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Electromagnetism Chapter 9: Steady Current Pablo Laguna Georgia Institute of Technology, USA Fall 2012 Notes based on textbook: Modern Electrodynamics by A. Zangwill Laguna Electromagnetism Let’s consider the class of problems where the source of E(r) is electric charge in steady motion; that is, current density j = j(r) which yield a current Z I = dS · j S From the continuity equation we have that ∂ρ +∇·j=∇·j=0 ∂t And the Maxwell equations still read ∇·E= ρ 0 and Laguna ∇×E=0 Electromagnetism The condition ∇ · j = 0 implies that the lines of current density are like the electric field lines. In some instance, we can define the convection current density as j(r) = ρ(r)υ(r) Current density in many systems obeys Ohm’s law, j = σE. Like P = 0 χE, Ohm’s law is a constitutive relation which describes the material-dependent response of a many-particle system to a specified stimulus. Here, it is the conductivity σ which carries the material-dependent information. Laguna Electromagnetism If dS k j (Filamentary Wire Limit) Z Z Z Z Z Z j d 3r = j (dS · d`) = (j · dS) d` = I d` Thus, for an arbitrary vector C(r) Z Z d 3 r j × C → I d` × C. Laguna Electromagnetism Since ∇ · j(r) = 0 j(r) = σ(r)E(r) then σ(r) ∇ · E(r) + E(r) · ∇σ(r) = 0 If σ = constant, and given that ρ(r) = 0 ∇ · E(r), we have that σ∇·E ρ σ 0 ρ = 0 = 0 = 0 Thus ∇2 ϕ = 0 Laguna Electromagnetism Matching Conditions Ohmic matter From ∇·j = 0 j = σE ∇×B = 0 we can derive n̂2 · [j1 − j 2 ] = 0 ∂ϕ1 ∂ϕ2 = σ2 σ1 ∂n S ∂n S or and n̂2 × [E1 − E2 ] = 0 or ϕ1 |S = ϕ2 |S Notice the similarity with the matching conditions at the interface between two dielectrics. If no current flows through a boundary surface separating an Ohmic medium from a non-conducting medium, ∂ϕ = 0. ∂n S Laguna Electromagnetism Electrical Resistance "! # 0 "n S ! #!A ! # !B "! # 0 "n S R= ϕA − ϕB . I where j is found from Z dS · j I= S and ϕ from ∇2 ϕ = 0 Laguna Electromagnetism An alternative to computing R= is to fix I from ϕA − ϕB I Z Z dS · j = σ dS ∂ϕ = ±I ∂n S S and we calculate the potential difference from the line integral of E from one electrode to the other. Therefore, RB A R= R RB d` · E 1 A R = × σ dS · j dS · E d` · E S S Consider calculating the resistance of two conducting electrodes are embedded in an infinite medium with conductivity σ RC = Laguna 0 σ Electromagnetism Joule Heating The inelastic collisions remove kinetic energy from the current-carrying charges. That energy is dissipated in the form of Joule heat. In an Ohmic medium, the moving charges have a constant drift velocity with no change in potential energy. From the 1st law of thermodynamics, the rate of Joule heating is equal to the rate the electric field does work on the current carrying charges: Z N d X dW = qk E(rk ) · rk (t) = d 3 r j · E dt dt k=1 But Z V Thus d 3r j · E = − Z V d 3 r ∇ · (jϕ) = − V Z dS · j ϕ = (ϕA − ϕB )I = I 2 R. S dW = I2R dt Laguna Electromagnetism Electromotive Force A steady current requires the presence of a source of energy to compensate for Joule heating. The source of energy must be non-electrostatic (chemical, thermal, gravitational, nuclear, or even electromagnetic) to maintain the current. The force associated with this source of energy is called electromotive force (EMF). One represents the effect of any source of EMF by a fictitious electric field E0 such that j(r) = σ[E(r) + E0 (r)]. Laguna Electromagnetism The voltage difference between two points 1 and 2 along a circuit as Z2 ϕ1 − ϕ2 = d` · E. 1 The EMF between the points 1 and 2 is Z2 E12 = d` · E0 . 1 Thus 1 σ Z2 Z2 d` · j = 1 d` · [E + E0 ] = ϕ1 − ϕ2 + E12 1 and I R12 = ϕ1 − ϕ2 + E12 H For a closed circuit, d` · E = 0, and then E = IR where R is the resistance of the entire circuit and the EMF of the entire circuit is I I 1 E = d` · E0 = d` · F. q Laguna Electromagnetism Kirchhoff’s Laws They are a consequence of ∇ · j = 0 and ∇ × E = 0 for filamentary wire circuits. From ∇ · j = 0 currents must flow into and out of each node in a circuit with no accumulation of charge. Thus X Ik = 0 (Kirchhoff’s current law). k with Ik the current on each segment. H For each closed loo, ∇ × E = 0 implies that d` · E = 0. Therefore, if In is the current which flows through resistor Rn , then X X Ek = I n Rn (Kirchhoff’s voltage law). k n Laguna Electromagnetism