FdM 1 Electromagnetism University of Twente Department Applied Physics First-year course on Part II: Magnetism: slides © F.F.M. de Mul

FdM 2 Magnetic Force on Current Wire eTeT dS dl=dl.e T | j | = dI/dS j = n q v (n = part./m 3 ) Lorentz force on one charge: F = q v x B ….. per unit of volume : f = nq v x B = j x B ….. on volume dV : dF = j x B. dV dF = j x B. dS.dl j = j.e T dF = j.dS. e T x B. dl dF = I. dl x B Suppose: total current = I ; cross section S variable B F = I.e T x B  dl = F = I. L. e T x B Straight conductor in homogeneous field: B

FdM 3 Hall effect G H P Q B I I d l a Suppose n charge carriers / m 3 F v B-field causes deviation of path of charge carriers Stationary case: F magn = F elec  q v B = q E Hall Build up of electric field E Hall between Q and P: Q+ ; P- With j = nqv = I/(ad) 

FdM 4 I Magnetic field of a line current P R Question: Determine B in P O dl=dz  erer Approach: Current line elements dl dl=dl.e T z r Integration over z from -  to +  Integration over  from 0 to  Calculation: e T x e r = e  ; tangential component only: dB dB ee

FdM 5 Magnetic field of a circular circuit Question: Determine B in P P R y I l a Approach: Current line elements dl r erer dl dl = dl. e T R r a dB P   dB y Calculation: e T x e r = 1 ; symmetry: y- component only: dB dB y

FdM 6 Magnetic field of a circular solenoid Radius: a ; Current I Length: L Coils: N, or per meter: n Approach: Solenoid = set of circular circuits ; and for each circuit: R is distance from circuit to P Each circuit: strip dy; current dI = n.dy.I Question: Determine B in P y P Result for L   : B =  0 n I e y Result independent of a, L P O y yPyP L a dy y R=y p -y R

FdM I1I1 I2I2 r 12 Magnetic forces between currents Question: determine F 1  2 e T1 e T2 e r1 e  1 Relations: F12F12 B12B12 Calculation

FdM 8 Why is the wire moved by Lorentz force ? (since inside the wire only the conduction electrons move) I I electron Conductor: - fixed ion lattice, - conduction electrons Hall effect: concentration of electrons (- charge) at one side of conductor Lattice ions feel a force F E upwards + FEFE B v FLFL -q This force (= F L ) is electrical !! B Magnetic field B  plane of drawing

FdM 9 Ampère’s Law (1) I Long thin straight wire; current I B ee r Question: Determine the “Circulation of B-field” along circle l l

FdM 10 Ampère’s Law (2) I B ee r l Question: Determine the “Circulation of B-field” along circuit c c Consequences: 1. More currents through c add up ; 2. Currents outside c do not contribute ; 3. Position of current inside c is not important. r B dl =r. d  dd

FdM 11 I Cylinder: radius a B-field from a thick wire Options for current: I: at surface II: in volume (suppose: homogeneous) Use Ampere-circuits (radius r):

FdM 12 Magnetic Induction of a Solenoid Radius: R ; Current: I Length: L >> R Coils: n per meter I Result: inside: B =  0 n I e z outside: B = 0 ezez erer ee Components: B z B r B  G B r : Gauss-box G:  total =0  top =  bottom   wall =0  B r =0 BrBr c c B  : Circuit c (radius r): Ampere: B .2  r=0  B  =0 BB 1 2 B z : Circuit 2: B(a)=B(b)=0 Circuit 1: Ampere: B z l=  0 n I l a b l BzBz

FdM 13 Symmetries for Ampere’s Law Wire,  long Plane,  extending Solenoid,  long Toroid, along core line

FdM 14 Plane layer L with current density j j L Magnetic Pressure Magnetic pressure: P = ½  0 j 2 = ½B e 2 /  0. Example: this situation is met at the wall of a (long) solenoid. Then the pressure is outward, thus maximizing the cross section area. Suppose we add an external field B e, with B e = B L, so that the total field behind the layer = 0 BeBe j Lorentz force on : (F L =I.L.e T  B): dF L = (j.db).dl  ½  0 j e z dF L = ½  0 j 2.db.dl e y dl db BeBe dF L B-field of the layer (circuit l ): 2 B L l =  0 j l  B L =  ½  0 j e z l z BLBL BLBL x y

FdM 15 Vectorpotential A r = f (x,y,z, x’,y’,z’ ) } erer P (x,y,z) r dv’dv’ Flow tube v’ x’,y’,z’ B = rot A Question: determine A in P from Biot-Savart for B j

FdM 16 Magnetic Dipole (1): Far Field rPrP r r’ O e rP dl P I  Far field: r’ << r P Monopole-term =0 Dipole-term Goal: expression in r P in stead of all r-values over circuit

FdM 17 Magnetic Dipole (2): Dipole Moment Assume: Y-axis along OP’ Assume: circular circuit, with radius R << r P P P’ y x z   r’ e rP rPrP O  Define: dipole moment: m = I. Area. e n = I  R 2.e n enen

FdM 18 Magnetization and Polarization L S enen Magnet = set of “elementary circuits” ; n per m 3 Total surface current = I tot Total magnetic moment = I tot S e n PolarizationMagnetization Dipole moment: p [Cm]Dipole moment: m [Am 2 ] Polarization P = np [Cm -2 ] = surface charge / m 2 Magnetization M = nm [Am -1 ] = surface current / m Def.: Magnetization M = magnetic moment / volume = surface current / length V=SL

FdM 19 Magnetic circuit: Torque and Energy Torque: moment:  = F. b sin  = I B l b sin  = I B S sin  Magnetic dipole moment: m = I S e n Torque: moment:  = m x B Potential energy: min for  = 0 ; max for  =   Potential energy: E pot = -MB. cos  = - m. B B m F F F F m B b l b.sin   I

FdM 20 Electret and Magnet E D H B Electret P E = (D-P)/  0 D =  0 E + P Magnet M H = B /  0 -M B =  0 H + M

FdM 21 B- and H-fields at interface Given: B 1 ;  1 ;  2 Question: Calculate B 2 B1B1 B2B2 11 22 1 2 Gauss box : B 1.Acos  1 - B 2.Acos  2 =0  B 1  = B 2  Circuit: : no I : H 1.Lsin  1 - H 2.Lsin  2 =0  H 1// = H 2// Needed: “Interface-crossing relations”: Relation H and B: B =  0  r H

FdM 22 Suppose: - toroid solenoid: R, L, N,  r ; - air gap,  r =1, width d <<R; g = gap ; m = metal I Core line; radius R; L = 2  R Toroid with air gap Question: Determine H g in gap Relations: H g d + H m (L-d) = N I B g =B m B g =  o H g ; B m =  0  r H m Result: d

FdM 23 Induction: conductor moves in field B v L1(t)L1(t) L 2 (t+dt ) Suppose: circuit L moves with velocity v through field B Question: Show equivalence: v.dt dl  Area = v dt dl sin  E n = non-electrostatic field Gauss box: top lid S 1 in L 1, bottom lid S 2 in L 2 S1S1 S2S2

FdM 24 Induction: Faraday’s Law V0V0 B c S I0I0 Static: V ind Dynamic: B c S I ind Suppose  changes  V ind  0  Non-electrostatic field E N Consequence: Let circuit c shrink, with keeping S. S c c S For closed surface :

FdM 25 Induction in rotating circuit frame Induction potential difference : (II) Using flux change: B ext enen b l  v v v enen   =  t ; v=  ½b (I) Using Lorentz force: =EN=EN

FdM 26 Electromagnetic brakes Why is a conducting wire decelerated by a magnetic field? Case I: switch open P Q v B Electrons feel F L FLFL P Q v B Case II: switch closed Potential difference V PQ : P -, Q + (= Hall effect) FLFL vLvL F L moves electrons: v L, which will act on positive metal ions: F ions. F ions FEFE F L counteracted by F E I F L2 which causes I and F L2, which causes electric field,

FdM 27 Coupled circuits (1): M and L 1 2 I1I1 Suppose: circuit 1 with current I 1 Part of flux from 1 will pass through 2 :  21  21 ~ I 1 Definition:  21 = M 21.I 1 M : coefficient of mutual induction : “Mutual Inductance” : [ H ] = [ NA -1 m -1.m 2.A -1 ] = [ NmA -2 ] Suppose: Circuit 2 has current as well: I 2 I2I2 Flux through 2:  2 =  21 +  22 = M 21 I 1 + M 22 I 2 L : coefficient of self-induction “(Self) Inductance”: M 22 = L 2 M and L are geometrical functions (shape, orientation, distance etc.)

FdM 28 Coupled circuits (2): toroid Question: determine M 21 Flux from 1 through S :  S = BS =  0  r N 1 I 1 S / L Cross section S N 1, I 1 N2N2 Core line: L rr Linked flux from 1 through 2:  21 = N 2  S =  0  r N 1 N 2 I 1 S / L Coefficient of mutual induction: M 21 =  0  r N 1 N 2 S / L This expression is symmetrical in 1 and 2: M 21 = M 12 This result is generally valid: M ij = M j i

FdM 29 Radii: a and b (a<b) Length: l Current: I ; choice: inside = upward I Coax cable : Self inductance Ampere: B-field tangential Flux through circuit: Self-inductance (coefficient of self-induction): per unit of length: (Compare with capacity of coax cable, per meter: Gauss surface B

FdM 30 Magnetic Field Energy x z y I Circuit c in XY-plane H H-field lines: circles around circuit. A A-surfaces: everywhere  H (or B) -field lines  through A: Magnetic energy: E m = ½ L I 2 = dA dl Compare: electric energy: E e =

FdM 31 Maxwell’s Fix of Ampere’s Law Changing B-field causes E-field Question: Does a changing E-field cause a B-field ? j Suppose: chargeing a capacitor using j L S1S1 L encloses S 1 : S2S2 L encloses S 2 : L encloses S 3 : S3S3 Volume enclosed by S 2 and S 3 : D