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STEADY CURRENTS Stationary charges produce electric fields that are constant in time; hence the term electrostatics is used. Steady currents produce magnetic fields that are constant in time; hence the term magnetostatics is used. By steady current we mean a flow of charge which has been going on forever, never increasing, never decreasing. Current Distribution If the charges are in motion, they constitute current distributions. Current Density Consider a tube of charge with density v . The charges are moving with a mean velocity u through a surface S whose surface normal is n̂ . Over a period t , the charges move a distance l u t The amount of charge that crosses the tube’s cross‐ section S in t is: q v v v u t.nˆS The corresponding current: I q v u .nˆ S t Let, 2 J v u ( A / m ) , the current density. The total current through a surface S is: I J . nˆ ds S ( A) If the surface S is a closed surface: I J . nˆ ds S ( A) For steady currents: J . nˆ ds 0 S This means that the net current through a closed surface is zero. The amount of current entering into S is exactly equal to the amount leaving it. So, steady currents establish a solenoidal vector field. i.e. . J 0 There are two types of electric currents caused by the motion of the free charges: 1) Convection Currents These currents are due to the motion of the positively and negatively charged particles in a vacuum or gas. 2)Conduction Currents Conduction currents in conductors or semiconductors are caused by drift motion of conduction electrons or holes. They obey the Ohm’s Law. Here, we will concentrate on conduction currents. Conduction Currents The drift velocity ue of electrons in a conducting material is related to the externally applied electric field E , ue e E Where e is the electron mobility (material property) in (m2 / V .s) . In a semiconductor, the current flow is due to the motion of both electrons and holes (+ve charges). uh h E h is the hole mobility Since, J vu v ue v uh e h J v e v h E e h Let v e v h the conductivity of the medium in Simens/meter (S/m) or mohos /meter ( / m ). In either case, e h J E ( A / m ) This is the Field Theory Expression of the Ohms Law. Note that, in a perfect dielectric with 0 , 2 J 0 and in a perfect conductor , E J 0 . is of order of 10 or 10 (S/m) for a good conductor. For semiconductors, i.e. for pure Germanium 2.2 (S/m). For insulators, the conductivity varies from 10 to 10 (S/m). 6 7 10 17 Oh hm’s Laaw for aa Lineaar, Hom mogeneous Conduc C ctor of Length h When a dc voltagee V is aapplied d betw ween the endss hich has cond ductivity .. In thiss of a cconducctor wh case the ele ectric field is not zerro becaause th he ductor iis not issolated d but iss wired d to a source o of cond electtromotive sou urce wh hich co ompiless the ch harges to move e and p preventts the eequilib brium to o be estab blished d. Thus electriic field exists to susttain thee curre ent flow w. As electron ns movve, theyy encounter some e forces called d resisttance. Based d on thee Ohm’’s Law, we will derivve the rresistan nce of the conductin ng medium of uniform cross‐section of area S and length . The applied electric field is uniform and its magnitude is: E V Since the conductor has uniform cross‐section, the magnitude of the current density is: J I I J .ds ) S (since s Both, the electric field intensity and the current density are in the direction of the current flow in the conductor. J E I V Substituting, S Define the resistance of the conductor, R V ( ) I S POWER DISSIPATION AND JOULE’S LAW As the free electrons drift inside a conductor there will be collisions between them and the atoms. The kinetic energy of the electron is converted into the thermal energy of the atom in a collision. Following this, the temperature of the conductor raises as the current flows. This heat increase is called the Joule’s heat. It is clear that to maintain an electric current inside a conductor, energy is to be transferred to the electrons continuously. This is dissipated as heat in the conductor. This dissipated power in a volume V of the conductor can be calculated as: P J .E dv v Watts. For a linear conductor: P E .E dv E 2 dv v v