LECTURE 17 Motional EMF Eddy Currents Self Inductance Reminder Magnetic Flux Faraday s Law ε = dφ B Flux through one loop Φ B = BAcosθ da Flux through N loops Φ B = NBAcosθ 1
Reminder How to Change Magnetic Flux in a Coil 1.! B changes: ΔΦ B Δt = N ΔB Δt Acosθ 3. θ changes: ΔΦ B Δt = NBA Δ cosθ Δt 2. A changes: ΔΦ B Δt = NB ΔA Δt cosθ 4. N changes: ΔΦ B = ΔN Δt Δt BAcosθ Unlikely Example Ideal Solenoid Within a Coil 120-turn coil of radius 2.4 cm and resistance 5.3 Ω Initial current in the solenoid is 1.5 A. Current is reduced to zero in 25 ms. Solenoid with radius 1.6 cm and n = 220 turns/cm 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=µ 0 n I Can easily find flux of this field through the N coils Example Ideal Solenoid Within a Coil ε = dϕ m ϕ m = N! B ˆnA = NBA No. of turns in the coil µ 0 ni NOT area enclosed by coil 3
Example Ideal Solenoid Within a Coil At point P 1 : B = µ 0 ni At point P 2 : 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 µ 0 ni i πr solenoid = 120 4π 10 7 220 turns cm ϕ m, i = 4.00 10 3 T m 2 ϕ m, f = 0 100 cm 1.5 π ( 0.016) 2 1 m dϕ m = ϕ m, f ϕ m, i 0.025 s = 4.00 10 3 0.025 T m 2 = 160 10 3 = ε s Current is given by Ohm's law I = ε R = 160 10 3 V 5.3 Ω = 30.2 10 3 A = 30.2 ma 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 d! B = 0, but da or d cosθ 0 ϕ m =! B ˆnA = B n A = B n lx dϕ m = B n l dx = B n lv ε = dϕ m = B n lv 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 = B n lv I = ε R = Blv 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: F L = IlB = lbv lb = l2 B 2 v R R Energy Conservation External Agent must exert equal but opposite force F R to move the loop with velocity v Agent does work at rate P P = F R v = l 2 B 2 v 2 R Energy is dissipated in circuit at rate P P ʹ = I 2 R = lbv 2 R = l 2 B 2 v 2 R R 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 x x x x x x x x x x x x F x x x x x x x x x x x x F v x x x x x x Flux x x x x x x increasing Induced current clockwise v Flux decreasing B field into the page 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 φ m = LI Inductance, L, depends on geometry of conductor Symbol for Inductor: Units: Henry = Webers/Ampere = Tm 2 /A Inductors Magnetic flux: Changing current: φ m = LI dφ m = L di = ε 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, ε L, appears in any coil in which the current is changing Close switch, current starts to flow B field produced within loop 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 di/ X X X X X X X X X X X X X X a b Self-Inductance in a Coil L = ϕ m I ϕ m = NBA = N µ 0 ni A = µ 0 N 2 IA l = µ 0 n 2 IAl since n = N l L = ϕ m = µ I 0 n 2 Al : geometric factors only, just like capacitance Note that Al = volume 12
Magnetic Energy in an Inductor Upon closing switch, S, apply Kirchoff s loop rule: ε IR L di = 0 Multiply through by I: power delivered by battery εi = I 2 R + LI di power dissipated by resistor power delivered to inductor Magnetic Energy in an Inductor Inductor stores magnetic energy If U m = energy in the inductor, du m U m = = LI di du = LI di m I du m = f LI di 0 U m = 1 2 LI f 2 Energy stored in an inductor 13
Magnetic Energy in a Solenoid B = µ 0 ni I = B µ 0 n L = µ 0 n 2 Al U m = 1 2 LI 2 = 1 2 µ B 0 n2 Al µ 0 n 2 = B2 2µ 0 Al Magnetic energy density u m = B2 2µ 0 Electric energy density u e = 1 2 ε 0 E 2 Both of these are general results 14
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