1 / 15

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)

floragreen
Download Presentation

Energy Density

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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)

More Related