Chapter29: Electromagnetic Induction Motional EMF’s Conductor Moving through a magnetic field Magnetic Force on charge carriers F v B Accumulation of charge = q × Balanced Electrostatic/Magnetic Forces qvB=qE Induced Potential Difference E = EL = vBL Generalized E v B l d = × ⋅d 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, dE = v ×B⋅dl (d) magnetic force on conducting bar, (e) mechanical force needed to maintain motion and (f) mechanical power necessary to maintain motion. p212c29: 3 Faraday disk dynamo = DC generator r r r dE = v × B ⋅ ld = ω rBdr R E = ω rBdr ∫0 dl=dr R2 = ω B 2 I ω B p212c29: 4 TYU A physics student carries a metal rod while walking east along the equator where the earth's magnetic field is horizontal and directed north. (A) Wow should the rod be held to get the maximum emf? 1) horizontally east-west 2) horizontally north-south 3) vertically (B) Wow should the rod be held to get zero emf? 1) horizontally east-west 2) horizontally north-south 3) vertically (C) What direction should the student walk if there is to be zero emf regardless of how the rod is held? 1) north or south 2) east or west p212c29: 5 Faraday’s Law of Induction Changing magnetic flux induces and EMF r r dΦ = B ⋅ Ad = BdA cos φ B r r Φ B = ∫ B ⋅ Ad r r Φ B = B ⋅ A = BA cos φ for a uniform magnetic field dΦ dΦ E −= B −= N B for N loops 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: 6 B A B A (increasing) (decreasing) E E A A E E B B (increasing) (decreasing) p212c29: 7 Conducting rails: changing flux from changing area dΦ = BdA = Blvdt I dΦ E = − = −Blv dt watch sign convention s + RHR dA=l vdt Lenz' s Law + induced currents p212c29: 8 Conducting rails: changing flux from changing area dΦ = BdA = Blvdt I dΦ E = − = −Blv dt watch sign convention s + RHR dA=l vdt Lenz' s Law + induced currents p212c29: 9 Simple Alternator Φ B = AB cos ωt dΦ θ = ω t E = − B dt B = −AB (−ω sin ωt) = ω AB sin ωt E ω AB sin ωt I = = R R ()ω AB sin ωt 2 P = I 2 R = elec R ω A2Bsin ωt M = IA = R ω Pmech =τω =| M × B |ω = ()MB sin θ ω ω A2Bsin ωt = Bsin ω t ω = Pelec R p212c29: 10 TYU F F A circular coil lies perpendicular to a uniform magnetic field. (A) As the coil is squeezed from state (a) to state (b), the induced emf in the coil is 1) clockwise 2) counter clockwise 3) zero (B) Once the coil is at its final squeezed from state (a) to state (b), the induced emf is 1) clockwise 2) counter clockwise 3) zero p212c29: 11 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 60 o 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? 60 o p212c29: 12 A determination of B vs r (lab experiment) I=I 0 sin(wt) µ0I µ0I 0 r B(,)r t = = sin( ω t) 2π r 2π r Φ ≈ B(,)r t Acoil µ I Φ = B(r,t) E = −ω NA 0 0 cos( ω t) coil 2π r 1 E ∝ coil r p212c29: 13 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: 14 Induced Electric Fields B (increasing) Changing magnetic Flux produces an EMF E E Induced Electric Field r r E = ∫ E ⋅ ld r r dΦ E ⋅ ld = − B ∫ dt Not an Electrostatic field Not Conservative! p212c29: 15 I’ induced E E B I’ induced increasing I I increasing r r ∫ E⋅ dl = 2π rE dΦ = B dt 1 dΦ E = B 2π r dt p212c29: 16 TYU (A) If you wiggle a magnet back and forth in your hand, are you generating an electric field? 1) yes 2) no (B) If so, is it conservative? 1) yes 2) no 3) none generated p212c29: 17 Displacement Current “Generalizing” displacement current for Ampere’s Law conduction current creates magnetic field Amperian loop with surface iC Ienc = i C Parallel Plate Capacitor Amperian loop with surface I = 0??? iC enc Parallel Plate Capacitor p212c29: 18 Define Displacement Current between plates so that i D = i C A q = Cv = ε Ed = εEA = εΦ d E dq dΦ i = = ε E ≡ i C dt dt D r r ∫ B ⋅ ld = µo (iC + iD ) d r r = µ iCo + µoε ∫ E ⋅ Ad r dt r Ed j = ε D dt changing Electric Fields can create Magnetic Fields! p212c29: 19 Maxwell’s Equations r r Q r r 4πρ ∫ E⋅ dA = enc ∇ ⋅E = ε ε o r r o r r B⋅ dA =0 ∇ ⋅B = 0 ∫ r r r r r r dΦ dE E ∇ ×BJ =µo() c + ε o ∫ B⋅ dl =µo() I c + ε o dt v dt r r r r dΦ dB E⋅ dl = − M ∇ ×E = − ∫ dt dt Lorentz Force Law r r r r F= q() E + v × B p212c29: 20 TYU (A) Which of these explains how a credit card reader works? (B) which of these describes how a steady current generates a magnetic field? r r Q 1) E ⋅dA = enc ∫ ε r r o 2) ∫ B⋅dA =0 r r dΦ 3) B ⋅ ld =µ (I + ε E ) ∫ o c o dt r r dΦ 4) E ⋅ ld = − M ∫ dt p212c29: 21 Superconductivity zero resistivity below T c (with no external magnetic field) with field above the critical field B c, no superconductivity at any temperature B Bc Normal Superconducting T Tc p212c29: 22 More magnetic fun with superconductivity: the Meissner Effect Within a superconducting material, B inside = 0! magnetic field lines are expulsed from superconductor Relative permeability K m = 0 internal field is reduced to 0 => Superconductor is the perfect diamagnetic material! p212c29: 23 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 T c and B c2 p212c29: 24.