Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Energy Density Work done to bring the charge in: qV But V itself is changing as you assemble unit, ie, V=V(q) Starts off at zero, ends at V, average V/2 So net energy stored per unit volume ½ rV Another way to think of ½ Only 1 member of each pair works against the other WE = ½ SiSj qiqj/4pe0Rij So wE = ½ rV W: total energy w: energy density 1 Energy Density WE = ½Q(V)V = ½CV2 C = eA/d, V = Ed WE = ½ eE2 Ad wE V Can show this wE is also ½ rV as expected 2 Energy Density B I WM = ½IFB(I) = ½LI2 L = mN2Sl, I = B/mN WM = ½B2/m Sl wM V Can show this wM is also ½ J.A as expected 3 Energy Density So electric energy is WE = ½QV, which gives energy density wE = ½rV Capital Letter for Total energy Small Letter for Energy density Equivalently, we have WE = ½CV2, which gives energy density wE = ½E.D (check for parallel plate capacitor, C = e.A/d and V = Ed) Magnetic energy is WB = ½IF, which gives energy density wB = ½J.A (recall I/S = J, S:area. Also, flux F = ∫B.dS = ∫(x A).dS = ∫A.dl) Equivalently, we have WB = ½LI2, which gives energy density wB = ½B.H Energy Density Is energy stored in charges or fields? Both viewpoints correct – just matter of bookkeeping Where energy ‘sits’ is debatable, but total energy same Integrate where charge r sits W = ∫½ rV dv’ OR Integrate where field E sits W = ∫½ eE2 dv’ 5 Energy Density Area S B WM = ½B2/m Sl I Here we have the field point of view ie, energy density B2/2m sitting in field lying inside solenoid occupying a volume Sl Let’s now use the charge/current point of view Energy density ½ J.A stored in volume of wires around solenoid 6 Energy Density B WM = ½B2/m Sl I S0 J = (I/S0) f Total wire volume V = Nl . 2pR . S0 #turns Length/turn Cross sectional area/turn 7 Energy Density J = (I/S0) f Total wire volume V = Nl . 2pR . S0 B I WM = ½B2/m Sl B = mNIz A = mNIr/2 f (Check B = x A) Set r = R because only those terms multiply with J WM = ½ J.AV = ½(I/S0).(mNIR/2).(Nl.2pR.S0) rewrite I in terms of B 8 = (B2/2m) x (pR2l) Revelation! Although different volumes to integrate over, result same in the end! 9 EM Energy Density H = E/Z0 Can show then that wE = w M H is characterized by m E is characterized by e Z0= m/e 10 EM Energy Density H = E/Z0 Can show then that wE = w M This makes sense! Comparison of E and H strengths is meaningless since they have different units and dimensions. But when comparing energy densities both fields contribute equally to an EM signal, as expected ! 11 Power flow Along propagation direction Charge current J = vr Energy current S = v(wE + wM) = b.(1/√me)(eE2) = bE2/Z0 = bEH = E x H 12 Power flow = EH<cos(wt)cos(wt+f)> = EHcos(f)/2 S = E x H : Poynting Vector (1885) Sav = ½ Re(E x H*) Pin = ∫S.dA 13 Energy Conservation From Maxwell’s equations can prove (Try it!) -.S = J.E + Incoming Power density w/t Power Dissipation Increasing EM Energy Density w = eE2/2 + B2/2m Shows energy conservation just like eqn of continuity gives charge conservation -.J = r/t 14 Loss: finite conductivity (r=0,J=sE) Power dissipation/Vol J=sE I=JA V=El I=(sA/l)V (Ohm’s Law) J.E = J2/s = I2R/lA V Joule’s Law R=l/sA 15