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Energy Density

Energy Density. Another way to think of ½ Only 1 member of each pair works against the other W E = ½ S i S j q i q j /4 pe 0 R ij So w E = ½ r V. Work done to bring the charge in: qV But V itself is changing as you assemble unit, ie, V=V(q)

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Energy Density

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  1. Energy Density 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 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 W: total energy w: energy density

  2. Energy Density V wE WE = ½Q(V)V = ½CV2 C = eA/d, V = Ed  WE = ½ eE2Ad Can show this wE is also ½ rV as expected

  3. Energy Density V wM B I WM = ½IFB(I) = ½LI2 L = mN2Sl, I = B/mN  WM = ½B2/mSl Can show this wM is also ½ J.A as expected

  4. Energy Density So electric energy is WE = ½QV, which gives energy densitywE =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

  5. Energy Density OR Integrate where field Esits W = ∫½ eE2dv’ Integrate where charge r sits W = ∫½ rV dv’ Is energy stored in charges or fields? Both viewpoints correct – just matter of bookkeeping Where energy ‘sits’ is debatable, but total energy same

  6. Energy Density B Area S WM = ½B2/mSl 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

  7. Energy Density J = (I/S0) f S0 B WM = ½B2/mSl I Total wire volume V = Nl . 2pR . S0 #turns Length/turn Cross sectional area/turn

  8. Energy Density B = mNIz A = mNIr/2 f (Check B =  x A) Set r = R because only those terms multiply with J B J = (I/S0) f Total wire volume V = Nl . 2pR . S0 I WM = ½B2/mSl WM = ½ J.AV = ½(I/S0).(mNIR/2).(Nl.2pR.S0) rewrite I in terms of B = (B2/2m) x (pR2l)

  9. Revelation! Although different volumes to integrate over, result same in the end!

  10. EM Energy Density H = E/Z0 Can show then that wE = wM H is characterized by m E is characterized by e Z0= m/e

  11. EM Energy Density H = E/Z0 Can show then that wE = wM 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 !

  12. Power flow Charge current J = vr Energy current S = v(wE + wM) = b.(1/√me)(eE2) = bE2/Z0 = bEH = E x H Along propagation direction

  13. Power flow S = E x H : Poynting Vector (1885) Sav = ½ Re(E x H*) Pin = ∫S.dA = EH<cos(wt)cos(wt+f)> = EHcos(f)/2 dA

  14. Energy Conservation -.S = J.E + w/t Incoming Power density Power Dissipation Increasing EM Energy Density w = eE2/2 + B2/2m -.J = r/t From Maxwell’s equations can prove (Try it!) Shows energy conservation just like eqn of continuity gives charge conservation

  15. Loss: finite conductivity(r=0,J=sE) V=El I=JA V R=l/sA Power dissipation/Vol J.E = J2/s = I2R/lA J=sE Joule’s Law I=(sA/l)V (Ohm’s Law)

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