Chapter29: Electromagnetic Induction Motional EMF’s Conductor Moving through a magnetic field Magnetic Force on charge carriers F = qv ×B Accumulation of charge Balanced Electrostatic/Magnetic Forces qvB=qE Induced Potential Difference E = EL = vBL Generalized d E = v ×B⋅dl p212c29: 1 With conduction rails: I F = qv ×B demonstrations with galvanometer moving conductor moving field! p212c29: 2 I Consider circuit at right with total circuit resistance of 5 Ω, B = .2 T, speed of conductor = 10 m/s, length of 10 cm. Determine: (a) EMF, (b) current, (c) power dissipated, (d) magnetic force on conducting bar, (e) mechanical force needed to maintain motion and (f) mechanical power necessary to maintain motion. dE = v ×B⋅dl p212c29: 3 Faraday disk dynamo = DC generator dE = v × B ⋅ dl = ω rBdr R E = ∫ ω rBdr dl=dr 0 R2 =ω B 2 I ω B p212c29: 4 Faraday’s Law of Induction Changing magnetic flux induces and EMF dΦ B = B ⋅ dA = BdA cos φ Φ B = ∫ B ⋅ dA Φ B = B ⋅ A = BA cos φ for a uniform magnetic field =− dΦ B dΦ B for N loops = −N dt dt Lenz’s Law: The direction of any magnetic induction effect (induced current) is such as to oppose the cause producing it. (opposing change = inertia!) p212c29: 5 B (increasing) A E B (decreasing) E A E B (increasing) A A E B (decreasing) p212c29: 6 Conducting rails: changing flux from changing area dΦ = BdA = Blvdt dΦ E=− = − Blv dt watch sign conventions + RHR Lenz' s Law + induced currents I dA=l vdt p212c29: 7 Simple Alternator θ=ωt B Φ B = AB cos ω t dΦ B dt = − AB(−ω sin ω t ) E =− = ω AB sin ω t I= ω E ω AB sin ω t R ( ω AB sin ω t )2 2 Pelec = I R = R 2 ω A B sin ω t M = IA = R Pmech = τω =| M × B | ω = (MB sin θ )ω = R = ω A2 B sin ω t R B sin ω t ω = Pelec p212c29: 8 Example: A coil of wire containing 500 circular loops with radius 4.00 cm is placed in a uniform magnetic field. The direction of the field is at an angle of 60o with respect to the plane of the coil. The field is decreasing at a rate of .200 T/s. What is the magnitude of the induced emf? 60o p212c29: 9 A determination of B vs r (lab experiment) I=I0 sin(wt) B(r , t ) = r µ0 I µ0 I 0 = sin(ω t ) 2π r 2π r Φ ≈ B (r , t ) Acoil Φ = B(r,t) Ecoil = −ω NA Ecoil ∝ µ0 I 0 cos(ω t ) 2π r 1 r p212c29: 10 Eddy Currents Induced currents in bulk conductors magnetic forces on induced currents energy losses to resistance ex: Conducting disk rotating through a perpendicular magnetic field Ferromagnets ~ Iron (Electrical Conductor) Constrain Eddy currents with insulating laminations p212c29: 11 Induced Electric Fields Changing magnetic Flux produces an EMF E B (increasing) E Induced Electric Field E = ∫ E ⋅ dl dΦ B E ⋅ d l = − ∫ dt Not an Electrostatic field Not Conservative! p212c29: 12 I’ induced E I’ induced E B I increasing I increasing E ∫ ⋅ dl = 2π rE dΦ B dt 1 dΦ B E= 2π r dt = p212c29: 13 Qenc E ∫ ⋅ dA = εo Maxwell’s Equations 4πρ ∇⋅ E = B ∫ ⋅ dA =0 dΦ E µ ε B ⋅ dl = ( I + ) o c o ∫ dt dΦ M E ⋅ dl = − ∫ dt εo ∇⋅ B = 0 dE ∇ × B = µ o ( Jc + ε o ) dt dB ∇× E = − dt Lorentz Force Law F = q ( E + v × B) p212c29: 14 Superconductivity zero resistivity below Tc (with no external magnetic field) with field above the critical field Bc, no superconductivity at any temperature B Bc Normal Superconducting Tc T p212c29: 15 More magnetic fun with superconductivity: the Meissner Effect Within a superconducting material, Binside = 0! magnetic field lines are expulsed from superconductor Relative permeability Km = 0 internal field is reduced to 0 => Superconductor is the perfect diamagnetic material! p212c29: 16 mechanical effects ferromagnetic or paramagnetic materials attracted to permanent magnet superconductor repels permanent magnet! Type-II superconductors small filaments of “normal phase” coexist within bulk superconductor some magnetic field lines penetrate material two critical fields field just begins to penetrate material Bc1 : magnetic field just begins to penetrate material Bc2 :bulk material ”goes normal” more practical for electromagnets: higher Tc and Bc2 p212c29: 17