1 Basic unit of a magnetic systemNot a bar magnet… .. but a current loop

2 Inside a bar magnet Microscopic current loops  “spin”Ferromagnetic Domains Unpaired spin on Fe Exchange Microscopic current loops  “spin” Origin: Quantum mechanics

3 In pics H1t H2t B1n Fields form loops Js B2n

4 Maths .D = r  x E = 0 .B = 0  x H = J  D.dS=q  E.dl = 0Differential eqns (Gauss’ law) Fields diverge, but don’t curl Defines Scalar potential E = -U .B = 0  x H = J Differential eqns (Gauss’ law) Fields curl, but don’t diverge Defines Vector potential B =  x A Integral eqns  D.dS=q  E.dl = 0  B.dS = 0  H.dl = I Integral eqns D1n-D2n=rs E1t-E2t = 0 Electrostatic bcs B1n-B2n = 0 H1t-H2t = Js Magnetic bcs

5 Ampere’s Law Analogue of Gauss’ law, except…It counts field lines circulating a linear loop instead of lines piercing out of an area. Recall Gauss Law for electrostatics: Total flux of D equals the total enclosed charge Ampere’s Law for magnetostatics: Total circulation of H equals the total enclosed current

6 An example of Ampere’s LawChoose Amperean loop (Solenoid)  B.dl = m0I B=0 along upper edge of loop B perpendicular to side edges of loops So only contribution to circulation is from bottom of loop, equal to Bl If n loops/length, then current enclosed by loop = nlI

7 The world gets a little Loopy !!!Single Loop Two loops (Helmholtz Coils) Many loops (Solenoid) Loops in a circle (Toroid) Should be able to find B’s using Ampere’s Law

8 The world gets a little Loopy !!!Single Loop Two loops (Helmholtz Coils) Many loops (Solenoid) Loops in a circle (Toroid) Or using Magnetic Poisson A = m0∫Jdv’/4pR’

9 Equations with sourcesxB = mJ .E = r/e So we use the equations with sources (or their integral versions due to Gauss or Ampere) to find their strengths for given charge/current densities

10 What about the other source-free equations?We use them implicitly when we guess directions for the diverging or curling lines We do use them explicitly when we define potentials Combined with the source eqns on the previous slide, they give Poisson’s equations .B = 0  x E = 0 xE = 0  E = -V .B = 0  B =  x A

11 For easy geometries Use Gauss’/Ampere’s laws to find fieldsand then integrate to find potentials

12 For complex geometriesUse Integral solution to Poisson’s equations (Coulomb/Biot-Savart’s laws) to find potentials and then take grad/curl to find fields.

13 For interfaces Solve on each side separately, and thenuse boundary conditions to match them at the interface

14 Ampere  Biot-Savart Biot-Savart’s Law Current  magnetic fieldB.2pr = m0I B.= m0I/2pr For a distributed current element B.= m0I∫dlxr/4pr3

15 ‹ I ‹ Biot-Savart example r dlxr ║ dB a z  a2 + z2 f dB = ________m0Iadf 4p( a2+z2) dB = ________ m0Idl 4p( a2+z2) dl dBz = ________ m0Iadf 4p( a2+z2) x _____ a ( a2+z2) I B=m0Ia2/2(a2+z2)3/2 z ‹ Multiply by # rings ndz in range dz and integrate over z from –∞ to ∞ to get result for solenoid B=m0nI

16 ∫ ‹ ‹ I Vector Potential example A = m0∫Jdv’/4pR’ m0Iadff A = ________4p( a2+z2) ‹ ∫ a z  a2 + z2 f dl Keep track of angular dep. of f ‹ f = ycosf - xsinf I 2. A = 0 on z-axis because cosf and sinf integrals zero. 3. But to get curl(A), need to go a bit off-center (r > 0). 4. Messy algebra, but Taylor expand around r to get A, then take curl. If you do it correctly, should get same B as previous page!

17 Self Inductance (Solenoid) Vcap = Q/CVind = d(BA)/dt [Lenz’s law, next chapter]

18 Self Inductance (Solenoid) C = Q/V (capacity to store charge)L = BA/I (capacity to store flux)

19 Self Inductance L “Flux Capacitor” !!!

20 From Ampere’s Law (Solenoid) B= m0nI L= BA/I = m0A/dSimilar to C= e0A/d

21 (Solenoid) L = (NBA)/I “Flux Capacitor”!

22 (Solenoid) L = (m0N2A)/l