Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Maxwell's equations wikipedia , lookup
Neutron magnetic moment wikipedia , lookup
Electrical resistance and conductance wikipedia , lookup
Magnetic field wikipedia , lookup
History of electromagnetic theory wikipedia , lookup
Magnetic monopole wikipedia , lookup
Electromagnetism wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Superconductivity wikipedia , lookup
LECTURE 17 • Motional EMF • Eddy Currents • Self Inductance Reminder Magnetic Flux • Faraday’s Law ε =− dΦB dt dA • Flux through one loop ΦB = BAcosθ • Flux through N loops ΦB = NBAcosθ 1 Reminder How to Change Magnetic Flux in a Coil ! ΔΦB ΔB 1. B changes: =N Acosθ Δt Δt 2. A changes: ΔΦB ΔA = NB cosθ Δt Δt 3. θ changes: Δ ⎡⎣cosθ ⎤⎦ ΔΦB = NBA Δt Δt 4. N changes: ΔΦB ΔN = BAcosθ Δt Δt Unlikely Example Ideal Solenoid Within a Coil 120-turn coil of radius 2.4 cm and resistance 5.3 Ω Solenoid with radius 1.6 cm and n = 220 turns/cm Initial current in the solenoid is 1.5 A. Current is reduced to zero in 25 ms. What is the current in the coil while the current is being reduced? 2 Example Ideal Solenoid Within a Coil • IDEAL B field inside the solenoid – Constant and parallel to its axis B=µ0n I • Can easily find flux of this field through the N coils Example Ideal Solenoid Within a Coil dϕ m dt! ϕ m = NB ⋅ n̂A = NBA ε =− No. of turns in the coil µ0nI NOT area enclosed by coil 3 Example Ideal Solenoid Within a Coil At point P1 : B = µ0nI At point P2 : B≅0 Ideal solenoid (closely wound, long): B = 0 outside the core of the solenoid ∴ relevant area for ϕ m = area enclosed by solenoid Example Ideal Solenoid Within a Coil 2 ϕ m, i = N ⋅ µ0nIi ⋅ π rsolenoid = 120 ⋅ 4π ×10−7 × 220 2 turns 100 cm × ×1.5 ⋅ π (0.016) cm 1m ϕ m, i = 4.00 ×10−3 T ⋅ m2 ϕ m, f = 0 dϕ m ϕ m, f − ϕ m, i 4.00 ×10−3 T ⋅ m2 = = = 160 ×10−3 = −ε dt 0.025 s 0.025 s Current is given by Ohm's law I= ε 160 ×10−3 V = = 30.2 ×10−3 A = 30.2 mA R 5.3 Ω 4 Example Ideal Solenoid Within a Coil What is the direction of the current in the coil? (A) Same direction as the current in the solenoid (B) Opposite the direction of the current in the solenoid Motional EMF • Definition – Any emf induced by the motion of a conductor in a magnetic field ! dB dt = 0, but dA or dt d⎡ ⎣cosθ ⎤⎦ ≠ 0 dt ! ϕ m = B ⋅ n̂A = Bn A = Bn ℓx dϕ m dx = Bn ℓ = Bn ℓv dt dt ε =− dϕ m = −Bn ℓv dt 5 Motional EMF • Definition – Any emf induced by the motion of a conductor in a magnetic field • Charge Separation – + charge experiences upward force qvB – Electric field vB, downward – Potential ΔV=vBL Calculation A thin conducting rod is pulled with velocity v along conducting rails that are connected by a resistor, R. A uniform magnetic field B is directed into the page. Surface S is increasing therefore flux is increasing. What will be the magnitude & direction of the induced current? Lenz Law ⇒ counterclockwise ε =− dϕ m = −Bn ℓv dt I= ε Bℓv = R R 6 Energy Conservation • What is the rate of work by the applied force? • The induced current gives rise to a net magnetic force F in the loop which opposes the motion: ⎛ ℓBv ⎞ ℓ 2B 2v FL = IℓB = ⎜ ⎟ ℓB = R ⎝ R ⎠ Energy Conservation • External Agent must exert equal but opposite force FR to move the loop with velocity v l 2B 2v 2 – Agent does work at rate P P = FRv = R • Energy is dissipated in circuit at rate P’ 2 ⎛ lBv ⎞ l 2B 2v 2 Pʹ = I R = ⎜ ⎟ R= R ⎝ R ⎠ 2 ⇒ P =P' 7 Eddy Currents • Previously we considered currents confined to a loop or a coil • Electrical equipment contains other metal parts which are often located in changing B fields – Currents in such parts are called Eddy currents • Swirl around Induced current anti-clockwise F x x x x x x x x v x x x x Flux increasing x x x x x x Induced current clockwise x x x x x x x x x x x xF x x x x x x Flux v B field into the page decreasing Eddy Currents Replace the loop with a solid Cu plate • The energy is dissipated by heating the metal – Conducting material near magnetic fields will heat up, sometimes a lot 8 Induction Heating Laboratory and industrial processes Cooking Eddy Currents • Relative motion between a B field and a conductor induces a current in the conductor • The induced current give rises to a net magnetic force, FM, which opposes the motion 9 Demo Eddy Currents ! ! ! • The force F = i ℓ × B opposes the motion – This “braking effect” is used for the brakes of trains Demo Reduce Eddy Currents Metal strips with insulating glue Cut slots into the metal 10 Inductance and Inductors • Capacitance: Q ∝V Q = CV • Coil carrying current, I, produces a magnetic field – Magnetic flux proportional to current – L = self-inductance of coil φm = LI • Inductance, L, depends on geometry of conductor • Symbol for Inductor: • Units: Henry = Webers/Ampere = Tm2/A Inductors φm = LI d φm dI Changing current: = L = −ε dt dt • Magnetic flux: • • Changing the current through an inductor induces an opposing voltage across the inductor – Acts like a voltage source in a circuit ΔV = −ε 11 Self-Inductance • An induced emf, εL, appears in any coil in which the current is changing • Close switch, current starts to flow – B field produced within loop dI/dt X XX X X XX XX X X XX X a b • Flux through loop increases as current increases – EMF is induced in loop opposing change in current flow – Opposes change in flux • Self-Induction – Changing current through a loop induces an opposing voltage in that same loop Self-Inductance in a Coil L= ϕm I ϕ m = NBA = N ⋅ µ0nI ⋅ A = µ0N 2IA N = µ0n 2IAℓ since n = ℓ ℓ ϕm = µ0n 2 Aℓ : geometric factors only, just like capacitance I Note that Aℓ = volume L= 12 Magnetic Energy in an Inductor • Upon closing switch, S, apply Kirchoff’s loop rule: ε − IR − L dI =0 dt • Multiply through by I: εI = I 2R + LI power delivered by battery dI dt power dissipated by resistor power delivered to inductor Magnetic Energy in an Inductor • Inductor stores magnetic energy If Um = energy in the inductor, dUm dI = LI ⇒ dUm = LI dI dt dt Um = ∫ dU Um = 1 2 LI 2 f m = ∫ If 0 LI dI Energy stored in an inductor 13 Magnetic Energy in a Solenoid B = µ0nI ⇒ I = B µ 0n L = µ0n 2 Al 2 ⎛ B ⎞ 1 1 B2 Um = LI 2 = µ0n 2 Al ⎜ = Al ⎟ 2 2 µ n 2 µ ⎝ 0 ⎠ 0 Magnetic energy density um = B2 2µ0 1 Electric energy density ue = ε 0E 2 2 Both of these are general results 14 15