated with the magnetic ﬁeld • Inductance, which is the ability of a structure to store energy in a magnetic ﬁeld The word ending “-statics” refers to the fact that these aspects of electromagnetic theory can be developed by assuming that the sources of the magnetic ﬁeld are time-invariant; we might say that magnetostatics is the study of the magnetic ﬁeld at DC. However, many aspects of magnetostatics are applicable at “AC” as well. 7.1 Comparison of Electrostatics and Magnetostatics [m0115] Students encountering magnetostatics for the ﬁrst time have usually been exposed to electrostatics already. Electrostatics and magnetostatics exhibit many similarities. These are summarized in Table 7.1. The elements of magnetostatics presented in this table are all formally introduced in other sections; the sole purpose of this table is to point out the similarities. The technical term for these similarities is duality. Duality also exists between voltage and current in electrical circuit theory. For more about the concept of duality, see “Additional Reading” at the end of this section. Additional Reading: • “Duality (electricity and magnetism)” on Wikipedia. • “Duality (electrical circuits)” on Wikipedia. Electromagnetics Vol 1. c⃝2018 S.W. Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics-vol-1	Electromagnetics_Vol1_Page_161_Chunk1462
7.2. GAUSS’ LAW FOR MAGNETIC FIELDS: INTEGRAL FORM 147 electrostatics magnetostatics Sources static charge steady current, magnetizable material Field intensity E (V/m) H (A/m) Flux density D (C/m2) B (Wb/m2=T) Material relations D = ǫE B = µH J = σE Force on charge q F = qE F = qv × B Maxwell’s Eqs. H S D · ds = Qencl H S B · ds = 0 (integral) H C E · dl = 0 H C H · dl = Iencl Maxwell’s Eqs. ∇· D = ρv ∇· B = 0 (differential) ∇× E = 0 ∇× H = J Boundary Conditions ˆn × [E1 −E2] = 0 ˆn × [H1 −H2] = Js ˆn · [D1 −D2] = ρs ˆn · [B1 −B2] = 0 Energy storage Capacitance Inductance Energy density we = 1 2ǫE2 wm = 1 2µH2 Energy dissipation Resistance Table 7.1: A summary of the duality between electrostatics and magnetostatics. 7.2 Gauss’ Law for Magnetic Fields: Integral Form [m0018] Gauss’ Law for Magnetic Fields (GLM) is one of the four fundamental laws of classical electromagnetics, collectively known as Maxwell’s Equations. Before diving in, the reader is strongly encouraged to review Section 2.5. In that section, GLM emerges from the “ﬂux density” interpretation of the magnetic ﬁeld. GLM is not identiﬁed in that section, but now we are ready for an explicit statement: Gauss’ Law for Magnetic Fields (Equation 7.1) states that the ﬂux of the magnetic ﬁeld through a closed surface is zero. This is expressed mathematically as follows: I S B · ds = 0 (7.1) where B is magnetic ﬂux density and S is a closed surface with outward-pointing differential surface normal ds. It may be useful to consider the units. B has units of Wb/m2; therefore, integrating B over a surface gives a quantity with units of Wb, which is magnetic ﬂux, as indicated above. GLM can also be interpreted in terms of magnetic ﬁeld lines. For the magnetic ﬂux through a closed surface to be zero, every ﬁeld line entering the volume enclosed by S must also exit this volume – ﬁeld lines may not begin or end within the volume. The only way this can be true for every possible surface S is if magnetic ﬁeld lines always form closed loops. This is in fact what we ﬁnd in practice, as shown in Figure 7.1. Following this argument one step further, GLM implies there can be no particular particle or structure that can be the source of the magnetic ﬁeld (because then that would be a start point for ﬁeld lines). This is one way in which the magnetic ﬁeld is very different from the electrostatic ﬁeld, for which every ﬁeld line begins at a charged particle. So, when we say that current (for example) is the source of the magnetic ﬁeld, we mean only that the ﬁeld coexists with current, and not that the magnetic ﬁeld is somehow attached to the current. Summarizing, there is no “localizable” quantity, analogous to charge for electric ﬁelds, associated with magnetic ﬁelds. This is just another way in which magnetic ﬁelds are weird!	Electromagnetics_Vol1_Page_162_Chunk1463
148 CHAPTER 7. MAGNETOSTATICS c⃝Youming / K. Kikkeri CC BY SA 4.0 Figure 7.1: Gauss’ Law for Magnetostatics applied to a two-dimensional bar magnet. For the surface S = SA, every ﬁeld line entering S also leaves S, so the ﬂux through S is zero. For the surface S = SB, every ﬁeld line within S remains in S, so the ﬂux through S is again zero. Summarizing: Gauss’ Law for Magnetic Fields requires that magnetic ﬁeld lines form closed loops. Further- more, there is no particle that can be identiﬁed as the source of the magnetic ﬁeld. Additional Reading: • “Gauss’ Law for Magnetism” on Wikipedia. • “Maxwell’s Equations” on Wikipedia. 7.3 Gauss’ Law for Magnetism: Differential Form [m0047] The integral form of Gauss’ Law (Section 7.2) states that the magnetic ﬂux through a closed surface is zero. In mathematical form: I S B · ds = 0 (7.2) where B is magnetic ﬂux density and S is the enclosing surface. Just as Gauss’s Law for electrostatics has both integral (Sections 5.5) and differential (Section 5.7) forms, so too does Gauss’ Law for Magnetic Fields. Here we are interested in the differential form for the same reason. Given a differential equation and the boundary conditions imposed by structure and materials, we may then solve for the magnetic ﬁeld in very complicated scenarios. The equation we seek may be obtained from Equation 7.2 using the Divergence Theorem (Section 4.7), which in the present case may be written: Z V (∇· B) dv = I S B · ds (7.3) Where V is the mathematical volume bounded by the closed surface S. From Equation 7.2 we see that the right hand side of the equation is zero, leaving: Z V (∇· B) dv = 0 (7.4) The above relationship must hold regardless of the speciﬁc location or shape of V. The only way this is possible is if the integrand is everywhere equal to zero. We conclude: ∇· B = 0 (7.5) The differential (“point”) form of Gauss’ Law for Magnetic Fields (Equation 7.5) states that the ﬂux per unit volume of the magnetic ﬁeld is al- ways zero.	Electromagnetics_Vol1_Page_163_Chunk1464
7.4. AMPERE’S CIRCUITAL LAW (MAGNETOSTATICS): INTEGRAL FORM 149 This is another way of saying that there is no point in space that can be considered to be the source of the magnetic ﬁeld, for if it were, then the total ﬂux through a bounding surface would be greater than zero. Said yet another way, the source of the magnetic ﬁeld is not localizable. Additional Reading: • “Gauss’ Law for Magnetism” on Wikipedia. 7.4 Ampere’s Circuital Law (Magnetostatics): Integral Form [m0019] Ampere’s circuital law (ACL) relates current to the magnetic ﬁeld associated with the current. In the magnetostatic regime, the law is (see also Figure 7.2): I C H · dl = Iencl (7.6) That is, the integral of the magnetic ﬁeld intensity H over a closed path C is equal to the current enclosed by that path, Iencl. Before proceeding to interpret this law, it is useful to see that it is dimensionally correct. That is, H (units of A/m) integrated over a distance (units of m) yields a quantity with units of current (i.e., A). In general, Iencl may be either positive or negative. The direction corresponding to positive current ﬂow must be correctly associated with C. The relationship follows the right-hand rule of Stokes’ Theorem (Section 4.9) summarized as follows. The direction of positive Iencl is the direction in which the ﬁngers of the right hand intersect any surface S bordered by C when the thumb of the right hand points in the direction of integration. The connection to Stokes’ Theorem is not a coincidence. See Section 7.9 for more about this. Iencl c⃝K. Kikkeri CC BY SA 4.0 Figure 7.2: Reference directions for Ampere’s Cir- cuital Law (Equation 7.6).	Electromagnetics_Vol1_Page_164_Chunk1465
150 CHAPTER 7. MAGNETOSTATICS Note that S can be any surface that is bounded by C – not just the taut surface implied in Figure 7.2. The integral form of Ampere’s Circuital Law for magnetostatics (Equation 7.6) relates the mag- netic ﬁeld along a closed path to the total cur- rent ﬂowing through any surface bounded by that path. ACL plays a role in magnetostatics that is very similar to the role played by the integral form of Gauss’ Law in electrostatics (Section 5.5). That is, ACL provides a means calculate the magnetic ﬁeld given the source current. ACL also has a similar limitation. Generally, symmetry is required to simplify the problem sufﬁciently so that the integral equation may be solved. Fortunately, a number of important problems fall in this category. Examples include the problems addressed in Sections 7.5 (magnetic ﬁeld of a line current), 7.6 (magnetic ﬁeld inside a straight coil), 7.7 (magnetic ﬁeld of a toroidal coil), 7.8 (magnetic ﬁeld of a current sheet), and 7.11 (boundary conditions on the magnetic ﬁeld intensity). For problems in which the necessary symmetry is not available, the differential form of ACL may be required (Section 7.9). Finally, be aware that the form of ACL addressed here applies to magnetostatics only. In the presence of a time-varying electric ﬁeld, the right side of ACL includes an additional term known as the displacement current (Section 8.9). Additional Reading: • “Maxwell’s Equations” on Wikipedia. • “Ampere’s Circuital Law” on Wikipedia. 7.5 Magnetic Field of an Inﬁnitely-Long Straight Current-Bearing Wire [m0119] In this section, we use the magnetostatic form of Ampere’s Circuital Law (ACL) (Section 7.4) to determine the magnetic ﬁeld due to a steady current I (units of A) in an inﬁnitely-long straight wire. The problem is illustrated in Figure 7.3. The wire is an electrically-conducting circular cylinder of radius a. Since the wire is a cylinder, the problem is easiest to work in cylindrical coordinates with the wire aligned along the z axis. Here’s the relevant form of ACL: I C H · dl = Iencl (7.7) where Iencl is the current enclosed by the closed path C. ACL works for any closed path, so to exploit the z y x a I ρ c⃝K. Kikkeri CC BY SA 4.0 Figure 7.3: Determination of the magnetic ﬁeld due to steady current in an inﬁnitely-long straight wire.	Electromagnetics_Vol1_Page_165_Chunk1466
7.5. MAGNETIC FIELD OF AN INFINITELY-LONG STRAIGHT CURRENT-BEARING WIRE 151 symmetry of the cylindrical coordinate system we choose a circular path of radius ρ in the z = 0 plane, centered at the origin. With this choice we have Iencl = I for ρ ≥a (7.8) For ρ < a, we see that Iencl < I. a steady (DC) current will be distributed uniformly throughout the wire (Section 6.4). Since the current is uniformly distributed over the cross section, Iencl is less than the total current I by the same factor that the area enclosed by C is less than πa2, the cross-sectional area of the wire. The area enclosed by C is simply πρ2, so we have Iencl = I πρ2 πa2 = I ρ2 a2 for ρ < a (7.9) For the choice of C made above, Equation 7.7 becomes Z 2π φ=0 H · ˆφ ρ dφ = Iencl (7.10) Note that we have chosen to integrate in the +φ direction. Therefore, the right-hand rule speciﬁes that positive Iencl corresponds to current ﬂowing in the +z direction, which is consistent with the direction indicated in Figure 7.3. (Here’s an excellent exercise to test your understanding. Change the direction of the path of integration and conﬁrm that you get the same result obtained at the end of this section. Changing the direction of integration should not change the magnetic ﬁeld associated with the current!) The simplest way to solve for H from Equation 7.10 is to use a symmetry argument, which proceeds as follows: • Since the distribution of current is uniform and inﬁnite in the z-dimension, H can’t depend on z, and so H · ˆz must be zero everywhere. • The problem is identical after any amount of rotation in φ; therefore, the magnitude of H cannot depend on φ. This is a form of radial symmetry. Since we determined above that H can’t depend on z either, it must be that the magnitude of H can depend only on ρ. • The radial symmetry of the problem also requires that H · ˆρ be equal to zero. If this were not the case, then the ﬁeld would not be radially symmetric. Since we determined above that H· ˆz is also zero, H must be entirely ±ˆφ-directed. From the above considerations, the most general form of the magnetic ﬁeld intensity can be written H = ˆφH(ρ). Substituting this into Equation 7.10, we obtain Iencl = Z 2π φ=0 h ˆφH(ρ) i · ˆφ ρ dφ = ρH(ρ) Z 2π φ=0 dφ = 2πρH(ρ) (7.11) Therefore, H(ρ) = Iencl/2πρ. Reassociating the known direction, we obtain: H = ˆφIencl 2πρ (7.12) Therefore, the ﬁeld outside of the wire is: H = ˆφ I 2πρ for ρ ≥a (7.13) whereas the ﬁeld inside the wire is: H = ˆφ Iρ 2πa2 for ρ < a (7.14) (By the way, this is a good time for a units check.) Note that as ρ increases from zero to a (i.e., inside the wire), the magnetic ﬁeld is proportional to ρ and therefore increases. However, as ρ continues to increase beyond a (i.e., outside the wire), the magnetic ﬁeld is proportional to ρ−1 and therefore decreases. If desired, the associated magnetic ﬂux density can be obtained using B = µH. Summarizing: The magnetic ﬁeld due to current in an inﬁnite straight wire is given by Equations 7.13 (outside the wire) and 7.14 (inside the wire). The mag- netic ﬁeld is +ˆφ-directed for current ﬂowing in the +z direction, so the magnetic ﬁeld lines form concentric circles perpendicular to and centered on the wire.	Electromagnetics_Vol1_Page_166_Chunk1467
152 CHAPTER 7. MAGNETOSTATICS I H c⃝Jfmelero CC BY SA 4.0 (modiﬁed) Figure 7.4: Right-hand rule for the relationship be- tween the direction of current and the direction of the magnetic ﬁeld. Finally, we point out another “right-hand rule” that emerges from this solution, shown in Figure 7.4 and summarized below: The magnetic ﬁeld due to current in an inﬁnite straight wire points in the direction of the curled ﬁngers of the right hand when the thumb of the right hand is aligned in the direction of current ﬂow. This simple rule turns out to be handy in quickly determining the relationship between the directions of the magnetic ﬁeld and current ﬂow in many other problems, and so is well worth committing to memory. 7.6 Magnetic Field Inside a Straight Coil [m0120] In this section, we use the magnetostatic integral form of Ampere’s Circuital Law (ACL) (Section 7.4) to determine the magnetic ﬁeld inside a straight coil of the type shown in Figure 7.5 in response to a steady (i.e., DC) current. The result has a number of applications, including the analysis and design of inductors, solenoids (coils that are used as magnets, typically as part of an actuator), and as a building block and source of insight for more complex problems. The present problem is illustrated in Figure 7.6. The coil is circular with radius a and length l, and consists of N turns (“windings”) of wire wound with uniform winding density. Since the coil forms a cylinder, the problem is easiest to work in cylindrical coordinates with the axis of the coil aligned along the z axis. To begin, let’s take stock of what we already know about the answer, which is actually quite a bit. The magnetic ﬁeld deep inside the coil is generally aligned with axis of the coil as shown in Figure 7.7. This can be explained using the result for the magnetic ﬁeld due to a straight line current (Section 7.5), in which we found that the magnetic ﬁeld follows a “right-hand rule.” The magnetic ﬁeld points in the direction of the ﬁngers of the right hand when the thumb of the right hand is aligned in the direction of current ﬂow. The wire comprising the coil is obviously not straight, but we can consider just one short segment of one turn and then sum the results for all such segments. When we consider this for a single turn of the coil, the situation is as shown in Figure 7.8. Summing the results for many loops, we see that the direction of the magnetic ﬁeld inside the coil must be generally in the by Zureks (public domain) Figure 7.5: A straight coil.	Electromagnetics_Vol1_Page_167_Chunk1468
7.6. MAGNETIC FIELD INSIDE A STRAIGHT COIL 153 Figure 7.6: Determination of the magnetic ﬁeld due to DC current in a coil. c⃝Geek3 CC BY SA 3.0 Figure 7.7: Magnetic ﬁeld lines inside a straight coil with closely-spaced windings. (Dotted circles repre- sent current ﬂowing up/out from the page; crossed cir- cles represent current ﬂowing down/into the page.) +ˆz direction when the current I is applied as shown in Figures 7.6 and 7.7. However, there is one caveat. The windings must be sufﬁciently closely-spaced that the magnetic ﬁeld lines can only pass through the openings at the end of the coil and do not take any “shortcuts” between individual windings. Figure 7.7 also indicates that the magnetic ﬁeld lines near the ends of the coil diverge from the axis of the coil. This is understandable since magnetic ﬁeld lines form closed loops. The relatively complex structure of the magnetic ﬁelds near the ends of the coil (the “fringing ﬁeld”) and outside of the coil make them relatively difﬁcult to analyze. Therefore, here we shall restrict our attention to the magnetic ﬁeld deep inside the coil. This restriction turns out to be of little consequence as long as l ≫a. Also, it is apparent from the radial symmetry of the coil that the magnitude of the magnetic ﬁeld cannot depend on φ. Putting these ﬁndings together, we ﬁnd I c⃝Chetvorno CC0 1.0 (modiﬁed) Figure 7.8: Magnetic ﬁeld due to a single loop. that the most general form for the magnetic ﬁeld intensity deep inside the coil is H ≈ˆzH(ρ). That is, the direction of H is ±ˆz and the magnitude of H depends, at most, on ρ. In fact, we will soon ﬁnd with the assistance of ACL that the magnitude of H doesn’t depend on ρ either. Here’s the relevant form of ACL: I C H · dl = Iencl (7.15) where Iencl is the current enclosed by the closed path C. ACL works for any closed path that encloses the current of interest. Also, for simplicity, we prefer a path that lies on a constant-coordinate surface. The selected path is shown in Figure 7.9. The beneﬁts of this particular path will soon become apparent. However, note for now that this particular choice is consistent with the right-hand rule relating the direction of C to the direction of positive I. That is, when I is positive, the current in the turns of the coil pass through the surface bounded by C in the same direction as the ﬁngers of the right hand when the thumb is aligned in the indicated direction of C. Let’s deﬁne N to be the number of windings in the coil. Then, the winding density of the coil is N/l (turns/m). Let the path length in the z direction be l′, as indicated in Figure 7.9. Then the enclosed current is Iencl = N l l′I (7.16) That is, the number of turns per unit length times	Electromagnetics_Vol1_Page_168_Chunk1469
154 CHAPTER 7. MAGNETOSTATICS z l' ρ2 ρ1 A D C B z1 z2 cylindrical form of coil c⃝K. Kikkeri CC BY SA 4.0 Figure 7.9: Selected path of integration. length gives number of turns, and this quantity times the current through the wire is the total amount of current crossing the surface bounded by C. For the choice of C made above, and taking our approximation for the form of H as exact, Equation 7.15 becomes I C [ˆzH(ρ)] · dl = N l l′I (7.17) The integral consists of segments A, B, C, and D, as shown in Figure 7.9. Let us consider the result for each of these segments individually: • The integral over segments B and D is zero because dl = ˆρdρ for these segments, and so H · dl = 0 for these segments. • It is also possible to make the contribution from Segment C go to zero simply by letting ρ2 →∞. The argument is as follows. The magnitude of H outside the coil must decrease with distance from the coil, so for ρ sufﬁciently large, H(ρ) becomes negligible. If that’s the case, then the integral over Segment C also becomes negligible. With ρ2 →∞, only Segment A contributes signiﬁcantly to the integral over C and Equation 7.17 becomes: N l l′I = Z z2 z1 [ˆzH(ρ1)] · (ˆzdz) = H(ρ1) Z z2 z1 dz = H(ρ1) [z2 −z1] (7.18) Note z2 −z1 is simply l′. Also, we have found that the result is independent of ρ1, as anticipated earlier. Summarizing: H ≈ˆzNI l inside coil (7.19) Let’s take a moment to consider the implications of this remarkably simple result. • Note that it is dimensionally correct; that is, current divided by length gives units of A/m, which are the units of H. • We have found that the magnetic ﬁeld is simply winding density (N/l) times current. To increase the magnetic ﬁeld, you can either use more turns per unit length or increase the current. • We have found that the magnetic ﬁeld is uniform inside the coil; that is, the magnetic ﬁeld along the axis is equal to the magnetic ﬁeld close to the cylinder wall formed by the coil. However, this does not apply close to ends of the coil, since we have neglected the fringing ﬁeld. These ﬁndings have useful applications in more complicated and practical problems, so it is worthwhile taking note of these now. Summarizing: The magnetic ﬁeld deep inside an ideal straight coil (Equation 7.19) is uniform and proportional to winding density and current. Additional Reading: • “Electromagnetic Coil” on Wikipedia. • “Solenoid” on Wikipedia.	Electromagnetics_Vol1_Page_169_Chunk1470
7.7. MAGNETIC FIELD OF A TOROIDAL COIL 155 7.7 Magnetic Field of a Toroidal Coil [m0049] A toroid is a cylinder in which the ends are joined to form a closed loop. An example of a toroidal coil is shown in Figure 7.10. Toroidal coils are commonly used to form inductors and transformers. The principal advantage of toroidal coils over straight coils in these applications is magnetic ﬁeld containment – as we shall see in this section, the magnetic ﬁeld outside of a toroidal coil can be made negligibly small. This reduces concern about interactions between this ﬁeld and other ﬁelds and structures in the vicinity. In this section, we use the magnetostatic form of Ampere’s Circuital Law (ACL) (Section 7.4) to determine the magnetic ﬁeld due to a steady (DC) current ﬂowing through a toroidal coil. The problem is illustrated in Figure 7.11. The toroid is circular with inner and outer radii a and b, respectively. The coil consists of N windings (turns) of wire wound with uniform winding density. This problem is easiest to work in cylindrical coordinates with the toroid centered on the origin in z = 0 plane. To begin, let’s take stock of what we already know about the answer, which is actually quite a bit. First, a review of Section 7.6 (“Magnetic Field Inside a Straight Coil”) is recommended. There it is shown that the magnetic ﬁeld deep inside a straight coil is aligned with axis of the coil. This can be explained c⃝Slick CC0 1.0 Figure 7.10: A toroidal coil used as a large-value in- ductor in the power supply of a wireless router. x y a b I H Figure 7.11: Geometry of a toroidal coil. using the result for the magnetic ﬁeld due to a straight line current (Section 7.5), in which we found that the magnetic ﬁeld follows a “right-hand rule” – The magnetic ﬁeld points in the direction of the curled ﬁngers of the right hand when the thumb of the right hand is aligned in the direction of current ﬂow. The wire comprising the coil is obviously not straight, but we can consider just one short segment of one turn and then sum the results for all such segments. When we do this, we see that the direction of the magnetic ﬁeld inside the coil must be in the +ˆφ direction when the current I is applied as shown in Figure 7.11. Also, because the problem is identical after any amount of rotation around the z axis, the magnitude of the magnetic ﬁeld cannot depend on φ. Putting these ﬁndings together, we ﬁnd that the most general form for the magnetic ﬁeld intensity inside or outside the coil is H = ˆφH(ρ, z). Here’s the relevant form of ACL: I C H · dl = Iencl (7.20) where Iencl is the current enclosed by the closed path C. ACL works for any closed path, but we need one that encloses some current so as to obtain a relationship between I and H. Also, for simplicity, we prefer a path that lies on a constant-coordinate surface. The selected path is a circle of radius ρ	Electromagnetics_Vol1_Page_170_Chunk1471
156 CHAPTER 7. MAGNETOSTATICS x y a b I ρ C Figure 7.12: Selected path of integration. centered on the origin in the z = z0 plane, as shown in Figure 7.12. We further require C to lie entirely inside the coil, which ensures that the enclosed current includes the current of all the windings as they pass through the hole at the center of the toroid. We choose the direction of integration to be in the +φ direction, which is consistent with the indicated direction of positive I. That is, when I is positive, the current in the windings of the coil pass through the surface bounded by C in the same direction as the curled ﬁngers of the right hand when the thumb is aligned in the indicated direction of C. In terms of the variables we have deﬁned, the enclosed current is simply Iencl = NI (7.21) Equation 7.20 becomes I C h ˆφH(ρ, z0) i · dl = NI (7.22) Now evaluating the integral: NI = Z 2π 0 h ˆφH(ρ, z0) i · ˆφ ρ dφ = ρH(ρ, z0) Z 2π 0 dφ = 2πρH(ρ, z0) (7.23) It is now clear that the result is independent of z0. Summarizing: H = ˆφ NI 2πρ inside coil (7.24) Let’s take a moment to consider the implications of this result. • Note that it is dimensionally correct; that is, current divided by the circumference of C (2πρ) gives units of A/m, which are the units of H. • We have found that the magnetic ﬁeld is proportional to winding density (i.e., number of windings divided by circumference) times current. To increase the magnetic ﬁeld you can either use more windings or increase the current. • Remarkably, we have found that the magnitude of the magnetic ﬁeld inside the coil depends only on ρ; i.e., the distance from the central (here, z) axis. It is independent of z. Summarizing: The magnetic ﬁeld inside a toroidal coil (Equa- tion 7.24) depends only on distance from the cen- tral axis and is proportional to winding density and current. Now let us consider what happens outside the coil. For this, we consider any path of integration (C) that lies completely outside the coil. Note that any such path encloses no current and therefore Iencl = 0 for any such path. In this case we have: I C H · dl = 0 (7.25) There are two ways this could be true. Either H could be zero everywhere along the path, or H could be non-zero along the path in such a way that the integral windings out to be zero. The radial symmetry of the problem rules out the second possibility – if H is radially symmetric and C is radially symmetric, then the sign of H · dl should not change over C. Therefore:	Electromagnetics_Vol1_Page_171_Chunk1472
7.8. MAGNETIC FIELD OF AN INFINITE CURRENT SHEET 157 The magnetic ﬁeld everywhere outside an ideal toroidal coil is zero. Note the caveat signaled by the use of the adjective “ideal.” In a practical toroidal coil, we expect there will be some leakage of magnetic ﬂux between the windings. In practice, this leakage can be made negligibly small by using a sufﬁciently high winding density and winding the wire on material on a toroidal form (a “core”) having sufﬁciently large permeability. The use of a high-permeability core, as shown in Figure 7.10, will dramatically improve the already pretty-good containment. In fact, the use of such a core allows the spacing between windings to become quite large before leakage becomes signiﬁcant. One ﬁnal observation about toroidal coils is that at no point in the derivation of the magnetic ﬁeld did we need to consider the cross-sectional shape of the coil; we merely needed to know whether C was inside or outside the coil. Therefore: The magnetic ﬁeld inside an ideal toroidal coil does not depend on the cross-sectional shape of the coil. Additional Reading: • “Toroidal Inductors and Transformers” on Wikipedia. 7.8 Magnetic Field of an Inﬁnite Current Sheet [m0121] We now consider the magnetic ﬁeld due to an inﬁnite sheet of current, shown in Figure 7.13. The solution to this problem is useful as a building block and source of insight in more complex problems, as well as being a useful approximation to some practical problems involving current sheets of ﬁnite extent including, for example, microstrip transmission line and ground plane currents in printed circuit boards. The current sheet in Figure 7.13 lies in the z = 0 plane and the current density is Js = ˆxJs (units of A/m); i.e., the current is uniformly distributed such that the total current crossing any segment of width ∆y along the y direction is Js∆y. To begin, let’s take stock of what we already know about the answer, which is actually quite a bit. For example, imagine the current sheet as a continuum of thin strips parallel to the x axis and very thin in the y dimension. Each of these strips individually behaves like a straight line current I = Js∆y (units of A). The magnetic ﬁeld due to each of these strips is determined by a “right-hand rule” – the magnetic ﬁeld points in the direction of the curled ﬁngers of the right hand when the thumb of the right hand is aligned in the direction of current ﬂow. (Section 7.5). It is apparent from this much that H can have no ˆy component, since the ﬁeld of each individual strip has z C Js Lz Ly y Figure 7.13: Analysis of the magnetic ﬁeld due to an inﬁnite thin sheet of current.	Electromagnetics_Vol1_Page_172_Chunk1473
158 CHAPTER 7. MAGNETOSTATICS no ˆy component. When the magnetic ﬁeld due to each strip is added to that of all the other strips, the ˆz component of the sum ﬁeld must be zero due to symmetry. It is also clear from symmetry considerations that the magnitude of H cannot depend on x or y. Summarizing, we have determined that the most general form for H is ˆyH(z), and furthermore, the sign of H(z) must be positive for z < 0 and negative for z > 0. It’s possible to solve this problem by actually summing over the continuum of thin current strips as imagined above.1 However, it’s far easier to use Ampere’s Circuital Law (ACL; Section 7.4). Here’s the relevant form of ACL: I C H · dl = Iencl (7.26) where Iencl is the current enclosed by a closed path C. ACL works for any closed path, but we need one that encloses some current so as to obtain a relationship between Js and H. Also, for simplicity, we prefer a path that lies on a constant-coordinate surface. A convenient path in this problem is a rectangle lying in the x = 0 plane and centered on the origin, as shown in Figure 7.13. We choose the direction of integration to be counter-clockwise from the perspective shown in Figure 7.13, which is consistent with the indicated direction of positive Js according to the applicable right-hand rule from Stokes’ Theorem. That is, when Js is positive (current ﬂowing in the +ˆx direction), the current passes through the surface bounded by C in the same direction as the curled ﬁngers of the right hand when the thumb is aligned in the indicated direction of C. Let us deﬁne Ly to be the width of the rectangular path of integration in the y dimension and Lz to be the width in the z dimension. In terms of the variables we have deﬁned, the enclosed current is simply Iencl = JsLy (7.27) Equation 7.26 becomes I C [ˆyH(z)] · dl = JsLy (7.28) Note that H · dl = 0 for the vertical sides of the path, since H is ˆy-directed and dl = ˆzdz on those sides. 1In fact, this is pretty good thing to try, if for no other reason than to see how much simpler it is to use ACL instead. Therefore, only the horizontal sides contribute to the integral and we have: Z +Ly/2 −Ly/2 ˆyH −Lz 2 · (ˆydy) + Z −Ly/2 +Ly/2 ˆyH +Lz 2 · (ˆydy) = JsLy (7.29) Now evaluating the integrals: H −Lz 2 Ly −H +Lz 2 Ly = JsLy (7.30) Note that all factors of Ly cancel in the above equation. Furthermore, H(−Lz/2) = −H(+Lz/2) due to (1) symmetry between the upper and lower half-spaces and (2) the change in sign between these half-spaces, noted earlier. We use this to eliminate H(+Lz/2) and solve for H(−Lz/2) as follows: 2H(−Lz/2) = Js (7.31) yielding H(−Lz/2) = +Js 2 (7.32) and therefore H(+Lz/2) = −Js 2 (7.33) Furthermore, note that H is independent of Lz; for example, the result we just found indicates the same value of H(+Lz/2) regardless of the value of Lz. Therefore, H is uniform throughout all space, except for the change of sign corresponding for the ﬁeld above vs. below the sheet. Summarizing: H = ±ˆyJs 2 for z ≶0 (7.34) The magnetic ﬁeld intensity due to an inﬁnite sheet of current (Equation 7.34) is spatially uni- form except for a change of sign corresponding for the ﬁeld above vs. below the sheet.	Electromagnetics_Vol1_Page_173_Chunk1474
7.9. AMPERE’S LAW (MAGNETOSTATICS): DIFFERENTIAL FORM 159 7.9 Ampere’s Law (Magnetostatics): Differential Form [m0118] The integral form of Amperes’ Circuital Law (ACL; Section 7.4) for magnetostatics relates the magnetic ﬁeld along a closed path to the total current ﬂowing through any surface bounded by that path. In mathematical form:I C H · dl = Iencl (7.35) where H is magnetic ﬁeld intensity, C is the closed curve, and Iencl is the total current ﬂowing through any surface bounded by C. In this section, we derive the differential form of this equation. In some applications, this differential equation, combined with boundary conditions associated with discontinuities in structure and materials, can be used to solve for the magnetic ﬁeld in arbitrarily complicated scenarios. A more direct reason for seeking out this differential equation is that we gain a little more insight into the relationship between current and the magnetic ﬁeld, disclosed at the end of this section. The equation we seek may be obtained using Stokes’ Theorem (Section 4.9), which in the present case may be written: Z S (∇× H) · ds = I C H · dl (7.36) where S is any surface bounded by C, and ds is the differential surface area combined with the unit vector in the direction determined by the right-hand rule from Stokes’ Theorem. ACL tells us that the right side of the above equation is simply Iencl. We may express Iencl as the integral of the volume current density J (units of A/m2; Section 6.2) as follows: Iencl = Z S J · ds (7.37) so we may rewrite Equation 7.36 as follows: Z S (∇× H) · ds = Z S J · ds (7.38) The above relationship must hold regardless of the speciﬁc location or shape of S. The only way this is possible for all possible surfaces in all applicable scenarios is if the integrands are equal. Thus, we obtain the desired expression: ∇× H = J (7.39) That is, the curl of the magnetic ﬁeld intensity at a point is equal to the volume current density at that point. Recalling the properties of the curl operator (Section 4.8) – in particular, that curl involves derivatives with respect to direction – we conclude: The differential form of Ampere’s Circuital Law for magnetostatics (Equation 7.39) indicates that the volume current density at any point in space is proportional to the spatial rate of change of the magnetic ﬁeld and is perpendicular to the mag- netic ﬁeld at that point. Additional Reading: • “Ampere’s circuital law” on Wikipedia. • “Boundary value problem” on Wikipedia.	Electromagnetics_Vol1_Page_174_Chunk1475
160 CHAPTER 7. MAGNETOSTATICS 7.10 Boundary Conditions on the Magnetic Flux Density (B) [m0022] In homogeneous media, electromagnetic quantities vary smoothly and continuously. At an interface between dissimilar media, however, it is possible for electromagnetic quantities to be discontinuous. Continuities and discontinuities in ﬁelds can be described mathematically by boundary conditions and used to constrain solutions for ﬁelds away from these interfaces. In this section, we derive the boundary condition on the magnetic ﬂux density B at a smooth interface between two material regions, as shown in Figure 7.14.2 The desired boundary condition may be obtained from Gauss’ Law for Magnetic Fields (GLM; Section 7.2): I S B · ds = 0 (7.40) where S is any closed surface. Let S take the form of cylinder centered at a point on the interface, and for which the ﬂat ends are parallel to the surface and perpendicular to ˆn, as shown in Figure 7.14. Let the radius of this cylinder be a, and let the length of the 2It may be helpful to note the similarity (duality, in fact) between this derivation and the derivation of the associated boundary condi- tion on D presented in Section 5.18. region 1 region 2 n a h h c⃝K. Kikkeri CC BY SA 4.0 Figure 7.14: Determination of the boundary condition on B at the interface between material regions. cylinder be 2h. From GLM, we have I S B · ds = Z top B · ds + Z side B · ds + Z bottom B · ds = 0 (7.41) Now let us reduce h and a together while (1) maintaining a constant ratio h/a ≪1 and (2) keeping S centered on the interface. Because h ≪a, the area of the side can be made negligible relative to the area of the top and bottom. Then, as h →0, we are left with Z top B · ds + Z bottom B · ds →0 (7.42) As the area of the top and bottom sides become inﬁnitesimal, the variation in B over these areas becomes negligible. Now we have simply: B1 · ˆn∆A + B2 · (−ˆn) ∆A →0 (7.43) where B1 and B2 are the magnetic ﬂux densities at the interface but in regions 1 and 2, respectively, and ∆A is the area of the top and bottom sides. Note that the orientation of ˆn is important – we have assumed ˆn points into region 1, and we must now stick with this choice. Thus, we obtain ˆn · (B1 −B2) = 0 (7.44) where, as noted above, ˆn points into region 1. Summarizing: The normal (perpendicular) component of B across the boundary between two material re- gions is continuous. It is worth noting what this means for the magnetic ﬁeld intensity H. Since B = µH, it must be that The normal (perpendicular) component of H across the boundary between two material re- gions is discontinuous if the permeabilities are unequal.	Electromagnetics_Vol1_Page_175_Chunk1476
7.11. BOUNDARY CONDITIONS ON THE MAGNETIC FIELD INTENSITY (H) 161 7.11 Boundary Conditions on the Magnetic Field Intensity (H) [m0023] In homogeneous media, electromagnetic quantities vary smoothly and continuously. At a boundary between dissimilar media, however, it is possible for electromagnetic quantities to be discontinuous. Continuities and discontinuities in ﬁelds can be described mathematically by boundary conditions and used to constrain solutions for ﬁelds away from these boundaries. In this section, we derive boundary conditions on the magnetic ﬁeld intensity H. To begin, consider a region consisting of only two media that meet at a smooth boundary as shown in Figure 7.15. The desired boundary condition can be obtained directly from Ampere’s Circuital Law (ACL; Section 7.4): I C H · dl = Iencl (7.45) where C is any closed path and Iencl is the current that ﬂows through the surface bounded by that path in the direction speciﬁed by the “right-hand rule” of Stokes’ theorem. Let C take the form of a rectangle centered on a point on the boundary as shown in Figure 7.15, perpendicular to the direction of current ﬂow at that location. Let the sides A, B, C, and D be perpendicular and parallel to the boundary. Let the length of the parallel sides be l, and let the length of region 1 A D w t n Figure 7.15: Determining the boundary condition on H at the smooth boundary between two material re- gions. the perpendicular sides be w. Now we apply ACL. We must integrate in a counter-clockwise direction in order to be consistent with the indicated reference direction for Js. Thus: I H · dl = Z A H · dl + Z B H · dl + Z C H · dl + Z D H · dl = Iencl (7.46) Now we let w and l become vanishingly small while (1) maintaining the ratio l/w and (2) keeping C centered on the boundary. In this process, the contributions from the B and D segments become equal in magnitude but opposite in sign; i.e., Z B H · dl + Z D H · dl →0 (7.47) This leaves Z A H · dl + Z C H · dl →Iencl (7.48) Let us deﬁne the unit vector ˆt (“tangent”) as shown in Figure 7.15. Now we have simply: H1 ·	Electromagnetics_Vol1_Page_176_Chunk1477
162 CHAPTER 7. MAGNETOSTATICS Eliminating the common factor of ∆l and arranging terms on the left: (H2 −H1) · ˆt = Js ·	Electromagnetics_Vol1_Page_177_Chunk1478
7.12. INDUCTANCE 163 7.12 Inductance [m0123] Current creates a magnetic ﬁeld, which subsequently exerts force on other current-bearing structures. For example, the current in each winding of a coil exerts a force on every other winding of the coil. If the windings are ﬁxed in place, then this force is unable to do work (i.e., move the windings), so instead the coil stores potential energy. This potential energy can be released by turning off the external source. When this happens, charge continues to ﬂow, but is now propelled by the magnetic force. The potential energy that was stored in the coil is converted to kinetic energy and subsequently used to redistribute the charge until no current ﬂows. At this point, the inductor has expended its stored energy. To restore energy, the external source must be turned back on, restoring the ﬂow of charge and thereby restoring the magnetic ﬁeld. Now recall that the magnetic ﬁeld is essentially deﬁned in terms of the force associated with this potential energy; i.e., F = qv × B where q is the charge of a particle comprising the current, v is the velocity of the particle, and B is magnetic ﬂux density (Section 2.5). So, rather than thinking of the potential energy of the system as being associated with the magnetic force applied to current, it is equally valid to think of the potential energy as being stored in the magnetic ﬁeld associated with the current distribution. The energy stored in the magnetic ﬁeld depends on the geometry of the current-bearing structure and the permeability of the intervening material because the magnetic ﬁeld depends on these parameters. The relationship between current applied to a structure and the energy stored in the associated magnetic ﬁeld is what we mean by inductance. We may fairly summarize this insight as follows: Inductance is the ability of a structure to store en- ergy in a magnetic ﬁeld. The inductance of a structure depends on the ge- ometry of its current-bearing structures and the permeability of the intervening medium. Note that inductance does not depend on current, which we view as either a stimulus or response from this point of view. The corresponding response or stimulus, respectively, is the magnetic ﬂux associated with this current. This leads to the following deﬁnition: L = Φ I (single linkage) (7.58) where Φ (units of Wb) is magnetic ﬂux, I (units of A) is the current responsible for this ﬂux, and L (units of H) is the associated inductance. (The “single linkage” caveat will be explained below.) In other words, a device with high inductance generates a large magnetic ﬂux in response to a given current, and therefore stores more energy for a given current than a device with lower inductance. To use Equation 7.58 we must carefully deﬁne what we mean by “magnetic ﬂux” in this case. Generally, magnetic ﬂux is magnetic ﬂux density (again, B, units of Wb/m2) integrated over a speciﬁed surface S, so Φ = Z S B · ds (7.59) where ds is the differential surface area vector, with direction normal to S. However, this leaves unanswered the following questions: Which S, and which of the two possible normal directions of ds? For a meaningful answer, S must uniquely associate the magnetic ﬂux to the associated current. Such an association exists if we require the current to form a closed loop. This is shown in Figure 7.16. Here C is the closed loop along which the current ﬂows, S is a surface bounded by C, and the direction of ds is deﬁned according to the right-hand rule of Stokes’ Theorem (Section 4.9). Note that C can be a closed loop of any shape; i.e., not just circular, and not restricted to lying in a plane. Further note that S used in the calculation of Φ can be any surface bounded by C. This is because magnetic ﬁeld lines form closed loops such that any one magnetic ﬁeld line intersects any open surface bounded by C exactly once. Such an intersection is sometimes called a “linkage.” So there we have it	Electromagnetics_Vol1_Page_178_Chunk1479
– we require the current I to form a closed loop, we measure the magnetic ﬂux through this loop using the sign convention of the right-hand rule, and the ratio is the inductance. Many structures consist of multiple such loops – the coil is of course one of these. In a coil, each winding	Electromagnetics_Vol1_Page_178_Chunk1480
164 CHAPTER 7. MAGNETOSTATICS I B ds c⃝K. Kikkeri CC BY SA 4.0 Figure 7.16: Association between a closed loop of current and the associated magnetic ﬂux. carries the same current, and the magnetic ﬁelds of the windings add to create a magnetic ﬁeld, which grows in proportion to the winding density (Section 7.6). The magnetic ﬂux density inside a coil is proportional to the number of windings, N, so the ﬂux Φ in Equation 7.58 should properly be indicated as NΦ. Another way to look at this is that we are counting the number of times the same current is able to generate a unique set of magnetic ﬁeld lines that intersect S. Summarizing, our complete deﬁnition for inductance is L = NΦ I (identical linkages) (7.60) An engineering deﬁnition of inductance is Equa- tion 7.60, with the magnetic ﬂux deﬁned to be that associated with a single closed loop of cur- rent with sign convention as indicated in Fig- ure 7.16, and N deﬁned to be the number of times the same current I is able to create that ﬂux. What happens if the loops have different shapes? For example, what if the coil is not a cylinder, but rather cone-shaped? (Yes, there is such a thing – see “Additional Reading” at the end of this section.) In this case, one needs a better way to determine the factor NΦ since the ﬂux associated with each loop of current will be different. However, this is beyond the scope of this section. An inductor is a device that is designed to exhibit a speciﬁed inductance. We can now make the connection to the concept of the inductor as it appears in elementary circuit theory. First, we rewrite Equation 7.60 as follows: I = NΦ L (7.61) Taking the derivative of both sides of this equation with respect to time, we obtain: d dtI = N L d dtΦ (7.62) Now we need to reach beyond the realm of magnetostatics for just a moment. Section 8.3 (“Faraday’s Law”) shows that the change in Φ associated with a change in current results in the creation of an electrical potential equal to −NdΦ/dt realized over the loop C. In other words, the terminal voltage V is +NdΦ/dt, with the change of sign intended to keep the result consistent with the sign convention relating current and voltage in passive devices. Therefore, dΦ/dt in Equation 7.62 is equal to V/N. Making the substitution we ﬁnd: V = L d dtI (7.63) This is the expected relationship from elementary circuit theory. Another circuit theory concept related to inductance is mutual inductance. Whereas inductance relates changes in current to instantaneous voltage in the same device (Equation 7.63), mutual inductance relates changes in current in one device to instantaneous voltage in a different device. This can occur when the two devices are coupled (“linked”) by the same magnetic ﬁeld. For example, transformers (Section 8.5) typically consist of separate coils that are linked by the same magnetic ﬁeld lines. The voltage across one coil may be computed as the time-derivative of current on the other coil times the mutual inductance. Let us conclude this section by taking a moment to dispel a common misconception about inductance. The misconception pertains to the following question. If the current does not form a closed loop, what is the inductance? For example, engineers sometimes refer	Electromagnetics_Vol1_Page_179_Chunk1481
7.13. INDUCTANCE OF A STRAIGHT COIL 165 to the inductance of a pin or lead of an electronic component. A pin or lead is not a closed loop, so the formal deﬁnition of inductance given above – ratio of magnetic ﬂux to current – does not apply. The broader deﬁnition of inductance – the ability to store energy in a magnetic ﬁeld – does apply, but this is not what is meant by “pin inductance” or “lead inductance.” What is actually meant is the imaginary part of the impedance of the pin or lead – i.e., the reactance – expressed as an equivalent inductance. In other words, the reactance of an inductive device is positive, so any device that also exhibits a positive reactance can be viewed from a circuit theory perspective as an equivalent inductance. This is not referring to the storage of energy in a magnetic ﬁeld; it merely means that the device can be modeled as an inductor in a circuit diagram. In the case of “pin inductance,” the culprit is not actually inductance, but rather skin effect (see “Additional References” at the end of this section). Summarizing: Inductance implies positive reactance, but posi- tive reactance does not imply the physical mech- anism of inductance. Additional Reading: • “Inductance” on Wikipedia. • “Inductor” on Wikipedia. • T.A. Winslow, “Conical Inductors for Broadband Applications,” IEEE Microwave Mag., Vol. 6, No. 1, Mar 2005, pp. 68–72. • “Skin Effect” on Wikipedia. 7.13 Inductance of a Straight Coil [m0124] In this section, we determine the inductance of a straight coil, as shown in Figure 7.17. The coil is circular with radius a and length l and consists of N windings of wire wound with uniform winding density. Also, we assume the winding density N/l is large enough that magnetic ﬁeld lines cannot enter or exit between windings but rather must traverse the entire length of the coil. Since the coil forms a cylinder, the problem is easiest to work in cylindrical coordinates with the axis of the coil aligned along the z axis. Inductance L in this case is given by (Section 7.12) L = NΦ I (7.64) where I is current and Φ is the magnetic ﬂux associated with one winding of the coil. Magnetic ﬂux in this case is given by Φ = Z S B · ds (7.65) where B is the magnetic ﬂux density (units of T = Wb/m2), S is the surface bounded by a single current loop, and ds points in the direction determined by the right hand rule with respect to the direction of positive current ﬂow. First, let’s determine the magnetic ﬁeld. The magnetic Figure 7.17: Determination of the inductance of a straight coil.	Electromagnetics_Vol1_Page_180_Chunk1482
166 CHAPTER 7. MAGNETOSTATICS ﬂux density deep inside the coil is (Section 7.6): B ≈ˆzµNI l (7.66) Is it reasonable to use this approximation here? Since inductance pertains to energy storage, the question is really what fraction of the energy is stored in a ﬁeld that is well-described by this approximation, as opposed to energy stored in the “fringing ﬁeld” close to the ends of the coil. If we make l sufﬁciently large relative to a, then presumably energy storage in the fringing ﬁeld will be negligible in comparison. Since the alternative leads to a much more complicated problem, we shall assume that Equation 7.66 is valid for the interior of the coil. Next, we determine Φ. In this case, a natural choice for S is the interior cross-section of the coil in a plane perpendicular to the axis. The direction of ds must be +ˆz since this is the direction in which the ﬁngers of the right hand point when the current ﬂows in the direction indicated in Figure 7.17. Thus, we have Φ ≈ Z S ˆzµNI l · (ˆzds) = µNI l Z S ds = µNI l A (7.67) where A is the cross-sectional area of the coil. Finally from Equation 7.64 we obtain L ≈µN 2A l (l ≫a) (7.68) Note that this is dimensionally correct; that is, permeability (units of H/m) times area (units of m2) divided by length (units of m) gives units of H, as expected. Also, it is worth noting that inductance is proportional to permeability and cross-sectional area, and inversely proportional to length. Interestingly the inductance is proportional to N 2 as opposed to N; this is because ﬁeld strength increases with N, and independently there are N ﬂux linkages. Finally, we note that the inductance does not depend on the shape of the coil cross-section, but only on the area of the cross-section. Summarizing: The inductance of a long straight coil is given ap- proximately by Equation 7.68. Again, this result is approximate because it neglects the non-uniform fringing ﬁeld near the ends of the coil and the possibility that magnetic ﬁeld lines escape between windings due to inadequate winding density. Nevertheless, this result facilitates useful engineering analysis and design. Additional Reading: • “Inductance” on Wikipedia.	Electromagnetics_Vol1_Page_181_Chunk1483
7.14. INDUCTANCE OF A COAXIAL STRUCTURE 167 7.14 Inductance of a Coaxial Structure [m0125] Let us now determine the inductance of coaxial structure, shown in Figure 7.18. The inductance of this structure is of interest for a number of reasons – in particular, for determining the characteristic impedance of coaxial transmission line, as addressed in Section 3.10. For our present purpose, we may model the structure as shown in Figure 7.18. This model consists of two concentric perfectly-conducting cylinders of radii a and b, separated by a homogeneous material having permeability µ. To facilitate analysis, let us place the +z axis along the common axis of the concentric cylinders, so that the cylinders may be described as the surfaces ρ = a and ρ = b. Below we shall ﬁnd the inductance by assuming a current I on the inner conductor and integrating over the resulting magnetic ﬁeld to obtain the magnetic ﬂux Φ between the conductors. Then, inductance can be determined as the ratio of the response ﬂux to the source current. Before we get started, note the derivation we are about to do is similar to the derivation of the capacitance of a coaxial structure, addressed in Section 5.24. The reader may beneﬁt from a review of that section before attempting this derivation. b a z ρ z=l I μ z=0 H Figure 7.18: Determining the inductance of coaxial line. The ﬁrst step is to ﬁnd the magnetic ﬁeld inside the structure. This is relatively simple if we may neglect fringing ﬁelds, since then the internal ﬁeld may be assumed to be constant with respect to z. This analysis will also apply to the case where the length l pertains to one short section of a much longer structure; in this case we will obtain the inductance per length as opposed to the total inductance for the structure. Note that the latter is exactly what we need for the transmission line lumped-element equivalent circuit model (Section 3.4). To determine the inductance, we invoke the deﬁnition (Section 7.12): L ≜Φ I (7.69) A current I ﬂowing in the +z direction on the inner conductor gives rise to a magnetic ﬁeld inside the coaxial structure. The magnetic ﬁeld intensity for this scenario was determined in Section 7.5 where we found H = ˆφ I 2πρ , a ≤ρ ≤b (7.70) The reader should note that in that section we were considering merely a line of current; not a coaxial structure. So, on what basis do we claim the ﬁeld for inside the coaxial structure is the same? This is a consequence of Ampere’s Law (Section 7.4): I C H · dl = Iencl (7.71) If in this new problem we specify the same circular path C with radius greater than a and less than b, then the enclosed current is simply I. The presence of the outer conductor does not change the radial symmetry of the problem, and nothing else remains that can change the outcome. This is worth noting for future reference: The magnetic ﬁeld inside a coaxial structure comprised of concentric conductors bearing cur- rent I is identical to the magnetic ﬁeld of the line current I in free space. We’re going to need magnetic ﬂux density (B) as opposed to H in order to get the magnetic ﬂux. This is simple since they are related by the permeability of the medium; i.e., B = µH. Thus: B = ˆφ µI 2πρ , a ≤ρ ≤b (7.72)	Electromagnetics_Vol1_Page_182_Chunk1484
168 CHAPTER 7. MAGNETOSTATICS Next, we get Φ by integrating over the magnetic ﬂux density Φ = Z S B · ds (7.73) where S is any open surface through which all magnetic ﬁeld lines within the structure must pass. Since this can be any such surface, we may as well choose the simplest one. The simplest such surface is a plane of constant φ, since such a plane is a constant-coordinate surface and perpendicular to the magnetic ﬁeld lines. This surface is shown as the shaded area in Figure 7.18. Using this surface we ﬁnd: Φ = Z b ρ=a Z l z=0 ˆφ µI 2πρ · ˆφ dρ dz = µI 2π Z l z=0 dz ! Z b ρ=a dρ ρ ! = µIl 2π ln b a (7.74) Wrapping up: L ≜Φ I = (µIl/2π) ln (b/a) I (7.75) Note that factors of I in the numerator and denominator cancel out, leaving: L = µl 2π ln b a (7.76) Note that this is dimensionally correct, having units of H. Also note that this is expression depends only on materials (through µ) and geometry (through l, a, and b). Notably, it does not depend on current, which would imply non-linear behavior. To make the connection back to lumped-element transmission line model parameters (Sections 3.4 and 3.10), we simply divide by l to get the per-unit length parameter: L′ = µ 2π ln b a (7.77) which has the expected units of H/m. Example 7.1. Inductance of RG-59 coaxial cable. RG-59 coaxial cable consists of an inner conductor having radius 0.292 mm, an outer conductor having radius 1.855 mm, and polyethylene (a non-magnetic dielectric) spacing material. Estimate the inductance per length of RG-59. Solution. From the problem statement, a = 0.292 mm, b = 1.855 mm, and µ ∼= µ0 since the spacing material is non-magnetic. Using Equation 7.77, we ﬁnd L′ ∼= 370 nH/m.	Electromagnetics_Vol1_Page_183_Chunk1485
7.15. MAGNETIC ENERGY 169 7.15 Magnetic Energy [m0127] Consider a structure exhibiting inductance; i.e., one that is able to store energy in a magnetic ﬁeld in response to an applied current. This structure could be a coil, or it could be one of a variety of inductive structures that are not explicitly intended to be an inductor; for example, a coaxial transmission line. When current is applied, the current-bearing elements of the structure exert forces on each other. Since these elements are not normally free to move, we may interpret this force as potential energy stored in the magnetic ﬁeld associated with the current (Section 7.12). We now want to know how much energy is stored in this ﬁeld. The answer to this question has relevance in several engineering applications. One issue is that any system that includes inductance is using some fraction of the energy delivered by the power supply to energize this inductance. In many electronic systems – in power systems in particular – inductors are periodically energized and de-energized at a regular rate. Since power is energy per unit time, this consumes power. Therefore, energy storage in inductors contributes to the power consumption of electrical systems. The stored energy is most easily determined using circuit theory concepts. First, we note that the electrical potential difference v(t) (units of V) across an inductor is related to the current i(t) (units of A) through the inductor as follows (Section 7.12): v(t) = L d dti(t) (7.78) where L (units of H) is the inductance. The instantaneous power associated with the device is p(t) = v(t)i(t) (7.79) Energy (units of J) is power (units of J/s) integrated over time. Let Wm be the energy stored in the inductor. At some time t0 in the past, i(t0) = 0 and Wm = 0. As current is applied, Wm increases monotonically. At the present time t, i(t) = I. Thus, the present value of the magnetic energy is: Wm = Z t0+t t0 p(τ)dτ (7.80) Now evaluating this integral using the relationships established above: Wm = Z t+t0 t0 v(τ)i(τ)dτ = Z t+t0 t0 L d dτ i(τ) i(τ)dτ = L Z t+t0 t0 d dτ i(τ) i(τ)dτ (7.81) Changing the variable of integration from τ (and dτ) to i (and di) we have Wm = L Z t+t0 t0 di dτ i dτ = L Z I 0 i di (7.82) Evaluating the integral we obtain the desired expression Wm = 1 2LI2 (7.83) The energy stored in an inductor in response to a steady current I is Equation 7.83. This energy increases in proportion to inductance and in pro- portion to the square of current. The long straight coil (Section 7.13) is representative of a large number of practical applications, so it is useful to interpret the above ﬁndings in terms of this structure in particular. For this structure we found L = µN 2A l (7.84) where µ is the permeability, N is the number of windings, A is cross-sectional area, and l is length. The magnetic ﬁeld intensity inside this structure is related to I by (Section 7.6): H = NI l (7.85)	Electromagnetics_Vol1_Page_184_Chunk1486
170 CHAPTER 7. MAGNETOSTATICS Substituting these expressions into Equation 7.83, we obtain Wm = 1 2 µN 2A l Hl N 2 = 1 2µH2Al (7.86) Recall that the magnetic ﬁeld inside a long coil is approximately uniform. Therefore, the density of energy stored inside the coil is approximately uniform. Noting that the product Al is the volume inside the coil, we ﬁnd that this energy density is Wm/Al; thus: wm = 1 2µH2 (7.87) which has the expected units of energy per unit volume (J/m3). The above expression provides an alternative method to compute the total magnetostatic energy in any structure. Within a mathematical volume V, the total magnetostatic energy is simply the integral of the energy density over V; i.e., Wm = Z V wm dv (7.88) This works even if the magnetic ﬁeld and the permeability vary with position. Substituting Equation 7.87 we obtain: Wm = 1 2 Z V µH2dv (7.89) Summarizing: The energy stored by the magnetic ﬁeld present within any deﬁned volume is given by Equa- tion 7.89. It’s worth noting that this energy increases with the permeability of the medium, which makes sense since inductance is proportional to permeability. Finally, we reiterate that although we arrived at this result using the example of the long straight coil, Equations 7.87 and 7.89 are completely general. 7.16 Magnetic Materials [m0058] As noted in Section 2.5, magnetic ﬁelds arise in the presence of moving charge (i.e., current) and in the presence of certain materials. In this section, we address these “magnetic materials.” A magnetic material may be deﬁned as a substance that exhibits permeability µ (Section 2.6) that is signiﬁcantly different from the permeability of free space µ0. Since the magnetic ﬂux density B is related to the magnetic ﬁeld intensity H via B = µH, magnetic materials may exhibit magnetic ﬂux density in response to a given magnetic ﬁeld intensity that is signiﬁcantly greater than that of other materials. Magnetic materials are also said to be “magnetizable,” meaning that the application of a magnetic ﬁeld causes the material itself to become a source of the magnetic ﬁeld. Magnetic media are typically metals, semiconductors, or heterogeneous media containing such materials. An example is ferrite, which consists of iron particles suspended in a ceramic. Magnetic media are commonly classiﬁed according to the physical mechanism responsible for their magnetizability. These mechanisms include paramagnetism, diamagnetism, and ferromagnetism. All three of these mechanisms involve quantum mechanical processes operating at the atomic and subatomic level, and are not well-explained by classical physics. These processes are beyond the scope of this book (but information is available via “Additional References” at the end of this section). However, it is possible to identify some readily-observable differences between these categories of magnetic media. Paramagnetic and diamagnetic materials exhibit permeability that is only very slightly different than µ0 and typically by much less than 0.01%. These materials exhibit very weak and temporary magnetization. The principal distinction between paramagnetic and diamagnetic media is in the persistence and orientation of induced magnetic ﬁelds. Paramagnetic materials – including aluminum, magnesium, and platinum – exhibit a very weak persistent magnetic ﬁeld, and the magnetic ﬁeld	Electromagnetics_Vol1_Page_185_Chunk1487
7.16. MAGNETIC MATERIALS 171 induced in the material is aligned in the same direction as the impressed (external) magnetic ﬁeld. Diamagnetic materials – including copper, gold, and silicon – do not exhibit a persistent magnetic ﬁeld, and the magnetic ﬁeld induced in the material is (counter to intuition!) aligned in the opposite direction as the impressed magnetic ﬁeld. The magnetization of paramagnetic and diamagnetic media is typically so weak that it is not often a consideration in engineering analysis and design. Paramagnetic and diamagnetic media exhibit per- meability only very slightly different than that of free space, with little or no magnetization. Ferromagnetic materials, on the other hand, exhibit permeability that can be many orders of magnitude greater than µ0. (See Appendix A.2 for some example values.) These materials can be readily and indeﬁnitely magnetized, thus, permanent magnets are typically comprised of ferromagnetic materials. Commonly-encountered ferromagnetic materials include iron, nickel, and cobalt. Ferromagnetic materials are signiﬁcantly non-linear (see deﬁnition in Section 2.8), exhibiting saturation and hysteresis. This is illustrated in Figure 7.19. In this plot, the origin represents a ferromagnetic material that is unmagnetized and in a region free of an external magnetic ﬁeld. The external magnetic ﬁeld is quantiﬁed in terms of H, plotted along the horizontal axis. As the external ﬁeld is increased, so to is B in the material, according to the relationship B = µH. Right away we see the material is non-linear, since the slope of the curve – and hence µ – is not constant. Once the external magnetizing ﬁeld H exceeds a certain value, the response ﬁeld B no longer signiﬁcantly increases. This is saturation. Once saturated, further increases in the external ﬁeld result do not signiﬁcantly increase the magnetization of the material, so there is no signiﬁcant increase in B. From this state of saturation, let us now reduce the external ﬁeld. We ﬁnd that the rate of decrease in B with respect to H is signiﬁcantly less than the rate that B originally increased with respect to H. In fact, B is still greater than zero even when H has been reduced to zero. At this point, the magnetization of B saturation H c⃝Ndthe (modiﬁed) CC BY SA 3.0 Figure 7.19: Non-linearity in a ferromagnetic material manifesting as saturation and hysteresis. the material is obvious, and a device comprised of this material could be used as a magnet. If we now apply an external ﬁeld in the reverse direction, we ﬁnd that we are eventually able to zero and then redirect the response ﬁeld. As we continue to decrease H (that is, increase the magnitude in the reverse direction), we once again reach saturation. The same behavior is observed when we once again increase H. The material is eventually demagnetized, remagnetized in the opposite direction and then saturated in that direction. At this point, it is apparent that a return to the start condition (H = B = 0; i.e., demagnetized when there is no external ﬁeld) is not possible. Hysteresis is the name that we apply to this particular form of non-linear behavior. Hysteresis has important implications in engineering applications. First, as identiﬁed above, it is an important consideration in the analysis and design of magnets. In applications where a ferromagnetic material is being used because high permeability is desired – e.g., in inductors (Section 7.12) and transformers (Section 8.5) – hysteresis complicates the design and imposes limits on the performance of the device. Hysteresis may also be exploited as a form of memory. This is apparent from Figure 7.19. If B > 0,	Electromagnetics_Vol1_Page_186_Chunk1488
172 CHAPTER 7. MAGNETOSTATICS then recent values of H must have been relatively large and positive. Similarly, If B < 0, then recent values of H must have been relatively large and negative. Furthermore, the most recent sign of H can be inferred even if the present value of H is zero. In this sense, the material “remembers” the past history of its magnetization and thereby exhibits memory. This is the enabling principle for a number of digital data storage devices, including hard drives (see “Additional Reading” at the end of this section). Summarizing: Ferromagnetic media exhibit permeability µ that is orders of magnitude greater than that of free space and are readily magnetizable. These mate- rials are also nonlinear in µ, which manifests as saturation and hysteresis. Additional Reading: • Section A.2 (“Permeability of Some Common Materials”) • “Magnetism” on Wikipedia. • “Ferrite (magnet)” on Wikipedia. • “Paramagnetism” on Wikipedia. • “Diamagnetism” on Wikipedia. • “Ferromagnetism” on Wikipedia. • “Magnetic hysteresis” on Wikipedia. • “Magnetic storage” on Wikipedia. • “Hard disk drive” on Wikipedia. [m0150]	Electromagnetics_Vol1_Page_187_Chunk1489
7.16. MAGNETIC MATERIALS 173 Image Credits Fig. 7.1: c⃝Youming / K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0018 fGLMBarMagnet (2).svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 7.2: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0019 fACL.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 7.3: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0119 fLSW.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 7.4: c⃝Jfmelero, https://et.wikipedia.org/wiki/Fail:Manoderecha.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/), modiﬁed. Fig. 7.5: Zureks, https://en.wikipedia.org/wiki/File:Solenoid-1.png, public domain. Fig. 7.7: c⃝Geek3, https://commons.wikimedia.org/wiki/File:VFPt Solenoid correct.svg, CC BY SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/). Fig. 7.8: Chetvorno, https://en.wikipedia.org/wiki/File:Magnetic ﬁeld of loop 3.svg, public domain via CC0 1.0, modiﬁed. Fig. 7.9: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0120 fPath.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 7.10: Slick, https://en.wikipedia.org/wiki/File: 3Com OfﬁceConnect ADSL Wireless 11g Firewall Router 2012-10-28-0869.jpg, public domain via CC0 1.0 (https://creativecommons.org/publicdomain/zero/1.0/deed.en). Fig. 7.14: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0022 fBoundary.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 7.16: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0123 fFluxCurrent.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 7.19: c⃝Ndthe, https://commons.wikimedia.org/wiki/File:HysteresisSVG.svg, CC BY SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/deed.en), modiﬁed by author.	Electromagnetics_Vol1_Page_188_Chunk1490
Chapter 8 Time-Varying Fields 8.1 Comparison of Static and Time-Varying Electromagnetics [m0013] Students encountering time-varying electromagnetic ﬁelds for the ﬁrst time have usually been exposed to electrostatics and magnetostatics already. These disciplines exhibit many similarities as summarized in Table 8.1. The principles of time-varying electromagnetics presented in this table are all formally introduced in other sections; the sole purpose of this table is to point out the differences. We can summarize the differences as follows: Maxwell’s Equations in the general (time- varying) case include extra terms that do not ap- pear in the equations describing electrostatics and magnetostatics. These terms involve time deriva- tives of ﬁelds and describe coupling between electric and magnetic ﬁelds. The coupling between electric and magnetic ﬁelds in the time-varying case has one profound consequence in particular. It becomes possible for ﬁelds to continue to exist even after their sources – i.e., charges and currents – are turned off. What kind of ﬁeld can continue to exist in the absence of a source? Such a ﬁeld is commonly called a wave. Examples of waves include signals in transmission lines and signals propagating away from an antenna. Additional Reading: • “Maxwell’s Equations” on Wikipedia. Electromagnetics Vol 1. c⃝2018 S.W. Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics-vol-1	Electromagnetics_Vol1_Page_189_Chunk1491
8.2. ELECTROMAGNETIC INDUCTION 175 Electrostatics / Time-Varying Magnetostatics (Dynamic) Electric & magnetic independent possibly coupled ﬁelds are... Maxwell’s Eqns. H S D · ds = Qencl H S D · ds = Qencl (integral) H C E · dl = 0 H C E · dl = −∂ ∂t R S B · ds H S B · ds = 0 H S B · ds = 0 H C H · ds = Iencl H C H · dl = Iencl+ R S ∂ ∂tD · ds Maxwell’s Eqns. ∇· D = ρv ∇· D = ρv (differential) ∇× E = 0 ∇× E = −∂ ∂tB ∇· B = 0 ∇· B = 0 ∇× H = J ∇× H = J+ ∂ ∂tD Table 8.1: Comparison of principles governing static and time-varying electromagnetic ﬁelds. Differences in the time-varying case relative to the static case are highlighted in blue. 8.2 Electromagnetic Induction [m0129] When an electrically-conducting structure is exposed to a time-varying magnetic ﬁeld, an electrical potential difference is induced across the structure. This phenomenon is known as electromagnetic induction. A convenient introduction to electromagnetic induction is provided by Lenz’s Law. This section explains electromagnetic induction in the context of Lenz’s Law and provides two examples. Let us begin with the example depicted in Figure 8.1, involving a cylindrical coil. Attached to the terminals of the coil is a resistor for which we may identify an electric potential difference V and current I. In this particular case, the sign conventions indicated for V and I are arbitrary, but it is important to be consistent once they are established. Now let us introduce a bar magnet as shown in Figure 8.1. The magnet is centered along the axis of coil, to the right of the coil, and with its north pole facing toward the coil. The magnet is responsible for the magnetic ﬂux density Bimp. We refer to Bimp as an impressed magnetic ﬁeld because this ﬁeld exists independently of any response that may be induced by interaction with the coil. Note that Bimp points to the left inside the coil. The experiment consists of three tests. We will ﬁnd in Y. Qin (modiﬁed) CC BY 4.0 Figure 8.1: An experiment demonstrating electro- magnetic induction and Lenz’s Law. two of these tests that current ﬂows (i.e., \|I\| > 0), and subsequently, there is an induced magnetic ﬁeld Bind due to this current. It is the direction of the current and subsequently the direction of Bind inside the coil that we wish to observe. The ﬁndings are summarized below and in Table 8.2. • When the magnet is motionless, we have the unsurprising result that there is no current in the coil. Therefore, no magnetic ﬁeld is induced, and the total magnetic ﬁeld is simply equal to Bimp. • When the magnet moves toward the coil, we observe current that is positive with respect to the reference direction indicated in Figure 8.1. This current creates an induced magnetic ﬁeld Bind that points to the right, as predicted by magnetostatic considerations from the right-hand	Electromagnetics_Vol1_Page_190_Chunk1492
176 CHAPTER 8. TIME-VARYING FIELDS rule. Since Bimp points to the left, it appears that the induced current is opposing the increase in the magnitude of the total magnetic ﬁeld. • When the magnet moves away from the coil, we observe current that is negative with respect to the reference direction indicated in Figure 8.1. This current yields Bind that points to the left. Since Bimp points to the left, it appears that the induced current is opposing the decrease in the magnitude of the total magnetic ﬁeld. The ﬁrst conclusion one may draw from this experiment is that changes in the magnetic ﬁeld can induce current. This was the claim made in the ﬁrst paragraph of this section and is a consequence of Faraday’s Law, which is tackled in detail in Section 8.3. The second conclusion – also associated with Faraday’s Law – is the point of this section: The induced magnetic ﬁeld – that is, the one due to the current induced in the coil – always opposes the change in the impressed magnetic ﬁeld. Generalizing: Lenz’s Law states that the current that is induced by a change in an impressed magnetic ﬁeld cre- ates an induced magnetic ﬁeld that opposes (acts to reduce the effect of) the change in the total magnetic ﬁeld. When the magnet moves, three things happen: (1) A current is induced, (2) A magnetic ﬁeld is induced (which adds to the impressed magnetic ﬁeld), and (3) the value of V becomes non-zero. Lenz’s Law does not address which of these are responding directly to the change in the impressed magnetic ﬁeld, and which of these are simply responding to changes in the other quantities. Lenz’s Law may leave you with the incorrect impression that it is I that is induced, and that Bind and V are simply responding to this current. In truth, the quantity that is induced is actually V . This can be veriﬁed in the above experiment by replacing the resistor with a high-impedance voltmeter, which will indicate that V is changing even though there is negligible current ﬂow. Current ﬂow is simply a response to the induced potential. Nevertheless, it is common to say informally that “I is induced,” even if it is only indirectly through V . So, if Lenz’s Law is simply an observation and not an explanation of the underlying physics, then what is it good for? Lenz’s Law is often useful for quickly determining the direction of current ﬂow in practical electromagnetic induction problems, without resorting to the mathematics associated with Faraday’s Law. Here’s an example: Example 8.1. Electromagnetic induction through a transformer. Figure 8.2 shows a rudimentary circuit consisting of a battery and a switch on the left, a voltmeter on the right, and a transformer linking the two. It is not necessary to be familiar with transformers to follow this example; sufﬁce it to say, the transformer considered here consists of two coils wound around a common toroidal core, which serves to contain magnetic ﬂux. In this way, the ﬂux generated by either coil is delivered to the other coil with negligible loss. The experiment begins with the switch on the left in the open state. Thus, there is no current and no magnetic ﬁeld apparent in the coil on the left. The voltmeter reads 0 V. When the switch is closed, what happens? Solution. Closing the switch creates a current in the coil on the left. Given the indicated polarity of the battery, this current ﬂows counter-clockwise through the circuit on the left, with current arriving in the left coil through the bottom terminal. Given the indicated direction of the winding in the left coil, the impressed magnetic ﬁeld Bimp is oriented counter-clockwise through the toroidal core. The coil on the right “sees” Bimp increase from zero to some larger value. Since the voltmeter presumably has input high impedance, negligible current ﬂows. However, if current were able to ﬂow, Lenz’s Law dictates that it would be induced to ﬂow in a counter-clockwise direction around the circuit on the right, since the induced magnetic ﬁeld Bind	Electromagnetics_Vol1_Page_191_Chunk1493
would then be clockwise-directed so as to oppose the increase in Bimp. Therefore, the potential measured at the bottom of the right coil would be higher than the potential at the top of the right coil. The	Electromagnetics_Vol1_Page_191_Chunk1494
8.2. ELECTROMAGNETIC INDUCTION 177 Magnet is ... \|Bimp\| in coil is ... Circuit Response Bind inside coil Motionless constant V = 0, I = 0 none Moving toward coil increasing V > 0, I > 0 Pointing right Moving away from coil decreasing V < 0, I < 0 Pointing left Table 8.2: Results of the experiment associated with Figure 8.1. + - + - + - - + V E. Bach CC0 1.0 (modiﬁed) Figure 8.2: Electromagnetic induction through a transformer. ﬁgure indicates that the voltmeter measures the potential at its right terminal relative to its left terminal, so the needle will deﬂect to the right. This deﬂection will be temporary, since the current provided by the battery becomes constant at a new non-zero value and Bind responds only to the change in Bimp. The voltmeter reading will remain at zero for as long as the switch remains closed and the current remains steady. Here are some follow up exercises to test your understanding of what is going on: (1) Now open the switch. What happens? (2) Repeat the original experiment, but before starting, swap the terminals on the battery. Finally, it is worth noting that Lenz’s law can also be deduced from the principle of conservation of energy. The argument is that if the induced magnetic ﬁeld reinforced the change in the impressed magnetic ﬁeld, then the sum magnetic ﬁeld would increase. This would result in a further increase in the induced magnetic ﬁeld, leading to a positive feedback situation. However, positive feedback cannot be supported without an external source of energy, leading to a logical contradiction. In other words, the principle of conservation of energy requires the negative feedback described by Lenz’s law. Additional Reading: • “Electromagnetic Induction” on Wikipedia. • “Lenz’s Law” on Wikipedia.	Electromagnetics_Vol1_Page_192_Chunk1495
178 CHAPTER 8. TIME-VARYING FIELDS 8.3 Faraday’s Law [m0055] Faraday’s Law describes the generation of electric potential by a time-varying magnetic ﬂux. This is a form of electromagnetic induction, as discussed in Section 8.2. To begin, consider the scenario shown in Figure 8.3. A single loop of wire in the presence of an impressed magnetic ﬁeld B. For reasons explained later, we introduce a small gap and deﬁne VT to be the potential difference measured across the gap according to the sign convention indicated. The resistance R may be any value greater than zero, including inﬁnity; i.e., a literal gap. As long as R is not inﬁnite, we know from Lenz’s Law (Section 8.2) to expect that a time-varying magnetic ﬁeld will cause a current to ﬂow in the wire. Lenz’s Law also tells us the direction in which the current will ﬂow. However, Lenz’s Law does not tell us the magnitude of the current, and it sidesteps some important physics that has profound implications for the analysis and design of electrical devices, including generators and transformers. The more complete picture is given by Faraday’s Law. In terms of the scenario of Figure 8.3, Faraday’s Law relates the potential VT induced by the time variation of B. VT then gives rise to the current identiﬁed in Lenz’s Law. The magnitude of this current is simply VT /R. Without further ado, here’s Faraday’s Law for V T + - R B Figure 8.3: A single loop of wire in the presence of an impressed magnetic ﬁeld. this single loop scenario: VT = −∂ ∂tΦ (single loop) (8.1) Here Φ is the magnetic ﬂux (units of Wb) associated with any open surface S bounded by the loop: Φ = Z S B · ds (8.2) where B is magnetic ﬂux density (units of T or Wb/m2) and ds is the differential surface area vector. To make headway with Faraday’s Law, one must be clear about the meanings of S and ds. If the wire loop in the present scenario lies in a plane, then a good choice for S is the simply the planar area bounded by the loop. However, any surface that is bounded by the loop will work, including non-planar surfaces that extend above and/or below the plane of the loop. All that is required is that every magnetic ﬁeld line that passes through the loop also passes through S. This happens automatically if the curve C deﬁning the edge of the open surface S corresponds to the loop. Subsequently, the magnitude of ds is the differential surface element ds and the direction of ds is the unit vector ˆn perpendicular to each point on S, so ds = ˆnds. This leaves just one issue remaining – the orientation of ˆn. This is sorted out in Figure 8.4. There are two possible ways for a vector to be perpendicular to a surface, and the direction chosen for ˆn will affect the sign of VT . Therefore, ˆn must be somehow related to the polarity chosen for VT . Let’s consider this relationship. Let C begin at the “−” terminal of VT and follow the entire perimeter of the loop, ending at the “+” terminal. Then, ˆn is determined by the following “right-hand rule:” ˆn points in the direction of the curled ﬁngers of the right hand when the thumb of the right hand is aligned in the direction of C. It’s worth noting that this convention is precisely the convention used to relate ˆn and C in Stokes’ Theorem (Section 4.9). Now let us recap how Faraday’s Law is applied to the single loop scenario of Figure 8.3: 1. Assign “+” and “−” terminals to the gap voltage VT .	Electromagnetics_Vol1_Page_193_Chunk1496
8.3. FARADAY’S LAW 179 V + - S n C Figure 8.4: Relationship between the polarity of VT and orientations of C and ˆn in the planar single-loop scenario. 2. The orientation of ˆn is determined by the right hand rule, taking the direction of C to be the perimeter of the loop beginning at “−” and ending at “+” 3. B yields a magnetic ﬂux Φ associated with the loop according to Equation 8.2. S is any open surface that intersects all the magnetic ﬁeld lines that pass through the loop (so you might as well choose S in a way that results in the simplest possible integration). 4. By Faraday’s Law (Equation 8.1), VT is the time derivative of Φ with a change of sign. 5. The current I ﬂowing in the loop is VT /R, with the reference direction (i.e., direction of positive current) being from “+” to “−” through the resistor. Think of the loop as a voltage source, and you’ll get the correct reference direction for I. At this point, let us reiterate that the electromagnetic induction described by Faraday’s Law induces potential (in this case, VT ) and not current. The current in the loop is simply the induced voltage divided by the resistance of the loop. This point is easily lost, especially in light of Lenz’s Law, which seems to imply that current, as opposed to potential, is the thing being induced. Wondering about the signiﬁcance of the minus sign in Equation 8.1? That is speciﬁcally Lenz’s Law: The current I that ends up circulating in the loop generates its own magnetic ﬁeld (“Bind” in Section 8.2), which is distinct from the impressed magnetic ﬁeld B and which tends to oppose change in B. Thus, we see that Faraday’s Law subsumes Lenz’s Law. Frequently one is interested in a structure that consists of multiple identical loops. We have in mind here something like a coil with N ≥1 windings tightly packed together. In this case, Faraday’s Law is VT = −N ∂ ∂tΦ (8.3) Note that the difference is simply that the gap potential VT is greater by N. In other words, each winding of the coil contributes a potential given by Equation 8.1, and these potentials add in series. Faraday’s Law, given in general by Equation 8.3, states that the potential induced in a coil is pro- portional to the time derivative of the magnetic ﬂux through the coil. The induced potential VT is often referred to as “emf,” which is a contraction of the term electromotive force – a misnomer to be sure, since no actual force is implied, only potential. The term “emf” is nevertheless frequently used in the context of Faraday’s Law applications for historical reasons. Previously in this section, we considered the generation of emf by time variation of B. However, Equation 8.1 indicates that what actually happens is that the emf is the result of time variation of the magnetic ﬂux, Φ. Magnetic ﬂux is magnetic ﬂux density integrated over area, so it appears that emf can also be generated simply by varying S, independently of any time variation of B. In other words, emf may be generated even when B is constant, by instead varying the shape or orientation of the coil. So, we have a variety of schemes by which we can generate emf. Here they are: 1. Time-varying B (as we considered previously in this section). For example, B might be due to a permanent magnet that is moved (e.g., translated or rotated) in the vicinity of the coil, as described in Section 8.2. Or, more commonly, B might be due to a different coil that bears a time-varying current. These mechanisms are collectively	Electromagnetics_Vol1_Page_194_Chunk1497
180 CHAPTER 8. TIME-VARYING FIELDS referred to as transformer emf. Transformer emf is the underlying principle of operation of transformers; for more on this see Section 8.5. 2. The perimeter C – and thus the surface S over which Φ is determined – can be time-varying. For example, a wire loop might be rotated or changed in shape in the presence of a constant magnetic ﬁeld. This mechanism is referred to as motional emf and is the underlying principle of operation of generators (Section 8.7). 3. Transformer and motional emf can exist in various combinations. From the perspective of Faraday’s Law, transformer and motional emf are the same in the sense in either case Φ is time-varying, which is all that is required to generate emf. Finally, a comment on the generality of Faraday’s Law. Above we have introduced Faraday’s Law as if it were speciﬁc to loops and coils of wire. However, the truth of the matter is that Faraday’s Law is fundamental physics. If you can deﬁne a closed path – current-bearing or not – then you can compute the potential difference achieved by traversing that path using Faraday’s Law. The value you compute is the potential associated with electromagnetic induction, and exists independently and in addition to the potential difference associated with the static electric ﬁeld (e.g., Section 5.12). In other words: Faraday’s Law indicates the contribution of elec- tromagnetic induction (the generation of emf by a time-varying magnetic ﬂux) to the potential dif- ference achieved by traversing a closed path. In Section 8.8, this insight is used to transform the static form of Kirchoff’s Voltage Law (Section 5.10) – which gives the potential difference associated with electric ﬁeld only – into the Maxwell-Faraday Equation (Section 8.8), which is a general statement about the relationship between the instantaneous value of the electric ﬁeld and the time derivative of the magnetic ﬁeld. Additional Reading: • “Faraday’s law of induction” on Wikipedia. 8.4 Induction in a Motionless Loop [m0056] In this section, we consider the problem depicted in Figure 8.5, which is a single motionless loop of wire in the presence of a spatially-uniform but time-varying magnetic ﬁeld. A small gap is introduced in the loop, allowing us to measure the induced potential VT . Additionally, a resistance R is connected across VT in order to allow a current to ﬂow. This problem was considered in Section 8.3 as an introduction to Faraday’s Law; in this section, we shall actually work the problem and calculate some values. This is intended to serve as an example of the application of Faraday’s Law, a demonstration of transformer emf, and will serve as a ﬁrst step toward an understanding of transformers as devices. In the present problem, the loop is centered in the z = 0 plane. The magnetic ﬂux density is B = ˆbB(t); i.e., time-varying magnitude B(t) and a constant direction ˆb. Because this magnetic ﬁeld is spatially uniform (i.e., the same everywhere), we will ﬁnd that only the area of the loop is important, and not it’s speciﬁc shape. For this reason, it will not be necessary to specify the radius of the loop or even require that it be a circular loop. Our task is to ﬁnd expressions for VT and I. V + - B Figure 8.5: A single loop of wire in the presence of an impressed spatially-uniform but time-varying mag- netic ﬁeld.	Electromagnetics_Vol1_Page_195_Chunk1498
8.4. INDUCTION IN A MOTIONLESS LOOP 181 To begin, remember that Faraday’s Law is a calculation of electric potential and not current. So, the approach is to ﬁrst ﬁnd VT , and then ﬁnd the current I that ﬂows through the gap resistance in response. The sign convention for VT is arbitrary; here, we have selected “+” and “−” terminals as indicated in Figure 8.5.1 Following the standard convention for the reference direction of current through a passive device, I should be directed as shown in Figure 8.5. It is worth repeating that these conventions for the signs of VT and I are merely references; for example, we may well ﬁnd that I is negative, which means that current ﬂows in a clockwise direction in the loop. We now invoke Faraday’s Law: VT = −N ∂ ∂tΦ (8.4) The number of windings N in the loop is 1, and Φ is the magnetic ﬂux through the loop. Thus: VT = −∂ ∂t Z S B · ds (8.5) where S is any open surface that intersects all of the magnetic ﬁeld lines that pass through the loop. The simplest such surface is simply the planar surface deﬁned by the perimeter of the loop. Then ds = ˆnds, where ds is the differential surface element and ˆn is the normal to the plane of the loop. Which of the two possible normals to the loop? This is determined by the right-hand rule of Stokes’ Theorem. From the “−” terminal, we point the thumb of the right hand in the direction that leads to the “+” terminal by traversing the perimeter of the loop. When we do this, the curled ﬁngers of the right hand intersect S in the same direction as ˆn. To maintain the generality of results derived below, we shall not make the substitution ˆn = +ˆz; nevertheless we see this is the case for a loop parallel to the z = 0 plane with the polarity of VT indicated in Figure 8.5. Taking this all into account, we have VT = −∂ ∂t Z S ˆbB(t) · (ˆnds) = − ˆb · ˆn ∂ ∂t Z S B(t) ds (8.6) 1A good exercise for the student is to repeat this problem with the terminal polarity reversed; one should obtain the same answer. Since the magnetic ﬁeld is uniform, B(t) may be extracted from the integral. Furthermore, the shape and the orientation of the loop are time-invariant, so the remaining integral may be extracted from the time derivative operation. This leaves: VT = − ˆb · ˆn ∂ ∂tB(t) Z S ds (8.7) The integral in this expression is simply the area of the loop, which is a constant; let the symbol A represent this area. We obtain VT = − ˆb · ˆnA ∂ ∂tB(t) (8.8) which is the expression we seek. Note that the quantity ˆb · ˆnA is the projected area of the loop. The projected area is equal to A when the the magnetic ﬁeld lines are perpendicular to the loop (i.e., ˆb = ˆn), and decreases to zero as ˆb · ˆn →0. Summarizing: The magnitude of the transformer emf induced by a spatially-uniform magnetic ﬁeld is equal to the projected area times the time rate of change of the magnetic ﬂux density, with a change of sign. (Equation 8.8). A few observations about this result: • As promised earlier, we have found that the shape of the loop is irrelevant; i.e., a square loop having the same area and planar orientation would result in the same VT . This is because the magnetic ﬁeld is spatially uniform, and because it is the magnetic ﬂux (Φ) and not the magnetic ﬁeld or shape of the loop alone that determines the induced potential. • The induced potential is proportional to A; i.e., VT can be increased by increasing the area of the loop. • The peak magnitude of the induced potential is maximized when the plane of the loop is perpendicular to the magnetic ﬁeld lines. • The	Electromagnetics_Vol1_Page_196_Chunk1499
induced potential goes to zero when the plane of the loop is parallel to the magnetic ﬁeld lines. Said another way, there is no induction unless magnetic ﬁeld lines pass through the loop.	Electromagnetics_Vol1_Page_196_Chunk1500

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