chapter 10 - time invarient currents in conductors

7/29/2019 Chapter 10 - Time Invarient Currents in Conductors

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Time-Invariant ElectricCurrent in Solid andLiquid Conductors

10.1 IntroductionThe term time-invariantelectriccurrent implies a steady, time-constantmotion of a velarge number of small charged particles. The term current is used because this motiis somewhat similar to the motion of a fluid. A typical example is the steady motionfree electrons inside a metallic conductor, but there are other types of time-invariacurren ts as well. What causes organize d mot ion of large numbers of electrons (other charges)? The answer is an electric field, which unlike in the electrostatic cadoes exist inside current-carrying conductors. Time-invariant currents are frequenalso called direct currents, abbreviated de.A domain in which currents exist is knowas the current field.Inside a metallic conductorwith no electric field present, a free electron (or aother type of free charge) moves chaotically in all directions, like a gas molecuIf there is an electric field inside the conductor, the electrons (negative charges) aaccelerated in the direction opposite to that of the local vector E. This acceleratmotion lasts until the electron collides with an atom .We can imagine that the electr

1


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140 CHAPTER 10

>(i;1)IRGCTloN TDthen stops, transfers the acquired kinetic energy to the atom, is again accelerated ithe oppositlE, and so on . So the electrons acquire an average "drift" vel ocity undethe influence of the field, and the result of th is organized motion is an electric curren

There are three important consequences of th is fact :1. In solid and liquid conductors, where the average path between two collision

is very short, the drift velocity is in the direction of the force, i.e., the chargefollow the lines of vectorE.2. Charges constantly lose the acquired kinetic ene rgy to the atoms they collid

with. This resu lts in a more vigorous vibration of the atoms, i.e., a higher temperature of the conductor. This means that in the case of an electric current iconductors, the energy of the electric field is constantly converted into heat. Thiheat is known as Joule's heat. It is also frequently called Joule's losses becauserepresents a loss of electric energy.

3. In the steady time-invariant state, the motion of electric charges is timeinvariant. The electric field driving th e charges must in turn be time-invarianand is therefore due to a time-constant distribution of charges. Such an electrifield is identical to the electros taticfield of charges distributed in the same manneThis is a conclus ion of extreme importance . All the concepts we derived fothe electrostatic field (scala r potential, vo ltage, etc.) are valid for time-invariancu rrents.Liquid conductors have pa irs of positive and negative ions,which move in op

posite directions under th e influence of the electric field .The electric curren t in liquiconductors is therefore made of two st reams of charged particles moving in opposite directions, but we have the same mech anism and the same effect of energy los(Joule's heat) of current flow as in the case of a solid conductor. There is an additionaeffect, however, known as electrolysis-ehemical changes in any liquid conductor thaalways accompany electric current.

In a class of materials called semiconductors , there are two types of chargcarriers -negatively charged electrons and positively charged holes . In this casethe electric ~ is due to both typ es of charges and depends very much on theiconcen trations. ~ c u R ~ G N i

In gases, elect ric current is also due to moving ions, but th e average path between two collisions is much longer than for solid and liquid conductors . The mechanism of current flow is therefore quite different.

In solid and liquid conductors the number of charges taking part in an electricurren t is extremely large. To understand th is, recall that a solid or l iquid containon the order of 1028 atoms pe r cubic meter. It is not easy to understand these hugnumber s. Perhaps it would help if we consider a volume of about (0.1 mm)3 (a cub0.1 mm on each side), wh ich is barely visible by the naked eye. Thi s tiny volumcontains about 1012 atoms, which is mo re than one hundred times the number ohumans on our planet! It is evident from thi s example that the term "electric currentis indeed approp riate.Questions and problems: Q10.1


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TIME-INVARIANTELECTRICCURRENT IN SOLID AND LIQUID CONDUCTORS

10.2 Current Density and Current Intensity: Point Form of Ohm'sand Joule's lawsElectric current in conductors is described by two quantities. The current density vtor,J, describes the organized motion of charged particles at a point. The currenttensity is a scalar that describes this motion in an integral manner, through a surfaLet a conductor have N free charges per unit volume, each carrying a chargeand having an average (drift) velocity v at a given point . The current density vecat this point is then defined as

J=NQv amperes per m2 (A/m2 ) . (10.1(Definition ofcurrentdensityforonekind ofcharge carriers

Note that this definition implies that the current densi ty vector of equal chargesopposite sign moving in opposite directions is the same. Of course, motion of dferent charges in opposite directions physically is different. However, experimeindicate that practically all effects (joule's heat, chemical effects, magnetic effects)an electric current depend on the product Qv, so it is convenient to adopt this defition for the current density vector.If there are several types of free charges inside a conductor, the current densis defined as a vector sum of the expression in Eq. (10.1). For example, let the currbe due to the motion of free charge carriers of charges QI, Q2, . . . ,Qn, moving wdrift velocities VI, v2, . . . , v n. Let there be NI, N2, . . . ,Nn of these charge carriers punit volume, respectively. The current density vector is then given by

(10(Definition of current densityfor several kindsofcharge carrie

The current intensity, I, through a surface is defined as the total amountcharge that flows through the surface during a: small time interval, divided by ttime interval. In counting this charge, opposite chargesmoving in opposite directioare added together. Thus

(Definition of current intensity through a surfaceI = dQthrough 5 in citdt (Cis = A). (10.3

This can also be expressed in terms of the current density vector, as follows.Consider a surface element dS of the surface S in Fig. 10.1. Let the drift velocof charges at dS be v, their charge Q, and their number per unit volumeN. During t


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142 CHAPTER 10

Figure 10.1 The current intensity I through S isequal to the flux of the current density vector Jthrough Stime interval dt the charges move by a distance v dt in the direction of v. Therefothe charge that crosses dS in dt is

dQthrough dSduring dt = dS v dt cos a NQ, (10where a is the angle between the velocity vector and the normal to the surface elment. The total charge through S during interval dt is obtained as a sum (integral)these elemental charges over the entire surface, and the current intensity is obtainby dividing this sum by dt. Noting that dS v cos aNQ = JdS cos a = J.dS, we obta

I = ~ J . dS (A ). (10.5(Definition of current intensity through a surface in terms of thecurrent densityvector

The unit for current is an ampere (A), equal to a coulomb pe r second (Cis). The unfor current density is A/m2.We now know that electric current in a conductor is produced by an electrfield. We also know that in solid and liquid conductors the vectors Jand E are in tsame direction. For most conductors, vector Jis a linear function of E,

J=aE [a - siemens pe r meter (S/m)]. (10.6[Point (local) formofOhm's law

Conductors for which (10.6) is val id are called linear conductors . The constant aknown as the conductivity of the conductor. The unit for conductivity is siemens pmeter (S/m).The reciprocal value of a is designated by p and is known as the resistivity. Tunit for resistivity is ohm meter (Q .m). Equation (10.6) can be written in the form


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TIME-INVARIANTELECTRIC CURRENT IN SOLIDAND LIQUIDCONDUCTORS 1

E = pJ [p - ohm meter (Q .m)]. (l0.7[Point (local) form ofOhm's Iato

Both Eqs. (10.6) and (10.7) are known as the pointform ofOhm's law for linear condutors because they give a relationship between the two field quantities at every poinside a conductor.For metallic conductors, conductivities range from about 10 MS/m (iron)abou t 60MS /m (silver). The conductivity of seawater is about 4 Sl m, that of grou(soil) is between 10- 2 and 10-4 Slm , and conductivities of good insulators are lthan about 10-12 S/ m .

Wehave already explained from a physical standpoint that there is a permantransformation of electric energy into heat in every current field. Let us now derthe expression, known as Joule's law in point form, for the volume density of powerthis energy transformation.

Let there be N charge carriers Q in the conductor, and let their local drift velity be v. The electric for ce on each charge is QE. The work done by the force whmoving the charge during a t ime interval d t is equal to QE . (v d t).The work donemoving all the N dv charges insi de a small volume dv is therefore

dAel.forces = QE . (v dt)Ndv = J. Edv dt (J). (l OIfwe divide this expression by dv dt,we get the desired power per unit volume (vume power density) - the electric power that is lost to heat:

dPJ P 2PJ = - = J .E = - = oE watts / rn'' (W/m3) .dv a (l0.9(Jou le's law in pointform

(10.

If we wish to determine the power of Joule's losses in a domain of space,just have to in tegrate the expression in Eq. (10.9) over that domain:PJ =1J'Edv watts (W).

(Joule's losses in a domain ofspaExample IO.I-Fuses. Electrical devices are frequently protected from excess ive curreby fuses, one type of which is sketched in Fig. 10.2 . The fuse conductor is made to be mu

thinner than the circuit conductors elsewhere. For example, let the radius of the circui t condtor be /1 times th at of the fuse. If a current of in tensity I exists in the circuit, the volume densof Joule's losses in the thin conductor section is /14 larger than those in the other section. Incase of excessive cu rrent, therefore, the thin conductor section melts long before the normsection is heated up . Whe n the fuse melts, it becomes an open circuit and doe s not allow a


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(10.1

(10.1

(10.1

144 CHAPTER 10

Figure 10 .2 A simple model of a fusefurther cur rent to flow and possibly damage the device protected by the fuse. Usually the thconductor is a meta l with a conductivity smaller than tha t of the thick wire.Questions andproblems: Q10.2 and Q10.3, P10.1 to P10.7

10.3 Current-Continuity Equation and Kirchhoff's Current lawExperiments tell us that electric charge cannot be crea ted or destroyed . This is knowas the lawofconservationofelectriccharge .The continuity equation is the ma thematicexpression of this law. Its general form is valid for time-varying currents, bu t it caeasily be specialized for time-invariant currents.Consider a closed surface 5 in a current field. Let J be the current densityfunction of coordinates and, in the general case considered here, of t ime). The deinition of current intensity applies to any surface, so it app lies to a closed surfacas well. The current intensity, i(t), through 5, with respect to the outward normal,given by Eq. (10.3):

i(t) = d q ( t O U ~ ; f S in dt .According to the law of conservation of electric charge, if some amount of chargleaves a closed surface, the charge of opposite sign inside the surface must increasby the same amount. So we can write Eq. (10.11)as

i(t) = _ d q ( t ) i n ~ ~ Sindt .The current intensity can also be written in the form of Eq. (10.5), anddq(t)inside S in dt = ~ 1 (t)dv

dt dt v P ,where v is the volume enclosed by 5. Recall that by convention we always adopt thoutward unit vector normal to a closed sur face. Thus Eq. (10.12) can be rewritten i


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TIME-INVARIANT ELECTRIC CURRENT IN SOLIDAND LIQUID CONDUCTORS 1

the form1. J. dS = - ~ { p(t )dv. (10.115 dt Jv

(General formof thecurrent continuity equation, where surface S mayvary in timThis is the current continuity equation. Note that we can imagine the surfaceto change in time, in which case the form of the continuity equation in Eq. (10.1

must be used. This, however, is needed only in rare instances. I f the surface 5 donot change in time, the time derivative acts on p only. Because p is a function of botime and space coordinates, the ordinary deriva tive needs to be replaced by a partderivative, and we obtain a much more important form of the current continuequation:

1. J .dS = - [ Bp(t) dv. (10.1515 Jv at (Curren t continuityequationfor a time-invariantsurfaceAlthough the current continuity equation is not a field equation, it is of fu

damental importance in the analysis of electromagnetic fields because only sourc(charges and currents) satisfying this equation can be real sources of the field.Now let the current field be constant in time, in which case the charge densis also constant in time. The partial derivative of p on the right side of Eq. (10.15)

then zero, and both forms of the current continuity equation reduce to

(10.16(Generalized Kirchhoff's current law

This equation tells us that in time-constant current fields the amount of charge thflows into a closed surface is exactly the same as that which flows ou t of it. Eqution (10.16) represents, in fact, the generalized form of the familiar Kirchhoff'S currelaw from circuit theory. Indeed, if a surface S encloses a node of a circuit, there acurrents only through the circuit branches, and Eq. (10.16) becomes

(10.1

We know that Kirchhoff's curren t law in this form is applied also to circuits wtime-vary ing currents. Considering the preceding discussion, it should be clear thin such cases it is only approximate. (Can you explain why?)


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146 CHAPTER10

Figure 10.3 Application of generalizedKirchhoff's current law to a node withfour wires

Example lO.2-Continuity equation applied to a circuit node. Let the surface 5 encloa node with four wires (Fig. 10.3), w ith de currents I I , 12, 13, and 14 , The vector J is nonzero onove r small areas of 5 where the wires go through the surface. There, the flux of J is simply thcurrent intensity in that wire, so that Eq. (10.16) yields - I I - 12+ 13 +14 = 0, which is wha t wwo uld ge t if we simply applied Kirchhoff's curren t law to the node. How are the signs of thcurrents de termined and what do they correspond to in Eq. (10.16)?Questions and problems: QI0.4 and QI0.5

10.4 Resistors: Ohm's and jou le's LawsA resistor is a resistive body wi th two equipotential contacts. A resis to r of genershape is shown in Fig. lOA. Assume that the material of the resis tor is linear. Wknow that the resistivity, p, for linear materials does not depend on the current density. Then the current density, J, is proportional at all poin ts of the resistor to thcurren t in tensity I th rough its terminals. Therefore, the electric field vector in th

Figure 10.4 A resis tor consists of a resistivebod y with two metallic (equipotential)con tacts (the resis tor terminals)


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TIME-INVARIANT ELECTRIC CURRENT IN SOLID AND LIQUID CONDUCTORS

resistor material, E = pJ, and the potential difference between its terminals are aproportional to the current intensity,

[R ohms (Q)], (10.where R is a constant. This equation is known as Ohm's law. Resistors for which tequation holds are called linear resistors. The constant R is called the resistance ofresistor. In some instances it can be computed starting from the defining formulaEq. (10.18),but it can always be measured. The unit for resistance is the ohm (Q).

The reciprocal of resistance is called the conductance, G. Its unit is calledsiemens (S).In the United States, sometimes themho (ohmbackwards) is used instebut this is not a legal SI uni t and we will not use it.

Example lO.3-Resistance of a straight wire segment. As an example of calculatingsistance, consider a straight wire of resistivity p, length 1, and cross-sectional area S. Letcurrent intensi ty in the wire be I. The current density vector is parallel to the wire axis, aits magnitude is J = l i S, The electric field vector is therefore also parallel to the wire aand its magnitude is E = pJ = pIIS. The potential difference between the ends of the wsegment is VI - Vz = EI = pIllS. So the resistance of the wire segment is

IR=p -5 (Q). (10

Consider now a resistor of resistance R. Let the current intensity in the resisbe I, and the voltage between its terminals V. During a time interval t, a charge eqto Q= It flows through the resistor. This charge is transported by electric forces frone end of the resistor to the other end. From the definition of voltage, the work doby electric forces is

AeLforces = QV = VIt 0) (10.Because of energy conservation, an energy equal in magnitude to this work is traformed into heat inside the resistor:

0) (10.Since the process of transformation of electric energy into heat is constant in time,power of this transformation of energy is W/ t, that is,

(W).V2p= VI=RP = -R (10.(Joule's laThis is the familiar Joule's lawfrom circuit theory. It is named after the British physicJames Prescott Joule (1818-1889), who established this law experimentally.Questions and problems: Q10.6 to Q10.9, P10.8 to PlO.14


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148 CHAPTER 1010.5 Electric Generators

We know that actual sources of the electric field are electric charges. We also knowthat we must remove some charges from a body, or put them on a body, in ordeto obtain excess electric charges on it. This obviously cannot be done by the electriforces themselves. Devices that do this must use some nonelectric energy to separatone type of charge from another. Such devices are known as electric generators.

An electric generator is sketched in Fig. 10.5. In a region of the generator therare nonelectric forces, known as impressed forces, that separate charges of differensigns on the two generator terminals, or electrodes, denoted in the figure by " -(negative) and "+" (positive). These forces can be diverse in nature. For example, ichemical batteries these are chemical forces ; in thermocouples these are forces duto different mobilities of charge carriers in the two conductors that make the connection; in large rotating generators these are magnetic forces acting on charges insidconductors.Impressed forces, by definition, act onl y on charges. Therefore they can alwaybe represented as a product of the charge on which they act and a vector quantit

that must have the dimension of the electric field strength:Fimpressed = QEi . (10.23The vector Ei is no t necessarily an electric field strength (although it does hav

the same dimension and unit). It is known as the impressedelectricfieldstrength.The region of space it exists in is known as the impressed electricfield.The impressed electrifield is a concept used often in electromagnetic theory. For example, in the analysis oradar wave scattering from a radar target, the radar wave is considered to be a fielimpressed on the target.

The electromotive force, or emf, of a genera tor is defined as the work done bimpressed forces in taking a unit charge through the generator, from its negative t

Figure 10.5 Sketch of an electric generator


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its positive terminal:e = {j+ Ei . dl} (definition of emf).

- through the gener ator(10

By simple reasoning, the emf of a generator can be expressed in terms ofvoltage between open-circuited generator terminals, as follows. Assume that the gerator is open-circuited (no current is flowing through the generator). Then at allpoints, the electric field strength due to separated charges and the impressed etric field strength must have equal magnitudes and opposite directions, i.e., E = -(otherwise free charges in the conducting material of the generator would move)we substitute this into Eq. (10.24) and exchange the integration limits, we get

e= { r- E . dl} = V+ - V_.J+ any path (10The electric field strength E is due to time-constant charges, i.e., it is an electrostfield. The line integral can therefore be taken along any path, including one outsthe generator (dashed line in Fig. 10.5). Th is means that we can measure the eof a generator by a voltmeter with the voltmeter leads connected in any way toterminals of an open-circuited generator. Wewill see that this is not the case for timvarying voltages.Because the generator is always made of a material with nonzero resistivsome energy is transformed into heat in the generator itself. Therefore, we desc ribgenerator by its internal resistance also. This is simply the resistance of the generain the absence of the impressed field.

Questions and problems: Q10.10 to QI0.12, PI0.IS and P10.1610.6 Boundary Conditions for Time-Invariant Currents

Current fields often exist in media of different conductivities separated by boundsurfaces. We now formulate boundary conditions for this case.Consider the boundary surface sketched in Fig. 10.6. If we apply the continuequation for time-invariant currents to the small, coinlike closed surface, we immeately see that the normal components of the current density vector in the two memust be the same:

or (10

(10r

We know that the electric field in a current field has the same properties aselectrostatic field. Therefore the same boundary condition for the tangential comnent of the vector E on a boundary applies:lIt ht


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150 CrLAPTERIO

Figure 10.6 Boundary surface between twoconducting media

I- ( .....\ .......... ' .X ;.;'..:l ... :J:/al\ r'it I-Figure 10.7 The normal component of vector Jin a conductor adjacent to an insulator is zero .

Example lO.4-Boundary conditions between a conductor and an in sulator. If onethe two media in Fig. 10.6, say, medium 1, is an insulator (e.g., air), wha t are the expressiofor the boundary conditions?We know that there can be no current in the insulator. Hence from Eq. (10.26) we coclude tha t the normal component of the current density vector in a conductor adjacent to thinsu lator must be zero . Tangential componen ts of the vector E are the same in the two medibut of course ht does not exist. Lines of vector J in such a case are sketched in Fig. 10.7.Questions and problems: QI0.13

10.7 Grounding Electrodes and an Image Method for CurrentsConsider an electrode (a conducting body) of arbitrary shape buried in a poorly coducting medium (for example, soil) near the flat boundary sur face between the pooconductor and some insu lator (for example, air) . Suppose that a current of intensiI is flowing from the electrode in to the conducting medium, and is supplied fromdistant current source through a thin insulated wire, as shown in Fig. 10.8.Accord into the boundary conditions, the current flow lines are tangential to the surface.


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TIME-INVARIANT ELECTRIC CURRENT IN SOLIDAND LIQUIDCONDUCTORS

,....--(---t- - - - - - - - - _ .

Figure 10.8 Image method for de currents flowing out of an electrode into a poor conductonear the interface with an insulator

Imagine that the entire space is filled with the poor conductor. If an image eltrode with the current of the same intensity flowing out of it is placed in the upper ha lthe space, as shown in the figure, the resulting current flow in both half spaces wbe tangential to the plane of symmetry (the former boundary surface). Therefore,current distribution in the lower half space is the same as in the actual case. Conquently, the influence of the boundary surface may be replaced by the image etrode, provided the curren t through the image electrode has the same intensitydirection as that th rough the real electrode . This method is very useful for analyzcurrent flow from electrodes buried unde r the surface of the ear th. Such electroare often used for grounding purposes.

Example lO.S-Hemispherical grounding electrode. Suppose that a hemispherical etrode ofvery high conductivity is buried in poorly conducting soil ofconductivity o , as shoin Fig. 10.9.Weare interested in

Figure 10.9 A hemispherical groundingelectrode


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152 CHAPTER 10

The resistance of such a grounding system The intensity of the electric field at all points on the earth 's sur face if a current I flowfrom the electrode into the earth What happens if a person in noninsulating shoes app roaches thi s grounding electrodeWe use the image method. Let the current I flow out of the hemispherical electrode. I fspace is filled with earth, the image is another hemispherical electrode, so we get a sphericelectrode with current 21 in a homogeneous conductingmedium. Due to symmetry, the currefrom the spherical electrodes is radial. Consequently, the current from the original hemisphecal electrode is also radial. The current density at a distance r from the center of the hemispheis therefore

21 IJ= -= - .41Tr2 21T r 2The magn itude of the electric field is

J IE= - = -- ,a 21Tar2so that the potential of the electrode with respect to a reference po in t at infinity is obtained a

100 100 I IVa = Ed r = - -dr = -- .a a 21Tar 2 21T aaIs it po ssible to de fine the resistance of this grounding sys tem? We need two terminafor a resistor, and the hemispherical electrod e has (seemingly) just one. However, the curre

must be collected somewhere at a distant point by a generator and returned to the electrothrough a wire. The distan t point is the other "resistor" contact. Usually this other contact islarge ground ing system of a power plant, with a large contact area w ith the ground , so that tprincipal contribution to the resistance of this resistor comes from the hemispherical electrodWe can therefore define the resistance of the hemispherical grounding electrode as

Va 1R= - = -- .I 21TaaThe potential at any point on the surface of the earth due to the current flow is

10 IVr = E(r )dr = --,r 21Tarand the potential difference between two points at a distance Sr = d apart is given by

I If lV= --- .21Tar 21Ta (r + d)Suppose a person approaches the grounding electrode when a large current of I1000A is flowing through it. The potential difference between his feet can be very large aneven fatal. For example, if a = 10- 2 S /m , r = 1 rn, and the person 's step is d = 0.75m long, t

potential difference between the two feet will be 6820 volts.


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TIME-INVARIANTELECTRIC CURRENT IN SOLIDAND LIQUID CONDUCTORSReal grounding electrodes are usually in the shape of a plate, a rod, or a tmetal mesh and are buried in the ground. The variation in the conductivity ofsoil has a large effect on the behavior of the grounding electrode. For that reasonconductivity around it is sometimes purposely increased by, for example, addingto the soil. The example of a ='spherical electrode is useful because it gives us

idea of the order of magnitude, bu t it is certainly not precise.Questions andproblems: QlO.14 to QI0 .17, PlO.17 to PlO.19

10.8 Chapter Summary1. The basic quantities that describe electric current are the current density vecJ, describing current flow at any point, and the current intensity, I, describcurrent flow through a surface.2. The current density vector J is most f requent ly a linear function of the loelectric field strength, J = oE (point form of Ohm's law), also expressedE = pJ, where a is the conductor conductivity, and p its resistivity.

3. The volume power density of transformation of electric energy into hea t in cductors is described by the point form of Joule's law, PI = J .E.4. The current continuity equation is a mathematical expression of the experimtal law of conservation of electric charges. Its form for time-invariant curreis just the Kirchhoff's current law of circuit theory.5. A resistor is an element, made of a resistive material, with two terminals, eequipotential. Ohm's and Joule's laws for resistors known from circuit theare derived from field theory.6. It is not possible to maintain a current field without devices known as elecgenerators, which convert some other form of energy into electric energy,

energy of separated electric charges. Electric generators are characterizedtheir electromotive force and internal resistance.7. Based on boundary conditions for current fields, an image method can be f

mulated for these fields , similar to that in electrostatics. However, in this cthe boundary conditions are satisfied by a "positive" image. The imagemethcan be used, for example, to determine the field and resistance of groundelectrodes.

QI0.l . What do you think is the main difference between the motion of a fluid and the moof charges constituting an electric current in conductors?QI0.2. Describe in your own words the mechanism of transformation of electric energy

heat in current-carrying conductors.QI0.3. Is Eq. (10.5) valid also for a closed surface, or must the surface be open? Explain.


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154 CHAPTER10QIO.4. A closed surface S is situa ted in the field of time-invariant currents. Wha t is the charge

that passes through S during a time interval d t?QIO.5. Is a current intensity on the order of 1A frequen t in engineering applications? Arecur rent densities on the order of 1mA /m 2 or 1kA / m2 frequent in engineering applications? Explain.QIO.6. What is the difference between linear and nonlinear resistors? Can you think of anexample of a nonlinear resistor?QIO.7. A wire of length I, cross-sectional area S, and resistivity p is made to meander verydensely. The lengths of the successive parts of the meander are on the order of th

wire radius . Is it possible to eva lua te the resistance of such a wire accurately usingEq. (10.19)? Explain.QIO.8. Explain in your own words the statement in Eq. (10.20).QIO.9. Assume tha t you made a resistor in the form of an uninsulated metal container (oneresistor contact) with a conducting liquid (e.g., tap water with a small amount of salt)

and a th in wire dipped into the liquid (the o ther resistor contact). If you change thelevel of water, but keep the length of the wire in the liqu id constant, will thi s producea substantial variation of the resistor resis tance? If you change the length of the wire inthe liquid, will this produce a substantial variation of the resis tor resistance? Explain

QIO.IO. What is meant by "none lectric forces" acting on electric charges inside electric generators?

QIO. l l . List a few types of electric generators, and exp lain the nonelectric (impressed) forceacting in them.QIO.12. Does the impressed electric field strength describe an electricfield?What is the unit othe impressed electric field strength?QIO.13. Prove that on a boundary surface in a time-invarian t curren t field, J lnonn = J 2nonn 'QIO.14. Where do you th ink the charges producing the electric field in the ground in Fig. 10.are located? (These charges cause the current flow in the ground .)QIO.15. Explain in your own words what is meant by the grounding resistance, and what thiresistor is physically.QIO.16. Is it possible to define the grounding resistance if the generator is not groundedExplain.QIO.17. Assume that a large current is flowing through the grounding electrode. Propose aleast three d ifferentways to approach the electrode with a minimum danger of electrishock.

PI O. I. Prove that the current in any homogen eous cylindrical conductor is distributed uniformlyover the conductor cross section.PIO.2. Uniformly distributed charged pa rticles are pla ced in a liquid dielectric. The number oparticles per unit volume is N = 109 m- 3, and each is cha rged with Q= 10-16 C. Calculatthe current density and the current magnitude obta ined when such a liquid moves witha velocity of v = 1.2m / s through a pipe of cross section area S = 1 cm-. Is this currenproduced by an electric field?


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PI0.3. A condu ctive wire has the shape of a hollow cylinder with inner radius a and ou terdi us b.A current I flows through the wire. Plot the current density as a function of radfe).If the conductivity of the wire is (J , what is the resistance of the wire per unit lengPlOA. A conducto r of radius a is connected to one with radius b. If a current I is flowing throuthe conductor, find the ratio of the current densities and of the densities of Joule's losseboth parts of the conductor if the conductivity for both parts is (J .PIO.5. The homogeneous dielectric inside a coaxial cable is no t perfect. Therefore, there is so

current, I = 50/-LA, flowing through the dielectric from the inner toward the outer cductor. Plot the current density inside the cable die lectric, i f the inner conductor radiua = 1mm, the ou ter radius b= 7mm , and the cable length I = 10m.

PI0.6. Find the expression for the current through the rectangular surface 5 in Fig. PlO.6 afunction of the surface wid th x.y

J

Figure PI 0.6 Calculating current intensity

PIO.7. Find the expression for the current intensity through a circular sur face 5 shownFig. PlO.7 for 0 < r < 00 .

sFigure PI 0.7 A coaxial cable


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156 CrLAPTERIOPt o.S. The resistivity of a wire segment of length 1 and cross-sectional area S varies along ilength as p(x) = Po(1 + x] I).Determine the wire segment resistance.PtO.9. Find the resistance between points 2 and 2' of the resistor shown in Fig. P10.9.

: /b b b

b-t12 21_ cl28I 8coppe r P2 copper(resistive) 2bFigure P10.9 An idealized resistor CROSS-fCTIo ...J Figure P1 0.10 Two inhomogeneous conductorsOf IVrOI IJ sa . .

pt a.t o. Show that the electric field is uniform in the case of both inhomogeneous conductorsFig. PlO.lO. Find the resistance per unit length of these conductors, the ratio of currentsthe two layers, and the ratio of Joule 's losses in the two layers.Pt O.H. The dielectric in a coaxial cable with inner radius a and outer radius b has a very larg e, bufinite, resistivity p. Find the conductance per unit length between the cable conductorSpecifically, find the conductance between the conductors of a cable L = 1km long wi

a= 1 ern, b = 3ern, and p = 1011 Q .m.PtOt 2. A lead battery is shown schematically in Fig. P10.12. The total surface area of the leaplates is S = 3.2dm - and the distance between the plates is d = 5mm. Find the approx

mate internal resistance of the battery, if the resistivity of the electrolyte is p = 0.016Q .m+

Figure P1 0.12 A lead battery


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TIME-INVARIANT ELECTRIC CURRENT IN SOLID AND LIQUID CONDUCTORS 1

PIO.13 . Calculate approximately the resistance between cross sections 1 and 2 of the nonunifostrip conductor sketched in Fig. PIO.13. The resistiv ity of the conductor is p, Why can tresistance be calculated only approximately?

c

x=op

d

Figure Pl 0.13 A nonuniform strip conductor

b

2

PlO.14. Calcul ate approximately the resis tance between cross sections 1 and 2 of the conical pof the conductor sketched in Fig. PIO.I4. The resis tivity of the conductor is p.2 2

2

Figure Pl 0.14 A conical conductor

2b

d

Figure P1 0.15 A resistor with animpressed field

PIO.IS. In the da rker shaded region of the very large conducting slab of conductivity a and pmit tivity Eo shown in Fig. PIO.IS, a uniform impressed field E, acts as indicated. The esurfaces of the slab are coated with a conductor of conductivity much grea ter than (J ' , aconnected by a wire of negligible resistance. Determine the current density, the electfield in tensity, and the charge density at all poin ts of the system . Ignore the fringing effePIO.16. Repeat prob lem PIO.IS assuming that the permittivity of the slab is 10 (different from ENote that in tha t case polarization charges are also present.


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158 CHAPTER 10PIO.17. Determine the resistance of a hemispherical grounding electrode of radius a if the grounis not homogeneous, bu t has a conductivity (J l for a < r < b, and (J 2 for r > b, where b >

and r is the distance from the gro undi ng electrode cen ter.PIO.IS. Determine the resistance between two hemispherical grounding electrodes of radii R1 anR2, which are a distance d (d R) , R2) apart. The ground conductivity is (J .PIO.19. A grounding sphere of radius a is buried at a depth d (d a) below the surface of th

ground of conductivity (J (Fig. PI0.19). Determine the points at the surface of the grounat which the electric field intensity is the largest. Determine the electric field intensitythese points i f the intensity of the current through the grounding sphere is 1.

image

d

d

electrode--1 r-

2a

E'0"

Figure P1 0.19 A deeply buried sphericalelectrode

chapter 10 - time invarient currents in conductors

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