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Static Electromagnetics, 2008 Spring Prof. C.W. Baek & H. Kim Chapter 5. Steady Electric Currents •Types of electric currents – Convection currents result from motion of electrons and/or holes in a vacuum or rarefied gas. (electron beams in a CRT, violent motion of charged particles in a thunder storm). Convection current , the result of hydrodynamic motion involving a mass transport, are not governed by Ohm’s law. – Electrolytic current are the result of migration of positive and negative ions. Electrolysis chemical decomposition +ion Electrolyte Usually a diluted salt solution -ion current - Conduction currents result from drift motion of electrons and/or holes in conductors and semiconductors. Atoms of the conducting medium occupy regular positions in a crystalline structure and do not move. Electrons in the inner shells are tightly bound to the nuclei are not free to move away. Electrons in the outermost shells do not completely fill the shells; they are valence or conduction electrons and are only very loosely bound to the nuclei. When an external E field is applied, an organized motion of the conduction electrons will result (e.g. electron current in a metal wire). The average drift velocity of the electrons is very low (10 -4~10-3 m/s) because of collision with the atoms, dissipating part of their kinetic energy as a heat. Chapter 5 : Steady Electric Currents Static Electromagnetics, 2008 Spring • Prof. C.W. Baek & H. Kim Steady current density – Electric currents : motion of free charges – Current density : current per unit area I dQ / dt [A]=[C/s] J dI / ds [A/m2 ] Consider a tube of charge with volume charge density v moving with a mean velocity u along the axis of the tube. Over a period t, the charges move a distance l = u t. The amount of charge that crosses the tube's cross-sectional surface s' in time t is therefore q ' v v v l s ' vus ' t If the charges are flowing through a surface s whose surface normal n̂ is not necessarily parallel to u, q v u st I q v u s J s t J u [C/m3 m/s=A/m 2 ] The total current flowing through an arbitrary surface S is I Chapter 5 : Steady Electric Currents : (volume) current density S J d s [A] Static Electromagnetics, 2008 Spring In vacuum, F qE me a u adt Prof. C.W. Baek & H. Kim Example 5-1. – In conductors and semiconductors, electrons and/or holes can not be accelerated due to the collision. – The drift velocity is proportional to the applied E field. u e E (e : electron mobility) J u [C/m3 m/s=A/m 2 ] J e e E E : point form of Ohm's law σ: conductivity (S/m) For semiconductors : e e h h ( e h ) cf) resistivity : 1 ( m) For metal, Ohmic media : material following Ohm's law V12 E E V12 / I J d s JS S J I /S V12 I RI S The resistance of a material having a straight length , uniform cross section area S, and conductivity : R () S J E Chapter 5 : Steady Electric Currents V12 I S Static Electromagnetics, 2008 Spring Prof. C.W. Baek & H. Kim •Electromotive force (emf) Static (conservative) electric field : For an ohmic material : 1 C E C Ed 0 J d 0 This equation tells us that a steady current cannot be maintained in the same direction in a closed circuit by an electrostatic field (Charge carriers collide with atoms and therefore dissipate energy in the circuit). This energy must come from a nonconservative field source for continuous current flow (e.g. battery, generator, thermocouples, photovoltaic cells, fuel cells, etc.). These energy sources, when connected in an electric circuit, provide a driving force to push a current in a circuit : impressed electric field intensity Ei . – EMF of a battery : the line integral of the impressed field intensity Ei from the negative to the positive electrode inside the battery. V = ò 1 ur r E i ×d l = - 2 ò 1 ur r E ×d l 2 ur r òÑE ×d l = C ò 2 ur r E ×d l + 1 ur r E ×d l = 0 2 Outside the source Inside the source ò 1 1 1 2 2 V E d 2 1 V Ei d 2 Chapter 5 : Steady Electric Currents 1 ur r E ×d l - ò 1 ur r E i ×d l = 0 2 Inside the source 1 V E i d E d V12 V1 V2 ò 2 (inside source, for emf ) Current flows from (-) to (+) inside source! Static Electromagnetics, 2008 Spring Prof. C.W. Baek & H. Kim • Kirchhoff's voltage law When a resistor is connected between terminal 1 and 2 of the battery to complete the circuit : the total electric field intensity (E + Ei) must be used in the point form of Ohm's law. J E Ei 1 J d C C E Ei E E d i C R J E d Ei d C 1 Ei d V 2 If the resistor has a conductivity , length V 1 C J d , and uniform cross section S, J = I / S. 1 I RI S – Kirchhoff's voltage law : Around a closed path in an electric circuit, the algebraic sum of the emf’s (voltage rises) is equal to the algebraic sum of the voltage drops across the resistances. V R I j j Chapter 5 : Steady Electric Currents k k k (V) Static Electromagnetics, 2008 Spring • Prof. C.W. Baek & H. Kim Equation of continuity and Kirchhoff's current law – Principle of conservation of charge : If a net current I flows across the surface out of (into) the region, the charge in the volume must decrease (increase) at a rate that equals the current. dQ d dv J dv dv S V V V dt dt t J (A/m 3 ) : Equation of continuity t I J ds For steady currents, =0 t and Jtherefore 0 S J ds 0 I j 0 (A) j – Kirchhoff's current law : Algebraic sum of all the dc currents flowing out of (into) a junction in an electric circuit is zero. For ac currents, 0 J 0 and therefore t Charge relaxation J ( E ) E t E / 0 t Chapter 5 : Steady Electric Currents S J ds 0 I j 0 (A) j Really? Quasi-static case (at low frequency =0) 0 e ( / ) t 3 (C/m ) For a good conductor (e.g. copper), = 1.5210-19 [s] : relaxation time decay to 1/e (36.8% value) Static Electromagnetics, 2008 Spring Prof. C.W. Baek & H. Kim • Power dissipation and Joule's law – Power dissipated in a conducting medium in the presence of an electrostatic field E Microscopically, electrons in the conducting medium moving under the influence of an electric field collide with atoms or lattice sites Energy is thus transmitted from the electric field to the atoms in thermal vibration. The work W done by an electric field E in moving a charge q a distance is w qE w qE u t 0 t dP pi E N i qi u i dv E J dv i i p lim dP EJ dv (W/m 3 ) : Power density under steady-current conditions For a given volume V, the total electric power converted into heat is P E J dv V (W) : Joule’s law In a conductor of a constant cross section, dv dsd, with d P Ed L S Jds VI Since V = RI, P I 2R Chapter 5 : Steady Electric Currents (W) measured in the direction J. Static Electromagnetics, 2008 Spring Prof. C.W. Baek & H. Kim [HW] Solve Example 5-3. • Boundary conditions for current density – For steady current density J in the absence of nonconservative energy sources Differential form J 0 J 0 Integral form J ds 0 S 1 J d 0 C (1) Normal component : the normal component of a divergenceless vector field is continuous. J1n J 2 n (A/m 2 ) (2) Tangential component : the tangential component of a curl-free vector field is continuous across an interface. J 1t 1 J 2t 2 The ratio of the tangential components of J at two sides of an interface is equal to the ratio of the conductivities. Chapter 5 : Steady Electric Currents Static Electromagnetics, 2008 Spring Prof. C.W. Baek & H. Kim • Resistance calculations We have calculated the resistance of a conducting medium having a straight length , uniform cross section area S, and conductivity . R S () This equation can not be used if the S of the conductor is not uniform How can we calculate the resistance? V E d LE d R L I J d s E d s S S Q S E d s cf) C V E d L RC – Procedures for resistance calculation (1) Choose an appropriate coordinate system for the given geometry. (2) Assume a potential difference V0 between the conductor material. (3) Find E within the conductor (by solving Laplace's equation and taking E V ). (4) Find the total current I from I J d s E d s S (5) Find resistance R by taking the ratio V0 / I. Chapter 5 : Steady Electric Currents S Static Electromagnetics, 2008 Spring Prof. C.W. Baek & H. Kim • Example 5-6 : Resistance of a conducting flat circular washer Sol. (1) Choose a coordinate system : CCS (2) Assume a potential difference V0. (3) Find E . d 2V 0 d 2 V c1 c2 V 0 at 0 V V0 at / 2 2V V E V ˆ ˆ 0 r r Boundary conditions are : V 2V0 , + V0 - (4) Find the total current I. 2 V0 J E ˆ r (5) Find R. I J ds S R Chapter 5 : Steady Electric Currents 2 V0 V0 I 2h ln b a h b a dr 2 hV0 b ln (at = /2 surface) r a