Types of electric currents

1. +ion -ion 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. Electrolyte Usually a diluted salt solution Electrolysis chemical decomposition 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.

2. Steady current density • Electric currents : motion of free charges • Current density : current per unit area • 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 • If the charges are flowing through a surface s whose surface • normal is not necessarily parallel to u, : (volume) current density The total current flowing through an arbitrary surface S is

3. In vacuum, 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. For metal, For semiconductors : cf) resistivity :  Ohmic media : material following Ohm's law : point form of Ohm's law σ:conductivity (S/m) The resistance of a material having a straight length , uniform cross section area S, and conductivity :

4. Electromotive force (emf) Static (conservative) electric field : 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 forcontinuous 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. For an ohmic material : Outside the source Inside the source Inside the source Current flows from (-) to (+) inside source!

5. 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. If the resistor has a conductivity , length , and uniform cross section S, J = I / 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. R

6. 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. • Kirchhoff's current law : Algebraic sum of all the dc currents flowing out of (into) a junction in an electric circuit is zero. : Equation of continuity For steady currents, and therefore For ac currents, and therefore Really? Quasi-static case (at low frequency =0) Charge relaxation  For a good conductor (e.g. copper),  = 1.5210-19 [s] : relaxation time decay to 1/e (36.8% value)

7. 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 Wdone by an electric field E in moving a charge q a distance is For a given volume V, the total electric power converted into heat is In a conductor of a constant cross section, , with measured in the direction J. Since V = RI, Power density under steady-current conditions Joule’s law

8. [HW] Solve Example 5-3. • Boundary conditions for current density • For steady current density J in the absence of nonconservative energy sources (1) Normal component : the normal component of a divergenceless vector field is continuous. (2) Tangential component : the tangential component of a curl-free vector field is continuous across an interface. Integral form Differential form The ratio of the tangential components of J at two sides of an interface is equal to the ratio of the conductivities.

9. Resistance calculations We have calculated the resistance of a conducting medium having a straight length , uniform cross section area S, and conductivity . This equation can not be used if the S of the conductor is not uniform  How can we calculate the resistance? • 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 ). (5) Find resistance R by taking the ratio V0 / I. (4) Find the total current I from

10. 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 . Boundary conditions are : (4) Find the total current I. (5) Find R.  + V0 - (at  = /2 surface)