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University of Twente Department Applied Physics First-year course on Electromagnetism Part II: Magnetism: slides © F.F.M. de Mul 1 FdM Magnetic Force on Current Wire dS Suppose: total current = I ; cross section S variable eT dl=dl.eT B | j | = dI/dS j = n q v (n = part./m3) 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.eT dF = j.dS. eT x B . dl dF = I . dl x B Straight conductor in homogeneous field: B FdM F = I.eT x B dl = F = I. L. eT x B 2 Hall effect l I G P Suppose n charge carriers / m3 I a d H B v Q F B-field causes deviation of path of charge carriers Build up of electric field EHall between Q and P: Q+ ; P- Stationary case: Fmagn = Felec q v B = q EHall IB With j = nqv = I/(ad) EHall vB nqad IB Hall potential : VHall VQ VP EHall . d nqa FdM 3 Magnetic field of a line current 0 I eT er Biot & Savart : dB dl 2 4 r dl=dl.eT er dl=dz Question: Determine B in P z O r e R P I (1) var z : dB dB 0 I 1 4 R 2 z 2 Integration over z from - to + Approach: Current line elements dl Calculation: eT x er = e ; tangential component only: dB 0 I sin dB dz 2 4 r R dz Result : 2 2 R z 0 I 1 Rd (2) var : dB sin 2 2 2 4 R sin sin FdM Integration over from 0 to 0 I BP eφ 2R 4 Magnetic field of a circular circuit 0 I eT er Biot & Savart : dB dl 2 4 r I l er dB r dl P dl = dl. eT FdM dBy dB r R Approach: Current line elements dl R a a Question: Determine B in P P dBy y Calculation: eT x er = 1 ; symmetry: y- component only: I 1 dBy 0 2 dl cos 4 l r dBy 0 I 1 a 2 a 4 a 2 R 2 a2 R2 0 I . a 2 BP 2 2 3 / 2 e y 2a R 5 Magnetic field of a circular solenoid Radius: a ; Current I Length: L Coils: N , or per meter: n y P Question: Determine B in P Approach: Solenoid = set of circular circuits ; and for each circuit: 0 I . a 2 BP 2 e 2 3/ 2 y 2a R R is distance from circuit to P L Each circuit: strip dy; current dI = n.dy.I dy L O R y yP FdM 0 R=yp-y a P B 0 .nI .dy.R 2 2 R2 a 2 3/ 2 ey Result for L : y B = 0 n I ey Result independent of6 a, L Magnetic forces between currents I1 Question: determine F12 I2 eT1 er1 r12 eT2 F12 B12 e 1 1 eT2 e 1 er1 FdM dFL I dl B ; dl dl.eT Calculation B1 2 0 I1 e 1 2r12 F12 I 2 .eT2 B12 .dl2 2 If L1 L2 Relations: 0 I B eθ ; eθ eT er 2R : F12 F21 F1 2 0 I1 I 2 L2 (e r1 ) 2 r12 7 Why is the wire moved by Lorentz force ? (since inside the wire only the conduction electrons move) Conductor: - fixed ion lattice, - conduction electrons I + + + + + + + + + + + + + + + + + + + + + I electron Magnetic field B plane of drawing FL B v B -q Hall effect: concentration of electrons (- charge) at one side of conductor Lattice ions feel a force FE upwards FdM This force (= FL ) is electrical !! -------FE + 8 Ampère’s Law (1) Long thin straight wire; current I I l Question: Determine the “Circulation of B-field” along circle l B r e B dl 0 I 0 I l B dl l 2r e e dl 2r 2r B dl I : Ampere 0 l 9 FdM Ampère’s Law (2) B dl I : Ampere I 0 l l B r dl d r B and again : B dl I 0 c =r. d FdM B dl 0 I c B dl c B.r.d 2r r.2 e c Question: Determine the “Circulation of B-field” along circuit c Consequences: 1. More currents through c add up ; 2. Currents outside c do not contribute ; 3. Position of current inside c 10 is not important. B-field from a thick wire Cylinder: radius a current : I j dS S I B dl 0 I 0 j dS l S Options for current: I: at surface II: in volume (suppose: homogeneous) Use Ampere-circuits (radius r): 0 I 2r r a B(r ).2r 0 I B(r ) r a ( I ) : B(r ).2r 0 B(r ) 0 0 I .r B (r ) 2a 2 r 2 ( II ) : B(r ).2r 0 I 2 a 11 FdM Magnetic Induction of a Solenoid ez I er G B r Br : Gauss-box G: total=0 B c c top =bottom wall=0 Br=0 B : Circuit c (radius r): Ampere: B .2r=0 B =0 Bz l 1 a b FdM e Radius: R ; Current: I Length: L >> R Coils: n per meter Components: Bz Br B 2 Bz : Circuit 2: B(a)=B(b)=0 Circuit 1: Ampere: Bz l=0 n I l Result: inside: B = 0 n I ez outside: B = 0 12 Symmetries for Ampere’s Law B dl 0 I 0 j dS l Wire, long Solenoid, long S Plane, extending Toroid, along core line 13 FdM Magnetic Pressure z y Plane layer L with current density j L x BL BL Be dFL l db Suppose we add an external field Be , with Be = BL , so that the total field behind the layer = 0 Lorentz force on : (FL=I.L.eT B): dFL= (j.db).dl ½0 j ez dFL= ½0 j2.db.dl ey j dl B-field of the layer (circuit l ): 2 BL l = 0 j l BL = ½0 j ez Be j Magnetic pressure: P = ½0 j2 = ½Be2 /0 . Example: this situation is met at the wall of a (long) solenoid. Then the pressure is outward, thus maximizing the cross section area.14 FdM Vectorpotential A Flow tube v’ dv’ r er P (x,y,z) j μ0 B 4 v' B = rot A x’,y’,z’ j er dv' 2 r μ0 4 v' Question: determine A in P from Biot-Savart for B r = f (x,y,z, x’,y’,z’ ) μ0 1 1 j dv' j.dv' r 4 v ' r 0 j j u.dv d (uv) v.du B dv' 4 v ' r r f ( x, y , z ) ; j f ( x ' , y ' , z ' ) j 0 0 B 4 FdM j dv' r v' } B 0 4 j v' r dv' 0 j A dv' in general, 4 v ' r 0 I 1 and A dl for a circuit 15 4 loop r Magnetic Dipole (1): Far Field r’ Far field: r’ << rP 1 r 1 rP erP rP r O I dl 1 2 rP r' r' 1 2 cos rP rP 0 I Thus : AP dl 4rP FdM 0 I A 4 dl r Goal: expression in rP in stead of all r-values over circuit 1 1 2 r rP r '2 2rP r '.cos P Monopole-term =0 r' 1 cos ..... rP 0 I r '.cos .dl 2 4rP Dipole-term 16 Magnetic Dipole (2): Dipole Moment 0 I AP r '.cos .dl 2 4rP z en P rP erP y O x r’ Ax Assume: circular circuit, with radius R << rP Assume: Y-axis along OP’ 0 R sin r '.cos er r ' sin R cos cos 0 P’ 0 I ( R sin cos )( R.d )( cos ) 2 4rP 0 I 2 0 I 2 2 R sin cos . d R sin . 2 2 4rP 4rP Ay Az 0 FdM R sin cos Define: dipole moment: m = I. Area. en = IR2.en 0 A m er 2 4rP 17 Magnetization and Polarization Magnet = set of “elementary circuits” ; n per m3 V=SL Total surface current = Itot Total magnetic moment = Itot S en Def.: Magnetization M = Polarization L en S magnetic moment / volume = surface current / length Magnetization Dipole moment: p [Cm] Dipole moment: m [Am2] Polarization P = np [Cm-2] = surface charge / m2 Magnetization M = nm [Am-1] = surface current / m 18 FdM Magnetic circuit: Torque and Energy b F F b.sin I l m B F B m F Torque: moment: = F . b sin = I B l b sin = I B S sin Magnetic dipole moment: m = I S en Torque: moment: = m x B Potential energy: min for = 0 ; max for = Potential energy: Epot = -MB. cos = - m . B 19 FdM Electret and Magnet P Electret E = (D-P)/0 D = 0 E + P E dl 0 D dS Q f 0 Magnet E D H B +++ --M H = B / 0 -M B = 0 H + M H dl I 0 B dS 0 f 20 FdM B- and H-fields at interface 1 1 B1 B2 Gauss box Circuit: 2 Given: B1 ; 1 ; 2 Question: Calculate B2 Needed: “Interface-crossing relations”: H dl I free and B dA 0 2 Relation H and B: B = 0 r H : B1.Acos 1 - B2.Acos 2 =0 B1 = B2 : no I : H1.Lsin 1 - H2.Lsin 2 =0 B1 1 A B2 1 A H1 tan 1 L H 2 tan 2 L H1// = H2// r1 tan 1 r 2 tan 2 21 FdM Toroid with air gap Core line; radius R; L = 2R I Relations: d Result: Suppose: - toroid solenoid: R, L, N, r ; - air gap, r =1 , width d <<R; g = gap ; m = metal Question: Determine Hg in gap H dl I B dA 0 f B 0 r H Hg d + Hm (L-d) = N I Bg =Bm Bg=oHg ; Bm =0r Hm Hg r NI Hg and H m L d ( r 1) r 22 FdM Induction: conductor moves in field B L1(t) S1 L2 (t+dt ) S2 v v.dt dl top bottom wall wall Gauss box: top lid S1 in L1, bottom lid S2 in L2 Area = v dt dl sin B dS 0 (t ) (t dt ) d(t ) B e [v.dt.dl.sin ] B (v.dt dl ) dt. (v B) dl n wall L d(t ) (v B) dl E n dl dt L L FdM Suppose: circuit L moves with velocity v through field B Question: Show equivalence: d Vind E dl B dS dt S L L En = non-electrostatic field 23 Induction: Faraday’s Law B S B V0 c Static: I0 E dl 0 c Vind S c Iind Dynamic: Suppose changes Vind 0 Non-electrostatic field EN d d Vind E N dl B dS dt dt S c Consequence: Let circuit c shrink, with keeping S. c For closed surface : c S S E c FdM N dl 0 B dS 0 S 24 Induction in rotating circuit frame v b v l Bext Bext v en = t ; v= ½b en Induction potential difference : (I) Using Lorentz force: Vind E N dl (v Bext ) dl =EN 2lvBext sin 2lvBext sin t (II) Using flux change: d d d Bext dS Bext [lb cos ] dt dt dt Bext lb sin t 2 Bext lv sin t Vind 25 FdM Electromagnetic brakes Why is a conducting wire decelerated by a magnetic field? Case I: switch open B v FL P Q FE B vL Q I Case II: switch closed v FL P FL moves electrons: vL , which causes I and FL2 , which causes electric field, FL2 Fions FdM Electrons feel FL Potential difference VPQ : P - , Q + (= Hall effect) FL counteracted by FE which will act on positive metal ions: Fions. 26 Coupled circuits (1): M and L 1 2 I1 I2 Suppose: circuit 1 with current I1 Part of flux from 1 will pass through 2 : 21 21 ~ I1 Definition: 21 = M21.I1 M : coefficient of mutual induction : “Mutual Inductance” : [ H ] = [ NA-1m-1.m2.A-1] = [ NmA-2 ] Suppose: Circuit 2 has current as well: I2 Flux through 2: 2= 21 + 22 = M21 I1 + M22 I2 M22 = L2 L : coefficient of self-induction “(Self) Inductance”: M and L are geometrical functions (shape, orientation, distance etc.) 27 FdM Coupled circuits (2): toroid N1 , I1 r N2 Question: determine M21 Cross section S Core line: L Flux from 1 through S : S = BS = 0 r N1 I1 S / L Linked flux from 1 through 2: 21 = N2 S = 0 r N1 N2 I1 S / L Coefficient of mutual induction: M21 = 0 r N1 N2 S / L This expression is symmetrical in 1 and 2: This result is generally valid: Mij = Mj i FdM M21 = M12 28 Coax cable : Self inductance I Radii: a and b (a<b) Length: l Current: I ; choice: inside = upward B Gauss surface 0 r I Ampere: B(r ) B-field tangential 2r Flux through circuit: 0 r I 0 r I . l b B B dS dl dr ln 2r 2 a S 0 a l b Self-inductance (coefficient of self-induction): per unit of length: b L 0 r ln (Compare with capacity of C 2 ln b 1 0 r 2 a a coax cable, per meter: 29 FdM Magnetic Field Energy Circuit c in XY-plane z H-field lines: circles around circuit. I H y A A-surfaces: everywhere H (or B) -field lines through A: B dA A x Magnetic energy: Em= ½ L I2 = 12 B dA H dl 12 ( B H )(dA dl ) 12 B H dv A c dl dA A c Compare: electric energy: Ee = V 1 2 D E dv v 30 FdM Maxwell’s Fix of Ampere’s Law Induction : En dl c d B dS dt S Changing B-field causes E-field Question: Does a changing E-field cause a B-field ? D L Suppose: chargeing a capacitor using j B dl j dS L encloses S1 : j 0 L B dl j dS L encloses S2 : S1 0 L S2 S3 S2 Volume enclosed j dS dQ f d D dS S dt dt by S2 and S3: S 2 L encloses S3 : 3 B dl 0 L FdM S1 d D dS dt S3 d In general : B dl j dS D dS dt S L S 31